A body of mass 3.0kg moves with a velocity 10m/s calculate the moment of the body

Answers

Answer 1

In this case, the body has a mass of  [tex]3.0[/tex] kg and is moving with a velocity of [tex]10 m/s[/tex].  the moment of the body is  [tex]30 kg m/s.[/tex]

What is the momentum?

The moment of a body (also known as its momentum) is defined as the product of the body's mass and velocity.

In this instance, the body weighs  3.0  kilogrammes and is travelling at a speed of 10 m/s.

The formula to calculate the moment of the body is:

momentum = mass x velocity

Plugging in the given values, we get:

momentum [tex]= 3.0 kg \times 10 m/s[/tex]

Solving the above expression, we get:

momentum = [tex]30 kg m/s[/tex]

Therefore, the moment of the body is [tex]30[/tex] kg m/s.

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Related Questions

1. In the picture above, how would the amount of kinetic energy in the third pendulum compare to the
amount of potential energy in the first and fourth pendulums?

Answers

The kinetic energy would be at its maximum compared to the potential energy which would be less. The potential energy of the first and the last pendulum would be at their maximum

What is kinetic energy?

Kinetic energy is the energy an object possesses due to its motion. It is defined as the energy that an object has by virtue of its motion, and is dependent on both its mass and velocity. The formula for calculating kinetic energy is 1/2 times the mass of the object times the square of its velocity, or KE = 1/2 mv^2.

This means that the kinetic energy of an object increases as its mass and velocity increase. For example, a heavy object moving at a high velocity will have a higher kinetic energy than a lighter object moving at a lower velocity. Kinetic energy is a scalar quantity, meaning it has no direction, only magnitude.

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An object has greater momentum if it has ?

Answers

The more massive or swifter an object is, the more momentum it has.

When a substance has more momentum?

A moving thing is more difficult to stop the more momentum it possesses. The amount of momentum an object possesses is influenced by its mass. For instance, while a car driving at the same speed as a baseball can be stopped, it cannot be caught. Because the car is heavier, it has more momentum.

What increases an object's momentum?

A moving object's mass and speed both affect its momentum. The greater the object's momentum and the more difficult it is to stop are directly proportional to its weight and speed.

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Consider a person of mass 59 kg. What would his mass be if he was composed entirely of neutron-star material of density 3 x 10^17 kg/m³? (Assume that his average density is 1000 kg/m³)
Express your answer using two significant figures.

Answers

If the human were fully made of neutron star matter, his mass would be roughly 1.8 x 10¹⁶ kg, to two significant numbers.

What elements make up a neutron star?

As most protons and electrons will have merged to become neutrons under the extremely dense conditions, neutron stars get their name because they are largely made of neutrons.

Assume the individual has a volume of 59,000 cm3, or about the same as someone with a density of 1000 kg/m3. When we convert this volume to m3, we obtain:

V = 59,000 cm³

V = 0.059 m³

If the individual were made entirely of matter from neutron stars, his mass would be:

m = ρV

m = (3 x 10¹⁷ kg/m³) (0.059 m³)

m = 1.77 x 10¹⁶ kg

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Need help I don't know if these are label correct but I need to know which letter on the diagram correctly labels amplitude of the wave and please label the rest
Pic attached below​

Answers

Explanation:

I  wavelength

J amplitude

K trough

L crest

M equal

1. What is the horizontal distance of the center of gravity of the system from the point where the ladder touches the ground?
2. What is the torque about the axis of rotation (point B) by taking the total weight of the person + ladder acting at the center of gravity?

Answers

To answer the first question, we need to determine the location of the center of gravity of the system. Assuming the person and ladder can be treated as a uniform object, the center of gravity will be located at the midpoint of the ladder.

Let's say the ladder is 10 feet long, so the midpoint would be 5 feet from either end. If we assume the ladder is resting at a 60 degree angle against a vertical wall, we can use trigonometry to determine the horizontal distance of the center of gravity from the point where the ladder touches the ground.
Using the sine function, we know that sin(60) = opposite/hypotenuse, so the opposite side (which is the vertical height of the ladder) is 10*sin(60) = 8.66 feet. Therefore, the horizontal distance from the center of gravity to the point where the ladder touches the ground is also 8.66 feet.
To answer the second question :

We need to calculate the torque about the axis of rotation (point B) by taking the total weight of the person + ladder acting at the center of gravity. The formula for torque is torque = force x distance.
The force is equal to the weight of the person + ladder, which we'll assume is 300 pounds. The distance is the horizontal distance we just calculated, which is 8.66 feet.

So the torque about point B would be 300 pounds x 8.66 feet = 2,598 Newton-meters (Nm) or 2,598 pound-feet (lb-ft).

