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RideAnS [48]
3 years ago
9

Name the force between the brakes and the wheel that does work to slow down the vehicle when the breaks are applied

Physics
1 answer:
Studentka2010 [4]3 years ago
5 0
Answer: Frictional force

Explanation: When a force is applied to the brakes, there is friction between the brakes and the wheel. This reduces the kinetic energy of the vehicle, slowing it down and causing the temperature of the brakes to increase.
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A stone of mass 150g is rotated in a horizontal circle at 10m/s which is attached to the end of a 1m long. what will be the acce
erastova [34]

force is mass multiply by acceleration so it will be 150 multiply by 10 is 1500N

8 0
3 years ago
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Given the 1-m stick shown below, which is held by a thread at its center. Block 1 is 15 N held at the 10 cm mark, while block 2
sattari [20]
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3 years ago
A batter hits a foul ball. The 0.140-kg baseball that was approaching him at 40.0 m/s leaves the bat at 30.0 m/s in a direction
lara31 [8.8K]
<h3>Answer</h3><h3>7 Ns</h3><h3>Explanation</h3>

Given in the question,

mass of foul ball = 0.140 kg

initial speed with which ball was hit with the bat = 30 m/s

final speed  = 40 m/s

According to the scenario the whole scene is making a right angle triangle

So, to the solve the question we will use pythagorus theorem

<h3>Hypotenuse² = base² + height²</h3>

Here,

Hypotenuse= Magnitude of impulse

Base = 1st change of momentum

height = 2nd change of momentum

 

1st impulse (1st change of momentum)

p = m(1)v(1) = (0.14 kg)(40.0 m/s) = 5.6 kg m / s = 5.6 N s

2nd impulse (2nd change of momentum)

p = m(2)v(2) = (0.14 kg)(30.0 m/s) = 4.2 kg m / s = 4.2 N s

Magnitude of impulse (hypotenuse of triangle)

impulse² = (5.6)² + (4.2)²

impulse² = 31.36 + 17.64

impulse² = 49

impulse² = √49

impulse = 7.0 N s

7 0
4 years ago
What are the advantages of using the metric system? SELECT ALL THAT APPLY
alexandr402 [8]

D. used by the entire scientific community

B.  more accurate system of measurement

3 0
3 years ago
A 10.0 g bullet moving at 300m/s is fired into a 1.00 kg block at rest. The bullet emerges (the bullet does not get embedded in
chubhunter [2.5K]

Answer:

v' = 1.5 m/s

Explanation:

given,

mass of the bullet, m = 10 g

initial speed of the bullet, v = 300 m/s

final speed of the bullet after collision, v' = 300/2 = 150 m/s

Mass of the block, M = 1 Kg

initial speed of the block, u = 0 m/s

velocity of the block after collision, u' = ?

using conservation of momentum

 m v + Mu = m v' + M u'

 0.01 x 300 + 0 = 0.01 x 150 + 1 x v'

v' = 0.01 x 150

v' = 1.5 m/s

Speed of the block after collision is equal to v' = 1.5 m/s

5 0
3 years ago
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