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zubka84 [21]
3 years ago
11

What does it depend on

Physics
1 answer:
gavmur [86]3 years ago
7 0

Electric force varies directly with the magnitude of each charge, and inversely with the distance between the charges.

but for short....

electrical force depends on the charge and distance
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A book on a 2-meter high shelf has a mass of 0.4 kg. What is its potential energy?
poizon [28]

Answer:

\boxed {\boxed {\sf 7.84 \ Joules}}

Explanation:

The formula for potential energy is:

PE=m*g*h

where <em>m </em>is the mass, <em>g</em> is the gravitational acceleration, and <em>h</em> is the height.

The mass of the book is 0.4 kilograms. The gravitational acceleration on Earth is 9.8 m/s². The height of the book is 2 meters.

m=0.4 \ kg \\g=9.8 \ m/s^2 \\h=2\ m

Substitute the values into the formula.

PE=(0.4 \ kg)(9.8 \ m/s^2)(2 \ m)

Multiply the first two numbers.

  • 0.4 kg*9.8 m/s²= 3.92 kg*m/s²
  • If we convert the units now, the problem will be much easier later on.
  • 1 kg*m/s² is equal to 1 Newton. So, our answer of 3.92 kg*m/s² is equal to 3.92 N

PE=(3.92 \ N )(2 \ m)

Multiply.

  • 3.92 N* 2 m=7.84 N*m
  • 1 Newton meter is equal to 1 Joule (this is why we converted the units).
  • Our answer is equal to<u> 7.84 Joules.</u>

PE=7.84  \ J

6 0
3 years ago
Read 2 more answers
Which process can transfer energy to a solid metal block
gladu [14]
Eneregy is the ability to to do work it takes work to heat someone heat is energy put the solid metal block in a campfire and heat it up tada wait, we need the options, nevermind
4 0
3 years ago
Read 2 more answers
Two permanent magnets produce a magnetic field pattern as shown below.
AlladinOne [14]

Answer:

S pole and S pole repelling

7 0
3 years ago
A 2.45-kg frictionless block is attached to an ideal spring with force constant 355 N/m. Initially the spring is neither stretch
ANTONII [103]

Answer:

A.    A = 0.913 m

B.    amax = 132.24m/s^2

C.    Fmax = 324.01N

Explanation:

When the block is moving at the equilibrium point , its velocity is maximum.

A. To find the amplitude of the motion you use the following formula for the maximum velocity:

v_{max}=A\omega          (1)

vmax = maximum velocity = 11.0 m/s

A: amplitude of the motion = ?

w: angular frequency = ?

Then, you have to calculate the angular frequency of the motion, by using the following formula:

\omega=\sqrt{\frac{k}{m}}           (2)

k: spring constant = 355 N/m

m: mass of the object = 2.54 kg

\omega = \sqrt{\frac{355N/m}{2.45kg}}=12.03\frac{rad}{s}

Next, you solve the equation (1) for A and replace the values of vmax and w:

A=\frac{v}{\omega}=\frac{11.0m/s}{12.03rad/s}=0.913m

The amplitude of the motion is 0.913m

B. The maximum acceleration of the block is given by:

a_{max}=A\omega^2 = (0.913m)(12.03rad/s)^2=132.24\frac{m}{s^2}

The maximum acceleration is 132.24 m/s^2

C. The maximum force is calculated by using the second Newton law and the maximum acceleration:

F_{max}=ma_{max}=(2.45kg)(132.24m/s^2)=324.01N

It is also possible to calculate the maximum force by using:

Fmax = k*A = (355N/m)(0.913m) = 324.01N

The maximum force exertedbu the spring on the object is 324.01 N

4 0
3 years ago
What belongs in the center section <br>​
MissTica

Answer:

The second one I think

Explanation:

B

8 0
3 years ago
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