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3 years ago

PLS HELP

Physics

1 answer:

Ahat [919]3 years ago

7 0

"This was the idea that non-living objects can give rise to living organisms."

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E<br> 3.6 What force is needed to give a mass of<br> 20 kg an acceleration of 5 m/s??

luda_lava [24]

Explanation:

Mass(m)= 20kg
Acceleration (a)= 5m/s²
Force(F)= ?

We know that,

F=ma
F=20×5
F=100N

Hence, the needed force is 100N.

6 0

1 year ago

Read 2 more answers

A baseball is released at rest from the top of the Washington Monument. It hits the ground after falling for 6 s. What was the h

alukav5142 [94]

Answer:

Total height (s) = 176.4 m

Explanation:

Given:

Initial velocity (u) = 0 m/s

Time taken (t) = 6 sec

Acceleration due to gravity = 9.8 m/s²

Find:

Total height (s)

Computation:

s = ut + [1/2]gt²

s = (0)(6) + [1/2][9.8][6²]

s = 176.4 m

Total height (s) = 176.4 m

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3 years ago

How to find the velocity using a graph

HACTEHA [7]

The slope of the line on a velocityversus time graph is equal to the acceleration of the object. If the object is moving with an acceleration of +4 m/s/s (i.e., changing its velocity by 4 m/s per second), then the slope of the line will be +4 m/s/s.

4 0

3 years ago

Calculate the centripetal force (in N) on the end of a 57 m (radius) wind turbine blade that is rotating at 0.3 rev/s. Assume th

enot [183]

Answer:

Check image.

Explanation:

This is what my solution is, not a professional tutor so take my answer with a grain of salt and check.

5 0

3 years ago

Suppose the Sun appeared to you 900 times dimmer than it does now. How far away from the Sun would you be? (A) 1/9 AU (B) 3 AU

Harrizon [31]

Answer:

The correct answer is option 'c': 30 AUs

Explanation:

For a spherical wave front emitted by sun with total energy 'E' the energy density over the surface when it is at a distance 'r' from the sun is given by

$e=\frac{E}{4\pi r^{2}}$

This energy per unit area is sensed by observer as intensity of the sun.

Let the initial intensity of sun at a distance $r_{1}$ be $e_{1}$

Thus if the sun becomes 900 times dimmer we have

$e'=\frac{e_{1}}{900}\\\\\frac{E}{4\pi r_{2}^{2}}=\frac{1}{900}\times \frac{E}{4\pi r_{1}^{2}}\\\\\Rightarrow r_{2}^{2}={r_{1}^{2}}\times 900\\\\\therefore r_{2}={r_{1}}\times {30}$

Thus the distance increases 30 times.

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3 years ago