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2 years ago

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A ball on the end of a string is revolving at a uniform rate in a vertical circle of radius 97.7 cm. If its speed is 3.74 m/s, a

nd its mass is 0.182 kg, calculate the tension (in newtons) in the string when the ball is at the bottom of the path.

1 answer:

noname [10]2 years ago

7 0

The tension in the string when the ball is at the bottom of the path is 2.61 Newtons.

<h3>Tension</h3>

A tension is simply referred to as a force along the length of a flexible medium such as strings, cable, ropes etc.

Tension in a string revolving can be determined using the expression;

T = mv² / r

Where m is mass of object, v is velocity and r is radius ( length of string )

Given the data in the question;

Mass of ball m = 0.182kg

Radius ( length of string ) r = 97.7cm = 0.977m

Velocity = 3.74m/s

Tension in the string; T = ?

To determine tension in the string, we substitute our given values into the expression above.

T = mv² / r

T = (0.182kg × (3.74m/s)²) / 0.977m

T = (0.182kg × 13.9876m²/s²) / 0.977m

T = (2.5457432kgm²/s²) / 0.977m

T = 2.61kgm/s²

T = 2.61N

Therefore, the tension in the string when the ball is at the bottom of the path is 2.61 Newtons.

Learn more about Tension here: brainly.com/question/14351325

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Answer:

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Explanation:

So, the very first thing we nee to do when solving this problem is draw a diagram that represents it. In the attached picture I show a diagram for each part of this problem.

part a)

So, for her to move in a direction directly opposite her starting point, the x-component of her velocity must be de same as the velocity of the river in the opposite direction. We can use this fact to find the angle we need. If we analize the triangle I drew in the diagram, we can ses that:

$cos \theta = \frac {V_{river}}{V_{boat}}$

When solving for theta, we get that:

$\theta =cos^{-1} ( \frac {V_{river}}{V_{boat}})$

so now we can substitute the corresponding values:

$\theta =cos^{-1} ( \frac {2.80km/hr}{5.60km/hr}})$

Which yields:

$\theta = 60^{o}$

but we are measuring the angle relative to the line perpendicular to the river, positive if down the river. So we need to subtract the angle from 270° so we get:

θ=270°-60°=210°

part b)

for part b, we need to find what the y-component for the velocity of the boat is for an angle of 210° as shown in the problem, so we get that:

$V_{y}=5.60km/hr*cos(210^{o})$

$V_{y}=-4.85km/hr$

The woman will head in a negative 5.60km distance from one side to the other, so we get that the time it takes her to go to the other side of the river is:

$t=\frac{y}{V_{y}}$

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part c)

In order to find the time it takes her to travel 2.80km down and up the river, we need to find the velocities she will have in both directions. First, down stream:

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and now up stream:

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$V_{us}=5.60km/hr-2.80km/hr=2.80km/hr$

Once we got these two velocities we will now need to find the time to take each trip:

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and the time up stream:

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so the total time will be:

$t_{ds}+t_{us}=0.333hr+1hr=1.333hr$

d) the time it takes the boat to go upstream and then downstream for the same distance is the same as the time we got on part c, since both times will be the same but they will come in different order, but their sum will be just the same:

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e) For her to cross the river faster, she must row in a 180° direction (this is in a direction straight accross the river) that way she will use all her velocity to move across the river. (Even though she will move a certain distance horizontally and will not reach a point opposite to the starting point.)

f) In order to find the time it takes her to get to the other side, we need to divide the distance into the velocity of the boat.

$t=\frac{d}{v_{boat}}$

$t=\frac{5.60km}{5.60km/hr}$

so

t= 1hr

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