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3 years ago

the table below shows the effect conditions on a car stopping distance. in witch condition was the car travelling most slowly

Physics

1 answer:

nikdorinn [45]3 years ago

7 0

Answer: Car 5

Explanation:

Higher speeds mean higher braking times so the car with the highest braking distance was travelling the the fastest and the car with the lowest braking distance was traveling the slowest.

Braking distance = Stopping distance - Thinking distance.

Car Braking distance

1 38

2 50 = 65 - 15

3 75

4 75

5 14 = 23 - 9

<em>Car 5 had the lowest braking distance so was going the slowest. </em>

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The force of friction always opposes the _______.

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A child bounces a 60 g superball on the sidewalk. The velocity change of the superball is from 22 m/s downward to 15 m/s upward.

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complete question:

A child bounces a 60 g superball on the sidewalk. The velocity change of the superball is from 22 m/s downward to 15 m/s upward. If the contact time with the sidewalk is 1/800 s, what is the magnitude of the average force exerted on the superball by the sidewalk

Answer:

F = 1776 N

Explanation:

mass of ball = 60 g = 0.06 kg

velocity of downward direction = 22 m/s = v1

velocity of upward direction = 15 m/s = v2

Δt = 1/800 = 0.00125 s

Linear momentum of a particle with mass and velocity is the product of the mass and it velocity.

p = mv

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I = pf − pi = ∆p

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let the upward velocity be the positive

Δp = mv2 - m(-v1)

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3 years ago

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Answer:

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Explanation:

The magnetic field at the center of the solenoid is given by;

B = μ(N/L)I

Where;

μ is permeability of free space

N is the number of turn

L is the length of the solenoid

I is the current in the solenoid

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$\frac{\delta B}{\delta t} = \frac{\mu N \frac{\delta i}{\delta t} }{L} \\\\\frac{\delta B}{\delta t} = \frac{4\pi *10^{-7} *600* \frac{5}{0.6} }{0.25}\\\\\frac{\delta B}{\delta t} =0.02514 \ T/s$

The induced emf in the shorter coil is calculated as;

$E = NA\frac{\delta B}{\delta t}$

where;

N is the number of turns in the shorter coil

A is the area of the shorter coil

Area of the shorter coil = πr²

The radius of the coil = 2.5cm / 2 = 1.25 cm = 0.0125 m

Area of the shorter coil = πr² = π(0.0125)² = 0.000491 m²

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the table below shows the effect conditions on a car stopping distance. in witch condition was the car travelling most slowly​

the table below shows the effect conditions on a car stopping distance. in witch condition was the car travelling most slowly