Answer:
The attraction is due to the induced charge.
Explanation:
When we approach a charged rod to a sheet, an induced load is produced in the sheet that is of the same magnitude as the rod of opposite sign, this is because the charges of different sign attract each other, this explains the initial attraction.
This induced load occurs if importing the plate load
The attraction is due to the induced charge.
In trial 1 of an experiment, a cart moves with a speed of vo on a frictionless, horizontal track and collides with another cart that is initially at rest. In trial 2, the setup is identical except the carts stick together during the collision. How does the speed of the two-cart system's center of mass change, if at all, during the collision in each trial
Answer:
1) elastic shock, the velocity of the center of mass does not change
2) inelastic shock, he velocity of the mass center change
Explanation:
The position of the center of mass of your system is defined by
[tex]x_{cm}[/tex] = [tex]\frac{1}{M} \sum x_i m_i[/tex]
in this case we have two bodies
x_{cm} = [tex]\frac{1}{M}[/tex] (x₁m₁ + x₂ m₂)
the velocity of the center of mass is
x_{cm} = dx_{cm} / dt = [tex]\frac{1}{M} ( m_1 \frac{dx_1}{dt} \ + m_2 \frac{dx_2}{dt} )[/tex]
x_{cm} = [tex]\frac{1}{M} ( m_1 v_1 + m_2 v_2 )[/tex]
where M is the total mass of the system.
Therefore to answer this question we have to find the velocity of the body after the collision.
Let's use momentum conservation, where the system is formed by the two bodies, so that the forces have been internal during the collision.
Let's solve each case separately.
2) inelastic shock
initial instant. Before the crash
p₀ = m₁ v₀ + 0
final instant. After the collision with the cars together
p_f = (m₁ + m₂) v
p₀ = p_f
m₁ v₀ = (m₁ + m₂) v
v = [tex]\frac{m_1}{m_1+m_2}[/tex] v₀
let's find the velocity of the center of mass
M = m₁ + m₂
initial.
[tex]v_{cm o}[/tex] = [tex]\frac{1}{m_1 +m_2}[/tex] (m₁ vo)
final
[tex]v_{cm f}[/tex] = [tex]\frac{1}{M} ( \frac{m_1}{m_1 + m_2} v_o )[/tex] ( v) = v
v_{cm f} = [tex]\frac{m_1}{M^2} v_o[/tex]
Let's find the ratio of the velocities of the center of mass
vcmf / vcmo = [tex]\frac{1}{M} = \frac{1}{m_1 +m_2}[/tex]
therefore the velocity of the mass center change
1) elastic shock
initial instant.
p₀ = m₁ v₀
final moment
p_f = m₁ v_{1f} + m₂ v_{2f}
p₀ = p_f
m₁ v₀ = m₁ v_{1f} + m₂ v_{2f}
m₁ (v₀ - v_{2f}) = m₂ v_{2f}
in this case the kinetic energy is conserved
K₀ = K_f
½ m₁ v₀² = ½ m₁ v_{1f}² + ½ m₂ v_{2f}²
m₁ (v₀² - v_{1f}²) = m₂ v_{2f}²
m₁ (v₀ + v_{1f}) (v₀ - v_{1f}) = m₂ v_{2f}
we write our system of equations
m₁ (v₀ - v_{1f}) = m₂ v_{2f} (1)
m₁ (v₀ - v_{1f}) (v₀ + v_{1f}) = m₂ v_{2f}²
we solve the system
v₀ + v_{1f} = v_{2f}
we substitute and look for the final speeds
v_{1f} = [tex]\frac{m_1 -m_2}{m1 +m2 } v_o[/tex]
v_{2f} = [tex]\frac{2 m_1}{m-1+m_2} vo[/tex]
now let's find the velocity of the center of mass
initial
[tex]v_{cm o}[/tex] = [tex]\frac{1}{M}[/tex] m₁ v₀
final
[tex]v_{cm f}[/tex] = [tex]\frac{1}{M}[/tex] (m₁ v_{1f} + m₂ v_{2f} )
v_{cm f} = [tex]\frac{1}{M}[/tex] [ [tex]m_1 \frac{m_2}{M}[/tex] + [tex]m_2 \frac{2 m_1}{M}[/tex] ] v₀
v_{cm f} = [tex]\frac{1}{M^2}[/tex] ( m₁² - m₁m₂ +2 m₁m₂) v₂
v_{cm f} = [tex]\frac{1}{M^2}[/tex] (m₁² + m₁ m₂) v₀
let's look for the relationship
v_{cm f} / v_{cm o} = [tex]\frac{1}{M}[/tex] M
v_{cm f} / v_{cm o} = 1
therefore the velocity of the center of mass does not change
we see in either case the velocity of the center of mass does not change.
