The visible portion of the electromagnetic spectrum occurs for wavelengths between.

Answer:

The farmer is moving at 10 m / 40 s = .4 m/s

2 min 20 sec = 140 sec

The farmer will have moved 140 * .4 = 56 m

To cover the x-axis twice he will have moved x = 10

56 - 50 = 6 m   and he will have moved a distance of 6 on the y-axis

S = (10^2 + 6^2)^1/2 = 11.7 m from the origin

Answer:

[tex]{ \rm{power, \: p = current \times p.d}} \\ { \rm{50 = 4.5 \times (current \times resistance)}} \\ { \rm{50 = 4.5 \times (4.5 \times r)}} \\ { \rm{resistance = \frac{50}{ {4.5}^{2} } }} \\ \\ { \rm{resistance = 2.5 \: ohms}}[/tex]

When the syringe is compressed the pressure spike to a value more than twice its initial value because the temperature of the gas increases.

Pressure is the force applied perpendicularly to an object's surface in relation to the area across which it is applied (symbol: p or P). Gauge pressure is the pressure when compared to the ambient pressure.

Various units are used to express pressure. The pound-force per square inch (psi), which is equal to one newton per square metre (N/m2) in the SI, is the standard unit of pressure in the imperial and U.S. customary systems. Some of these results derive from dividing a unit of force by a unit of area. The atmosphere, which is equal to this pressure, and the torr, which is defined as 1760 of this, are two other ways to describe pressure.

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Answer:

gogle it

Explanation:

i cant help

An astronaut could 6 times higher on the moon than on earth if his takeoff speed is the same in both locations.

From lunar circle, space explorers guided cameras through the window of their space apparatus toward catch photographs of the moon's surface. The nearest look we've had at the moon came from the sendoff of NASA's Apollo program during the 1960s. Somewhere in the range of 1967 and 1972, a progression of missions handled the primary men on the moon. At the point when the following space explorer to arrive at the moon strolls on the lunar surface in 2024, she'll confront radiation levels multiple times higher than on Earth. While Apollo mission space travelers. The prompt surface was extremely fine, as best we could see peering down from 15 feet. Off somewhere out there was an exceptionally clear skyline, perhaps with a rock.

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Option(c) is Kilometer is correct. Kilometer unit for length is greater than a meter.

Worldwide adoption of the International System of Units (SI) as a system of measurement units is widespread. For convenience, this updated version of the metric system is based on the number 10. The metric prefixes or the SI prefixes are a collection of prefixes that have been established. According on the prefixes, the unit is either a multiple or a fraction of base ten. It enables the zeros of very small or very high numbers to be reduced, for example, from 7,500,000 Joules to 1 nanometer and from 0.000000001 meter to 7.5 Megajoules, respectively. Additionally, some SI prefixes have a group of symbols before the unit symbol.

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The acceleration of Mr. Henderson Porsche is 54545.45 mi/hr².

Velocity is the directional speed of an object, body, or particle in motion as an indication of its rate of change in position as observed from a particular frame of reference and as measured by a particular standard of time.

We know that acceleration of a body is Change in Velocity per unit time.

[tex]a=\frac{v-u}{t}[/tex]

Here ,

v is the Final Velocity,

u is the Initial Velocity and;

t is the time taken.

In the given question ,

The Porsche accelerated from 0 to 60 miles per hour in four seconds.

u = 0 mi/hr

v = 60 mi/hr

t = 4 sec

[tex]t=\frac{4}{3600}{~}hr\\[/tex]

[tex]t=\frac{1}{900} {~}hr[/tex]

t  = 0.0011 hr

Then,

[tex]a=\frac{v-u}{t}[/tex]

[tex]a=\frac{60-0}{0.0011}[/tex]

[tex]a=54545.45 {~}mi/hr^{2}[/tex]

Hence, the acceleration of the Porsche is 54545.45 mi/hr².

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The answers for question numbers 3 and 4 is correct. For question number 9, the average speed is 2.5 mi /hr and for question number 10 the speed is 5 yards / s.

