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Neporo4naja [7]
2 years ago
9

A boy pulls his toy on a smooth horizontal surface with a rope inclined at 60 degrees to the horizontal. If the effective force

pulling the toy along the horizontal surface is 5N, calculate the tension in the rope

Physics
2 answers:
lana66690 [7]2 years ago
6 0
I'm thinking that you're supposed to divide. So you would divide 5 into 60 and get 12
myrzilka [38]2 years ago
6 0

As per the question the rope is inclined at an angle of 60 degree with the horizontal.

Let the tension produced in the rope is T.

The effective force which pulls the toy along horizontal direction is 5 N.

Resolving T into two components we get-

[1]\ \ Tsin\theta  in vertical upward direction.

[2]\ \ Tcos\theta in horizontal direction.

The vertical component is balanced by the weight of the body i.e

                            Tsin\theta=mg

Hence the toy moves due to the horizontal component.

                   Hence\ Tcos\theta=5 N

                         Tcos60= 5N

                          T= \frac{5}{cos60}

                                   = 5*2 N   [  cos60 = \frac{1}{2} ]

                                   = 10 N

Hence the tension produced in the rope is 10 N.

                         

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taurus [48]
According to Newton laws of motion, 
F = m*a
Here, m = 1,560 Kg
a = 1.30 m/s²

Substitute their values, 
F = 1,560 * 1.30
F = 2028 N ~ 2030 N  [ Closest value ]

In short, Your Answer would be Option C

Hope this helps!
6 0
3 years ago
Notice that all the initial spring potential energy was transformed into gravitational potential energy. If you compressed the s
Nostrana [21]

<u><em>The  question doesn't provide enough data to be solved, but I'm assuming some magnitudes to help you to solve your own problem</em></u>

Answer:

<em>The maximum height is 0.10 meters</em>

Explanation:

<u>Energy Transformation</u>

It's referred to as the change of one energy from one form to another or others. If we compress a spring and then release it with an object being launched on top of it, all the spring (elastic) potential energy is transformed into kinetic and gravitational energies. When the object stops in the air, all the initial energy is now gravitational potential energy.

If a spring of constant K is compressed a distance x, its potential energy is

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When the launched object (mass m) reaches its max height h, all that energy is now gravitational, which is computed as

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We have then,

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\displaystyle h=\frac{Kx^2}{2mg}

We have little data to work on the problem, so we'll assume some values to answer the question and help to solve the problem at hand

Let's say: x=0.2 m (given), K=100 N/m, m=2 kg

Computing the maximum height

\displaystyle h=\frac{(100)0.2^2}{2(2)(9.8)}

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8 0
2 years ago
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Answer:

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We have to find the root mean square speed of the particle

Which is given by v_{rms}=\sqrt{\frac{3RT}{m}}=\sqrt{\frac{3\times 8.314\times 300}{5\times 10^{-16}}}=38.68\times 10^8m/sec

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5 0
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olchik [2.2K]

Answer:

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3 0
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Answer:

Option (B)

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Thus, the correct answer is option (B).

6 0
3 years ago
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