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PtichkaEL [24]
4 years ago
10

The amount of power required to move an object can be increased without changing the

Physics
1 answer:
VLD [36.1K]4 years ago
3 0

Answer:

Yes it is possible to increase the power with out changing the amount of work.

Explanation:

The power is defined by the amount of power divided by the time. This time is the one needed to do the work. We can understand this issue by analyzing an example with numeric values.

Work = 500 [J]

Time = 5 [s]

Power will be:

Power=\frac{500}{5} \\Power=100 []watt]\\

Now if we change the time to 2 seconds:

Power = 500 [J]/2[s]\\Power = 250 [watt]\\

As we can see, the power was increased without the need to change the work.

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I give a ball a push on an acclivity. The "start velocity" is on 7m/s. The time it took the ball to get back to me was 10 second
Degger [83]

Answer:

7÷10

Explanation:

initial velocity=7m/s

final velocity=0m/s

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4 years ago
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An athlete is working out in the weight room. He steadily holds 50 kilograms above his head for 10 seconds. Which statement is t
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Answer: the true statement form the given statements is “the athletes is not doing any work because he does not move weight”

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3 years ago
In a race, a car travels 60 times around a 3.6km track. This takes 2.4 hours. What is the average speed of the car?​
Finger [1]

Answer:

multiply

Explanation:

multiply

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3 years ago
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A proton moves through a region of space where there is a magnetic field B⃗ =(0.64i+0.40j)T and an electric field E⃗ =(3.3i−4.5j
fenix001 [56]

Answer:

F = (8.35 \times 10^{-16})\hat i - (12.12 \times 10^{-16})\hat j +(1.35 \times 10^{-16})\hat k

Explanation:

When a charge is moving in constant magnetic field and electric field both then the net force on moving charge is vector sum of force due to magnetic field and electric field both

so first the force on the moving charge due to electric field is given by

\vec F_e = q\vec E

\vec F_e = (1.6 \times 10^{-19})(3.3 \hat i - 4.5 \hat j) \times 10^3

\vec F_e = (5.28 \times 10^{-16}) \hat i - (7.2 \times 10^{-16}) \hat j

Now force on moving charge due to magnetic field is given as

\vec F_b = q(\vec v \times \vec B)

\vec F_b = (1.6 \times 10^{-19})((6.6 \hat i+2.8 \hat j−4.8 \hat k) \times 10^3 \times (0.64 \hat i + 0.40 \hat j) )

\vec F_b = (4.22 \times 10^{-16})\hat k - (2.87 \times 10^{-16})\hat k - (4.92 \times 10^{-16})\hat j + (3.07 \times 10^{-16}) \hat i

\vec F_b = (3.07\times 10^{-16})\hat i - (4.92 \times 10^{-16})\hat j + (1.35 \times 10^{-16})\hat k

Now net force due to both

F = F_e + F_b

F = (8.35 \times 10^{-16})\hat i - (12.12 \times 10^{-16})\hat j +(1.35 \times 10^{-16})\hat k

7 0
3 years ago
Unless indicated otherwise, assume the speed of sound in air to be v = 344 m/s. You have a stopped pipe of adjustable length clo
faltersainse [42]

Answer:

Length of pipe = 0.057 meter

Explanation:

Speed of a transverse wave on a string

v = \sqrt{\frac{F}{\mu} }

where F is the tension in string and \mu is the mass per unit length

Thus,

\mu = \frac{m}{L}

Substituting the given values we get -

\mu = \frac{7.25 * 10^{-3}}{0.62}\\mu = 0.0117 \frac{Kg}{m}

Speed of a transverse wave on a string

v = \sqrt{\frac{4510}{0.0117} } \\v = 620.86 \frac{m}{s}

For third harmonic wave , frequency is equal to

f = \frac{nv}{2L}

Substituting the given values, we get -

f = \frac{3 * 620.86}{2 * 0.62} \\f = 1502.08

Length of pipe

L = \frac{nv}{4 f}

Substituting the given values we get

n = 1 for first harmonic wave

L = \frac{344* 1}{4*1502.08} \\L = 0.057

Length of pipe = 0.057 meter

5 0
4 years ago
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