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vichka [17]
3 years ago
15

A 2.1W iPod is used for 30 minutes. How much energy does it use? (3

Physics
2 answers:
dusya [7]3 years ago
8 0

Answer:

Energy used, E = 3780 Joules

Explanation:

It is given that,

Power, P = 2.1 W

Time, t = 30 minutes = 1800 seconds

We need to find the energy used by the iPod. The product of power and time is called energy used i.e.

E = P × t

E=2.1\ W\times 1800\ s

E = 3780 Joules.

So, the energy used by the iPod is 3780 Joules. Hence, this is the required solution.

shutvik [7]3 years ago
3 0
It us 200 energy if it the right awnser


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two ice skaters push each other, and move in opposite directions. the mass of one is 75 kg, and his speed is 2 m/s. if the mass
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momentum conservation

75x2 = 30 x her speed

75x2/30=150/30=5m/s

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Which quantity measures the maximum distance an object on a spring moves from the equilibrium position during oscillation?
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If the temperature of the conductor is increased, the electrons’ speeds decrease
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3 years ago
A 55 newton force applied on an object moves the object 10 meters in the same direction as the force. What is the value of work
kifflom [539]

Answer: Option D: 5.5×10²Joules

Explanation:

Work done is the product of applied force and displacement of the object in the direction of force.

W = F.s = F s cosθ

It is given that the force applied is, F = 55 N

The displacement in the direction of force, s = 10 m

The angle between force and displacement, θ = 0°

Thus, work done on the object:

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3 0
2 years ago
Using Hooke's law, F spring=k delta x, find the distance a spring with an elastic constant of 4 N/cm will stretch if a 2 newton
pishuonlain [190]

Hello!

Using Hooke's law, F spring=k delta x, find the distance a spring with an elastic constant of 4 N/cm will stretch if a 2 newton force is applied to it.

Data:

Hooke represented mathematically his theory with the equation:

F = K * Δx  

On what:

F (elastic force) = 2 N

K (elastic constant) = 4 N/cm

Δx (deformation or elongation of the elastic medium or distance from a spring) = ?

Solving:


F = K * \Delta{x}

2\:N = 4\:N/cm*\Delta{x}

4\:N/cm*\Delta{x} = 2\:N

\Delta{x} = \dfrac{2\:\diagup\!\!\!\!\!N}{4\:\diagup\!\!\!\!\!N/cm}

simplify by 2

\Delta{x} = \dfrac{2}{4}\frac{\div2}{\div2}

\boxed{\boxed{\Delta{x} = \dfrac{1}{2}\:cm}}\Longleftarrow(distance)\end{array}}\qquad\checkmark

Answer:

B.) 1/2 cm

_______________________

I Hope this helps, greetings ... Dexteright02! =)

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