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3 years ago

9

Newton's first law of motion states that an object at rest or moving with constant velocity will remain that way until it is act

ed on by an outside force true or false

1 answer:

Step2247 [10]3 years ago

6 0

No that is the 3rd law

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What is the purpose of the overrunning clutch in the starter drive?

nadya68 [22]

<span>"prevent the engine from over speeding the armature"
hopes this help :) :D :)</span>

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4 years ago

An 85-kg man plans to tow a 109 000-kg airplane along a runway by pulling horizontally on a cable attached to it. Suppose that h

Lelu [443]

Answer:

The greatest acceleration the man can give the airplane is 0.0059 m/s².

Explanation:

Given that,

Mass of man = 85 kg

Mass of airplane = 109000 kg

Distance = 9.08

Coefficient of static friction = 0.77

We need to calculate the greatest friction force

Using formula of friction

$F=\mu mg$

Where, m = mass of man

g = acceleration due to gravity

Put the value into the formula

$F = 0.77\times85\times9.8$

$F= 641.41\ N$

We need to calculate the acceleration

Using formula of newton's second law

$F = ma$

$a=\dfrac{F}{m}$

Put the value into the formula

$a=\dfrac{ 641.41}{109000}$

$a=0.0059\ m/s^2$

Hence, The greatest acceleration the man can give the airplane is 0.0059 m/s².

3 0

3 years ago

Two equally charged tiny spheres of mass 1.0 g are placed 2.0 cm apart. When released, they begin to accelerate away from each o

zhannawk [14.2K]

Answer:

The magnitude of the charge on each sphere is 0.135 μC

Explanation:

Given that,

Mass = 1.0

Distance = 2.0 cm

Acceleration = 414 m/s²

We need to calculate the magnitude of charge

Using newton's second law

$F= ma$

$a=\dfrac{F}{m}$

Put the value of F

$a=\dfrac{kq^2}{mr^2}$

Put the value into the formula

$414=\dfrac{9\times10^{9}\times q^2}{1.0\times10^{-3}\times(2.0\times10^{-2})^2}$

$q^2=\dfrac{414\times1.0\times10^{-3}\times(2.0\times10^{-2})^2}{9\times10^{9}}$

$q^2=1.84\times10^{-14}$

$q=0.135\times10^{-6}\ C$

$q=0.135\ \mu C$

Hence, The magnitude of the charge on each sphere is 0.135μC.

7 0

3 years ago

Read 2 more answers

a car travels at 15 m/s for 10 s. It then speeds up with a constant acceleration of 2.0 m/s? for 15 s. At the end of this time,

Firlakuza [10]

V = u + at where u is initial velocity (15 m/s), a is acceleration (2m/s^2) and t is time (15 seconds)

V = 15 + 2 X 15

V = 45 m/s

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3 years ago

Cheetahs can accelerate to a speed of 20.0 m/s in 2.50 s and can continue to accelerate to reach a top speed of 29.5 m/s. Assume

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Answer:

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3 years ago