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Two pure tones Cs and Gs, with frequencies from the Pythagorean diatonic scale, are sounded simultaneously. Find
a) the frequencies of the three combination tones and
b) the notes on the Pythagorean scale to which these tones belong.

Answers

The combination tones correspond to the notes Bb, D, and Cs on the Pythagorean diatonic scale.

What is Frequency?

Frequency is the number of occurrences of a repeating event per unit of time. In other words, it is the rate at which a wave oscillates or completes one cycle. The unit of frequency is hertz (Hz), which is equivalent to one cycle per second.

When two pure tones with frequencies f1 and f2 are sounded simultaneously, several additional frequencies, known as combination tones, can be produced. The three most important combination tones are:

The sum tone, which has a frequency equal to the sum of the two original frequencies: f1 + f2

The difference tone, which has a frequency equal to the difference between the two original frequencies: |f1 - f2|

The octave tone, which has a frequency twice that of the lower of the two original frequencies: 2f1 or 2f2

In this case, we have two pure tones Cs and Gs with frequencies from the Pythagorean diatonic scale. We need to first determine the frequencies of these two tones. According to the Pythagorean tuning system, the frequency ratios for Cs and Gs are:

Cs:G = 9:8

Cs:fundamental = 2:1 (assuming Cs is one octave above the fundamental)

Gs:fundamental = 3:1 (assuming Gs is one octave and a fifth above the fundamental)

Let's assume that the fundamental frequency is f0. Then we can write:

Cs = 2f0 * (9/8) = 9f0/4

Gs = 4f0 * (3/2) * (9/8) = 27f0/8

a) To find the combination tones, we need to apply the equations above. The sum tone has a frequency of:

f1 + f2 = Cs + Gs = (9f0/4) + (27f0/8) = 45f0/8

The difference tone has a frequency of:

|f1 - f2| = |Cs - Gs| = |(9f0/4) - (27f0/8)| = 9f0/8

The octave tone has a frequency of:

2f1 = 2Cs = 9f0/2 = 9f0

Therefore, the three combination tones have frequencies of 45f0/8, 9f0/8, and 9f0.

b) To determine the notes on the Pythagorean scale to which these tones belong, we need to find the closest notes on the scale to each of the combination tones. The Pythagorean scale is based on a series of perfect fifths, so we can use the frequency ratios of 3:2 to determine the frequency of each note relative to the fundamental frequency f0.

The closest notes on the Pythagorean scale to the combination tones are:

45f0/8 is closest to the note Bb, which has a frequency of 3f0/2

9f0/8 is closest to the note D, which has a frequency of 9f0/8

9f0 is closest to the note Cs, which has a frequency of 9f0/4

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The compression ratio of a petrol engine is 20.0 to 1; that is, air in a cylinder is compressed adiabatically to 1/20.0 of its initial volume.
(a) If the initial pressure is 1.01× 10^5 and the initial temperature is 20℃, find the final pressure and the temperature after adiabatic compression.
(b) How much work does the gas do during the compression if the initial volume of the cylinder is 1.00 = 1.00 ×10^−3^3. Use the values = 20.8 /. and
= 1.400 for air.
(c) Hence, find the change in internal energy of the air.

Answers

The final temperature is 390 K and the final pressure is 6.46 x [tex]10^{6}[/tex] Pa. The work done by the petrol during compression is 7.20 x 10² J. The air's internal energy changed by -7.20 x 10² J.

Once adiabatic compression has occurred, determine the final pressure and temperature.

The final pressure and temperature can be calculated using the adiabatic compression equation:

To start, here is the last volume:

Vf = Vi / 20.0 = 1.00 × [tex]10^{-3}[/tex] / 20.0 = 5.00 × [tex]10^{-5}[/tex] m³

We may then determine the final pressure by:

1.01 105 Pa * Pf = Pi * (V i / V f) (1.00 × [tex]10^{-3}[/tex] m³ / 5.00 × [tex]10^{-5}[/tex] m³)1.4 = 6.46 × [tex]10^{6}[/tex] Pa

Final temperature = (1 mol × 8.31 J/mol K) / (6.46 106 Pa × 5.00 × [tex]10^{-5}[/tex] m³) = 390 K

The final pressure is therefore 6.46 106 Pa, and the final temperature is 390 K.

How much effort is put forth by the petrol during compression?

The equation: can be used to determine the work performed by the gas during compression.

W = (γ / (γ - 1)) × P i × V i × (1 - (1 / r(γ - 1)))

where P i and V i are the initial pressure and volume, r is the compression ratio (20.0), and is the adiabatic index (1.4 for air).

