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Dafna11 [192]
3 years ago
7

52.887 in significant figures

Physics
1 answer:
Anastasy [175]3 years ago
5 0

Answer:

There are 5 significant figures in 52.887

5, 2, 8, 8, 7 --> all of these numbers are considered significant.

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A 2.00 kg cat is in a 97.00 kg elevator. What force on the elevator cable would be needed to lower the cat/elevator pair with an
umka21 [38]

The magnitude of force on the elevator cable that would be needed to lower the cat/elevator pair is 198 Newton.

<u>Given the following data:</u>

  • Mass of cat = 2 kg
  • Mass of elevator = 97 kg
  • Acceleration = 2 m/s^2

To determine the magnitude of force on the elevator cable that would be needed to lower the cat/elevator pair, we would apply Newton's Second Law of Motion:

First of all, we would calculate the total mass of the cat/elevator pair.

Total \;mass=2 + 97

Total mass = 99 kilograms

Mathematically, Newton's Second Law of Motion is given by this formula;

Force = mass \times acceleration

Substituting the given parameters into the formula, we have;

Force = 99 \times 2

Net force = 198 Newton

Read more here: brainly.com/question/24029674

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3 years ago
What is shown in the diagram?
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Answer:

an electromagnet

Explanation:

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3 years ago
A bowler who always left the same three pins standing could be considered a(n) ____ bowler.
RideAnS [48]
A bowler who always left the same 3 pins standing could be considered a C. Precise bowler as from bowling countless number of times he has observed the same amount of pins knocked down each time.
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4 years ago
The best rebounders in basketball have a vertical leap (that is, the vertical movement of a fixed point on their body) of about
nadya68 [22]

Answer:

a) 4.45 m/s

b) 0.9 seconds

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration due to gravity = 9.81 m/s²

v^2-u^2=2as\\\Rightarrow -u^2=2as-v^2\\\Rightarrow u=\sqrt{v^2-2as}\\\Rightarrow u=\sqrt{0^2-2\times -9.81\times 1}\\\Rightarrow u=4.45\ m/s

a) The vertical speed when the player leaves the ground is 4.45 m/s

v=u+at\\\Rightarrow t=\frac{v-u}{a}\\\Rightarrow t=\frac{0-4.45}{-9.81}\\\Rightarrow t=0.45\ s

Time taken to reach the maximum height is 0.45 seconds

s=ut+\frac{1}{2}at^2\\\Rightarrow 1=0t+\frac{1}{2}\times 9.81\times t^2\\\Rightarrow t=\sqrt{\frac{1\times 2}{9.81}}\\\Rightarrow t=0.45\ s

Time taken to reach the ground from the maximum height is 0.45 seconds

b) Time the player stayed in the air is 0.45+0.45 = 0.9 seconds

6 0
3 years ago
Can anyone pls help me with this I’ll appreciate it. I’m trying to study for a test but I don’t know the answer to this.
forsale [732]

Answer:

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Explanation:

7 0
3 years ago
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