All categories

3 years ago

How high above the ground would a 10 kg object needs to be to have the same GPE as the 30 kg object in the example

Physics

1 answer:

Black_prince [1.1K]3 years ago

4 0

Answer:

The 10 kg mass would have to be at a height 3 times that of the 30 kg mass to have the same gravitational potential energy as the 30 kg mass.

Explanation:

Gravitational potential energy, U = mgh where m = mass of object, g = acceleration due to gravity = 9.8 m/s² and h = height of object above the ground.

Now, let U' = gravitational potential energy of 10 kg mass = m'gh' where m' = 10 kg and h' = height of 10 kg mass above the ground.

So, U' = 10gh'

Also, let U" = gravitational potential energy of 30 kg mass = m"gh" where m" = 30 kg and h" = height of 30 kg mass above the ground.

So, U" = 30gh"

Since both object are supposed to have the same potential energy at a given height of the 10 kg mass,

U' = U"

10gh' = 30gh"

h' = 30gh"/10g

h' = 3h"

<u>So, the 10 kg mass would have to be at a height 3 times that of the 30 kg mass to have the same gravitational potential energy as the 30 kg mass.</u>

You might be interested in

When writing a summary of current research, you should _____.

tangare [24]

Answer:

Explanation:

A is completely necessary, in fact citing your sources is important. So false.

B is correct, unreliable bad research is not good to use

C is incorrect because not everything may be truthful

D is not necessary because it may be disturbing to the reader.

Hope it helped! hit the brainliest button if it's there! <3

3 0

3 years ago

Read 2 more answers

In a vehicle traveling at 20 miles per hour, the force of your car impacting a surface is ______ times as great as at 10 MPH. Tw

bogdanovich [222]

Answer:

Explanation:

7 0

3 years ago

The membrane of the axon of a nerve cell is a thin cylindrical shell of radius r = 10-5 m, length L = 0.32 m, and thickness d =

maksim [4K]

Answer:

5.3 x 10⁻⁹ C

Explanation:

r = radius of cylindrical shell = 10⁻⁵ m

L = length = 0.32 m

A = area

Area is given as

A = 2πrL

A = 2 (3.14) (10⁻⁵) (0.32)

A = 20.096 x 10⁻⁶ m²

d = separation = 10⁻⁸ m

$k_{appa}$ = dielectric constant = 4

Capacitance is given as

$Q=\frac{k_{appa}\epsilon _{o}A}{d}$ eq-1

V = Potential difference across the membrane = 74 mV = 0.074 Volts

Q = magnitude of charge on each side

Magnitude of charge on each side is given as

Q = CV

using eq-1

$Q=\frac{k_{appa} \epsilon _{o}AV}{d}$

Inserting the values

$Q=\frac{4 (8.85\times 10^{-12})(20.096\times 10^{-6})(0.074)}{10^{-8}}$

Q = 5.3 x 10⁻⁹ C

4 0

3 years ago

Suppose you push horizontally with precisely enough force to make the block start to move, and you continue to apply the same am

Tcecarenko [31]

Answer:

Incomplete question,

This is the complete question

Suppose you push horizontally with precisely enough force to make the block start to move, and you continue to apply the same amount of force even after it starts moving. Find the acceleration ~a of the block after it begins to move. Express your answer in terms of some or all of the variables µs, µk, and m, as well as the acceleration due to gravity g.

Explanation:

Let the force that make the object to start moving be F,

Frictional force is opposing the motion, the body has to overcome two frictional forces acting in the opposite direction of the motion.

Also, weight and normal reaction are acting in vertical direction, the weight is acting downward while the reaction is acting upward.

Weight of the object is given as

W=mg

Analyzing the vertical motion i.e y-axis.

ΣF = ma

since the body is not moving upward, the a=0

N-W=0

Then, N=W

So, N=mg

So, from friction law

Fr=µN

For static

Fs=µsN

For kinetic or dynamic

Fk= µkN

Using newton law

Along x-axis

Before the body start moving we can get the Force and since the force is the same use to start the block in motion.

Then,

ΣF = ma

Since at static the body is not moving then, a=0

F-Fs=0

F=Fs

Since, Fs=µsN

F=Fs=µsN

Then, the force to keep the body in motion too is F=µsN

Now analyses when the body is in motion

ΣF = ma

F-Fk=ma

ma=F - Fk

Substituting F=µsN and Fk=µkN

ma=µsN - µkN

ma=N(µs - µk)

Since N=mg

Then, ma=mg(µs - µk)

m cancels out, then

a=g(µs - µk)

Then the acceleration of the body is given as "a=g(µs - µk)"

5 0

3 years ago

How far will a free falling object fall in 8.7 secs if it started from rest? Remember acceleration is negative for free fall. Do

sleet_krkn [62]

Answer:

h~=371.26m

Explanation:

when an object falls we use the equations of accelerated motion. There is only one that gives distance.

$x = ut + \frac{1}{2} a {t}^{2}$

Since we have no initial velocity (started from rest) we can get rid of the (ut) term

where a we substitute g (gravitational acceleration, constant for given heights and almost 9.81m/s^2).

$h = \frac{1}{2} g {t}^{2} = \frac{1}{2} \times 9.81 \times {8.7}^{2} = 371.26m$

4 0

3 years ago