Inputting the values provided yields:

W = (1.4 / (1.4 - 1)) × 1.01 × [tex]10^{5}[/tex] Pa × 1.00 × [tex]10^{-3}[/tex] m³ × (1 - (1 / 20.0(1.4 - 1))) = 7.20 × 10² J

Calculate the change in internal energy.

The first law of thermodynamics can be used to compute the change in internal energy:

ΔU = Q - W

where W is the work performed by the petrol and Q is the heat contributed to the system, which in this instance is adiabatic.

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What mass of water at 27.0°C must be allowed to come to thermal equilibrium with a 1.95-kg cube of aluminum initially at 150°C to lower the temperature of the aluminum to 60.0°C? Assume any water turned to steam subsequently recondenses.
kg

Answers

To solve this problem, we can use the equation:

Q_aluminum = -Q_water

where Q_aluminum is the heat lost by the aluminum cube and Q_water is the heat gained by the water. We can also use the specific heat capacity of aluminum and water to calculate the heat lost or gained:

Q = m x c x ΔT

where Q is the heat, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature.

First, let's calculate the heat lost by the aluminum cube:

Q_aluminum = m_aluminum x c_aluminum x ΔT_aluminum
Q_aluminum = 1.95 kg x 0.91 J/g°C x (150°C - 60°C)
Q_aluminum = 311.22 kJ

Next, let's calculate the heat gained by the water:

Q_water = m_water x c_water x ΔT_water
Q_water = m_water x 4.18 J/g°C x (27.0°C - 60.0°C)
Q_water = -m_water x 4.18 J/g°C x 33.0°C
Q_water = -1381.14 m_water J

Since Q_aluminum = -Q_water, we can set these equations equal to each other and solve for the mass of water:

311.22 kJ = -1381.14 m_water J
m_water = -311.22 kJ / (-1381.14 J/g°C x 33.0°C)
m_water = 0.666 kg

Therefore, a mass of 0.666 kg of water at 27.0°C must be allowed to come to thermal equilibrium with the aluminum cube to lower its temperature to 60.0°C.

A 2.8 kg block slides along a frictionless surface at 1.1 m/s . A second block, sliding at a faster 4.8 m/s , collides with the first from behind and sticks to it. The final velocity of the combined blocks is 2.3 m/s.

What was the mass of the second block?

Answers

Conservation of momentum is a major law of physics which states that the momentum of a system is constant if no external forces are acting on the system. It is embodied in Newton’s First Law or The Law of Inertia.the mass of the second block is 1.1Kg.

principle of momentum conservation

M1u1 plus M2u2 equals M1v1 and M2V2.

As all collisions were elastic in nature and no energy loss through friction, heat, etc. was taken into account, theoretic calculations alone cannot guarantee that there was a complete transfer of energy.

Consider the scenario where a football with mass M2 is lying on the ground and a bowling ball with mass M1 is hurled at the football at a velocity of

The formula is: (2.8 kg * 1.1 m/s) + (m2 * 4.8 m/s) = (2.3 kg + m2). 2.3 m/s

The formula is 2.8 J + (4.8 m/s m2) = 4.8 J + (2.3 m/s m2).

4.8 m/s m2 = 2.8 J plus (2.3 m/s m2)

4.8 m2 = 2.8 + 2.3 m2

2.3 m2 on each side of the equation

2.5 m = 2.8 m = 2.8 / 2.5\sm = 1.1kg

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A runner is sprinting at 3 m/s. But 40 seconds later they are sprinting at 3.8 m/s. What is the runner’s acceleration?

Answers

Question:

A runner is sprinting at 3 m/s. But 40 seconds later they are sprinting at 3.8 m/s. What is the runner’s acceleration?

Answer:

We can use the following formula to calculate the acceleration:

a = (vf - vi) / t

where:

a = acceleration

vf = final velocity

vi = initial velocity

t = time

In this case, the initial velocity (vi) is 3 m/s, the final velocity (vf) is 3.8 m/s, and the time (t) is 40 seconds.

So, we can plug these values into the formula and solve for the acceleration:

a = (3.8 m/s - 3 m/s) / 40 s

a = 0.8 m/s^2

Therefore, the runner's acceleration is 0.8 m/s__2__.

If all of the dimensions of the block double (to become 20 cm wide, 8 cm tall, and 6 cm deep), what happens to the resistance of electric current along each axis?

