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mylen [45]
3 years ago
13

A truck contains both hydraulic brakes (reliability 0.96) and mechanical brakes (reliability 0.99). What is the probability of s

topping the truck (i) while moving at a high speed, assuming both systems must work to avoid hitting a vehicle stalled in the roadway, and (ii) while moving at a low speed, assuming either system will stop the truck in time if they function properly
Physics
1 answer:
Oksana_A [137]3 years ago
8 0

Answer:

i) 0.9504

ii) 0.0452

Explanation:

Given data: reliability of hydraulic brakes= 0.96

reliability of mechanical brakes = 0.99

So the probability of stopping the truck = 0.96×0.99= 0.9504

At low speed

case: A works and B does not

= 0.96×(1-0.99) = 0.0096

case2 : B works and A does not

= 0.99×(1-0.96) = 0.0396

Therefore, probality of stopping = 0.0096+0.0396 = 0.0492

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Answer:

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Explanation:

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2 years ago
a 0.25 kg arrow is moving at 5 m/s and hits a 0.10 kg apple. The apple sticks to the arrow, and both move to the right together.
MatroZZZ [7]

Answer:

The arrow-apple combo will move with a velocity 3.6 m/s to the right.

Explanation:

Use the law of conservation of momentum to solve this problem. In this case the law can be written as follows:

m_{arrow}v_{arrow}+m_{apple}\cdot 0= (m_{arrow}+m_{apple})v_{both}

from which the desired velocity can be isolated:

\frac{m_{arrow}v_{arrow}}{m_{arrow}+m_{apple}}= v_{both}\\v_{both} = \frac{0.25kg\cdot 5\frac{m}{s}}{0.35kg}=3.6\frac{m}{s}

The arrow-apple combo will move with a velocity 3.6 m/s to the right.

3 0
3 years ago
Car A (mass 1100 kg) is stopped at a traffic light when it is rear­ended by car B(mass 1400 kg). Both cars then slide with locke
viva [34]

Answer:

Part a)

v_a = 3.94 m/s

Part b)

v_b = 3.35 m/s

Part C)

v_b = 6.44 m/s

Part d)

Due to large magnitude of friction between road and the car the momentum conservation may not be valid here as momentum conservation is valid only when external force on the system is zero.

Explanation:

Part a)

As we know that car A moves by distance 6.1 m after collision under the frictional force

so the deceleration due to friction is given as

a = -\frac{F_f}{m}

a = -\frac{\mu mg}{m}

a = - \mu g

now we will have

v_f^2 - v_i^2 = 2ad

0 - v_i^2 = 2(-\mu g)(6.1)

v_a = \sqrt{(2(0.13)(9.81)(6.1)}

v_a = 3.94 m/s

Part b)

Similarly for car B the distance of stop is given as 4.4 m

so we will have

v_b = \sqrt{2(0.13)(9.81)(4.4)}

v_b = 3.35 m/s

Part C)

By momentum conservation we will have

m_1v_{1i} = m_1v_{1f} + m_2v_{2f}

1400 v_b = 1100(3.94) + 1400(3.35)

v_b = 6.44 m/s

Part d)

Due to large magnitude of friction between road and the car the momentum conservation may not be valid here as momentum conservation is valid only when external force on the system is zero.

3 0
3 years ago
__________ (marketing research) information is collected from questions (measurements) that are free from systematic or statisti
tatiyna

Answer: reliable

Explanation:

Reliable (marketing research) information is collected from questions (measurements) that are free from systematic or statistical error. An absence of systematic error implies that the respondents (i.e., the sampled people) who answer questions actually understand what the questions were asking.

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3 years ago
Directions: Fill in the blank with the correct word. Choose from the list of possible answers in the word bank below:
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Answer:

1).atoms (3). mixture. (5). Element

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Explanation:

(7). Homogeneous (8). Heterogeneous

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3 years ago
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