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shtirl [24]
3 years ago
13

A tennis player receives a shot with the ball (0.0600 kg) traveling horizontally at 22.0 m/s and returns the shot with the ball

traveling horizontally at 35.5 m/s in the opposite direction. (assume the initial direction of the ball is in the −x direction.) (a) what is the impulse delivered to the ball by the tennis racket?
Physics
1 answer:
morpeh [17]3 years ago
8 0
I need an answer choice
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(2) Put 5kg mass at left side (at 2m). This is fixed throughout the experiment! (3) Try to balance by putting 5kg mass at the ri
dmitriy555 [2]

We can conclude that as the mass on the right increases, the distance of the mass towards the right decreases. Also when the two masses balance, the net torque is zero.

<h3>What is torque</h3>

The torque experienced by an object a given position is the product of the applied force and the perpendicular distance of the object.

When 5 kg mass is at 2 m on the left, another 5 kg at 2 m on the right will balance it.

\tau _{net} = (2 \times 5 \times 9.8)  - (2 \times 5 \times 9.8)\\\\\tau _{net} = 0

<h3>Position of 10 kg mass on the right</h3>

Apply principle of moment

F_1r_1 = F_2r_2\\\\(m_1gr_1) = (m_2gr_2)\\\\r_2 = \frac{m_1gr_1}{m_2g} \\\\r_2 = \frac{m_1 r_1}{m_2} \\\\r _2 = \frac{5 \times 2}{10} \\\\r_2 = 1 \ m

<h3>Net torque</h3>

\tau_{et} = m_2gr_2 - m_1gr_1\\\\\tau_{et} = (10 \times 9.8 \times 1) - (5 \times 9.8 \times 2)\\\\\tau_{et} = 0

<h3>Position of the 20 kg mass</h3>

r_2 = \frac{5 \times 2}{20} \\\\r_2 = 0.5 \ m

<h3>Net torque</h3>

\tau_{et} = m_2gr_2 - m_1gr_1\\\\\tau_{et} = (20 \times 9.8 \times 0.5) - (5 \times 9.8 \times 2)\\\\\tau_{et} = 0

Thus, we can conclude that as the mass on the right increases, the distance of the mass towards the right decreases. Also when the two masses balance, the net torque is zero.

Learn more about principles of moment here: brainly.com/question/26117248

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2 years ago
What do repeated trails in an experiment allow scientists to do?
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Answer:

Explanation:

When we do multiple trials of the same experiment, we can make sure that our results are consistent and not altered by random events. Multiple trials can be done at one time. If we were testing a new fertilizer, we could test it on lots of individual plants at the same time.

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Where is the near point of an normal eye when accidentally wear a contact lens with a power of +2.0 diopters?
Lerok [7]

Answer:

The near point of an eye with power of +2 dopters, u' = - 50 cm

Given:

Power of a contact lens, P = +2.0 diopters

Solution:

To calculate the near point, we need to find the focal length of the lens which is given by:

Power, P = \frac{1}{f}

where

f = focal length

Thus

f = \frac{1}{P}

f = \frac{1}{2} = + 0.5 m

The near point of the eye is the point distant such that the image formed at this point can be seen clearly by the eye.

Now, by using lens maker formula:

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u = object distance = 25 cm = 0.25 m = near point of a normal eye

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Now,

\frac{1}{u'} = \frac{1}{f} - \frac{1}{u}

\frac{1}{u'} = \frac{1}{0.5} - \frac{1}{0.25}

\frac{1}{u'} = \frac{1}{f} - \frac{1}{u}

Solving the above eqn, we get:

u' = - 0.5 m = - 50 cm

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