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Y_Kistochka [10]

3 years ago

8

A 2-kg box sits on a horizontal table. the force of friction between the box and the table is 10 n. the box is pushed to the rig

ht with an applied horizontal force of 20 n. what is the acceleration of the box?

1 answer:

dangina [55]3 years ago

6 0

By Newton's 2nd law of motion, F = ma, where F is force, m is mass, and a is acceleration.

Rearranging this equation to find acceleration would give us:
a = F/m

The horizontal force to the right is 10N, because the box is pushed to the right with a force of 20N, and the friction force of 10N opposes that, so:
20N - 10N = 10N

The mass is 2kg.

Putting these values into the equation gives us:
a = F/m
= 10/2
= 5ms^-2

The acceleration of the box is 5ms^-2

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Answer:

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Explanation:

First we convert 2 cycles/s to angular velocity knowing that each circle has an angle of 2π

$\omega = 2 * 2\pi = 4\pi rad/s$

Then we calculate the moment of inertia of the cylindrical shell, assuming there's no mass inside the wheel (only at the rim):

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So the kinetic energy of this is

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How far from the earth must a body be along a line toward the sun so that the sun’s

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Answer:

r = 1.63×10^5 mi

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This simplifies to

Ms/(Rs - r)^2 = Me/r^2

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3 years ago

A transverse wave on a string is described by the following wave function.y = (0.090 m) sin (px/11 + 4pt)(a) Determine the trans

alukav5142 [94]

Explanation:

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Now, we will compare above equation with the standard form of transeverse wave equation,

y = $A sin(kx + \omega t)$

where, A is the amplitude = 0.09 m

k is the wave vector = $\frac{\pi}{11}$

$\omega$ is the angular frequency = $4\pi$

x is displacement = 1.40 m

t is the time = 0.16 s

Now, we will differentiate the equation with respect to t as follows.

The speed of the wave will be:

v(t) = $\frac{dy}{dt}$

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a(t) = $\frac{dv}{dt}$

a(t) = $-A \omega 2sin(kx + \omega t)$

a(t) = $-(0.09 m)(4 \pi)2 sin(\frac{\pi \times 1.4}{11} + 4\pi \times 0.16)$

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Hence, the value of transverse wave is 0.84 m/s and the value of acceleration is 9.49 $m/s^{2}$ .

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T = $\frac{2 \pi}{\omega}$

T = $\frac{2 \pi}{4 \pi}$

= 0.5 sec

Now, we will calculate the speed of propagation of wave as follows.

v = $\frac{\lambda}{T}$

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3 years ago

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xenn [34]

Answer:

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4 years ago