Answers

Answer:

Assuming the block has uniform resistivity throughout, if all of the dimensions of the block double, then the resistance of electric current along each axis will increase by a factor of 8. This is because the resistance of a material is dependent on its dimensions, and specifically on its length, area, and resistivity. When the dimensions of the block double, its length, width, and height all double, which means that the overall length of the path the current must take through the material increases by a factor of 2+2+2=6. Since resistance is directly proportional to length, the overall resistance of the block increases by a factor of 6. Additionally, since the area of the block's cross-section increases by a factor of 4 (2 x 2), the overall resistance decreases by a factor of 4. Therefore, the overall effect is that the

A glass windowpane in a home is 0.62 cm thick and has dimensions of 1.5 m × 2.3 m. On a certain day, the indoor temperature is 26°C and the outdoor temperature is 0°C.
(a) What is the rate at which energy is transferred by heat through the glass?
W

(b) How much energy is lost through the window in one day, assuming the temperatures inside and outside remain constant?
J

Answers

lass windowpane in a home is 0.62 cm thick and has dimensions of 1.5 m × 2.3 m. On a certain day, the indoor temperature is 26°C and the outdoor temperature is 0°C.the amount of energy lost through the window in one day is approximately 37.9 MJ.

We can use the formula for heat transfer through a material, which is:

Q = k * A * (T1 - T2) / d

where Q is the rate of heat transfer, k is the thermal conductivity of the material, A is the surface area of the material, d is the thickness of the material, T1 is the temperature on one side of the material, and T2 is the temperature on the other side of the material.

(a) We first need to find the thermal conductivity of glass. According to engineeringtoolbox.com, the thermal conductivity of glass is approximately 1.05 W/(m*K). We convert the temperatures to Kelvin:

T1 = 26°C + 273.15 = 299.15 K

T2 = 0°C + 273.15 = 273.15 K

Plugging in the values:

Q = (1.05 W/(m*K)) * (1.5 m * 2.3 m) * (299.15 K - 273.15 K) / (0.62 cm / 100 cm/m)

Q = 438.37 W

So the rate at which energy is transferred by heat through the glass is 438.37 W.

(b) We can convert the rate of heat transfer to energy over time by using the formula:

E = Q * t

where E is the energy, Q is the rate of heat transfer, and t is the time. Assuming 24 hours in a day:

E = 438.37 W * 24 h * 3600 s/h

E = 37,910,899.2 J

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A marble rolls off a tabletop 1.1m high and hits the floor at a point 2.1m away from the tables edge in the horizontal direction.

A. How long (in seconds) is the marble (in m/s) in the air?

B. What is the speed of the marble (in m/s) when it leaves the tables edge?

C. What is the speed (in m/s) when it hits the floor?

Answers

To solve this problem, we can use the equations of motion for an object in free fall with constant acceleration due to gravity (g = 9.81 m/s^2), which are:

y = vit + (1/2)gt^2 (Equation 1)
v = vi + gt (Equation 2)

where y is the vertical displacement, vi is the initial vertical velocity (which is zero when the marble leaves the table), t is the time, and v is the final velocity.

A. To find the time the marble is in the air, we can use Equation 1 with y = 1.1 m and vi = 0:

1.1 = (1/2)gt^2

Solving for t, we get:

t = sqrt(2*1.1/g) = 0.47 seconds

Therefore, the marble is in the air for 0.47 seconds.

B. To find the speed of the marble when it leaves the table, we can use Equation 2 with vi = 0 and t = 0.47 seconds:

v = gt = 9.810.47 = 4.61 m/s

Therefore, the speed of the marble when it leaves the table is 4.61 m/s.

C. To find the speed of the marble when it hits the floor, we need to find the horizontal and vertical components of its velocity. The horizontal component is constant, since there is no horizontal acceleration. The vertical component can be found using Equation 2 with vi = 0 and t = 0.47 seconds:

vy = gt = 9.810.47 = 4.61 m/s

The horizontal component of the velocity can be found from the distance traveled in the horizontal direction (2.1 m) and the time in the air (0.47 seconds):

vx = 2.1/0.47 = 4.47 m/s

The speed of the marble when it hits the floor is the magnitude of its velocity, which can be found using the Pythagorean theorem:

v = sqrt(vx^2 + vy^2) = sqrt(4.47^2 + 4.61^2) = 6.35 m/s

Therefore, the speed of the marble when it hits the floor is 6.35 m/s.

2. What gravitational force does the moon produce on the Earth if their centers are 3.88 x10^8 m apart and the moon has a mass of 7.43 x 10^22 kg?

Answers

About 1.98 x 10²⁰ Newtons of gravitational force are exerted by the moon on the Earth.

The moon is located 3.84 x 10⁸ metres away from the earth.

The angle subtended if observed from two diametrically opposed places on the Earth. The moon is 3.84 10⁸ metres away from the Earth. The angle subtended at the moon, when seen from two diametrically opposed places on Earth, is 1° 54′.

where F is the gravitational force, G is the gravitational constant (6.6743 x 10⁻¹¹ N*m²/kg²), m1 and m2 are the masses of the two objects, and r is the distance between their centers.

