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Ipatiy [6.2K]
3 years ago
11

You are working out on a rowing machine. Each time you pull the rowing bar (which simulates the oars) toward you, it moves a dis

tance of 1.3 m in a time of 0.96 s. The readout on the display indicates that the average power you are producing is 77 W. What is the magnitude of the force that you exert on the handle?
Physics
1 answer:
tangare [24]3 years ago
5 0

Answer:

56.86153 N

Explanation:

t =Time taken

F = Force

Power

P=\frac{W}{t}\\\Rightarrow W=P\times t\\\Rightarrow W=77\times 0.96\\\Rightarrow W=73.92\ Joules

Work done

W=F\times s\\\Rightarrow F=\frac{W}{s}\\\Rightarrow F=\frac{73.92}{1.3}\\\Rightarrow F=56.86153\ N

The magnitude of the force that is exerted on the handle is 56.86153 N

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3 years ago
⦁ A certain resistor is required to dissipate 0.25 W, what standard rating should be used?
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Answer:A

Explanation:

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3 years ago
A spring on a horizontal surface can be stretched and held 0.2 m from its equilibrium position with a force of 16 N. a. How much
nekit [7.7K]

Answer:

a   W_{3.5} = 490 \  J

b  W_{2.5} =  250 \  J

Explanation:

Generally the force constant is mathematically represented as

       k  = \frac{F}{x}

substituting values given in the question

=>   k  = \frac{16}{0.2}

=>   k  =  80 \ N /m

Generally the workdone  in stretching the spring 3.5 m is mathematically represented as

       W_{3.5} =  \frac{1}{2}  *  k  *  (3.5)^2

=>     W_{3.5} =  \frac{1}{2}  *  80  *  (3.5)^2

=>    W_{3.5} = 490 \  J

Generally the workdone  in compressing the spring 2.5 m is mathematically represented as

        W_{2.5} =  \frac{1}{2}  *  k  *  (2.5)^2

=>      W_{2.5} =  \frac{1}{2}  *  80 *  (2.5)^2

=>       W_{2.5} =  250 \  J

5 0
2 years ago
A package of mass 5 kg sits on an airless asteroid of mass 7.6 × 1020 kg and radius 8.0 × 105 m. We want to launch the package i
Effectus [21]

Answer:

s =  1.7 m

Explanation:

from the question we are given the following:

Mass of package (m) = 5 kg

mass of the asteriod (M) = 7.6 x 10^{20} kg

radius = 8 x 10^5 m

velocity of package (v) = 170 m/s

spring constant (k) = 2.8 N/m

compression (s) = ?

Assuming that no non conservative force is acting on the system here, the initial and final energies of the system will be the same. Therefore  

• Ei = Ef

• Ei = energy in the spring + gravitational potential energy of the system

• Ei = \frac{1}{2}ks^{2} + \frac{GMm}{r}

• Ef = kinetic energy of the object

• Ef = \frac{1}{2}mv^{2}  

• \frac{1}{2}ks^{2} + (-\frac{GMm}{r}) = \frac{1}{2}mv^{2}  

• s = \sqrt{\frac{m}[k}(v^{2}+\frac{2GM}{r})}

s = \sqrt{\frac{5}[2.8 x 10^5}(170^{2}+\frac{2 x 6.67 x10^{-11} x 7.6 x 10^{20}}{8 x 10^5})}

s =  1.7 m

7 0
3 years ago
What happens to balloon filled with air when it goes very high attitude from surface of earth why​
vlada-n [284]

Answer:

The balloon will continue to expand and eventually burst.

Explanation:

Simply, the reason for this is because the density of the atmosphere decreases gradually as you increase in altitude closer to space. This means that the air on the outside of the balloon can't provide enough pressure over the surface of the balloon in order to counteract the gas on the inside of the balloon from expanding.

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2 years ago
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