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alexira [117]
3 years ago
7

Before railroad were invented, goods often traveled along canals, with mules pulling barges from the bank. If a mule is exerting

a 12,000N force for 10km, and the rope connecting the mule to the barge is at a 20 degree angle from the direction of travel, how much work did the mole do on the barge?
A. 12MJ
B. 11MJ
C. 4.1MJ
D. 6MJ
Physics
1 answer:
morpeh [17]3 years ago
8 0

Answer:

W = 112.76MJ

Explanation:

the work is:

W = F_xD

where F_x is the force executed in the direction of the displacement and the d the displacement.

so:

W = 12000Ncos(20)(10000)

we use the cos of the angule because it give us the proyection in the axis x of the force, that means the force in the direction of the displacement.

W = 112.76MJ

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3 years ago
A pendulum completes 2 oscillation in 5s.
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Answer:

<h2>a) 2.5 s</h2><h2>b) 1.55 m</h2><h2 />

Explanation:

<h3>Given :-</h3><h2>Completes 2 oscillation in 5s.</h2><h2>A) Time period = ?</h2><h2 /><h2>t \:  = \frac{total \: time \: }{total \: no \: of \: oscillation}</h2><h2 /><h2>\frac{5}{2}  = 2.5s</h2>

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<h2>B) If g = 9.8 ms−², find its length.</h2><h2>t \:  = 2\pi  \frac{ \sqrt{l} }{g }</h2><h3>= 2.5s </h3><h3>2\pi \frac{ \sqrt{l} }{g}</h3><h3>Length = ? </h3><h2>\frac{ {t}^{2} }{4\pi {}^{2} }  \times g</h2><h3 /><h3>\frac{2.5}{ {4}^{2} }  \times 9.8 = 1.55m</h3>

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2 years ago
A pendulum oscillates 50 times in 6 seconds. Find its time period and frequency? ​
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4 0
3 years ago
A uranium and iron atom reside a distance R = 37.50 nm apart. The uranium atom is singly ionized; the iron atom is doubly ionize
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Answer:

r=15.53 nm

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Explanation:

Lets take electron is in between iron and uranium

Charge on electronq_1= -1.602\times 10^{-19}C

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Charge on uraniumq_3= 1.602\times 10^{-19}C

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F=K\dfrac{q_1 q_2}{r^2}  

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K\dfrac{q_1 q_2}{r^2}=K\dfrac{q_1 q_3}{(37.5-r)^2}  

K\dfrac{q_1q_2}{(37.5-r)^2}=K\dfrac{q_1 q_3}{r^2}  

q_2= 2\timesq_1,q_3=q_1

\dfrac{q_1\times 2\timesq_1}{r^2}=\dfrac{q_1\times q_1}{(37.5-r)^2}

So r=15.53 nm

So force

F=9\times 10^9\dfrac{1.602\times 10^{-19}\times 1.602\times 10^{-19}}{(15.53\times 10^{-9})^2}  

F=9.57\times 10^{-13}N

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