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3 years ago

A ball is thrown vertically upward. What is its acceleration right before it hits the earth?

2 answers:

7 0

<span>The entire time the ball is in the air, its acceleration is 9.8 m/s2 down provided this occurs on the surface of the Earth. Note that the acceleration can be either 9.8 m/s2 or -9.8 m/s2.
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Levart [38]3 years ago

4 0

It increases. Because it hasn't hit the ground yet.

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A wire has a current that flows to the right.

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D) circles around the wire that go in at the bottom

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2 years ago

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A kayaker moves 22 meters northward, then 18 meters southward , and finally 24 meters northward what is the distance and magnitu

Elina [12.6K]

Answer:

Distance = 64 metres

Displacement = 28 metres

Direction = northward

Explanation:

Given that a kayaker moves 22 meters northward, then 18 meters southward , and finally 24 meters northward

Distance covered is only about the magnitude of the length covered.

Distance = 22 + 18 + 24 = 64 metres

The direction will be considered when calculating the displacement

Let northward be positive and southward be negative

Displacement = 22 - 18 + 24 = 28 metres

Since the displacement is positive, the direction of the motion is northward.

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2 years ago

Increase the mass of the beanbag its gravitational potential energy will

kotegsom [21]

The gravitational potential energy will increase

Explanation:

The gravitational potential energy (GPE) of an object is the energy possessed by the object due to its position in a gravitational field.

Near the Earth's surface, the GPE of an object is given by

$GPE=mgh$

where

m is the mass of the object

g is the acceleration of gravity

h is the heigth of the object above the ground

From the equation, we see that the GPE is directly proportional to the mass: therefore, if the mass increases, the GPE will increase as well.

So, for the beanbag in this problem, when its mass increases, the GPE will increase as well.

Learn more about gravitational potential energy:

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2 years ago

You are camping with two friends, Joe and Karl. Since all three of you like your privacy, you don't pitch your tents close toget

aleksklad [387]

Answer:

35.7 m

Explanation:

Let

$\mid A\mid=18.5 m$

$\mid B\mid=41 m$

We have to find the distance between Joe's and Karl'e tent.

$A_x=Acos\theta$

$A_y=Asin\theta$

Substitute the values then we get

$A_x=18.5cos23^{\circ}=17 m$

$A_y=18.5sin 23^{\circ}=7.2 m$

$B_x=41cos37.5^{\circ}=32.5 m$

$B_y=41sin37.5^{\circ}=-24.96 m$

Because vertical component of B lie in IV quadrant and y-inIV quadrant is negative.

By triangle addition of vector

$B=A+C$

$C=B-A$

$C_x=B_x-A_x=32.5-17=15.5 m$

$C_y=B_y-A_y=-24.96-7.2=-32.16\approx=-32.2 m$

$\mid C\mid=\sqrt{C^2_x+C^2_y}$

$\mid C\mid=\sqrt{(15.5)^2+(-32.2)^2}=35.7 m$

Hence, the distance between Joe's and Karl's tent=35.7 m

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2 years ago

2. Which of the following accurately describes the effects of temperature changes on solids? A. If the temperature of a solid is

balandron [24]

<span>B. If the temperature of a solid is increased, the solid expands and its volume increases.</span>

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3 years ago

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