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AVprozaik [17]
2 years ago
13

When an objects total mechanical energy is conserved and it is dropped from rest how does the objects initial potential energy c

ompare with its final kinetic energy?
Physics
1 answer:
frozen [14]2 years ago
8 0

Answer:Magnitude of potential energy before falling will be equal to the magnitude of the final kinetic energy, energy is only converted from potential to kinetic

Explanation: law of conservation of energy says that energy can neither be created or destroyed but Change from one form to another.so nothing will happen to the size of the potential energy when it changes to kinetic energy

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(8c7p26) During spring semester at MIT, residents of the parallel buildings of the East Campus dorms battle one another with lar
likoan [24]

Answer: 1175 J

Explanation:

Hooke's Law states that "the strain in a solid is proportional to the applied stress within the elastic limit of that solid."

Given

Spring constant, k = 102 N/m

Extension of the hose, x = 4.8 m

from the question, x(f) = 0 and x(i) = maximum elongation = 4.8 m

Work done =

W = 1/2 k [x(i)² - x(f)²]

Since x(f) = 0, then

W = 1/2 k x(i)²

W = 1/2 * 102 * 4.8²

W = 1/2 * 102 * 23.04

W = 1/2 * 2350.08

W = 1175.04

W = 1175 J

Therefore, the hose does a work of exactly 1175 J on the balloon

7 0
3 years ago
What is the wavelength of microwaves with a frequency of 3x10^10 Hz?​
RUDIKE [14]

Answer:

0.01 m

Explanation:

Since the speed of light is 3.0×10^8 m/s

Use the equation,

Wavelength = speed ÷ frequency

Wavelength = 3.0×10^8 ÷ 3×10^10

Wavelength = 0.01m

5 0
2 years ago
An equilibrium constant is not changed by a change in pressure <br> a. True<br> b. False
Umnica [9.8K]
Hi There! :)


An equilibrium constant is not changed by a change in pressurea. True
b. False

False! :P
7 0
3 years ago
Read 2 more answers
A copper wire 1.0 meter long and with a mass of .0014 kilograms per meter vibrates in two segments when under a tension of 27 Ne
Furkat [3]

Answer:

the frequency of this mode of vibration is 138.87 Hz

Explanation:

Given;

length of the copper wire, L = 1 m

mass per unit length of the copper wire, μ = 0.0014 kg/m

tension on the wire, T = 27 N

number of segments, n = 2

The frequency of this mode of vibration is calculated as;

F_n = \frac{n}{2L} \sqrt{\frac{T}{\mu} } \\\\F_2 = \frac{2}{2\times 1} \sqrt{\frac{27}{0.0014} }\\\\F_2 = 138.87 \ Hz

Therefore, the frequency of this mode of vibration is 138.87 Hz

7 0
3 years ago
Automotive technology
viktelen [127]

lol what is this question

5 0
2 years ago
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