We know that,

v = d / t

where,

v = Speed

d = Distance

t = Time

1 ) v = 360 / 6 = 60 km / h

2 ) v = 120 / 3 = 40 mi / h

3 ) v = 18 / 6 = 3 m / s

4 ) v = 1000 / 20 = 50 m / min

5 ) t = 6pm - 5 pm = 1 hr

    v = 2.5 / 1 = 2.5 mi / hr

6 ) v = 1.5 / 0.33 = 4.5 mi / hr ( Since 20 min = 20 / 60 = 0.33 hr )

   d = 4.5 * 1 = 4.5 mi

7 ) d = 20 * 1 = 20 mi ( Since 60 min = 1 hr )

8 ) d = 60 * 2 = 120 mi

9 ) Distance per lap = 0.5 mi

     Total laps = 10

     Total distance = 10 * 0.5 = 5 mi

     v = 5 / 2 = 2.5 mi / hr

10 ) v = 100 / 20 = 5 yards / s

Therefore, the answer for:

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The Big Bang Theory is an example of the universe originated from the explosion and expansion of all matter and energy

The Big Bang Theory is the way that astronomers explain how the universe began as a tiny, dense, fireball that exploded 13.8 billion years ago. It includes Albert Einstein's general theory of relativity along with standard theories of fundamental particles.

The theory was born of the statement that other galaxies are moving away from our own at countless speed in all directions, as if they had all been forced by an ancient explosive force.

The solar system is the gravitationally bound system of the sun, the objects that orbit it or travel around the sun. It consists of an average star (sun), the planets (Mercury, Venus, Earth, Mars, Jupiter, Saturn, Uranus, Neptune and Pluto) and their moons, dwarf planets and countless asteroids, comets, and other small icy objects.

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The car's acceleration would be for a mass of 4 kg is 3 m/s²

Given the information below:

12 Newtons of force.

Increasing speed Equals

By using Newton's Second Law of Motion, we can determine the mass of the car:

According to Newton's second law of motion, force is equal to the rate at which momentum changes. Force is defined as mass times acceleration for a constant mass.

force= mass* acceleration

This formula provides Newton's Second Law of Motion mathematically;

12= mass*3

mass=12/ 3

mass=4 kilogram

By changing the parameters in the formula, we obtain;

= 4 kilos of mass.

As a result, the vehicle with a mass of 4 kg would accelerate at 3 m/s².

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The work done by the spring force while it is compressed to the length of 6m is 0.58 J.

The definition of work done includes both the forces applied to the body and the total displacement of the body. When we apply force "F" to a block, the body moves with some acceleration, or, additionally, its speed increases or decreases depending on the direction of the force. The system's kinetic energy changes as speed increases or decreases. Since we are aware that energy cannot be created or destroyed, it must be changed into another form. This perspective refers to it as completed work. When negative energy is finished, the energy declines, and when positive energy is finished, the energy rises.

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The thing that happens when you push or pull a water tube is that the force makes the tube to speed up, slow down, or be in one place.

It is very hard when pulling  water and if a person need to consistently pull air out of the tube before water makes it to the pump, it is very  difficult and some pumps may or cannot do it.

In regards to pushing, when a person is pushing above the Center of Gravity, one is pushing the said object into the ground and thus is increasing the rate of friction.

If a person is pulling you are said to be pulling above the Center of Gravity and lowering friction.

So, The thing that happens when you push or pull a water tube is that the force makes the tube to speed up, slow down, or be in one place.

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The car’s velocity when it passed the checkpoint was of: 60 m/s

The formula and procedure we will use to solve this exercise is:

vi = [(2 *x)/ t] - vf

Where:

Information about the problem:

Applying the initial velocity formula we have:

vi = [(2 *x)/ t] - vf

vi = [(2 * 225m)/5 s ] - 30 m/s

vi = [(450 m)/5 s ] - 30 m/s

vi = 90 m/s - 30 m/s

vi = 60 m/s

It is a physical quantity that indicates the displacement of a mobile per unit of time, it is expressed in units of distance per time, for example (miles/h, km/h).

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Answer:

48.9 m/s

Explanation:

I am ignoring air resistance in this:

horizontal velocity stays constant, so we need to find vertical velocity

v_0y = 0 m/s, t = 1.23s, a = 9.81 m/s^2 <- (g)

v_y = v_0y + at = 0 + 9.81 * 1.23 = 12.066 m/s

the magnitude v = sqrt(v_x ^ 2 + v_y ^2) = sqrt(47.4^2 + 12.066^2) = 48.9m/s

Answer:

V = 49.2 m/s

Explanation:

Given:

Vₓ  = 47.4 m/s

t = 1.23 s

___________

V - ?

Vy =  Voy + g·t = 0 + 9.8·1.23  ≈ 12,1 m/s

V = √( (Vx)² + (Vy)² ) = √ ( 47,7² + 12,1²) = 49.2 m/s

The block's third dimension is 2.77 cm.

The mass of the osmium block is 1 kg.

