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Svetach [21]
2 years ago
15

2. A car accelerates down the road. What is the "reaction" to the tires pushing on the road?

Physics
1 answer:
Artist 52 [7]2 years ago
7 0
The answer would be friction
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What is the acceleration of a 0.30-kg volleyball when a player uses a force of 42 N to spike the ball?
quester [9]

Answer:

The acceleration will be 140 meter per second

Explanation:

Force F = mass m × acceleration a

If F = 42 N and m = 0.30 kg

Then  acceleration a = F/m

a = 42/0.30

a = 140 m/s

5 0
3 years ago
The equation for the chemical reaction shown is not balanced. What number should replace the question mark to balance this equat
Fudgin [204]
The equation is balanced as it is.
7 0
3 years ago
If a small sports car collides head-on with a massive truck, which vehicle experiences the greater impact force? Which vehicle e
garri49 [273]

Answer:

Small sports car.

Explanation:

Lets take

mass of the small car = m

mass of the truck = M

As we know that when car collide with the massive truck then due to change in the moment of the car both car as well as truck will feel force.We also know that from Third law of Newton's ,it states that every action have it reaction with same magnitude but in the opposite direction.

Therefore

F = m a

a=Acceleration of the car

a=\dfrac{F}{m}

F= M a'

a'=Acceleration of the massive truck

a'=\dfrac{F}{M}

Here given that M > m that is why a > a'

Therefore car will experiences more acceleration.

5 0
3 years ago
the half-life of carbon-14 is 5,730 years. After 11,460 year, how much of original carbon-14 remains?
Inessa [10]

Via the half-life equation:

A_{final}=A_{initial}(\frac{1}{2})^{\frac{t}{h}}

Where the time elapse is 11,460 year and the half-life is 5,730 years.

A_{final}=A_{initial}(\frac{1}{2})^\frac{11460}{5730} \\\\A_{final}=A_{initial}\frac{1}{4} \\\\A_{final}=\frac{1}{4}A_{initial}

Therefore after 11,460 years the amount of carbon-14 is one fourth (1/4) of the original amount.

6 0
2 years ago
A 200 N trash can is pulled across the sidewalk by a person at constant speed by a force of 75 N. What is the coefficient of fri
Pavel [41]

Answer:

μ = 0.375

Explanation:

F = Applied force on the trash can = 75 N

W = weight of the trash can = 200 N

f = frictional force acting on trash can

Since the trash can moves at constant speed, force equation for the motion of can is given as

F - f = 0

75 - f = 0

f = 75 N

μ = Coefficient of friction

frictional force is given as

f = μ W

75 = μ (200)

μ = 0.375

5 0
2 years ago
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