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Citrus2011 [14]
3 years ago
13

The effect of gravity on a falling object can be modeled by a ball dropped from different heights. What is a limitation of this

model?
A. Not all objects bounce, even though balls do.

B. Some balls float in water, while others sink.

C. The ball can be dropped from varying heights.

D. Friction with air also affects the fall of the object.

Correct answer is D
Physics
2 answers:
Delvig [45]3 years ago
5 0

Answer:

D.

Explanation:

try it..  *lil uzi vert's voice*

Lelechka [254]3 years ago
4 0

Answer:

D

Explanation:

Just did it on A P E X

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As shown in the picture above, a sled of mass 67 kilograms is pulled
Scorpion4ik [409]

Explanation:

F net of sled = Tension force by rope - Kinetic friction between ground.

F normal of sled = mg = (67kg)(9.81kg/m^2) = 657.27N.

Kinetic friction = 0.18 (I cannot see the value) * Normal force of sled = 0.18 * 657.27N = 118.31N

So F net of sled = 800N - 118.31N = 681.69N.

(I cannot see what the question is asking for, please check on your own!)

3 0
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When we get out of the bed on a very cold morning, we feel that the air of the room is cold. But when we come back after staying
S_A_V [24]

Answer:

This is because the air outside is always cooler than the air inside, so after staying outside your body adapts to the cold air, when you come back inside, the cold air is still in you which makes the room seem warmer.

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3 years ago
The phases of the moon occur because the moon revolves around the earth.<br><br> True<br><br> False
Ira Lisetskai [31]

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7 0
3 years ago
"What is the magnifying power of an astronomical telescope using a reflecting mirror whose radius of curvature is 5.9 m and an e
notka56 [123]

Answer:

The Magnifying power of a telescope is M = 109.26

Explanation:

Radius of curvature R = 5.9 m = 590 cm

focal length of objective f_{objective} = \frac{R}{2}

⇒ f_{objective} = \frac{590}{2}

⇒ f_{objective} = 295 cm

Focal length of eyepiece f_{eyepiece} = 2.7 cm

Magnifying power of a telescope is given by,

M = \frac{f_{objective} }{f_{eyepiece} }

M = \frac{295}{2.7}

M = 109.26

therefore the Magnifying power of a telescope is M = 109.26

4 0
3 years ago
pls help quick. the number line shows the starting and ending velocities for ball 1 what's the change in velocity of ball 1 calc
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Answer:

The starting velocity for ball 1 is 1.00 meter/second. Its ending velocity is 0.25 meter/second.

The change in velocity for ball 1 is 0.25 – 1.00 = -0.75 meter/seconds

7 0
3 years ago
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