In this case, we are calculating the force that the moon produces on the Earth, so we can set m1 to the mass of the Earth (5.97 x 10²⁴ kg).

F = 6.6743 x 10⁻¹¹ * (5.97 x 10²⁴ kg) * (7.43 x 10²² kg) / (3.88 x 10⁸ m)²

Simplifying the calculation, we get:

F = 1.98 x 10²⁰ N

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From 1998 to 2005,the percentage of mothers with infants who were in the workforce

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From 1998 to 2005, the percentage of mothers with infants in the workforce grew considerably. In 1998, 59.5% of mothers with infants were in the labor force and by 2005, that number grew to 68.8%, an increase of 9.3%, according to data from the Bureau of Labor Statistics.

What is Bureau of Labor Statistics?

The United States Department of Labor has a division called the Bureau of Labor Statistics (BLS). It functions as a key agency of the U.S. Federal Statistics System and is the primary fact-finding body for the federal government of the United States in the broad field of labour economics and statistics. The American public, the U.S. Congress, other Federal agencies, State and local governments, industry and labour representatives, and the general public are all served by the BLS's collection, processing, analysis, and dissemination of crucial statistical data. The BLS also conducts research on the income levels families require to sustain an acceptable standard of living and acts as a statistical resource for the US Department of Labor.

The largest overall increase was in mothers with infants aged 6 to 11 months. In 1998, 57.7% of this group was in the labor force, but by 2005, that increased to 68.3%.

There was also an increase among mothers with infants aged 0 to 5 months, though it was much smaller. In 1998, 55.4% worked, with that number increasing to 60.2% in 2005. Working mothers with infants aged 12 to 17 months experienced the smallest increase from 1998 to 2005, with the labor force participation rate reaching 71.7%, a 4.4 percentage point increase from 1998.

The fact that more and more mothers are opting to enter the workforce may indicate that the social expectation of mothers’ roles is slowly evolving. The increased availability of daycare, more flexible job positions, and changes in child-rearing practices may have contributed to the increase in mothers entering the workforce.

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An object has a mass of 50kg. On earth, the weight of the object is almost 500 newtons but the object floats in space. Why does an object that is so difficult to lift on earth float on space ?

Answers

The object will continue to float in a stationary position until acted upon by a force, such as a push or pull.

What is Weight?

Weight is the force with which an object is attracted towards the Earth or any other celestial body due to gravity. The weight of an object can be calculated by multiplying its mass with the acceleration due to gravity. In the SI system of units, weight is measured in Newtons (N).

An object that has a mass of 50kg has a weight of nearly 500 newtons on Earth due to gravity. However, in space, where there is no significant gravity, the object will not experience any weight or force pushing it down. The object will float because there is no force acting upon it to cause it to move in a particular direction. This is due to the absence of gravity, which is the force responsible for the weight of an object.

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How strong would a magnetic field need to be in order to make a particle with a mass of 4 x 10−11 kg and a charge of 8 nC move in a circular path with a speed of 400 m/s and a radius of 0.5 m?

Answers

As a result, a particle with a mass of 4 x 10-11 kg and a charge of 8 nC needs a magnetic field of 1.0 Tesla to move in a circular path with a speed of 400 m/s and a radius of 0.5 m.

When is the strongest magnetic pressure acting on a charged particle in a magnetic area?

Thus, when a charged particle moves at a 90° angle to the field, the magnetic force acting on it is greatest.

In a magnetic subject, what is the pressure on a moving charge?

What force does a moving object experience in a magnetic field? As a charge moves through an electric and magnetic field, the Lorentz force acts on it. The total of magnetic and electric forces must be what it is.

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10/10=x =35/56
x=?

help please

Answers

The answer is 6

Mark as brainy please

The escape speed from the surface of Planet Zoroaster is 12.0km/s. The planet has no atmosphere. A meteor far away from the planet moves at speed 5.0km/s on a collision course with Zoroaster. How fast is the meteor going when it hits the surface of the planet.

Answers

Answer:

The escape speed of a planet is the minimum speed that an object needs to attain to escape the gravitational pull of the planet and not fall back. Since the meteor's speed is less than the escape speed of Planet Zoroaster, it will not escape and will crash into the planet.

To find the final speed of the meteor when it hits the surface of the planet, we can use the principle of conservation of energy. At a great distance from the planet, the meteor has only kinetic energy. As it approaches the planet, its potential energy increases due to the planet's gravitational attraction, while its kinetic energy decreases due to the planet's gravitational deceleration.

At the moment of impact, all of the meteor's kinetic energy will be converted into other forms of energy (such as heat and sound) upon hitting the surface. Therefore, we can equate the initial kinetic energy of the meteor to the sum of its potential energy and its final kinetic energy just before impact.

