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Paraphin [41]
3 years ago
14

What are the potential out comes of force

Physics
1 answer:
AysviL [449]3 years ago
5 0

Answer:

F in the definition of potential energy is the force exerted by the force field, e.g., gravity, spring force, etc. The potential energy U is equal to the work you must do against that force to move an object from the U=0 reference point to the position r.

Explanation:

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On the moon what would be the force of gravity acting on an object that has a mass of 7 kg?
Rudiy27

Answer:

0

Explanation:

4 0
4 years ago
9. What could cause scientist to change their mind?
MakcuM [25]
A if it is one answer and if it’s multiple choice A and D
6 0
2 years ago
The image below shows two opposite forces acting on a rope, what can we say is true about the affect of the forces on the rope?
Tamiku [17]
It would be D) the rope is pulled to the right. This is because their is a greater force in that direction.
8 0
3 years ago
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At an accident scene on a level road, investigators measure a car's skid mark to be 98 m long. It was a rainy day and the coeffi
shutvik [7]

Answer:

The speed of the car was 28 m/s.

Explanation:

Hi there!

The initial kinetic energy of the car, KE, is equal to the negative work, W, done by friction to bring the car to stop. Let´s write the work-energy theorem:

W = ΔKE = final kinetic energy - initial kinetic energy

In this case, the final kinetic energy is zero, then:

W = - initial kinetic energy

Since the work done by friction is negative (the work is done in opposite direction to the movement of the car), then:

Wfr = initial kinetic energy

The work done by friction is calculated as follows:

Wfr = Fr · d

Where:

Fr = friction force.

d = distance.

The friction force is calculated as follows:

Fr = N · μ

Where:

N = normal force.

μ = coefficient of friction

Since the only vertical forces acting on the car are the weight of the car and the normal force, and the car is not being accelerated in the vertical direction, the normal force has to be equal to the weight of the car (with opposite sign).

Then the friction force can be written as follows:

Fr = m · g · μ

Where:

m = mass of the car.

g =  acceleration due to gravity (9.8 m/s²)

The work done by friction will be:

W = m · g · μ · d

The equation of kinetic energy is the following:

KE =  1/2 · m · v²

Where

m = mass of the car.

v = speed.

Then:

W = KE

m · g · μ · d = 1/2 · m · v²

2 · g · μ · d = v²

2 · 9.8 m/s² · 0.40 · 98 m = v²

v = 28 m/s

The speed of the car was 28 m/s.

7 0
4 years ago
X-rays with an energy of 400 keV undergo Compton scattering with a target. If the scattered X-rays are detected at \theta = 30^{
marissa [1.9K]

Answer:

37.91594 keV

Explanation:

E_i = Incident energy = 400 keV

θ = 30°

h = Planck's constant = 4.135×10⁻¹⁵ eV s = 6.626×10⁻³⁴ J s

Incident photon wavelength

\lambda_i=\frac{hc}{E_i}\\\Rightarrow \lambda_i=\frac{4.135\times 10^{-15}\times 3\times 10^8}{400\times 10^3}\\\Rightarrow \lambda_i=3.101\times 10^{-12}\ m

Difference in wavelength

\Delta \lambda=\frac{h}{m_ec}(1-cos\theta)\\\Rightarrow \Delta \lambda=\frac{6.626\times 10^{-34}}{9.11\times 10^{-31}\times 3\times 10^8}(1-cos30)\\\Rightarrow \Delta \lambda=3.248\times 10^{-13}\ m

\lambda_f=\lambda_i+\Delta \lambda\\\Rightarrow \lambda_f=3.101\times 10^{-12}+3.248\times 10^{-13}\\\Rightarrow \lambda_f=3.426\times 10^{-12}

Final photon wavelength

\lambda_f=\frac{hc}{\lambda_f}\\\Rightarrow E_f=\frac{4.135\times 10^{-15}\times 3\times 10^8}{3.426\times 10^{-12}}\\\Rightarrow E_f=362084.06\ eV = 362.08406\ keV

Energy of the recoiling electron

\Delta E=E_i-E_f\\\Rightarrow \Delta E=400-362.08406=37.91594\ keV

Energy of the recoiling electron is 37.91594 keV

8 0
4 years ago
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