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disa [49]
3 years ago
10

What is the magnitude of the velocity after it hits the ground?

Physics
1 answer:
Digiron [165]3 years ago
7 0

Answer:

Multiply the time by the acceleration due to gravity to find the velocity when the object hits the ground. If it takes 9.9 seconds for the object to hit the ground, its velocity is (1.01 s)*(9.8 m/s^2), or 9.9 m/s.

Explanation:

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A set of charged plates 0.00262 m apart has an electric field of 155 N/C between them. What is the potential difference between
mylen [45]

Answer: The potential difference between the plates = 0.4061V

Explanation:

Given that the

Electric field strength E = 155 N/C

Distance d = 0.00262 m

From the definition of electric field strength, is the ratio of potential difference V to the distance between the plates. That is

E = V/d

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155 = V/0.00262

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A stuntman of mass 48 kg is to be launched horizontally out of a spring-
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The velocity of the stuntman, once he has left the cannon is 5 m/s.

The right option is O A. 5 m/s

The Kinetic energy of the stuntman is equal to the elastic potential energy of the spring.

<h3 /><h3>Velocity: </h3>

This is the ratio of displacement to time. The S.I unit of Velocity is m/s.  The velocity of the stuntman can be calculated using the formula below.

⇒ Formula:

  • mv²/2 = ke²/2
  • mv² = ke².................. Equation 1

⇒ Where:

  • m = mass of the stuntman
  • v = velocity of the stuntman
  • k = force constant of the spring
  • e = compression of the spring

⇒ Make v the subject of the equation

  • v = √(ke²/m)................. Equation 2

From the question,

⇒ Given:

  • m = 48 kg
  • k = 75 N/m
  • e = 4 m

⇒ Substitute these values into equation 2

  • v = √[(75×4²)/48]
  • v = √25
  • v = 5 m/s.

Hence, The velocity of the stuntman, once he has left the cannon is 5 m/s.

The right option is O A. 5 m/s

Learn more about velocity here: brainly.com/question/10962624

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