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kipiarov [429]
3 years ago
8

What happens during the main sequence stage of a star?

Physics
1 answer:
Goryan [66]3 years ago
5 0
We the human earths rotate round him
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a ball is thrown striaght up in the air and then falls back to earth. if the downward fall takes 2.2s, how fast is the ball trav
lapo4ka [179]

The velocity of the ball when it strikes the ground, given the data is 21.56 m/s

<h3>Data obtained from the question</h3>

From the question given above, the following data were obtained:

  • Time to reach ground from maximum height (t) = 2.2 s
  • Initial velocity (u) = 0 m/s
  • Acceleration due to gravity (g) = 9.8 m/s²
  • Final velocity (v) =?

<h3>How to determine the velocity when the ball strikes the ground</h3>

The velocity of the ball when it strikes the ground can be obtained as illustrated below:

v = u + gt

v = 0 + (9.8 × 2.2)

v = 0 + 21.56

v = 21.56 m/s

Thus, the velocity of the ball when it strikes the ground is 21.56 m/s

Learn more about motion under gravity:

brainly.com/question/22719691

#SPJ1

5 0
1 year ago
What does popular sovereignty
Lana71 [14]

I'm a little confused by your question.

If you mean what is popular sovereignty, it is the belief that the governments is made and sustained by the consent of the people that it governs.

6 0
3 years ago
Pls give me an answer
wolverine [178]

Answer:

IV what is it's potential energy at the maximum height

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2 years ago
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Nikitich [7]

Answer:b speed with displacement

Explanation:

8 0
3 years ago
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A model airplane with a mass of 0.741kg is tethered by a wire so that it flies in a circle 30.9 m in radius. The airplane engine
liq [111]

(a) 24.6 Nm

The torque produced by the net thrust about the center of the circle is given by:

\tau = F r

where

F is the magnitude of the thrust

r is the radius of the wire

Here we have

F = 0.795 N

r = 30.9 m

Therefore, the torque produced is

\tau = (0.795 N)(30.9 m)=24.6 N m

(b) 0.035 rad/s^2

The equivalent of Newton's second law for a rotational motion is

\tau = I \alpha

where

\tau is the torque

I is the moment of inertia

\alpha is the angular acceleration

If we consider the airplane as a point mass with mass m = 0.741 kg, then its moment of inertia is

I=mr^2 = (0.741 kg)(30.9 m)^2=707.5 kg m^2

And so we can solve the previous equation to find the angular acceleration:

\alpha = \frac{\tau}{I}=\frac{24.6 Nm}{707.5 kg m^2}=0.035 rad/s^2

(c) 1.08 m/s^2

The linear acceleration (tangential acceleration) in a rotational motion is given by

a=\alpha r

where in this problem we have

\alpha = 0.035 rad/s^2 is the angular acceleration

r = 30.9 m is the radius

Substituting the values, we find

a=(0.035 rad/s^2)(30.9 m)=1.08 m/s^2

4 0
3 years ago
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