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kaheart [24]
4 years ago
14

Consider a particle on which several forces act, one of which is known to be constant in time: Fi = 3.00 i +4.00 ) N. As a resul

t, the particle moves along a straight path from a Cartesian coordinate of 0.00 m, 0.00 m) to (5.00 m, 6.00 m). What is the work done by Fi?
Physics
1 answer:
Leviafan [203]4 years ago
4 0

Given that.

F=3•i+4•j

And it from point (0,0)m to (5,6)m

dx=final position - initial position

dx=(5,6)-(0,0)

dx=(5,6)m

dx=5•i +6•j

Work done by the force is give by

W = F•dx

W=F•dx

Note that i•i=j•j=1 and i•j=j•i=0

Then,

W=(3i+4j)•(5i+6j)

Therefore,

W=3i•(5i+6j)+4j•(5i+6j)

W=15i•i+18i•j+20j•i+24j•j

W=15+0+0+24

W=39J

Then the work done by the force is 39 Joules

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A string with a length of 4.00 m is held under a constant tension. The string has a linear mass density of \mu=0.000600~\text{kg
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Answer:

T=245.76N

Explanation:

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f_{n+1}-f_n=(n+1)f-nf=nf+f-nf=f

Which for our values means (we do not need the value of <em>n</em>, that is, which harmonics are the frequencies given):

f=f_{n+1}-f_n=480Hz-400Hz=80Hz

Now we turn to the formula for the vibration frequency of a string (for the fundamental harmonic):

f=\frac{1}{2L} \sqrt{\frac{T}{\mu}}

So the tension is:

T=\mu(2Lf)^2

Which for our values is:

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Two cars are traveling in the same direction down a highway at 65 miles per hour. What is the relative velocity of the second ca
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Answer:

5 hours

Explanation:

Let the required time be x hours. The time will be the same for both cars.

The cars will cover different distances because they are travelling at different speeds.

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The distance travelled by the faster car = 58×x miles.

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8x=40

x=5 hours

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A second method:

The difference in the distances is 40 miles

The difference in the speeds is #8mph.

The time to make up the 40 miles= \frac{40}{8}=5 hours

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