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3 years ago

- What is the total distance traveled by the car

Physics

1 answer:

slega [8]3 years ago

7 0

*Graph is attached with the answer*

Answer:

Total distance traveled by the car is 40 m

Explanation:

It can be seen from the graph that for the first 4 second car is moving with uniform acceleration and for next 2 second it has constant velocity

So, we solved this question in 2 parts

Part 1:

initial velocity = v₁ = 0 m/s

final velocity = v₂ = 10 m/s

time = t = 4 s

acceleration = a = (v₂ - v₁)/t

a = 10/4 = 2.5 m/s²

Using 3rd equation of motion

2as = v₂² - v₁²

2(2.5)s = 10²

s₁ = 100/5

s₁ = 20 m

Part 2:

s₂ = v₂t

s₂ = 10 × 2

s₂ = 20 m

Total distance = S = s₁ + s₂

= 20 + 20

S = 40 m

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A rock hits a window and stops in 0.15 seconds. The net force on the rock is 58N during the collision. What is the magnitude of

nlexa [21]

Answer:

The change in momentum is $\Delta p = 0.7 \ kg\cdot m \cdot s^{-1}$

Explanation:

From the question we are told that

The time taken for the stone to stop is $\Delta t = 0.15 \ seconds$

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$I = F * \Delta t$

Substituting values

$I = 58 * 0.15$

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3 years ago

A block rides on a piston that is moving vertically with simple harmonic motion. (a) If the SHM has period 2.68 s, at what ampli

fredd [130]

Answer:

Part a)

$A = 1.78 m$

Part b)

$f = 2 rev/s$

Explanation:

Part A)

As we know that time period of the motion is given as

$T = 2.68 s$

so we have

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

$\omega = \frac{2\pi}{2.68}$

$\omega = 2.34 rad/s$

now at the point of maximum amplitude the force equation when Normal force is about to zero is given as

$mg = m\omega^2 A$

so we have

$A = \frac{g}{\omega^2}$

$A = \frac{9.81}{2.34^2}$

$A = 1.78 m$

Part b)

Now if the amplitude of the SHM is 6.23 cm

and now at this amplitude if object will lose the contact then in that case again we have

$mg = m\omega^2 A$

$g = \omega^2 (0.0623)$

$\omega = 12.5 rad/s$

so now we have

$2\pi f = 12.5$

$f = 2 rev/s$

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2 years ago

How high was a brick dropped from if if falls in 2.5 seconds?

jenyasd209 [6]

Using the kinematic equation d = V_0 * t + 1/2 * a * t^2, where d is height you can rewrite this to be d = 1/2*g*t^2 or 4.9t^2
g = a because this is a free fall
d = 1/2 * 9.81m/s^2 * 2.5^2
d = 30.65625m
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7 0

2 years ago