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3 years ago

10,000 W of power is used to move a spoon for 10 s. How much work is done on the spoon?

Physics

1 answer:

bixtya [17]3 years ago

6 0

Answer:

<em>The work done is 100,000 Joule</em>

Explanation:

<u>Work and Power </u>

Work is the amount of energy transferred by a force.

If both the force F and displacement s are parallel, then we can use the formula:

W = F.s

Power is the amount of energy transferred or converted per unit of time. In the SI, the unit of power is the watt, equal to one joule per second.

The power can be calculated as:

$\displaystyle P=\frac {W}{t}$

Where W is the work and t is the time.

We know the power needed to move a spoon is P=10,000 Watt and it took a time of t=10 s. The work done on the spoon is calculated by solving for W:

W = Pt

W = 10,000 Watt * 10 sec

W = 100,000 Joule

The work done is 100,000 Joule

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1 point

s2008m [1.1K]

Answer:

The person has no displacement

Explanation:

The given parameters are

The location of the person = The equator

The distance covered in one revolution = Total distance around the body

The total distance around the Earth = The circumference of the Earth = 40.075 kilometres

The total distance moved by the person standing at the equator during the Earths complete revolution = 40,075 kilometres

The initial location of the person in relation to a fixed point in space outside Earth at the start of the revolution = x km

The final location of the person in relation to the fixed point in space outside Earth at the completion of the revolution = x km

The displacement = Change in position = Final location - Initial location

∴ The displacement = x km - x km = 0 km.

5 0

4 years ago

Two charged particles are a distance of 1.62 m from each other. One of the particles has a charge of 7.10 nc, and the other has

k0ka [10]

Answer:

A. F=107.6nN

B. Repulsive

Explanation:

According to coulombs law, the force between two charges is express as

F=(Kq1q2) /r^2

If the charges are of similar charge the force will be repulsive and if they are dislike charges, force will be attractive.

Note the constant K has a value 9*10^9

Hence for a charge q1=7.10nC=7.10*10^-9, q2=4.42*10^-9 and the distance r=1.62m

If we substitute values we have

F=[(9×10^9) ×(7.10×10^-9) ×(4.42×10^-9)] /(1.62^2)

F=(282.4×10^-9)/2.6244

F=107.6×10^-9N

F=107.6nN

B. Since the charges are both positive, the force is repulsive

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3 years ago

Scientific work is currently underway to determine whether weak oscillating magnetic fields can affect human health. For example

nlexa [21]

Answer:

The maximum emf that can be generated around the perimeter of a cell in this field is $1.5732*10^{-11}V$

Explanation:

To solve this problem it is necessary to apply the concepts on maximum electromotive force.

For definition we know that

$\epsilon_{max} = NBA\omega$

Where,

N= Number of turns of the coil

B = Magnetic field

$\omega =$ Angular velocity

A = Cross-sectional Area

Angular velocity according kinematics equations is:

$\omega = 2\pi f$

$\omega = 2\pi*61.5$

$\omega =123\pi rad/s$

Replacing at the equation our values given we have that

$\epsilon_{max} = NBA\omega$

$\epsilon_{max} = NB(\pi (\frac{d}{2})^2)\omega$

$\epsilon_{max} = (1)(1*10^{-3})(\pi (\frac{7.2*10^{-6}}{2})^2)(123\pi)$

$\epsilon_{max} = 1.5732*10^{-11}V$

Therefore the maximum emf that can be generated around the perimeter of a cell in this field is $1.5732*10^{-11}V$

6 0

4 years ago

Weightlessness is experienced by an astronaut in space. This means that the astronaut's muscles have to be stronger to move his

just olya [345]

The answer is false. The speed of the astronaut cancels out the force of gravity, causing a 'stationary freefall'. While under these effects, it is not required for an astronaut to 'strengthen' his body.

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patriot [66]

Answer:

i think its ancestor

Explanation:

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