Ask question

Ask question

All categories

3 years ago

7

which object has the most gravitational potential energy? A. an 8 kg book at a height of 3m B. an 5 kg book at a height of 3 m C

. A 5 kg book at a height of 2 m D. a book at a height of 2 m

2 answers:

astra-53 [7]3 years ago

6 0

I think A is the correct answer because its high is more higher compared to the others, and the mass really does not matter, to know the gravitational potential energy, we need to know how high the object is located because gravity does not show any favor to an object that has more mass or an object that doesnt

amid [387]3 years ago

5 0

Answer: an 8 kg book at a height of 3m

Explanation:

Potential energy is the energy possessed by an object by virtue of its position.

$P.E=m\times g\times h$

m= mass of object = 200 kg

g = acceleration due to gravity = $10ms^{-2}$

h = height of an object

1. an 8 kg book at a height of 3m

$P.E=8kg\times 10ms^{-2}\times 3m=240Joules$

2. an 5 kg book at a height of 3 m

$P.E=5kg\times 10ms^{-2}\times 3m=150Joules$

3. an 5 kg book at a height of 2 m

$P.E=5kg\times 10ms^{-2}\times 2m=100Joules$

4. a book at a height of 2 m

$P.E=0kg\times 10ms^{-2}\times 3m=0Joules$

Thus a 8 kg book at a height of 3 m has most potential energy.

You might be interested in

An object has an average acceleration of +6.18 m/s2 for 0.266 s . At the end of this time the object's velocity is +9.90 m/s .

Archy [21]

At the start of the 0.266 s, the object's speed was 8.26 m/s.

The question can only be talking about speed, not velocity.

4 0

3 years ago

Which describes why people on earth can see light from the stars in the sky that are so far away?

Ugo [173]

Yes, C is correct. It self explains itself as we know light travels through a vacuum ( doesn't need a medium) and light is a type of electromagnetic wave.

7 0

3 years ago

4. A ball is thrown vertically upward from the ground with a velocity of 30m/s. (a) how long will it take to rise to the highest

yarga [219]

All the answers are:

a) The time that will it take to rise to the highest point is 3.06 seconds.

b) The ball will rise to a height of 45.87 meters.

c) The time at which the ball will have a velocity of 10 m/s upward is 2.04 seconds.

The time when the ball has 10 m/s downward is 1.02 seconds.

d) The displacement of the ball will be zero at 6.12 seconds.

e) The time when the magnitude of the ball's velocity is equal to half its velocity of projection is 1.53 seconds.

f) The ball's displacement is equal to half the maximum height to which it rises after 0.90 seconds.

g) In each moment (upward and downward) the magnitude of the acceleration is the value of g (9.81 m/s²) and is a vector in the negative y-direction.

Let's calculate the values for each case.

a) At the highest point, the final velocity is 0, so we can use the following equation.

$v_{f}=v_{i}-gt$ (1)

Where:

v(i) is the initial velocity
v(f) is the final velocity
g is the acceleration due to gravity (9.81 m/s²)

We know that v(i) = 30 m/s.

$0=30-9.81t$

Solve it for t:

$t=3.06\: s$

Hence, the time is 3.06 s.

b) At the highest point, the final velocity is 0, so we can use the following equation.

$v_{f}^{2}=v_{i}^{2}-2gh$ (2)

$0=v_{i}^{2}-2gh$

We know that the initial velocity is 30 m/s.

$0=30^{2}-2gh$

Solving it for h we have:

$h=\frac{30^{2}}{2*9.81}$

$h=45.87 \: m$

Then, the height is 45.87 m.

c) Using equation (1) we can find the time (t).

$10=30-(9.81t)$

So, the time elapsed to get 10 m/s is:

$t_{upward}=2.04\: s$

We know the upward time is equal to the downward time. So the time from v=10 m/s to v=0 m/s will be.

$t_{upward}=2.04+t$

$t=1.02\: s$

This is the time when the ball has 10 m/s downward.

Therefore, the time upward is 2.04 s, and the time downward is 1.02 s.

d) It will be when the ball returns to the ground.

$t=2t_{upward}$

$t=2*3.06$

$t=6.12\: s$

The displacement will be zero after 6.12 s.

e) Here we need to find the time when v(f) is 15 m/s

$15=30-gt$

$t=\frac{15}{9.81}$

$t=1.53\: s$

The time when the v(f) is 15 m/s is 1.53 s.

f) Here, we need to find t when h = 45.87/2 m = 22.94 m

We can use the next equation:

[tex]h=v_{i}t-0.5gt^{2}/tex]

[tex]22.94=30t-0.5*9.81*t^{2}/tex]

Solving this quadratic equation, t will be:

[tex]t=0.90\: s/tex]

Hence, the ball's displacement is equal to half the maximum h, at 0.90 s.

g) In each moment the magnitude of the acceleration is the value of g (9.81 m/s²) and is a vector in the negative y-direction.

Learn more about vertical motion here:

brainly.com/question/13966860

I hope it helps you!

3 0

2 years ago

The law of conservation of energy states:

Alborosie

Answer:

In physics and chemistry, the law of conservation of energy states that the total energy of an isolated system remains constant; it is said to be conserved over time. ... For instance, chemical energy is converted to kinetic energy when a stick of dynamite explodes.

6 0

3 years ago

The position of a 55 g oscillating mass is given by x(t)=(2.0cm)cos(10t), where t is in seconds. determine the velocity at t=0.4

NNADVOKAT [17]

The position of the mass is given by (in cm):
$x(t)=2 \cos (10 t)$
The velocity is the derivative of the position:
$v(t) = \frac{dx(t)}{dt} =-10\cdot 2 \sin (10t)=-20 \sin (10t)$
Substituting t=0.40 s, we can find the velocity at this time:
$v(0.40 s)= -20 \sin (10 \cdot 0.4)=15 cm/s=15 \cdot 10^{-2}m/s$

5 0

3 years ago

Read 2 more answers