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Setler79 [48]
3 years ago
8

An initially motionless test car is accelerated uniformly to 115 km/h in 8.83 seconds before striking a simulated deer. The car

is in contact with the faux fawn for 0.995 seconds, after which the car is measured to be traveling at 60.0 km/h. What is the magnitude of the acceleration of the car before the collision?
Physics
1 answer:
goblinko [34]3 years ago
3 0

We know the equation of motion v = u+ at, where v is the final velocity, u is the initial velocity, a is the acceleration and t is the time taken.

In this case Final velocity before collision = 115 km/hr = 115*5/18 = 31.94 m/s

Time taken by car to reach this velocity = 8.83 seconds

Initial velocity = 0 m/s

v = u +at

31.94 = 0 + a*8.83

a = 3.62 m/s^2

So acceleration of car just before collision = 3.62 m/s^2

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Write a hypothesis for Part II of the lab, which is about the relationship described by F = ma. In the lab, you will use a toy c
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Answer:

F=ma is the relationship where, F is force, m is mass and a is acceleration.

Newton's second law states that  the unbalanced force applied to the object accelerates the object which is directly proportional to the force and inversely to the mass.

If we apply force to a toy car then It will accelerate.

This is how Newton's second law of motion is verified.

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What has a larger capacitance, an aluminum sphere with a 10 cm diameter or one with a 100 cm diameter? Question 16 options: 10 c
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\boxed{\sf C=\dfrac{Q}{V}}

But

\boxed{\sf \Delta V_{R_2\to R_1}={\displaystyle{\int}^{R_1}_{R_2}}dV=-{\displaystyle{\int}^{R_1}_{R_2}}\dfrac{kQ}{r^2}dR}

  • Hence higher the radius lower the voltage
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<h3>100cm diameter having aluminium sphere has a larger capacitance</h3>
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A ball is attached to a string of length 3 m to make a pendulum. The pendulum is placed at a location that is away from the Eart
Musya8 [376]

1) 0.61 m/s^2

2) 13.9 s

Explanation:

1)

The acceleration due to gravity is the acceleration that an object in free fall (acted upon the force of gravity only) would have.

It can be calculated using the equation:

g=\frac{GM}{r^2} (1)

where

G is the gravitational constant

M=5.98\cdot 10^{24} kg is the Earth's mass

r is the distance of the object from the Earth's center

The pendulum in the problem is at an altitude of 3 times the radius of the Earth (R), so its distance from the Earth's center is

r=4R

where

R=6.37\cdot 10^6 m is the Earth's radius

Therefore, we can calculate the acceleration due to gravity at that height using eq.(1):

g=\frac{GM}{(4R)^2}=\frac{(6.67\cdot 10^{-11})(5.98\cdot 10^{24})0.}{(4\cdot 6.37\cdot 10^6)^2}=0.61 m/s^2

2)

The period of a simple pendulum is the time the pendulum takes to complete one oscillation. It is given by the formula

T=2\pi \sqrt{\frac{L}{g}}

where

L is the length of the pendulum

g is the acceleration due to gravity at the location of the pendulum

Note that the period of a pendulum does not depend on its mass.

For the pendulum in this problem, we have:

L = 3 m is its length

g=0.61 m/s^2 is the acceleration due to gravity (calculated in part 1)

Therefore, the period of the pendulum is:

T=2\pi \sqrt{\frac{3}{0.61}}=13.9 s

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3 years ago
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