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irinina [24]
3 years ago
14

A boy flies a kite with the string at a 30 degree angle to the horizontal. The tension in the string is 4.5N .

Physics
1 answer:
sp2606 [1]3 years ago
6 0
How much work in J does the string do on the boy if the boy stands still? 

<span>answer: None. The equation for work is W = force x distance. Since the boy isn't moving, the distance is zero. Anything times zero is zero </span>
<span>--------------------------------------... </span>
<span>How much work does the string do on the boy if the boy walks a horizontal distance of 11m away from the kite? </span>

<span>answer: might be a trick question since his direction away from the kite and his velocity weren't noted. Perhaps he just set the string down and walked away 11m from the kite. If he did this, it is the same as the first one...no work was done by the sting on the boy. </span>

<span>If he did walk backwards with no velocity indicated, and held the string and it stayed at 30 deg the answer would be: </span>
<span>4.5N + (boys negative acceleration * mass) = total force1 </span>
<span>work = total force1 x 11 meters </span>
<span>--------------------------------------... </span>

<span>How much work does the string do on the boy if the boy walks a horizontal distance of 11m toward the kite? </span>

<span>answer: same as above only reversed: </span>
<span>4.5N - (boys negative acceleration * mass) = total force2 </span>
<span>work = total force2 x 11 meters</span>
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<span>The correct option is C. Gravity, and the complete sentence is: "The force of gravity is the force at which the Earth attracts another object towards itself". In fact, the force of gravity between two objects is given by
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A particle moves in a straight line with the velocity function v ( t ) = sin ( w t ) cos 3 ( w t ) . find its position function
Sunny_sXe [5.5K]

Integrating the velocity equation, we will see that the position equation is:

$f(t)=\frac{\cos ^3(\omega t)-1}{3}

<h3>How to get the position equation of the particle?</h3>

Let the velocity of the particle is:

$v(t)=\sin (\omega t) * \cos ^2(\omega t)

To get the position equation we just need to integrate the above equation:

$f(t)=\int \sin (\omega t) * \cos ^2(\omega t) d t

$\mathrm{u}=\cos (\omega \mathrm{t})

Then:

$d u=-\sin (\omega t) d t

\Rightarrow d t=-d u / \sin (\omega t)

Replacing that in our integral we get:

$\int \sin (\omega t) * \cos ^2(\omega t) d t$

$-\int \frac{\sin (\omega t) * u^2 d u}{\sin (\omega t)}-\int u^2 d t=-\frac{u^3}{3}+c$

Where C is a constant of integration.

Now we remember that $u=\cos (\omega t)$

Then we have:

$f(t)=\frac{\cos ^3(\omega t)}{3}+C

To find the value of C, we use the fact that f(0) = 0.

$f(t)=\frac{\cos ^3(\omega * 0)}{3}+C=\frac{1}{3}+C=0

C = -1 / 3

Then the position function is:

$f(t)=\frac{\cos ^3(\omega t)-1}{3}

Integrating the velocity equation, we will see that the position equation is:

$f(t)=\frac{\cos ^3(\omega t)-1}{3}

To learn more about motion equations, refer to:

brainly.com/question/19365526

#SPJ4

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Explanation:

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