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NemiM [27]
3 years ago
13

An elevator manufacturing company is stress-testing a new elevator in an airless test shaft. The elevator is traveling at an unk

nown velocity when the cable snaps. The elevator falls 1.10 meters before hitting the bottom of the shaft. The elevator was in free fall for 0.900 seconds. Determine its velocity when the cable snapped. As usual, up is the positive direction.
Physics
1 answer:
devlian [24]3 years ago
4 0

Answer:

3.192 m/s

Explanation:

t = Time taken = 0.900 seconds

u = Initial velocity

v = Final velocity

s = Displacement = 1.1 meters

a = Acceleration due to gravity = 9.81 m/s²

s=ut+\frac{1}{2}at^2\\\Rightarrow u=\frac{s-\frac{1}{2}at^2}{t}\\\Rightarrow u=\frac{1.1-\frac{1}{2}\times 9.81\times 0.9^2}{0.9}\\\Rightarrow u=-3.192\ m/s

Velocity of the elevator when it snapped is 3.192 m/s

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Answer:

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Explanation:

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3 years ago
At any point P, the total electric field due to a group of point charges equals the vector sum of the electric fields of the ind
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3 years ago
A rocket takes off from Earth's surface, accelerating straight up at 69.2 m/s2. Calculate the normal force (in N) acting on an a
goldenfox [79]

According to Newton's 3rd law, there will be equal and opposite force on the astronaut which is  -6048 N

<h3>What does Newton's third law say ?</h3>

The law state that in every action, there will be equal and opposite reaction.

Given that a rocket takes off from Earth's surface, accelerating straight up at 69.2 m/s2. We are to calculate the normal force (in N) acting on an astronaut of mass 87.4 kg, including his space suit.

Let us first calculate the force involved in the acceleration of the rocket by using the formula

F = ma

Where mass m = 87.4 kg, acceleration a = 69.2 m/s2

Substitute the two parameters into the formula

F = 87.4 x 69.2

F = 6048.08 N

According to the Newton's 3rd law, there will be equal and opposite force on the astronaut.

Therefore, the normal force acting on the astronaut is -6048 N approximately

Learn more about forces here: brainly.com/question/12970081

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7 0
2 years ago
The windshield of a speeding car hits a hovering insect. Consider the time interval from just before the car hits the insect to
HACTEHA [7]

Answer:

A. False

B True

C. False

D.False

E. True

F. False

G. False

H. False

I. True

Explanation:

A. False: The system being analyzed consists of the bug and the car.  These are the two bodies involved in the collision.

B. True: The system being analyzed consists of the bug and the car

C. False: The magnitudes of the change in velocity are different from the car and the bug. The velocity of the bug changes from 0 to the velocity of the car, while there is no noticeable change in the velocity of the car

D.False: There is barely any change in the momentum of the car since the mass of the bug is very small.

E. True: Since the mass of the bug is small, and was initially at rest, the magnitude of the change in monentum will be large because the new velocity will be that of the car.

F. False: The system being analyzed consists of the bug and the car. Those are the two bodies involved in the collision

G. False: The car barely changes in velocity since the mass of the bug is small.

H. False: The car barely changes in momentum because the collision does not affect its speed so much. on the other hand the momentum change of the bug is large since its mass is small.

I. True: The bug which was initially at rest will begin moving with the velovity of the speeding car, while the car barely changes in its velocity

5 0
3 years ago
Please help me with this one
katen-ka-za [31]

Both of the pictures show how one picture is I believe in fall and the other one is in the forest and maybe in spring or summer. Also, the two pictures show 2 different animals. And maybe hunting for 2 different foods for them to survive.

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3 years ago
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