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3 years ago

Do electrons have more room to move along a thick or thin wire?

Physics

1 answer:

PolarNik [594]3 years ago

3 0

Answer:

Got it from g00gle: Similarly, electrons flow more easily through a thick wire than they flow through a thin wire of the same material. Resistance is greater in a longer wire because the charges travel farther. As temperature increases, a metal's resistance increases because electrons collide more often.

Explanation:

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A crate of mass 190 kg sits on a horizontal floor. The coefficient of static friction between the crate and the floor is 0.4, an

Oxana [17]

Answer:

Explanation:

Mass of 190kg

Coefficient of static friction is 0.4

Coefficient of kinetic friction 0.36

Horizontal force= 500N

Taking g=9.81m/s^2.

The weight of the body my

W=190×9.81=1863.91N

There is a normal acting on the body which is equal to the weight

N=W=1863.91N

Frictional force(fr) is acting on the body and it is opposite the horizontal force.

The minimum force to be overcome before the object can start to move is Fr = μsN

Fr= μsN. μs=0.4

Fr= 0.4×1863.91

Fr=745.56N.

Since the horizontal force (500N) is not up to the minimum force to make the object move, then the force of 500N the body is still at rest.

Then the frictional force at that time is equal to the horizontal force

Therefore

Functional force = 500N

b. Mass of asteroid is

M=2000kg

Asteroid velocity at a particular instant is,

U=(-1.30x10^4, 4.20x10^4, 0)m/s

Magnitude of U is

U=√(-1.30×10^4)^2 +(4.2×10^4)^2+0

U=√1.933E9

U=4.39×10^4m/s

Position of the asteroid from the centre of the earth is,

R= (6.00x10^6, 10.00x10^6, 0)m.

The magnitude of the radius is

R = √(6.00x10^6)^2+ (10.00x10^6)^2+ 0^2

R=√3.6E13+10E13+0

R=√13.6E13

R=1.17E7m

R^2=13.6E13m

The mass of the earth is

Me=5.97x10^24 kg

The momentum of the asteroid after time, t=1.5×10^3s

Given that G=6.67x10^-11Nm^2/kg^2

Momentum is

Mv-Mu=Ft

There the new momentum will be

Mv=Ft+Mu

Now we the to find the force the earth exert on the asteroid by using

F=GMMe/R^2

F=6.67E-11 ×2000× 5.97E24 /13.6E13

F=7.964E17/13.6E13

F=5855.88N

The new momentum

Mv= Mu+Ft

Mv= 2000(4.39E4)+5855.88(1.5E3)

Mv=9.66E7kgm/s

The new momentum is 9.66×10^7 Kgm/s

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3 years ago

A capacitor is made by taking two sheets of aluminum foil, each 0.022 mm thick and placing between them a sheet of paper which c

olasank [31]

The answer is 18 nf in this equation

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3 years ago

a bowling ball is pushed with a force of 22.0 n and acclerates at 5.5m/s2. what mass of of the bwling ball

inn [45]

A force can cause a resting object to move, or it can accelerate a moving object by changing the objects speed or direction

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3 years ago

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The parking brake on a 1200kg automobile has broken, and the vehicle has reached a momentum of 7800kg.m/s. What is the velocity

liraira [26]

Answer:

Velocity of the vehicle (v) = 6.5 m/s

Explanation:

Mass of automobile (m) = 1200 kg

Momentum of vehicle (p) = 7800 kg m/s

By using formula of momentum:

$\rm p = mv \\ \rm 7800 = 1200 \times v \\ \rm v = \frac{7800}{1200} \\ \rm v = 6.5 \: m {s}^{ - 1}$

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3 years ago

A 0.453 kg pendulum bob passes through the lowest part of its path at a speed of 2.58 m/s. What is the tension in the pendulum c

harkovskaia [24]

Answer with Explanation:

Mass of pendulum bob, m=0.453 kg

Speed, $v_1=$ 2.58 m/s

a.r=75.1 cm= $75.1\times 10^{-2}m$ =0.751 m

$1cm=10^{-2} m$

Tension in the pendulum cable is given by

Tension=Centripetal force+force due to gravity

$T=\frac{mv^2}{r}+mg$

Where $g=9.8 m/s^2$

Substitute the values

$T=\frac{0.453(2.58)^2}{75.1\times 10^{-2}}+0.453\times 9.8$

$T=8.45 N$

b.When the pendulum reaches its highest point,then

Final velocity, $v_2=0$

According to law of conservation of energy

$mgh_1+\frac{1}{2}mv^2_1=mgh_2+\frac{1}{2}mv^2_2$

$gh_1+\frac{1}{2}v^2_1=gh_2+\frac{1}{2}v^2_2$

$h_1=0$

Substitute the values

$9.8\times 0+\frac{1}{2}(2.58)^2=9.8\times h_2+\frac{1}{2}(0)^2$

$3.3282=9.8h_2$

$h_2=\frac{3.3282}{9.8}=0.34 m$

The angle mad by cable with the vertical= $cos\theta=\frac{0.751-0.34}{0.751}=0.55$

$\theta=cos^{-1}(0.55)=56.6^{\circ}$

c.When the pendulum reaches at highest point then

Acceleration, a=0

Therefore, the tension in the pendulum cable

$T=mgcos\theta$

Substitute the values

$T=0.453\times 9.8cos56.6$

$T=2.4 N$

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4 years ago