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denis-greek [22]
3 years ago
9

Car 1 brakes to a stop on a dry road. Car 2 does the same thing, at the same speed, on a road where it has been raining. Explain

how the following will be different for car 2: a) braking distance, b) thinking distance and c) stopping distance.
Physics
1 answer:
Zepler [3.9K]3 years ago
8 0

Answer: car 1 is going how fast

Explanation: no need to answer without speed I won't know distance.

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The Law of Conservation of energy states that the total energy of an isolated system remains constant-it is said to be the conserved over time. It applies by us burning calories because humans can lose or gain weight.
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How do mountain glaciers and continental glaciers differ in terms of dimensions, thickness and patterns of movement?
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How do mountain glaciers and continental glaciers differ in terms of dimensions, thickness and patterns of movement?

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4 years ago
The 20-g bullet is travelling at 400 m/s when it becomes embedded in the 2-kg stationary block. The coefficient of kinetic frict
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Answer:

The distance the block will slide before it stops is 3.3343 m

Explanation:

Given;

mass of bullet, m₁ = 20-g = 0.02 kg

speed of the bullet, u₁ =  400 m/s

mass of block, m₂ = 2-kg

coefficient of kinetic friction,  μk = 0.24

Step 1:

Determine the speed of the bullet-block system:

From the principle of conservation of linear momentum;

m₁u₁ + m₂u₂ = v(m₁ + m₂)

where;

v is the speed of the bullet-block system after collision

(0.02 x 400) + (2 x 0) = v (0.02 + 2)

8 = v (2.02)

v = 8/2.02

v = 3.9604 m/s

Step 2:

Determine the time required for the bullet-block system to stop

Apply the principle of conservation momentum of the system

v(m_1+m_2) -F_kt = v_f(m_1 +m_2)\\\\v(m_1+m_2) -N \mu_kt = v_f(m_1 +m_2)\\\\v(m_1+m_2) -g(m_1 +m_2) \mu_kt = v_f(m_1 +m_2)\\\\3.9604(2.02)-9.8(2.02)0.24t = v_f(2.02)\\\\8 - 4.751t = 2.02v_f\\\\3.9604 - 2.352t = v_f

when the system stops, vf = 0

3.9604 -2.352t = 0

2.352t = 3.9604

t = 3.9604/2.352

t = 1.684 s

Thus, time required for the system to stop is 1.684 s

Finally, determine the distance the block will slide before it stops

From kinematic, distance is the product of speed and time

S = \int\limits {v} \, dt \\\\S = \int\limits^t_0 {(3.9604-2.352t)} \, dt\\\\ S = 3.9604t - 1.176t^2

Now, recall that t = 1.684 s

S = 3.9604(1.684) - 1.176(1.684)²

S = 6.6693 - 3.3350

S = 3.3343 m

Thus, the distance the block will slide before it stops is 3.3343 m

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3 years ago
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Oduvanchick [21]

Answer:

5 I think

Explanation:

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