The dimensions of the rectangular osmium block are 4 cm × 4 cm.

Density is defined as the mass of the object by the volume of the object.

density = mass / volume

Now, the volume of a rectangle is:

Volume = length × breadth × height

And now let's assume that the block's height is y.

Then,

Volume = length × breadth × height

Volume = 4 cm × 4 cm × y cm

Osmium now has a density of ρ = 22.57 g/cm³.

Therefore, the density of the rectangular osmium will be:

density = mass / volume

[tex]\rho = \frac{1{~}kg}{4{~}cm {~}\times {~}4{~}cm {~}\times {~} y}[/tex]

[tex]22.57 {~}g/cm^{3}=\frac{1000{~}g}{4{~}cm \times 4{~}cm \times y}[/tex]

[tex]16{{~}cm^{2} \times y{~} =\frac{1000{~}g}{22.57{~} g/cm^{3}}[/tex]

16y cm² = 44.31 cm³

Dividing each side of the equation by 16 cm²,

( 16y cm² ÷ 16 cm²) = ( 44.31 cm³ ÷ 16 cm² )

y = 2.77 cm

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The complete question is mentioned below:

The density of osmium is 22.57 g/cm³. If a 1.00kg rectangular block of osmium has two dimensions of 4.00 cm x 4.00 cm, calculate the third dimension of the block.

When an emission line of helium from a galaxy is shifted toward red, it means that the galaxy is moving away from the viewer and is denoted as option D.

This refers to the systems of stars and interstellar matter that make up the universe which are gravitationally bound together and are made up of four main components which are:

When an emission line of helium from a galaxy is shifted toward red, there is a displacement to a longer wavelengths and this alteration is referred to as a doppler shift.

Moving away from the view means that the galaxy will be viewed in the red spectrum of the visible light which is a part of the electromagnetic spectrum.

This is therefore the reason why option D was chosen as the most appropriate choice.

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The magnitude of the electric field is 0 N/C.

We choose a Gaussian sphere of radius r > R concentric with the solid sphere.

Apply Gauss's law from Equation (1):

[tex]\Phi_{E} = \oint E\cdot dA = \frac{q_{in}}{\in_{0} }[/tex]

Since the electric field vector E is parallel to the area vector dA of the curved part the dot product can be the simple product EdA:

                                 [tex]\oint E\cdot dA = \frac{q_{in}}{\in_{0} }[/tex]

since the electric field is constant we can put E out of the integral:

                                [tex]E \oint dA = \frac{q_{in}}{\in_{0} }[/tex]

                                    [tex]EA = \frac{q_{in}}{\in_{0} }[/tex]

Substitute for A from Equation (2) and for [tex]q_{in}[/tex] from Equation (3)

recognizing that the enclosed charge by the Gaussian sphere is less than Q:

                              [tex]E(4\pi r^{2}) = \frac{pVgaussian}{\in_{0} }[/tex]

Substitute for V gaussian (volume of the Gaussian surface) from Equation (4):

                                          [tex]E(4\pi r^{2}) = \frac{\rho (\frac{4}{3} \pie r^{2})} {\in_{0} }[/tex]

                                           [tex]E = \frac{\rho r}{3\in_{0} }[/tex]

Where p is the charge per unit volume of the solid sphere which is equal to the total charge Q of the sphere divided by the volume V of the sphere :

                             [tex]E = \frac{(\frac{Q}{V} )r}{3\in_{0} }[/tex]

                                [tex]E = \frac{Qr}{(\frac{4}{3} \pi R^{3})3\in_{0} }[/tex]

                                [tex]E = \frac{Qr}{4\pi \in_{0}R^{3} }[/tex]

Substitute numerical values:

                                                  [tex]E = \frac{(26\times10^{-6})(0)}{4\pi(8.8542 \times10^{-12})(0.4)^{3}}[/tex]

                                 [tex]E = 0 N/C[/tex]

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The horizontal component of the magnetic field is 12.6 μT.

The magnetic influence on moving electric currents, electric charges, and magnetic materials is described by a magnetic field, which is a vector field. When a charge moves through a magnetic field, a force that is perpendicular to both its own velocity and the magnetic field operates on it.

The horizontal component of the Earth's magnetic field is perpendicular to the axis of a circular coil with five turns and a diameter of D = 30.0 cm that is vertically orientated.

A coil current of I = 0.600 A causes a horizontal compass to deflect 45.0° from magnetic north when it is positioned in the coil's center.

Let B be the magnetic field and R be the radius of the circular coil.

