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Readme [11.4K]
3 years ago
10

The maximum mass the can be hung vertically from a string without breaking the string is 10kg. A length of this string that is 2

m long is used to rotate a 0.5kg object in a circle on a frictionless table with the string horizontal. The maximum speed that the mass can attain under these conditions without the string breaking is most nearly:
Answer is 20 m/s
Please show me how I can get the answer. Thank you.
Physics
1 answer:
vichka [17]3 years ago
0 0

Answer:

Explanation:

Think about circular motion when answering this question. First determine the maximum force that can be applied on the string. F = mg so F = (10)(10) = 100 N. Then determine the centripetal acceleration of the .5 kg mass, a = F/m so a = 100/.5 = 200 m/s².  On the equation sheet, use equation a(centripetal acceleration) = v²/r so 200 = v²/2 therefore v = 20 m/s. Hope this helps!

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At a certain instant a particle is moving in the +x direction with momentum +8 kg m/s. During the next 0.13 seconds a constant f
jeka94

Answer:

The momentum of the particle at the end of the 0.13 s time interval is 7.12 kg m/s

Explanation:

The momentum of the particle is related to force by the following equation:

Δp = F · Δt

Where:

Δp =  change in momentum = final momentum - initial momentum

F = constant force.

Δt = time interval.

Let´s calculate the x-component of the momentum after the 0.13 s:

final momentum - 8 kg m/s = -7 N · 0.13 s

final momentum = -7 kg m/s² · 0.13 s + 8 kg m/s

final momentum = 7.09 kg m/s

Now let´s calculate the y-component of the momentum vector after the 0.13 s. Since the particle wasn´t moving in the y-direction, the initial momentum in this direction is zero:

final momentum = 5 kg m/s² · 0.13 s

final momentum = 0.65 kg m/s

Then, the mometum vector will be as follows:

p = (7.09 kg m/s,  0.65 kg m/s)

The magnitude of this vector is calculated as follows:

|p| = \sqrt{(7.09 kg m/s)^{2} + (0.65 kg m/s)^{2}} = 7.12 kg m/s

The momentum of the particle at the end of the 0.13 s time interval is 7.12 kg m/s

4 0
3 years ago
How to represent milligram in kilogram by standard formula?
Anettt [7]

Answer:

0.000001 kg

Explanation:

because 1 kg equal 1,000,000 milligrams

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4 0
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Factors that affect acceleration due to gravity.<br>​
hram777 [196]

Answer

Factors Affecting Acceleration Due to Gravity

 

(1) Position on the planetary surface, it is higher at the axis (poles),

(2) Mass of the planet, higher with planets of larger masses

(3) Distance between object and the centre of planet, the larger the distance, the smaller the acceleration due to gravity.

Mark me as Brainlist if this answer is helpful for you

8 0
3 years ago
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What real-world examples show no work begin done? Can you think of examples other than resisting the force of gravity?
irga5000 [103]
Oh my gosh !  Resisting the force of gravity always DOES involve doing work.
If no work is being done, then you're NOT resisting the force of gravity.

Example:

-- ball rolling on the floor . . . no work
-- ball rolling up a ramp . . . work being done
-- ball rolling down a ramp . . . work being done, BY gravity

6 0
3 years ago
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When landing after a spectacular somersault, a 45.0 kg gymnast decelerates by pushing straight down on the mat. Calculate the fo
STALIN [3.7K]
So the force that the gymnast with a mass m=45 kg, has to exert against the ground to stop if her acceleration is a=8*g, where g=9.81 m/s², can be obtained from the Newtons second law: F=m*a, where F is the force, m is the mass and a is acceleration. 

F=m*a=m*8*g=45*8*9.81=3531.6 N. 

So the force that a gymnast has to exert on the mat in order to stop is F=3531.6 N.  
6 0
3 years ago
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