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Fed [463]
4 years ago
8

A weightlifter works out at the gym each day. Part of her routine is to lie on her back and lift a 43 kg barbell straight up fro

m chest height to full arm extension, a distance of 0.55 m . Part A: How much work does the weightlifter do to lift the barbell one time? Part B: If the weightlifter does 23 repetitions a day, what total energy does she expend on lifting, assuming a typical efficiency for energy use by the body? Part C: How many 490 Calorie donuts can she eat a day to supply that energy
Physics
1 answer:
Tasya [4]4 years ago
5 0

Answer:

A. 231.77 J

B. 5330.71 J

C. 46 donuts

Explanation:

A. To lift the barbell once, she will have to extend it the full length of her arm. The work done will then be:

W = F * d

Where the force is the weight of the barbell.

F = m * g

Hence, the work done in lifting the barbell is:

W = m * g * d

W = 43 * 9.8 * 0.55

W = 231.77 J

B. If she does 23 repetitions, the total energy she expend will be equal to the Potential energy when the barbell is lifted multiplied by 23:

E = 23 * m * g * d

E = 23 * 231.77

E = 5330.71 J

C. 1 Joule = 4.184 calories

5330.71 Joules = 5330.71 * 4.184 = 22303.69

If 1 donut contains 490 calories, the number of donuts she will need will be:

N = 22303.69/490 = 45.5 donuts or 46 donuts

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lesantik [10]

Answer:

The radius of the cable is 0.0083 m or 8.3 mm.

Explanation:

The resistance of copper cable of 1 meter length will be given by

R_{cable} = \frac{\rho \times l}{a}    ....     (i)

where the resistivity of copper is \rho = 1.7 \times 10^{-8} \Omega.m , and l is the length of the wire which is considered to be 1m, and a is the cross sectional area of the wire in m^{2}.

From the formula of power we know that, P = I^2 R    ....    (ii)

Therefore 2 W/m  = (160)^2 \times R     ....     (iii)  

where  the resistance,R, actually means the resistance of the cable per meter.

Therefore R ( resistance of cable per meter)

= \frac{2}{160^2}  = 7.812 \times 10^{-5} ohms / meter.       ....    (iv)

Therefore from (i)

7.812 \times 10^{-5} = \frac{1.7 \times 10^{-8} \times 1}{a}  = \frac{1.7 \times 10^{-8} \times 1}{\pi r^{2} }       .....     (v)

where cross sectional area of the cable, a  = \pi r^2,

where r is the radius of the cable, and length of cable,l = 1m.

Therefore r  = \sqrt{\frac{ 1.7 \times 10^{-8}}{\pi \times 7.812 \times 10^{-5} } }  =  0.0083m = 8.3 mm

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3 years ago
How are frequency and period related to each other? A. They are the same for any given wave B. They have the same magnitude but
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Either of them is. 1/(the other one). That's 'C' .
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Explanation:

The acceleration of an object is the rate if change in velocity of the object. It is calculated as

a=\frac{v-u}{t}

where

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u is the initial velocity

t is the time elapsed

For an object moving in a straight line at constant acceleration, there are several equations that can be used to find the acceleration: they are called suvat equations. They are the following:

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Learn more about acceleration:

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U=1/4π∈₀×(q²/r)+1/4π∈₀×(q²/r)+1/4π∈₀×(q²/r)

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Substitute given values

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