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ipn [44]
3 years ago
13

A vw beetle goes from 0 to 60 mi/h with an acceleration of 2.35 m/s^2.

Physics
1 answer:
Vlad1618 [11]3 years ago
6 0
Part a.
u = 0, the initial velocity
v = 60 mi/h, the final velocity
a = 2.35 m/s², the acceleration.

Note that
1 m = 1609.34 m.
Therefore
v = (60 mi/h)*(1609.34 m/mi)*(1/3600 h/s) = 26.822 m/s
Use the formula
v = u + at
(26.822 m/s) = (2.35 m/s²)*(t s)
t = 26.822/2.35 = 11.4 s

Answer: 11.4 s

Part b.
We already determined that v = 60 mi/h = 26.822 m/s.
t = 0.6 s
Therefore
(26.822 m/s) = (a m/s²)*(0.6 s)
a = 26.822/0.6 = 44.7 m/s²

Answer:  44.7 m/s²
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Answer:

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Explanation:

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2. Conner flips a coin up in the air (to determine if he or his sister needs to do the dishes) at an upward velocity of 4.00 m/s
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2 years ago
A brick is dropped from a big scaffold. What is its velocity after 4.0s?
Flura [38]

The velocity of the brick is 39.2 m/s downward

Explanation:

The motion of the brick is a free fall motion, since the object is affected only by the force of gravity. Therefore, it has a uniformly accelerated motion towards the ground, with constant acceleration of g=9.8 m/s^2.

So, we can find its velocity using the suvat equation:

v=u+at

where

v is the final velocity

u is the initial velocity

a is the acceleration

t is the time

For the brick in this problem (taking downward as positive direction)

u = 0 (it is dropped from rest)

a=g=9.8 m/s^2

Therefore, its velocity after t = 4.0 s is:

v=0+(9.8)(4.0)=39.2 m/s

Downward, because the sign is positive.

Learn more about free fall motion:

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3 years ago
After the box comes to rest at position x₁, a person starts pushing the box, giving it a speed v₁. When the box reaches position
Vesna [10]

Answer:

W_p = \frac{1}{2}mv_1^2

Explanation:

As we know that box is initially at rest

So we will have

v_i = 0

now as it will be displaced from initial position to final position then final speed of the box is reached to

v_f = v_1

now we know by work energy theorem that work done by all the forces on the box will be equal to the change in kinetic energy

So here we will have

W_p = \frac{1}{2}m(v_f^2 - v_i^2)

W_p = \frac{1}{2}m(v_1^2 - 0)

W_p = \frac{1}{2}mv_1^2

8 0
3 years ago
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