Initial kinetic energy = 1/2 * m * v1^2

where m is the mass of the meteor and v1 is its initial speed.

Potential energy at the surface of the planet = -G * M * m / R

where G is the gravitational constant, M is the mass of the planet, m is the mass of the meteor, and R is the radius of the planet.

Final kinetic energy just before impact = 1/2 * m * v2^2

where v2 is the final speed of the meteor just before impact.

We can set these equal and solve for v2:

1/2 * m * v1^2 = -G * M * m / R + 1/2 * m * v2^2

Simplifying and solving for v2, we get:

v2 = sqrt(2 * G * M / R + v1^2)

Plugging in the given values, we get:

v2 = sqrt(2 * 6.6743 × 10^-11 m^3 kg^-1 s^-2 * M / 5 × 10^6 m + (5.0 km/s)^2)

where M is the mass of Planet Zoroaster.

Without knowing the mass of Planet Zoroaster, we cannot determine the exact value of v2. However, we can use the given escape speed to find the mass of the planet:

escape speed = sqrt(2 * G * M / R)

=> M = R * escape speed^2 / (2 * G)

Plugging in the given values, we get:

M = 5 × 10^6 m * (12.0 km/s)^2 / (2 * 6.6743 × 10^-11 m^3 kg^-1 s^-2) = 3.599 × 10^25 kg

Now we can calculate the final speed of the meteor:

v2 = sqrt(2 * 6.6743 × 10^-11 m^3 kg^-1 s^-2 * 3.599 × 10^25 kg / 5 × 10^6 m + (5.0 km/s)^2) ≈ 12.032 km/s

Therefore, the meteor will be moving at a speed of approximately 12.032 km/s when it hits the surface of Planet Zoroaster.

In Part I, the independent variable, the one that is intentionally manipulated, is . In Part II, the independent variable changes to . The dependent variable, the one you measure the response in, is the same for Parts I and II. For both parts of the lab, the dependent variable is .

Answers

Kinetic energy is the dependent variable for both Parts, while mass is the independent variable for Part I and speed is the independent variable for Part II.

The independent variable is what, exactly?

The variable in an experiment that is not altered by the experimental process is known as the independent variable. In contrast, the variable we must quantify and which is altered by the experimental circumstances is the dependent variable.

Is the experiment's manipulating the independent variable?

An experiment's manipulated variable, also referred to as an independent variable, is a component that you can alter to observe how other factors react. The three categories of factors in an experiment are as follows: Variable that has been altered and controlled based on the trial.

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Question:

In Part I, the independent variable, the one that is intentionally manipulated, is

In Part II, the independent variable changes to

The dependent variable, the one you measure the response in, is the same for Parts I and II. For both parts of the

lab, the dependent variable is

easy physics. HELP
if a cart is carrying 100kg of mass, at what rate will the cart accelerate if 200 N of force are applied? (a=F/m)

a. 20m/s *2
b. 50 m/s *2
c. 2m/s *2
d. 100 m/s *2

Answers

Answer:

c) 2 m/s *2

Step-by-step explanation:

Using the formula a = F/m, where F is the force applied and m is the mass of the cart:

a = F/m = 200 N / 100 kg = 2 m/s^2

Therefore, the cart will accelerate at a rate of 2 m/s^2.

Acceleration can be described as...
A. Change in velocity over time.
B. Change in position over time.

Answers

Can be described as A.

Answer:

A

Explanation:

because it is defined as the rate of change of velocity.

A proton (+1.6 × 10−19 C) moves 20 cm on a path in the direction of a uniform electric field of strength 2.4 N/C. How much work is done on the proton by the electrical field?

Answers

W=Fd

F=Eq

W=Eqd

E = 2.4 N/C

q = 1.6×10⁻¹⁹ C

d = 20 cm = 0.2 m

W = 2.4(1.6×10⁻¹⁹)(0.2)

W = 7.68×10⁻²⁰ J

a small metallic sphere has a net charge q1= -2.5 μ c it is held in a stationary position by means of insulating supports. a second small metallic sphere with a net charge q2= -7.8 μ c and a mass of 1.5 g, when the 2 spheres are at a distance of 0.8 m from each other, q2 moves towards q1 with a speed of 22 m/ s. A ¿What is the speed of q2 when the spheres are 0.4 m apart? B. ¿How close does q1 get to q2?​

Answers

The speed of q2 when the spheres are 0.4 m apart would be 238.9 m/s.The distance between q1 and q2 could be 0.109 m.

Conservation of energy problem

Let's first find the initial electrostatic potential energy of the system. The electrostatic potential energy between two point charges q1 and q2 separated by a distance r is given by:

U = k * q1 * q2 / r

where k is the Coulomb constant, which has a value of approximately 9 x 10^9 N m^2/C^2.

