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2 years ago

A 50-kg sprinter accelerates from 0 to 11 m/s in 3.0 s. What is the power output for this rapid start?

Physics

1 answer:

Citrus2011 [14]2 years ago

3 0

Answer:

1008.33 W.

Explanation:

Power: This can be defined as the rate at which energy is consumed or dissipated. The S.I unit of power is Watt (W).

P = E/t.......................... Equation 1

Where P = Energy, E = Energy, t = time.

But from the question,

E = 1/2mΔv²................... Equation 2

where m = mass of the sprinter, Δv = change in velocity of the sprinter = final velocity - initial velocity.

Substitute equation 2 into equation 1

P = 1/2mΔv/t....................... Equation 3

Given: m = 50 kg, Δv = 11-0 = 11 m/s, t = 3.0 s.

Substitute into equation 3

P = 1/2(50)(11)²/3

P = 25(121)/3

P = 1008.33 W.

Thus the power output = 1008.33 W.

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A lens of focal length 15.0 cm is held 10.0 cm from a page (the object ). Find the magnification .

nevsk [136]

Answer:

Magnification, m = 3

Explanation:

It is given that,

Focal length of the lens, f = 15 cm

Object distance, u = -10 cm

Lens formula :

$\dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{f}$

v is image distance

$\dfrac{1}{v}=\dfrac{1}{f}+\dfrac{1}{u}\\\\\dfrac{1}{v}=\dfrac{1}{15}+\dfrac{1}{(-10)}\\\\v=-30\ cm$

Magnification,

$m=\dfrac{v}{u}\\\\m=\dfrac{-30}{10}\\\\m=3$

So, the magnification of the lens is 3.

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1 year ago

A uniform rod of mass M and length L is free to swing back and forth by pivoting a distance x from its center. It undergoes harm

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Answer:

The moment of inertia is 0.7500 kg-m².

Explanation:

Given that,

Mass = 2.2 kg

Distance = 0.49 m

If the length is 1.1 m

We need to calculate the moment of inertia

Using formula of moment of inertia

$I=\dfrac{1}{12}ml^2+mx^2$

Where, m = mass of rod

l = length of rod

x = distance from its center

Put the value into the formula

$I=\dfrac{1}{12}\times2.2\times(1.1)^2+2.2\times(0.49)^2$

$I=0.7500\ kg-m^2$

Hence, The moment of inertia is 0.7500 kg-m².

5 0

1 year ago

A housefly walking across a surface may develop a significant electric charge through a process similar to frictional charging.

Umnica [9.8K]

Answer:

Number of electrons, $n=3.87\times 10^8\ electrons$

Explanation:

Given that,

Charge on the fly, $q=62\ pC=62\times 10^{-12}\ C$

Let n is the number of electrons lose to the surface it is walking across. It is case of quantization of electric charge. It is given by :

$q=ne$

$n=\dfrac{q}{e}$

$n=\dfrac{62\times 10^{-12}}{1.6\times 10^{-19}}$

n = 387500000 electrons

$n=3.87\times 10^8\ electrons$

So, there are $3.87\times 10^8$ electrons. Hence, this is the required solution.

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1 year ago

Concerned with citizen complaints of price gouging during past hurricanes, Florida's state government passes a law setting a pri

natta225 [31]

Answer:idk

Explanation:idk

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1 year ago

A double slit apparatus is held 1.2 m from a screen. [___/4] (a) When red light (λ = 600 nm) is sent through the double slit, th

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A. To solve for part A, we use the formula of Young’s double slit equation:

x = λ L m / d

Where,

x = distance between adjacent dark lines or fringes = 12.5 cm = 0.125 m

λ = wavelength of light = 600 × 10^-9 m

L = distance from the two sources of light to the screen = 1.2 m

m = number of fringes = 10 (tenth) – 1 (first) = 9

d = separation of slits = unknown

Rearranging the equation in terms of d and plugging in the values:

d = λ L m / x

d = (600 × 10^-9 m) (1.2 m) (9) / 0.125 m

d = 5.184 × 10^-5 m

d = 51.84 μm

B. The formula for path difference is:

path difference = (2 m + 1) (λ / 2)
path difference = (2 * 9 + 1) (600 × 10^-9 m / 2)

path difference = 5.7 × 10-6 m
path difference = 5.7 μm

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1 year ago