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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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Peter’s body supplies a force of 500 N to run up a 10-m hill in 10 s. How much power is involved in Peter’s run up the hill? Exp
shutvik [7]

Answer: 500 Watts

Explanation:

Power P is the speed with which work W is done. Its unit is Watts (W), being 1 W=\frac{1 Joule}{1 s}.

Power is mathematically expressed as:

P=\frac{W}{t} (1)

Where t is the time during which work W  is performed.

On the other hand, the Work W done by a Force F refers to the release of potential energy from a body that is moved by the application of that force to overcome a resistance along a path.  It is a scalar magnitude, and its unit in the International System of Units is the Joule (like energy). Therefore, 1 Joule is the work done by a force of 1 Newton when moving an object, in the direction of the force, along 1 meter (1J=(1N)(1m)=Nm  ).

When the applied force is constant and the direction of the force and the direction of the movement are parallel, the equation to calculate it is:  

W=(F)(d) (2)

In this case, we have the following data:

F=500 N

d=10 m

t=10 s

So, let's calculate the work done by Peter and then find how much power is involved:

From (2):

W=(500 N)(10 m) (3)

W=5000 J (4)

Substituting (4) in (1):

P=\frac{5000 J}{10 s} (5)

Finally:

P=500 W

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A) 8.11 m/s

For a satellite orbiting around an asteroid, the centripetal force is provided by the gravitational attraction between the satellite and the asteroid:

m\frac{v^2}{(R+h)}=\frac{GMm}{(R+h)^2}

where

m is the satellite's mass

v is the speed

R is the radius of the asteroide

h is the altitude of the satellite

G is the gravitational constant

M is the mass of the asteroid

Solving the equation for v, we find

v=\sqrt{\frac{GM}{R+h}}

where:

G=6.67\cdot 10^{-11} m^3 kg^{-1}s^{-2}

M=1.40\cdot 10^{16}kg

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h=6.00 km = 6000 m

Substituting into the formula,

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B) 11.47 m/s

The escape speed of an object from the surface of a planet/asteroid is given by

v=\sqrt{\frac{2GM}{R+h}}

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