Substituting the given values, we get:

U = (9 x 10^9 N m^2/C^2) * (-2.5 x 10^-6 C) * (-7.8 x 10^-6 C) / 0.8 mU = 2.284 J

At a separation of 0.4 m, the electrostatic potential energy of the system is:

U' = (9 x 10^9 N m^2/C^2) * (-2.5 x 10^-6 C) * (-7.8 x 10^-6 C) / 0.4 m

U' = 9.136 J

The change in potential energy is therefore:

ΔU = U' - U = 6.852 J

This change in potential energy is equal to the kinetic energy of sphere q2, which we can calculate using:

KE = (1/2) * m * v^2

where m is the mass of sphere q2 and v is its velocity.

To find the speed of q2 when the spheres are 0.4 m apart, we can rearrange the above equation and substitute the known values:

v = sqrt(2 * KE / m) = sqrt(2 * ΔU / m) = sqrt(2 * 6.852 J / 0.0015 kg) = 238.9 m/s

Therefore, the speed of sphere q2 when the spheres are 0.4 m apart is approximately 238.9 m/s.

To find how close q1 gets to q2, we can use the conservation of energy principle again. At the closest point of approach, all of the initial potential energy has been converted into kinetic energy, so we can equate the two:

(1/2) * m * v^2 = U

Solving for the separation r, we get:

r = k * q1 * q2 / (2 * KE)

Substituting the known values, we get:

r = (9 x 10^9 N m^2/C^2) * (-2.5 x 10^-6 C) * (-7.8 x 10^-6 C) / (2 * 2.284 J)

r = 0.109 m

Therefore, the closest separation between q1 and q2 is approximately 0.109 m.

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(a)
A copper block with a mass of 1.6 kg initially slides over a rough horizontal surface with a speed of 6.6 m/s. Friction slows the block to rest. While slowing to rest, 85.0% of the kinetic energy of the block is absorbed by the block itself as internal energy. What is the temperature increase of the block? (Enter your answer in degrees Celsius.)
°C
(b)
What happens to the remaining energy?
It becomes chemical energy.
It is absorbed by the horizontal surface on which the block slides.
It is so minute that it doesn't factor into the equation.
It vanishes from the universe.

Answers

Hey, Misha! I see you're working on a physics problem for college. I'd be happy to help you out!

(a) To solve for the temperature increase of the copper block, we can use the equation:

ΔE = mcΔT

Where ΔE is the change in internal energy of the block, m is the mass of the block, c is the specific heat capacity of copper, and ΔT is the change in temperature.

First, we need to find the initial kinetic energy of the block:

KE = 1/2mv^2 = 1/2(1.6 kg)(6.6 m/s)^2 = 35.1 J

Next, we need to find the internal energy absorbed by the block:

ΔE = 0.85(KE) = 0.85(35.1 J) = 29.8 J

Finally, we can solve for ΔT:

ΔT = ΔE/(mc) = (29.8 J)/(1.6 kg)(0.385 J/kg°C) ≈ 47°C

Therefore, the temperature increase of the copper block is approximately 47°C.

(b) The remaining energy is converted into thermal energy and dissipated into the surroundings as heat. It does not vanish from the universe, but rather it is dispersed into the environment.

when light strikes a green opaque object the green wavelength of light is .....

1. reflected
2. absorbed
3. transmitted




while all other wavelengths of visible light are......

1. reflected
2. absorbed
3. transmitted

Answers

When light strikes a green opaque object, the green wavelength of light is absorbed, while all other wavelengths of visible light are reflected.

Why does a green object appear green to our eyes?

A green object appears green to our eyes because it selectively absorbs all wavelengths of visible light except for the green wavelength of light, which is reflected back to our eyes.

When light strikes a green opaque object, the green wavelength of light is absorbed by the object. This means that the green light is not reflected or transmitted, but rather it is absorbed by the object.

What happens to the energy of the absorbed light when it is absorbed by an opaque object?

When light is absorbed by an opaque object, the energy of the absorbed light is converted into other forms of energy, such as thermal energy. This is because the absorbed light energy causes the atoms in the object to vibrate, which in turn generates heat energy.

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Microwaves travel at the speed of light 3.00x108 m/s. When the frequency of microwaves is 9.00x109Hz, what is their wavelength?​

Answers

The wavelength of microwaves with a frequency of 9.00x10^9 Hz is approximately 0.0333 meters or 33.3 millimeters.

What is the speed of microwaves and what is their frequency?

Microwaves are a type of electromagnetic radiation with a frequency range of around 300 MHz to 300 GHz. They are used for various applications, including communication, cooking, and medical imaging.

