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Flura [38]
1 year ago
12

If the same elevator accelerates downwards

Physics
1 answer:
PilotLPTM [1.2K]1 year ago
6 0

The upward force exerted by the elevator floor on the passenger is 7.6 times mass of the passenger.

<h3>What is force?</h3>

A force in physics is an effect that has the power to alter an object's motion. An object with mass can change its velocity, or accelerate, as a result of a force. An obvious way to describe force is as a push or a pull. A force is a vector quantity since it has both magnitude and direction.

Given in question elevator accelerates downwards with an acceleration of 1.2 m/s². There is acceleration due to gravity working downward which is 9.8 m/s².

So, net acceleration due to gravity is given as 9.8 - 1.2 = 7.6 m/s²

The upward force exerted by the elevator floor on the passenger is mass times acceleration, 7.6 times mass of passenger.

To learn more about force refer to the link:

brainly.com/question/13191643

 #SPJ1

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Answer: 500 N

Explanation:

The formula to find the force exerted by a mass, we may use F = mg, where g, the gravity, and a, the acceleration, can be interchangeable in the formula.

1) F = 50 x 10
2) F = 500 N

Hope this helps, brainliest would be appreciated :)
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3 years ago
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Answer:

Gravity is the force by which a planet or other body draws objects toward its center. The force of gravity keeps all of the planets in orbit around the sun.

<em><u>Please mark as brainliest</u></em>

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3 years ago
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A screen is placed 1.20m behind a single slit. The central maximum in the resulting diffraction pattern on the screen is 1.40cm
andrew11 [14]

Answer:

2.8 cm

Explanation:

y_1 = Separation between two first order diffraction minima = 1.4 cm

D = Distance of screen = 1.2 m

m = Order

Fringe width is given by

\beta_1=\dfrac{y_1}{2}\\\Rightarrow \beta_1=\dfrac{1.4}{2}\\\Rightarrow \beta_1=0.7\ cm

Fringe width is also given by

\beta_1=\dfrac{m_1\lambda D}{d}\\\Rightarrow d=\dfrac{m_1\lambda D}{\beta_1}

For second order

\beta_2=\dfrac{m_2\lambda D}{d}\\\Rightarrow \beta_2=\dfrac{m_2\lambda D}{\dfrac{m_1\lambda D}{\beta_1}}\\\Rightarrow \beta_2=\dfrac{m_2}{m_1}\beta_1

Distance between two second order minima is given by

y_2=2\beta_2

\\\Rightarrow y_2=2\dfrac{m_2}{m_1}\beta_1\\\Rightarrow y_2=2\dfrac{2}{1}\times 0.7\\\Rightarrow y_2=2.8\ cm

The distance between the two second order minima is 2.8 cm

8 0
3 years ago
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

Substitute E and d into the above formula

155 = V/0.00262

Cross multiply

V = 155 × 0.00262

V = 0.4061 V

The potential difference between the plates is 0.4061 V

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