4 years ago

PLEASE HELP AND HURRY

Physics

1 answer:

andreyandreev [35.5K]4 years ago

7 0

C
I took the tests earlier hope this helps

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A box is moving along the x-axis and its position varies in time according to the expression:

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Answer:

38.4 m/s

Explanation:

a) at t = 3.2s. $x = 6 * 3.2^2 = 61.44 m$

b) at t = 3.2 + Δt. $x = 6*(3.2 + \Delta t)^2$

c) As Δt approaches 0. We can find the velocity by the ratio of Δx/Δt

$v = \frac{\Delta x}{\Delta t} = \frac{x_2 - x_1}{\Delta t}$

$v = \frac{6*(3.2 + \Delta t)^2 - 61.44}{\Delta t}$

$v = \frac{6(3.2^2 + 6.4\Delta t + \Delta t^2) - 61.44}{\Delta t}$

$v = \frac{61.44 + 38.4\Delta t + \Delta t^2 - 61.44}{\Delta t}$

$v = \frac{\Delta t(38.4 + \Delta t)}{\Delta t}$

$v = 38.4 + \Delta t$

As Δt approaches 0, v = 38.4 + 0 = 38.4 m/s

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3 years ago

A toy cannon launches a 46-g golf ball straight up into the air with a kinetic energy of 6.8 J. What must the

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Answer : The correct option is, (C) 17 m/s

Explanation :

Formula used :

$K.E=\frac{1}{2}mv^2$

where,

K.E = kinetic energy = 6.8 J

m = mass of object = 46 g = 0.046 kg (1 kg = 1000 g)

v = velocity

Now put all the given values in the above formula, we get:

$K.E=\frac{1}{2}mv^2$

$6.8J=\frac{1}{2}\times 0.046kg\times v^2$

$6.8kg.m^2/s^2=\frac{1}{2}\times 0.046kg\times v^2$

$v=17.19m/s\approx 17m/s$

Therefore, the ball's velocity be as it leaves the cannon is, 17 m/s

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