Then the horizontal component of the Earth's magnetic field is given as:

[tex]B(h) = B(coil) = \frac{\mu_{0} NI}{2R}[/tex]

[tex]B(h) =\frac{4 \pi \times 10^{-7}\times(5)\times(0.6)}{0.3}[/tex]

B(h) = 12.6 μT

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Answer:

•So he stays healthy.

•He has energy and time to do it.

A balloon clings to a wall after it is negatively charged by rubbing because the wall is positively charged.

Yes, opposite charges attract each other. When a positive charge and a negative charge interact with each other, their forces act from the direction of positive to the direction of negative charge. As a result opposite charges attract each other while on the other hand, similar charges repel each other because their forces move in the opposite direction so they repel each other.

So we can conclude that a balloon clings to a wall after it is negatively charged by rubbing because the wall is positively charged.

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The resultant force is 10.77 pounds.

We need to know about vector to solve this problem. Force is included to vector because it has magnitude and also direction. The resultant force can be determined using cosine law. It can be written as

F² = F1² + F2² - 2F1 . F2 cos(θ⁰)

From the question above, we know that:

F1 = 10 pounds

F2 = 14 pounds

θ = 50°

By substituting the parameter, we get

F² = F1² + F2² - 2F1 . F2 cos(θ⁰)

F² = 10² + 14² - 2. 10 . 14 cos(50⁰)

F² = 100 + 196 - 280 . 0.64

F² = 100 + 196 - 280 . 0.64

F² = 116.02

F = 10.77 pounds

Hence, the resultant force is 10.77 pounds.

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The work required to pump all the water over the top of the tank of radius 5 m and depth 2 m is 5.12 x 10^6 J.

The work required to pump the water over the top of the tank is equal to the gravitational potential energy of the water inside the tank.

The potential energy is given by the following equation,

U= mgh

Here, m is the mass of water present inside the tank.

We know that, Density( Mass/Volume

Hence, m= Density x volume

So, the equation becomes

U= ρ x V x g x h

Where V is the volume of hemispherical tank= 2/3π x r^3

Putting the given values in the above equation,

U= 1000 x 2/3 x 3.14 x 53 x 9.8 x 2 = 5.12 x 10^6 J

Hence, the work required to pump all the water over the top of a tank of radius 5 m and a depth of 2 m is 5.12 x 10^6 J.

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The number of molecules in the given volume is 5.33 x 10⁻¹² moles.

The number of molecules of the gas is calculated as follows;

PV = nRT

n = PV/RT

where;

1 torr = 133 Pa

1 x 10⁻¹⁰ torr =  ?

= 1.33 x 10⁻⁸ Pa

n = (1.33 x 10⁻⁸ x 1) / (8.314 x 300)

n = 5.33 x 10⁻¹² moles

Thus, the number of molecules in the given volume is 5.33 x 10⁻¹² moles.

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Answer: The diaphragm

Explanation:

The diaphragm is a thin dome-shaped muscle which separates the thoracic cavity (lungs and heart) from the abdominal cavity (intestines, stomach, liver, etc.)

The speed of the wave in the rod is

[tex]The speed of the wave in the rod is $v = \sqrt{\frac{Y}{\rho}} = \sqrt{\frac{20\times 10^{10}\ \mathrm{N/m^2}}{7.86\times 10^3\ \mathrm{kg/m^3}}} = 5044\ \mathrm{m/s}$[/tex]

5044m/s

A sound wave is the pattern of disruption brought on by the movement of energy moving through a medium as it propagates away from the source of the sound (such as air, water, or any other liquid or solid substance). Pressure waves are produced when an item vibrates, such as a ringing phone, and these waves are known as sound waves. The surrounding medium's particles are disturbed by the pressure wave, and those particles disrupt the particles next to them, and so on. Like ocean waves, the disturbance's pattern causes outward movement in all directions. Usually in all directions and with decreased intensity as it gets further away from the source, the wave transmits the sound energy through the medium.

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Answer:

8.33 m/s^2

Explanation:

The equation for force is Force = mass * acceleration. The force is 500 N, and the mass is 60 kg, so substituting those into the equation you get that 500 = 60 * acceleration. Divide 500 by 60 to get the acceleration, which is 8.33 m/s^2.

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The average velocity or displacement of a particle for the first time interval is Δs / Δt = 6 cm/s.

Solution:

As we know that displacement is calculated in centimeters and the unit of time is second.

