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KengaRu [80]
1 year ago
15

The kinetic energy of an object is increased by a factor of 4 . By what factor is the magnitude of its momentum changed?(a) 16(b

) 8(c) 4 (d) 2(e) 1
Physics
1 answer:
maw [93]1 year ago
4 0

The kinetic energy of an object is increased by a factor of 4 . By what factor is the magnitude of its momentum changed: 2.

<h3>What is kinetic energy?</h3>
  • A particle or an item that is in motion has a sort of energy called kinetic energy. An item accumulates kinetic energy when work, which involves the transfer of energy, is done on it by exerting a net force.
  • Kinetic energy comes in five forms: radiant, thermal, acoustic, electrical, and mechanical.
  • The energy of a body in motion, or kinetic energy (KE), is essentially the energy of all moving objects. Along with potential energy, which is the stored energy present in objects at rest, it is one of the two primary types of energy.
  • Explain that a moving object's mass and speed are two factors that impact the amount of kinetic energy it will possess.

The kinetic energy of an object is increased by a factor of 4 . By what factor is the magnitude of its momentum changed: 2.

To learn more about kinetic energy, refer to:

brainly.com/question/25959744

#SPJ4

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A uniform electric field exists in the region between two oppositely charged plane parallel plates. A proton is released from re
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Answer:

a) 17.33 V/m

b) 6308 m/s

Explanation:

We start by using equation of motion

s = ut + 1/2at², where

s = 1.2 cm = 0.012 m

u = 0 m/s

t = 3.8*10^-6 s, so that

0.012 = 0 * 3.8*10^-6 + 0.5 * a * (3.8*10^-6)²

0.012 = 0.5 * a * 1.444*10^-11

a = 0.012 / 7.22*10^-12

a = 1.66*10^9 m/s²

If we assume the electric field to be E, and we know that F =qE. Also, from Newton's law, we have F = ma. So that, ma = qE, and E = ma/q, where

E = electric field

m = mass of proton

a = acceleration

q = charge of proton

E = (1.67*10^-27 * 1.66*10^9) / 1.6*10^-19

E = 2.77*10^-18 / 1.6*10^-19

E = 17.33 V/m

Final speed of the proton can be gotten by using

v = u + at

v = 0 + 1.66*10^9 * 3.8*10^-6

v = 6308 m/s

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A 5.00 kg mass is placed on top of a vertical spring, which compresses a distance of 3.13 cm. Calculate the force constant (in N
77julia77 [94]

Answer:

<h2>1567.09 N/m</h2>

Explanation:

Step one:

given data

mass m=5kg

compression x= 3.13cm to m= 0.0313m

<em>According to Hooke's law, provided the elastic limit of an elastic material is not exceeded the extension e is directly proportional to the applied force</em>

F=ke

where

k= spring constant in N/m

e= extension/compression in

Step two:

assume g= 9.81m/s^2

F=mg

F=5*9.81

F=49.05N

substitute in the expression F=ke

49.05=k*0.0313

k=49.05/0.0313

k=1567.09 N/m

<u>The force constant (in N/m) of the spring is 1567.09 N/m</u>

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