Describe Kinetic Energy and Potential Energy (in your own words please!!)
Answer:
Energy stored in an object due to its position is Potential Energy. · Energy that a moving object has due to its motion is Kinetic Energy.
Explanation:
Connecting math to physics
Answer:
wat
Explanation:
A material through which electricity cannot flow is called:
a conductor
an insulator
an electrode
a wet cell
Answer:
el conductor
Explanation:
gracias por los puntitoss
Answer:
conductor
Explanation:
You are given three pieces of wire that have different shapes (dimensions). You connect each piece of wire separately to a battery. The first piece has a length L and cross-sectional area A. The second is twice as long as the first, but has the same thickness. The third is the same length as the first, but has twice the cross-sectional area. Rank the wires in order of which carries the most current (has the lowest resistance) when connected to batteries with the same voltage difference.
Rank the wires from most current (least resistance) to least current (most resistance).
a. Wire of Lenght L and area A
b. Wire of Lenght 2L and area A
c. Wire of Lenght L and area 2A
Answer:
The answer is below
Explanation:
The resistance of a wire is directly proportional to the length of the wire and inversely proportional to its area. The resistance (R) is given by:
[tex]R=\frac{\rho L}{A}\\\\where\ L=length \ of\ wire,A=cross\ sectional\ area, \rho=resistivity\ of\ wire.[/tex]
Let us assume that all the wires have the same resistivity.
a) Wire of Length L and area A
[tex]R_1=\frac{\rho L}{A}[/tex]
b) Wire of Length 2L and area A
[tex]R_2=\frac{\rho *2L}{A}=2R_1[/tex]
C) Wire of Length L and area 2A
[tex]R_3=\frac{\rho L}{2A}=\frac{1}{2}R_1[/tex]
Therefore the wire of least resistance is R3 and R2 has the highest resistivity.
R₃ < R₁ < R₂
Therefore, the ranking of the wires from most current (least resistance) to least current (most resistance) is:
R₃ < R₁ < R₂
Help me !!!
What is the velocity of a ball with
a momentum of -4.5 kg*m/s and a
mass of 0.45 kg?
Answer:−4.05
Explanation:
why would the bulb not light?
are you a dmbss? the bulb and wire must be connected to both end
what is the angle between 3i-2j-3k and the negative x axis
Answer:
Um its the vbuck card on the 3 thrid level
Explanation:
Bc its a vbuck card you know sihdg;aig
A carmaker has designed a car that can reach a maximum acceleration of 12 meters/second2. The car’s mass is 1,515 kilograms. Assuming the same engine is used, what should the car’s mass be if the carmaker wants to reach an acceleration of 15 meters/second2? Use F = ma.
A.
1,212 kg
B.
1,335 kg
C.
1,466 kg
D.
1,515 kg
E.
1,894 kg
Answer:
A: 1,212 kg
Explanation:
Do you believe you can create a Controlled experiment without an Observational Study? Why or Why not. Include scientific evidence to support your response. PLEASE HELP I BEG YOU.
Answer:
No, it is very unlikely to perform a controlled experiment, because you need to observe the amount or anything from something. Consider someone on the busy street of a New York neighborhood asking random people that pass by how many pets they have, then taking this data and using it to decide if there should be more pet food stores in that area.
A ball 12 m in 4 seconds and then 2.5 seconds later it rolls 8 m in 2 seconds what is its acceleration
Answer:
If it accelerates at 20 m/s2 for a period of 22 seconds, what is its final velocity? ... How fast is the ball falling after 5 seconds? v = v0 + gt v = 0 + 10(5) v = 50 m/s. 4. ... + ½ 2.5(15)2 x = 281 m. 5. What is the total displacement of the car in question 2? ... 8. A base jumper falls until he reaches a speed of 200 m/s
Explanation:
6. A 25 g sample of iron (initially at 800.00°C) is dropped into 200 g of water (initially at
30.00°C). The final temperature of the system is 40.22°C. Find the specific heat of iron.
