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Snezhnost [94]
4 years ago
5

Which factor affects elastic potential energy but not gravitational potential energy?

Physics
2 answers:
julsineya [31]4 years ago
6 0

Explanation:

The energy stored in an elastic objects as a result of deformation is called elastic potential energy. The energy stored in a spring is given by :

E=\dfrac{1}{2}kx^2

Where

k = spring constant

x = compression or stretching in an spring

While gravitational potential energy is given by :

PE = mgh

where

m = mass

g = acceleration due to gravity

h = height from ground

So, the factor affecting elastic potential energy but not gravitational potential energy is " spring constant ".

anyanavicka [17]4 years ago
3 0

Answer:

spring constant  or C

Explanation:

if this helped please leave a thanks and go friend me if it did.

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At the point of fission, a nucleus of 235U that has 92 protons is divided into two smaller spheres, each of which has 46 protons
Gwar [14]

Answer:

Force = 3481.1 N.

Explanation:

Below is an attachment containing the solution.

7 0
3 years ago
Upon being struck by 240-nm photons, a metal ejects electrons with a maximum kinetic energy of 2.97 eV. What is the work functio
Neko [114]

Answer:

Work Function = 3.53 x 10⁻¹⁹ J = 2.2 eV

Explanation:

The work function of the metal metal can be found as follows:

Energy of Photon = Work Function + K.E

hc/λ = Work Function + K.E

Work Function = hc/λ - K.E

where,

h = Plank's Constant = 6.625 x 10⁻³⁴ J.s

c = speed of light  = 3 x 10⁸ m/s

λ = wavelength of photons = 240 nm = 2.4 x 10⁻⁷ m

K.E = Maximum Kinetic Energy = (2.97 eV)(1.6 x 10⁻¹⁹ J/1 eV) =  4.752 x 10⁻¹⁹ J

Therefore,

Work Function = (6.625 x 10⁻³⁴ J.s)(3 x 10⁸ m/s)/(2.4 x 10⁻⁷ m) - 4.752 x 10⁻¹⁹ J

Work Function = 8.281 x 10⁻¹⁹ J - 4.752 x 10⁻¹⁹ J

<u>Work Function = 3.53 x 10⁻¹⁹ J = 2.2 eV</u>

4 0
3 years ago
You break a piece of Styro foam packing material, and it releases lots of little spheres whose electric charge makes them stick
stich3 [128]

Answer:

The magnitude of charge on each sphere is q=2.50\times 10^{-8}\ C.

Explanation:

Given that,

Force of repulsion between the charges, F = 22 mN

The distance between spheres, r = 16 mm = 0.016 m

It is mentioned that both the spheres carry equal charges. The force between charges is given by :

F=\dfrac{kq^2}{r^2}

q=\sqrt{\dfrac{Fr^2}{k}}

q=\sqrt{\dfrac{22\times 10^{-3}\times (0.016)^2}{9\times 10^9}}

q=2.50\times 10^{-8}\ C

So, the magnitude of charge on each sphere is q=2.50\times 10^{-8}\ C. Hence, this is the required solution.

3 0
3 years ago
The manipulated variable is plotted on the horizontal axis
tatuchka [14]
<span>True.

Manipulated variable or also called the controlled variables are variables in which you regulate. Manipulate or as said control. By this you want to certain the outcome of a certain experiment. Making it close to which you want it or desired it to be, possibly. Characteristics could be: </span>
<span><span>1.       </span>You are able to have govern or a certain variable is controllable.</span> <span><span>
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8 0
4 years ago
Maria does 500 Newton-meters of work moving books from one table to another. The total force is 100 Newtons. How far were the bo
Mazyrski [523]

The displacement of the book is 5 meters

Explanation:

The work done by a force when moving an object is given by:

W=Fd cos \theta

where :

F is the magnitude of the force

d is the displacement

\theta is the angle between the direction of the force and the displacement

In this problem, we have:

W = 500 J is the work done by Maria

F = 100 N is the force applied by Maria

\theta=0^{\circ}, assuming the direction of the force is parallel to the displacement

So we can rearrange the equation to solve for d, the displacement:

d=\frac{W}{F cos \theta}=\frac{500}{(100)(cos 0)}=5 m

So, the book was moved 5 meters.

Learn more about work:

brainly.com/question/6763771

brainly.com/question/6443626

#LearnwithBrainly

5 0
3 years ago
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