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3 years ago

A 13.3 kg box sliding across the ground

Physics

1 answer:

adelina 88 [10]3 years ago

5 0

Answer:

0.25

Explanation:

According to newtons law of motion

\sum F_x = ma

F_f = ma

nR = ma

nmg = ma

ng = a

n = a/g

g is the acceleration due to gravity

Given

a = 2.42m/s²

g = 9.8m/s²

Substitute into the formula;

n = 2.42/9.8

n = 0.25

Hence the coefficient of kinetic friction is 0.25

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Energy Density = 1/2 × ε(0) × (V/d)^2

V = 100, d = 0.01, ε(0) = 8.85 x 10^-12

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3 years ago

What happens to the frequency of a wave of you increase the speed of the wave ?

exis [7]

Nothing happens. The frequency is determined at the source,
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3 years ago

Convert the following:

Alika [10]

Explanation:

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3 years ago

Read 2 more answers

| A T-ball with a mass of 0.6 kg travels in the

r-ruslan [8.4K]

Answer:

$|I|=6\ Kg.m/s$

$F=120\ N$

Explanation:

Impulse and Momentum

They are similar concepts since they deal with the dynamics of objects having their status of motion changed by the sudden application of a force. The momentum at a given initial time is computed as

$p_o=m.v_o$

When a force is applied, the speed changes to $v_1$ and the new momentum is

$p_1=m.v_1$

The change of momentum is

$\Delta p=p_1-p_0=m(v_1-v_o)$

The impulse is equal to the change of momentum of an object and it's defined as the average net force applied times the time it takes to change the object's motion

$I=F.t=\Delta p$

Part 1

The T-ball initially travels at 10 m/s and then suddenly it's stopped by the glove. The final speed is zero, so

$\Delta p=0.6\ Kg(0-10\ m/s)=-6\ Kg.m/s$

The impulse is

$I=\Delta p$

$I=-6\ Kg.m/s$

The magnitude is

$|I|=6\ Kg.m/s$

Part 2

The force can be computed from the formula

$I=F.t$

The direction of the impulse the T-ball receives is opposite to the direction of the force exerted by the ball on the glove, thus $I_b=6\ kg.m/s$

$\displaystyle F=\frac{I}{t}=\frac{6\ kg.m/s}{0.05\ s}$

$\boxed{F=120\ N}$

7 0

3 years ago

A block of mass 9.2kg rests on a slope an angle of 26.9∘ relative to the horizontal. What is the size of the contact force norma

Olegator [25]

Answer:

80.4 N

Explanation:

As the block is at rest on the slope, it means that all the forces acting on it are balanced.

We are only interested in the forces that act on the block along the direction perpendicular to the slope. Along this direction, we have two forces acting on the block:

- The normal reaction N (contact force), upward

- The component of the weight of the block, $mg cos \theta$ , downward, where m is the mass of the block, g is the gravitational acceleration and $\theta$ is the angle of the incline

Since the block is in equilibrium along this direction, the two forces must balance each other, so they must be equal in magnitude:

$N=mg cos \theta$

And by substituting the numbers into the equation, we find the size of the contact force normal to the slope:

$N=(9.2 kg)(9.8 m/s^2)(cos 26.9^{\circ})=80.4 N$

6 0

3 years ago