90
Answer:
[tex]c=0.45\ J/g^{\circ} C[/tex]
Explanation:
Given that,
A 25 g sample of iron (initially at 800.00°C) is dropped into 200 g of water (initially at 30.00°C). The final temperature of the system is 40.22°C.
We need to find the specific heat of iron.
It can be calculated as:
Cooler water gains = hot metal loses
mc∆T = - mc∆T
Put all the values,
[tex]200g(4.184\ J/g^{\circ} C)(T_f-T_i) = -25g(c)(T_f-T_i) \\\\200g(4.184 )( 40.22-30.00) = -25\times (c)\times (40.22-800.00)\\\\8552.096 = 18994.5c\\\\c=\dfrac{8552.096 }{18994.5}\\\\c=0.45\ J/g^{\circ} C[/tex]
So, the specific heat of iron is [tex]0.45\ J/g^{\circ} C[/tex]
What is the value of the angle of inclination of the slide?
Answer and I will give you brainiliest
Answer:
63°
Explanation:
90-27 =63
I am not completely sure
how much heat energy is needed to raise the temperature of 2.0 kg of concrete from 10c to 30c
what is the difference between heat and temperature
A disk with radius R and uniform positive charge density s lies horizontally on a tabletop. A small plastic sphere with mass M and positive charge Q hovers motionless above the center of the disk, suspended by the Coulomb repulsion due to the charged disk.
Required:
a. What is the magnitude of the net upward force on the sphere as a function of the height z above the disk?
b. At what height h does the sphere hover?
Answer:
a. F = Qs/2ε₀[1 - z/√(z² + R²)] b. h = (1 - 2mgε₀/Qs)R/√[1 - (1 - 2mgε₀/Qs)²]
Explanation:
a. What is the magnitude of the net upward force on the sphere as a function of the height z above the disk?
The electric field due to a charged disk with surface charge density s and radius R at a distance z above the center of the disk is given by
E = s/2ε₀[1 - z/√(z² + R²)]
So, the net force on the small plastic sphere of mass M and charge Q is
F = QE
F = Qs/2ε₀[1 - z/√(z² + R²)]
b. At what height h does the sphere hover?
The sphere hovers at height z = h when the electric force equals the weight of the sphere.
So, F = mg
Qs/2ε₀[1 - z/√(z² + R²)] = mg
when z = h, we have
Qs/2ε₀[1 - h/√(h² + R²)] = mg
[1 - h/√(h² + R²)] = 2mgε₀/Qs
h/√(h² + R²) = 1 - 2mgε₀/Qs
squaring both sides, we have
[h/√(h² + R²)]² = (1 - 2mgε₀/Qs)²
h²/(h² + R²) = (1 - 2mgε₀/Qs)²
cross-multiplying, we have
h² = (1 - 2mgε₀/Qs)²(h² + R²)
expanding the bracket, we have
h² = (1 - 2mgε₀/Qs)²h² + (1 - 2mgε₀/Qs)²R²
collecting like terms, we have
h² - (1 - 2mgε₀/Qs)²h² = (1 - 2mgε₀/Qs)²R²
Factorizing, we have
[1 - (1 - 2mgε₀/Qs)²]h² = (1 - 2mgε₀/Qs)²R²
So, h² = (1 - 2mgε₀/Qs)²R²/[1 - (1 - 2mgε₀/Qs)²]
taking square-root of both sides, we have
√h² = √[(1 - 2mgε₀/Qs)²R²/[1 - (1 - 2mgε₀/Qs)²]]
h = (1 - 2mgε₀/Qs)R/√[1 - (1 - 2mgε₀/Qs)²]
The distance to the other planets in the Solar System from the Earth depends on where they are in their orbit. The closest that Mars gets to the Earth is 0.5 AU. At this time it is observed to have an angular diameter of 18 arcseconds. The furthest it gets from Earth is 2.5 AU. What would its angular diameter be for this observation
Answer:
3.6 arcsec
Explanation:
angular diameter = diameter / distance
diameter is constant
so angular diameter ∝ 1 / distance
angular diameter = k / distance
For first case ,
18 = k / .5
for second case let angular diameter be D .
D = k / 2.5
dividing ,
D / 18 = .5 / 2.5 = 1 / 5
D = 18 / 5 = 3.6 arcsec
3.6 arcsec is the answer .