The speed of microwaves is the same as the speed of light, which is approximately 3.00x10^8 m/s. This means that microwaves travel at a very high speed and can cover long distances in a short amount of time.

How does the wavelength of microwaves change with their frequency?

The wavelength of microwaves is inversely proportional to their frequency. This means that as the frequency of microwaves increases, their wavelength decreases.

In other words, higher frequency microwaves have shorter wavelengths, and vice versa. For example, the wavelength of microwaves with a frequency of 9.00x10^9 Hz is approximately 0.0333 meters or 33.3 millimeters.

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Two iron nails hang from a bar magnet. Which diagram shows the magnetic poles induced in the nails?​

Answers

Answer:

option A

Explanation:similar pole repel each other and opposite pole attract each other.other options (exceptA) shows that similar pole is attracting ,it is not podsible.

The thermal conductivities of human tissues vary greatly. Fat and skin have conductivities of about 0.20 W/m · K and 0.020 W/m · K respectively, while other tissues inside the body have conductivities of about 0.50 W/m · K. Assume that between the core region of the body and the skin surface lies a skin layer of 1.0 mm, fat layer of 0.50 cm, and 3.2 cm of other tissues.
(a) Find the R-factor for each of these layers, and the equivalent R-factor for all layers taken together, retaining two digits.
Rskin
m2 · K/W
Rfat
m2 · K/W
Rtissue
m2 · K/W
R
m2 · K/W

(b) Find the rate of energy loss when the core temperature is 37°C and the exterior temperature is 0°C. Assume that both a protective layer of clothing and an insulating layer of unmoving air are absent, and a body area of 2.0 m2.

Answers

Rounded to two digits, the equivalent R-factor for all layers is 0.36 m2·K/W. Rounded to two digits, the rate of energy loss is approximately 219 W.

Why do people use thermal conductivity?

A crucial factor is thermal conductivity because it determines temperature gradients both during material development and inside of devices.

(a) The formula for the R-factor is:

R = thickness / thermal conductivity

For the skin layer:

Rskin = 0.001 m / 0.020 W/m·K = 0.05 m2·K/W

For the fat layer:

Rfat = 0.050 m / 0.20 W/m·K = 0.25 m2·K/W

For the other tissues:

Rtissue = 0.032 m / 0.50 W/m·K = 0.064 m2·K/W

To find the equivalent R-factor for all layers taken together, we need to add the individual R-factors together:

R = Rskin + Rfat + Rtissue

R = 0.05 m2·K/W + 0.25 m2·K/W + 0.064 m2·K/W

R = 0.364 m2·K/W

Rounded to two digits, the equivalent R-factor for all layers is 0.36 m2·K/W.

(b) The formula for the rate of energy loss is:

P = A * (Tcore - Texterior) / R

Converting the temperatures to kelvins:

Tcore = 37 + 273.15 = 310.15 K

Texterior = 0 + 273.15 = 273.15 K

Substituting the given values:

P = 2.0 m2 * (310.15 K - 273.15 K) / 0.36 m2·K/W

P = 219.44 W

Rounded to two digits, the rate of energy loss is approximately 219 W.

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An object of mass 0.35kg oscillates in SHM with an amplitude of 140mm
and a frequency of 0.60Hz.
Calculate or find:
i) Maximum kinetic energy of the object.
ii) Maximum potential energy of the object.
iii) Potential and kinetic energy at the mid-way point between the centre and the
extremity of the motion.

Answers

Answer:

The equation for the total mechanical energy of an object in SHM is:

E = 1/2 kA^2

where E is the total mechanical energy, k is the spring constant, and A is the amplitude.

To solve the problem, we need to find the spring constant of the oscillator:

f = 1/T

where f is the frequency and T is the period.

T = 1/f = 1/0.60 = 1.67 s

The angular frequency of the oscillator is:

ω = 2πf = 2π/T = 3.76 rad/s

The spring constant of the oscillator is:

k = mω^2 = 0.35 x (3.76)^2 = 4.97 N/m

i) The maximum kinetic energy of the object is equal to the maximum potential energy, which is:

Emax = 1/2 kA^2 = 1/2 x 4.97 x (0.14)^2 = 0.012 J

ii) The maximum potential energy of the object is the same as the maximum kinetic energy, which is:

Emax = 1/2 kA^2 = 1/2 x 4.97 x (0.14)^2 = 0.012 J

iii) At the mid-way point between the centre and the extremity of the motion, the displacement of the oscillator is half the amplitude, which is 70 mm or 0.07 m. At this point, the kinetic energy is zero, and the potential energy is:

E = 1/2 kx^2 = 1/2 x 4.97 x (0.07)^2 = 0.012 J

Therefore, the total mechanical energy at this point is also 0.012 J.

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