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Over [174]
3 years ago
15

3. Automobile companies often test the safety of cars by putting them through crash tests to observe the integrity of the passen

ger compartment. If 1100kg car is sent toward a cement wall with a speed of15 m/s, an impact force of 185,000 N stops the car, how long does it take before the car is brought to a stop?​
Physics
1 answer:
Nadya [2.5K]3 years ago
6 0

Answer:

Time taken for car to stop = 0.89 seconds (Approx.)

Explanation:

Given:

Mass of car = 1100 kg

Speed of car = 15 m/s

Impact force = 185,000 N

Find:

Time taken for car to stop

Computation:

Change in momentum of car = M(v) - M(u)

Change in momentum of car = 1100(0) - 1100(15)

Change in momentum of car = -16,500

Time taken for car to stop = I Change in momentum of car I / Impact force

Time taken for car to stop = I-16,500I / 185,000

Time taken for car to stop = 0.89 seconds (Approx.)

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A 100-lb child stands on a scale while riding in an elevator. What does the scale read while the elevator slows to stop at the l
Lelechka [254]

Answer: A 100-lb child stands on a scale while riding in an elevator. Then, the scale reading approaches to 100lb, while the elevator slows to stop at the lowest floor

Explanation: To find the correct answer, we need to know more about the apparent weight of a body in a lift.

<h3>What is the apparent weight of a body in a lift?</h3>
  • Consider a body of mass m kept on a weighing machine in a lift.
  • The readings on the machine is the force exerted by the body on the machine(action), which is equal to the force exerted by the machine on the body(reaction).
  • The reaction we get as the weight recorded by the machine, and it is called the apparent weight.
<h3>How to solve the question?</h3>
  • Here we have given with the actual weight of the body as 100lbs.
  • This 100lb child is standing on the scale or the weighing machine, when it is riding .
  • During this condition, the acceleration of the lift is towards downward, and thus, a force of ma .
  • There is also<em> mg </em>downwards and a normal reaction in the upward direction.
  • when we equate both the upward force and downward force, we get,

                             ma=mg-N\\N=mg-ma    i.e. during riding the scale reads a weight less than that of actual weight.

  • When the lift goes slow and stops the lowest floor, then the acceleration will be approaches to zero.

Thus, from the above explanation, it is clear that ,when the elevator moves to the lowest floor slowly and stops, then the apparent weight will become the actual weight.

Learn more about the apparent weight of the body in a lift here:

brainly.com/question/28045397

#SPJ4

7 0
1 year ago
What is the mass of basswood wing with these dimensions: 26.9 cm x 5.5 cm x 0.15 cm?
Softa [21]
The answer is 22.1925 I hope you get right let me know I got my answer because I multiply it all together.
6 0
3 years ago
Read 2 more answers
You are given the melting points of three unknown substances and are asked to predict which one is an ionic compound. You would
nikitadnepr [17]
The answer is <span>highest</span>
4 0
3 years ago
A 45.0-kg sample of ice is at 0.00°C. How much heat is needed to melt it? For water, Lf=334 kJ/kg and Lv=2257 kJ/kg 
Aleonysh [2.5K]

Heat required to change the phase of ice is given by

Q = m* L

here

m = mass of ice

L = latent heat of fusion

now we have

m = 45 kg

L = 334 KJ/kg

now by using above formula

Q = 45 * 334 * 10^3

Q = 1.5 * 10^7 J

In KJ we can convert this as

Q = 1.5 * 10^4 kJ

so the correct answer is D option

7 0
3 years ago
Read 2 more answers
A 10 m long uniform beam weighing 100 N is supported by two ropes at the ends. If a 400 N person sits at 2.0 m from one end of t
Naddik [55]

Answer:

T1 = 130N, T2 = 370N

Explanation:

In order for the system to be at rest, the sum of all forces must be zero and the torque around a point on the beam must be zero.

1. forces:

Let tension in rope 1 be T1 and in rope 2 be T2:

ma = T1 + T2 - 100N - 400N = 0

(1) T1 + T2 = 500N

2. torque around the center point of the beam:

τ = r x F = 5*T1 + 3*400N - 5*T2 = 0

(2) T1 - T2 = -240N

Solving both equations:

T1 = 130N

T2 = 370N

3 0
3 years ago
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