An 80-kg firefighter slides down a fire pole. After 1.3 seconds of sliding, the firefighter is sliding at a velocity of 6.5 m/s, straight down the pole. Once this velocity is reached, the firefighter grips the pole so that the force of friction exerted by the firefighter's hands on the pole is equal to the force of gravity. At this point what is the downward acceleration of the firefighter
Answer:
a= 0
Explanation:
In the vertical direction, if the friction force (directed upward) is equal to the force of gravity (downward) this means that no net force is acting on the firefighter.According to Newton's 2nd Law, if no net force is present, the acceleration in this direction is just zero, as follows:[tex]F_{net} = m*a = 0 (1)[/tex]
⇒ a = 0
Cara is building a model of the solar system, which includes the Sun. She plans to include a written description to provide details about each piece in her model. In order for her model to be realistic, which of the following should she include in her representation of the Sun?
Answer:
she should write about how big is it and what the sun looks and how far away is it from earth.
Cara is building a model of the solar system, which includes the Sun. She plans to include a written description to provide details about each piece in her model. In order for her model to be realistic, which of the following should she include in her representation of the Sun?
She should show that sunspots can ve seen as white areas on the Sun's surface.
She should explain that the Sun is made up of gaseous layers that surround an iron core.
She should show that the Sun revolves around the planets, determining the length of the year.
She should explain that the Sun rotates, even though different parts rotate at different rates.
Answer:
She should explain that the Sun is made up of gaseous layers that surround an iron core.
A monk is sitting atop a mountain in complete rest in meditation. What is the kinetic Energy of the monk? (assume mass of 65 kg and the mountain's height was 1000 m)
Answer:
no kinetic energy
hope this helps! :-D
Explanation:
the monk is not moving
An SUV is accelerated from rest to a speed v in a time interval t. Neglecting air resistance effects and assuming the engine is operating at its maximum power rating when accelerating, determine the time interval for the SUV to accelerate from rest to a speed 2v.
a. 2t
b. 4t
c. 2.5t
d. 3t
Answer:
A. [tex]2\cdot t[/tex]
Explanation:
The SUV accelerates uniformly. In this case, speed ([tex]v[/tex]) is directly proportional to time ([tex]t[/tex]). That is:
[tex]v \propto t[/tex] (1)
[tex]v = k\cdot t[/tex]
Where [tex]k[/tex] is the proportionality constant.
Then, we eliminate this constant by creating the following relationship:
[tex]\frac{v_{1}}{t_{1}} = \frac{v_{2}}{t_{2}}[/tex] (2)
If we know that [tex]v_{1} = v[/tex], [tex]t_{1} = t[/tex] and [tex]v_{2} = 2\cdot v[/tex], then the time interval for the SUV to accelerate from rest to a speed [tex]2\cdot v[/tex] is:
[tex]\frac{v}{t} = \frac{2\cdot v}{t_{2} }[/tex]
[tex]t_{2} = 2\cdot t[/tex]
Hence, correct answer is A.
Define the following soil conservation technique. Make sure to include pro’s and con’s of this method.
(Terraces)
Answer: Terraces on moderate to steep irregular slopes pro- ... sure of infertile or toxic soils. ... Following are terms used to define distances mea- ... the soil in the entire field will be disturbed to con-.
Explanation:
1. A SUV along with 5 passengers has a mass of 3500 kg. It has a driving force of 2500 N directed along west on a perfectly horizontal road. The surface of the road exerts a resistance force of 500 N due east. At the same time a high wind is blowing a force of 500 N due east in the opposite direction of the car's drive force. Does the car has any acceleration
Answer:
The net acceleration of the SUV is 0.429 meters per square second due west.
Explanation:
Statement is incomplete. Description is presented below:
A SUV along with 5 passengers has a mass of 3500 kg. It has a driving force of 2500 N directed along west on a perfectly horizontal road. The surface of the road exerts a resistance force of 500 N due east. At the same time a high wind is blowing a force of 500 N due east in the opposite direction of the car's drive force. Does the car has any acceleration? If yes, then what are the magnitude and direction of the car's acceleration?