The average velocity for the first interval [1,2] is given

Δs / Δt = s (t2) - s (t) / t2 - t1

Δs / Δt = 2sin2  π  + 3cos 2 π -  ( 2sin π + 3cos π ) / 2 - 1

Δs / Δt = 2(0) + 3(1) - 2(0) - 3 (-1) / 1

Δs / Δt = 6 cm/s

Thus the average velocity or displacement of a particle for the first time interval is Δs / Δt = 6 cm/s

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The complete question is:

The displacement of a particle moving back and forth along a line is given by the following equation s(t) = 2sin π t + 3cos π t. Estimate the instantaneous velocity of the particle when t = 1

The dot product of velocity vector and acceleration vector at maximum height is equal to zero.

[tex]$\overrightarrow{v}.\overrightarrow{a}= 0[/tex]

We have a football player who kicked a football in order to do a goal.

We have to determine the relationship between the direction of the velocity and acceleration vectors when the football reaches its maximum height.

Projectile motion is the two - dimensional motion of an object thrown or projected into the air such that it moves under the influence of acceleration of gravity.

According to the question -

Assume that the velocity with which the projectile was fired is u m/s.

Therefore, it can be resolved into two components as -

[tex]\overrightarrow{u} =[/tex] u(x) + u(y) = u cosθ [tex]a_{x}[/tex] + u sinθ [tex]a_{y}[/tex].

Now, when the projectile reaches the maximum height, it will no longer cover any vertical height but it will keep moving in the horizontal direction. Therefore - the vertical component of the velocity will become zero. Therefore -

[tex]\overrightarrow{u} =[/tex] u(x) + u(y) = u cosθ [tex]a_{x}[/tex] + 0 = u cosθ [tex]a_{x}[/tex]

Now, refer to the figure attached for reference -

At the maximum height, the velocity vector is in the horizontal direction and the vector for the acceleration due to gravity is vertically downwards.

Therefore - the acceleration vector and velocity vector will have an angle of 90° between them. Assume -

Velocity at maximum height = v = u cosθ [tex]a_{x}[/tex]

and acceleration = a = g = 9.8 m/[tex]s^{2}[/tex]

Therefore -

The dot product of velocity vector and acceleration vector is equal to zero.

[tex]\overrightarrow{v}.\overrightarrow{a}= 0[/tex]

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The relativistic time it takes the proton to travel the milky way is 9.877 × 10⁻⁷ s

Since the question has to do with relativistic motion, we need to find the relativistic time it takes the proton to travel the milky way. First, we use the equation for relativistic kinetic energy, we have

K = mc(γ - 1) where

So, making γ subject of the formula, we have

γ  = K/mc + 1

Substituting the values of the variables into the equation, we have

γ  = K/mc + 1

γ  = 1.602J/(1.6726 × 10⁻²⁷ kg × 3 × 10⁸ m/s) + 1

γ  = 1.602J/(5.0178 × 10⁻¹⁹ J) + 1

γ  = 0.3193 × 10¹⁹ + 1

γ  = 3.193 × 10¹⁸

Now since the proper diameter of the milky way d = 10⁵ ly, since the length moved by the proton is going to be contracted, the actual distance moved by the proton is

d' = d/γ

= 10⁵ ly/3.193 × 10¹⁸

= 0.3132 × 10⁻⁻¹³ ly

= 0.3132 × 10⁻⁻¹³ × 9.46 × 10¹⁵ m

= 2.963 × 10² m

Next we need to find the velocity of the proton from

γ = 1/[√(1 - (v/c)²]

3.193 × 10¹⁸ = 1/[√(1 - (v/c)²]

√(1 - (v/c)² = 1/3.193 × 10¹⁸

√(1 - (v/c)² = 0.3132 × 10¹⁸

1 - (v/c)² = (0.3132 × 10⁻¹⁸)²

1 - (v/c)² = 0.0981 × 10⁻³⁶

(v/c)² = 1 - 0.0981 × 10⁻³⁶

(v/c)² = 1 - 0.0000000000000000000000000000000000981

(v/c)² = 0.9999999999999999999999999999999999019

(v/c)² ≅ 0.99999

(v/c) = √0.99999

v/c = 0.999994999987

v = 0.999994999987c

So, the time it takes the proton to travel the milky way is t = d/v

= 2.963 × 10² m/0.999994999987c

= 2.963 × 10² m/2.99998499996 × 10⁸ m/s

= 0.9877 × 10⁻⁶ s

= 9.877 × 10⁻⁷ s

So, the time it takes the proton to travel the milky way is 9.877 × 10⁻⁷ s

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