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2 years ago

11

Bill has a mass of 85 kg and is skating west. He increases his speed from 3 m/s to 5 m/s by applying a force for 3 seconds. What

force did he apply?

1 answer:

likoan [24]2 years ago

4 0

F = ma
F = 85×(5-3)÷3
F = 85×(2÷3)
F = 85×0.667
F = 56.67N

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A positively-charged particle is released near the positive plate of a parallel plate capacitor. a. Describe its path after it i

il63 [147K]

Answer:

a. The electric field lines are linear and perpendicular to the plates inside a parallel-plate capacitor, and always from positive plate to the negative plate. If a positive charge is released near the positive plate, then<em> it will follow a linear path towards the negative plate under the influence of electrostatic force, F = Eq</em>, where q is the charge of the particle. The electric field inside a parallel plate capacitor is constant and equal to

This can be calculated by Gauss' Law.

A positive charge always follow the electric field lines when released. Another approach is that the positive plate repels the positive charge and negative plate attracts the positive charge. Therefore, the positive charge follows a path towards the negative charge.

b. The particle moves from the higher potential to the lower potential. <em>The direction of motion is the same as the direction of the force that moves the particle, so the work done on the particle by that force is positive.</em>

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2 years ago

A plane wall with constant properties is initially at a uniform temperature To. Suddenly, the surface at x = L is exposed to a c

Rzqust [24]

Answer:

The distribution is as depicted in the attached figure.

Explanation:

From the given data

The plane wall is initially with constant properties is initially at a uniform temperature, To.
Suddenly the surface x=L is exposed to convection process such that T∞>To.
The other surface x=0 is maintained at To

Uniform volumetric heating q' such that the steady state temperature exceeds T∞.

Assumptions which are valid are

There is only conduction in 1-D.
The system bears constant properties.
The volumetric heat generation is uniform

From the given data, the condition are as follows

<u>Initial Condition</u>

At t≤0

$T(x,0)=T_o$

This indicates that initially the temperature distribution was independent of x and is indicated as a straight line.

<u>Boundary Conditions</u>

<u>At x=0</u>

<u /> $T(0,t)=T_o$ <u />

This indicates that the temperature on the x=0 plane will be equal to To which will rise further due to the volumetric heat generation.

<u>At x=L</u>

<u /> $-k\frac{\partial T}{\partial x}]_{x=L}=h[T(L,t)-T_{\infty}]$ <u />

This indicates that at the time t, the rate of conduction and the rate of convection will be equal at x=L.

The temperature distribution along with the schematics are given in the attached figure.

Further the heat flux is inferred from the temperature distribution using the Fourier law and is also as in the attached figure.

It is important to note that as T(x,∞)>T∞ and T∞>To thus the heat on both the boundaries will flow away from the wall.

3 0

2 years ago

A penny is placed on a rotating turntable. Where on the turntable does the penny require the largest centripetal force to remain

Artyom0805 [142]

Answer:

m = mass of the penny

r = distance of the penny from the center of the turntable or axis of rotation

w = angular speed of rotation of turntable

F = centripetal force experienced by the penny

centripetal force "F" experienced by the penny of "m" at distance "r" from axis of rotation is given as

F = m r w²

in the above equation , mass of penny "m" and angular speed "w" of the turntable is same at all places. hence the centripetal force directly depends on the radius .

hence greater the distance from center , greater will be the centripetal force to remain in place.

So at the edge of the turntable , the penny experiences largest centripetal force to remain in place.

Explanation:

5 0

2 years ago

A cannon Is fired from the edge of a small clip the height of the clear is 80.0 at the cannonball inspired with a perfectly hori

Nutka1998 [239]

Answer:

The cannonball fly horizontally before it strikes the ground, S = 323.25 m

Explanation:

Given data,

The height of the cliff, h = 80 m

The horizontal velocity of the cannonball, Vₓ = 80 m/s

The range of the cannon ball with initial vertical velocity is zero is given by the formula,

$S=\frac{V_{x}\sqrt{2gh}}{g}$

$S=\frac{80\sqrt{2\times9.8\times80}}{9.8}$

S = 323.25 m

Hence, the cannonball fly horizontally before it strikes the ground, S = 323.25 m

7 0

2 years ago

Read 2 more answers

Solve A and B using energy considerations.

Alisiya [41]

<span>Use the kinematic equation vf^2 = vi^2 + 2ad where;
vf = ?
vi = 0 m/s
a = 9.8 m/s^2
d1 = 10 m
d2 = 25 m

final velocity at the ground (d1): vf = sqrt(2)(9.8)(10) = 14 m/s
final velocity to the bottom of the cliff (d2): vf = sqrt(2)(9.8)(25) = 22.14 m/s
</span>

7 0

2 years ago