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Lemur [1.5K]
2 years ago
13

A stuntman of mass 48 kg is to be launched horizontally out of a spring-

Physics
1 answer:
Damm [24]2 years ago
6 0

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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In both the camera and the __________, light enters a narrow opening and is projected onto a photosensitive surface. Group of an
Citrus2011 [14]

Answer: The HUMAN EYE

Explanation:

The human eye is made up of different parts which ranges from controlling the amount of light that enters the eye to the focusing of the image that is formed. The camera is a device which is both mechanically and electronically operated which shares a number of similarities with the eye.

In the human eye, the IRIS helps to regulate the amount of rays passing through the pupil to the lens by either contracting or dilating in light or dark environment respectively. While in the camera, the DIAPHRAGM controls the amount of light entering the camera.

The PUPIL serves as the passage for light into the eye while in the camera, the APERTURE does the same.

The photosensitive surface in the eye is the YELLOW SPOT while in the camera, the photosensitive surface is the PHOTOGRAPHIC FILM.

5 0
3 years ago
A rocket-launching vehicle is moving forward at a constant velocity of 5 m/s. a cannon on the vehicle shoots a shell straight up
Readme [11.4K]
For any object thrown upwards where only the force of gravity is acting upon it, uses the following formula for the maximum height attained.

H= v²/2g, where g = 9.81 m/s²

There are two information of velocities are given. However, we use the 20 m/s information because this is the launch velocity. Hence, the solution is as follows:

H = (20 m/s)²/2(9.81 m/s²)
<em>H = 20.4 m</em>
5 0
3 years ago
A circular bird feeder 19.0 cm in radius has rotational inertia 0.130 kg·m2. It's suspended by a thin wire and is spinning slowl
soldi70 [24.7K]

Answer:

N₂=20.05 rpm

Explanation:

Given that

R= 19 cm

I=0.13 kg.m²

N₁ = 24.2 rpm

\omega_1=\dfrac{2\pi \times 24.2}{60}\ rda/s

ω₁= 2.5 rad/s

m= 173 g = 0.173 kg

v=1.2 m

Initial angular momentum L₁

L₁ =  Iω₁  - m v r       ( negative sign because bird coming opposite to motion of the wire motion)

Final linear momentum L₂

L₂=  I₂ ω₂

 I₂ = I + m r²

The is no any external torque that is why angular momentum will be conserve

L₁ = L₂

Iω₁  - m v r =  I₂ ω₂

Iω₁  - m v r =  ( I + m r²) ω₂

Now by putting the all values

Iω₁  - m v r =  ( I + m r²) ω₂

0.13 x 2.5 - 0.173 x 1.2 x 0.19 =  ( 0.13 + 0.173 x  0.19²) ω₂

0.325  - 0.0394 = 0.136 ω₂

ω₂ = 2.1 rad/s

\omega_2=\dfrac{2\pi \times N_2}{60}

N₂=20.05 rpm

3 0
3 years ago
Find the radius Rrigel of the star Rigel, the bright blue star in the constellation Orion that radiates energy at a rate of 2.7×
Zinaida [17]

Answer:

r = 1.61 x 10^{11} m

Explanation:

energy radiated (H) = 2.7 x 10^31 W

surface temperature (T)  = 11,000 k

assuming ε = 1 and taking σ = 5.67 x 10^{-8} W/m^{2}.K^{4}

we can find the radius of the star from the equation below

H = A x  ε x σ x T^{4}              

where area (A) = 4 x π x r^{2} (assuming it is a sphere)

therefore  the equation becomes

H = 4 x π x r^{2} x  ε x σ x T^{4}  

2.7 x 10^31  = 4 x π x r^{2} x  1 x 5.67 x 10^{-8}  x (11,000)^{4}

r = \sqrt{\frac{2.7 x 10^31}{4 x π x 1 x 5.67 x 10^{-8}  x (11,000)^{4}} }

r = 1.61 x 10^{11} m

       

4 0
3 years ago
What is the standard measurement system used by scientists around the world?
sleet_krkn [62]

Answer:

Hey mate....

Explanation:

This is ur answer....

<h3><em>The International System of Units (SI), commonly known as the metric system, is the international standard for measurement.</em></h3>

Hope it helps!

Brainliest pls!

Follow me! ◇

5 0
3 years ago
Read 2 more answers
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