According to Newton's Laws of Motion, the SUV will accelerate if and only if net acceleration is different of zero. Let suppose as positive the direction of driving force ([tex]F[/tex]), measured in newtons:
[tex]\Sigma F = F - R -f = F_{net}[/tex] (1)
Where:
[tex]R[/tex] - Resistance force, measured in newtons.
[tex]f[/tex] - Wind force, measured in newtons.
[tex]F_{net}[/tex] - SUV net force, measured in newtons.
If we know that [tex]F = 2500\,N[/tex], [tex]R = 500\,N[/tex] and [tex]f = 500\,N[/tex], then net force experimented by the SUV is:
[tex]F_{net} = 2500\,N-500\,N-500\,N[/tex]
[tex]F_{net} = 1500\,N[/tex]
The car has acceleration.
By definition of force for systems with constant mass, we calculate the acceleration of the vehicle below:
[tex]a_{net} = \frac{F_{net}}{m}[/tex] (2)
Where [tex]m[/tex] is the mass of the SUV, measured in kilograms.
If we know that [tex]F_{net} = 1500\,N[/tex] and [tex]m = 3500\,kg[/tex], then the net acceleration of the car is:
[tex]a_{net} = \frac{1500\,N}{3500\,kg}[/tex]
[tex]a_{net} = 0.429\,\frac{m}{s^{2}}[/tex]
The net acceleration of the SUV is 0.429 meters per square second due west.
If the mass of the book is 50 sliding with acceleration 1.2 m/s ^ 2 then the friction force is
364N
185N
173N
73N
Answer and I will give you brainiliest
why no tempature can be lower than 0 kelvin
Answer:
At zero kelvin (minus 273 degrees Celsius) the particles stop moving and all disorder disappears. Thus, nothing can be colder than absolute zero on the Kelvin scale. Physicists have now created an atomic gas in the laboratory that nonetheless has negative Kelvin values.
Explanation:
what is the difference between mass and weight
Answer:
The mass of an object is a measure of the object's inertial property, or the amount of matter it contains. The weight of an object is a measure of the force exerted on the object by gravity, or the force needed to support it. The pull of gravity on the earth gives an object a downward acceleration of about 9.8 m/s2.
Answer:
Explanation:
The mass is essentially "how much stuff" is in an object. ... Weight: There is a gravitational interaction between objects that have mass. If you consider an object interacting with the Earth, this force is called the weight. The unit for weight is the Newton (same as for any other force).
Certain neutron stars (extremely dense stars) are believed to be rotating at about 500 rev/s. If such a star has a radius of 17 km, what must be its minimum mass so that material on its surface remains in place during the rapid rotation
Answer:
7.22 × 10²⁹ kg
Explanation:
For the material to be in place, the gravitational force on the material must equal the centripetal force on the material.
So, F = gravitational force = GMm/R² where M = mass of neutron star, m = mass of object and R = radius of neutron star = 17 km
The centripetal force F' = mRω² where R = radius of neutron star and ω = angular speed of neutron star
So, since F = F'
GMm/R² = mRω²
GM = R³ω²
M = R³ω²/G
Since ω = 500 rev/s = 500 × 2π rad/s = 1000π rad/s = 3141.6 rad/s = 3.142 × 10³ rad/s and r = 17 km = 17 × 10³ m and G = universal gravitational constant = 6.67 × 10⁻¹¹ Nm²/kg²
Substituting the values of the variables into M, we have
M = R³ω²/G
M = (17 × 10³ m)³(3.142 × 10³ rad/s)²/6.67 × 10⁻¹¹ Nm²/kg²
M = 4913 × 10⁹ m³ × 9.872 × 10⁶ rad²/s²/6.67 × 10⁻¹¹ Nm²/kg²
M = 48,501.942 × 10¹⁵ m³rad²/s² ÷ 6.67 × 10⁻¹¹ Nm²/kg²
M = 7217.66 × 10²⁶ kg
M = 7.21766 × 10²⁹ kg
M ≅ 7.22 × 10²⁹ kg
A Car is moving at a speed of 20 m/s. How Much Distance it will cover in 1 min? Express the answer in km.
Answer:
d=20m/sx60s=1200m=1200/1000Km=1.2km
Explanation:
HELP ! ILL MARK BRAINLIEST HELP ASAP
Answer:
A
Explanation:
In 5 minutes, they went 10 miles at both 2, 3, and 4 checkpoints. The bus then starts to speed up.
Hope this helps!