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klio [65]
3 years ago
6

Question 6.2 Calculate the reading on A2

Physics
1 answer:
Natali [406]3 years ago
5 0

Answer:

2.25A

Explanation:

that is the answer

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1. Alexandra and Rachel are on a train that sounds a whistle at a constant frequency as
exis [7]

Answer: the answer would be C trust me i took the test if its not that its b

hope that helps

Explanation: i took the test

6 0
3 years ago
Read 2 more answers
Describe an environment in which the sound of a ringing cell phone could not be transmitted, and explain why the ringing wouldn'
Sveta_85 [38]

Answer:

In a vacuum

Explanation:

Sound is a type of mechanical waves. Mechanical waves are waves that propagate through the oscillation of the particles in a medium, which can be either gas, liquid or solid.

A sound wave in air, for instance, is simply produced by the oscillations of the air particles back and forth along the direction of motion of the wave.

Given this definition, it is clear that mechanical waves (and so, sound waves as well) cannot be transmitted if there is no medium: therefore, they cannot be transmitted in a vacuum. So, the sound of the ringing bell would not be present in a vacuum.

8 0
3 years ago
What are three kinds of forces that affect an object’s motion? What effects do these forces have?
Harman [31]

Answer:

Different forces (including magnetism, gravity, and friction) can affect motion

4 0
2 years ago
A stuntman with a mass of 80.5 kg swings across a moat from a rope that is 11.5 m. At the bottom of the swing the stuntman's spe
goldenfox [79]

Answer:

  • No
  • 5.49 m/s

Explanation:

The net force required to accelerate the stuntman in a circular arc of radius 11.5 m will be ...

  F = mv²/r . . . . where this m is the mass being accelerated, v is the tangential velocity, and r is the radius.

Here, the net force needs to be ...

  F = (80.5 kg)(8.45 m/s)²/(11.5 m) . . . . . where this m is meters

  ≈ 499.8175 kg·m/s² = 499.8 N

Gravity exerts a force on the stuntman of ...

  F = mg = (80.5 kg)(9.8 m/s²) = 788.9 kg·m/s² = 788.9 N

Then the tension required in the rope/vine is ...

  499.8 N+788.9 N= 1288.7 N

This is more than the capacity of the rope, so we do not expect the stuntman to make it across the moat.

_____

The allowed net force for centripetal acceleration is ...

  1000 N -788.9 N = 211.1 N

Then the allowed velocity is ...

  211.1 = 80.5v²/11.5

  30.16 = v² . . . .  multiply by 11.5/80.5

  5.49 = v . . . . . . take the square root

The maximum speed the stuntman can have is 5.49 m/s.

_____

<em>Comment on crossing the moat</em>

The kinetic energy at the bottom of the swing translates to potential energy at the end of the swing. At the lower speed, the stuntman cannot rise as high, so will traverse a shorter arc. At 8.45 m/s, the moat could be about 16.8 m wide; at 5.49 m/s, it can only be about 11.5 m wide.

5 0
2 years ago
What’s the answer?? middle question
juin [17]
The new volume = 3 x 52.6 that’s because as the pressure decreases by 1/3 the volume increases x3
6 0
2 years ago
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