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2 years ago

A ball is projected into the air. Where is potential energy the greatest? A) A. B) B. C) C. D) A. and C.

Physics

2 answers:

Ksju [112]2 years ago

5 0

The point with the greatest potential energy is B.

The potential energy of an object is depends on the relative distance between the object and the ground and its mass. The higher the object is from the ground the greater the potential energy posses.

Potential energy (P.E) = mgh

Where m is the mass, g is the gravity and h is the height from the ground to where the object is.

Since the mass and gravity is constant in this case, only the height will determine the point with the greatest P.E and that point is B.

aev [14]2 years ago

4 0

Answer:

B) B

Explanation:

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An acorn falls from a tree. Its velocity just before it hits the ground is 28.2 m/s, downward. (acceleration of gravity is 9.81m

ss7ja [257]

Answer:

12.74 ms^-1 download

Explanation:

v=28.2, a=9.81

start from rest u=0

v=u+at=0+(9.81)t=28.2

t=2.875...

it reach 1.4 second before hitting the ground:

t=1.4, u=0, a=9.81

v=u+at=0+(9.81)(1.4)=12.74

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3 years ago

Anyone help me please?

choli [55]

Answer:

I believe that its B my apologies if its wrong/

Explanation:

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3 years ago

It is easier to roll a stone up a sloping road than to lift it vertical upwards because.

kvasek [131]

Answer:

Because the force one needs to apply is reduced.

Explanation:

Work can be expressed in terms of force and distance as follows,

Work = Force × distance

To be more precise, the selection of these two parameters should be in a manner that the distance is what the line of action of the force traces.
It is obvious that for a given amount of work when one parameter is changed the other changes to counteract the change of the first one.
Imagine the situation shown in the attached figure. The amount of work that needs to be done during the lifting of the mass to a height of $\small h$ vertically is $E = mg\times h = mgh$ .
On the slope, however, one needs to push the object an $L$ distance up to get the object to that $h$ height ultimately.
The work he does on the slope is $mgh = F\times L$ and since $L > h$ , $F < mg$ .
The force that needs to be applied is less than that applied in the vertical lift. Hence it feels easier. #SPJ4

3 0

1 year ago

If i try to fail and succeed which one did I do

Novay_Z [31]

You've failed because you failing becomes a statement rather than it becoming fact or what actually happened.

7 0

2 years ago

A very strong, but inept, shot putter puts the shot straight up vertically with an initial velocity of 11.0 m/s. how long does h

noname [10]

The shot putter should get out of the way before the ball returns to the launch position.

Assume that the launch height is the reference height of zero.
u = 11.0 m/s, upward launch velocity.
g = 9.8 m/s², acceleration due to gravity.

The time when the ball is at the reference position (of zero) is given by
ut - (1/2)gt² = 0
11t - 0.5*9.8t² = 0
t(11 - 4.9t) = 0
t = 0 or t = 4.9/11 = 0.45 s

t = 0 corresponds to when the ball is launched.
t = 0.45 corresponds to when the ball returns to the launch position.

Answer: 0.45 s

7 0

2 years ago