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BabaBlast [244]
3 years ago
12

A ball of mass m, attached to the end of a horizontal cord, is rotated in a circle of radius r on a frictionless horizontal surf

ace. If the cord will break when the tension in it exceeds F, what is the maximum speed the ball can have? Express your answer in terms of the given quantities.
Physics
1 answer:
kirza4 [7]3 years ago
8 0

Answer:v=\sqrt{\frac{FL}{m}}

Explanation:

Given

Ball of mass m

maximum Bearable Tension in string is F

Let length of the cord be L m and moving at a speed of v m/s

Here Tension will Provide Centripetal Force

T=Centripetal Force

F=T=\frac{mv^2}{L}

v=\sqrt{\frac{FL}{m}}

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<span>We know that pressure is the force applied into a surface, in our case the wall of the room, so then first we will calculate the surface of this wall: S = 2.2 * 3.2 = 7.04 m2 Then we also know the atmospheric pressure in normal conditions is 1 atm. That is the same 1 atm = 101325 Pascals or 101325 N/m2 Now we need to use the formula : P = F/S where P is pressure, F is force and S is surface to calculate the force: F = P * S = 101325 * 7.04 = 713,328 Newtons Conclusion: the force acts on the wall due the air inside the room is 713,328 N</span>
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3 years ago
Which lists types of materials from most conductive to least conductive?
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Superconductor, conductor, semiconductor, insulator
7 0
3 years ago
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Can anybody write a short poem about friction
Luden [163]
Answer:

you will be the clouds
and I will be the sky.
you will be the ocean
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8 0
2 years ago
Con lắc lò xo có độ cứng k = 100N/m được gắn vật có khối lượng m=0.1kg, kéo vật ra khỏi vị trí cân bằng 1 đoạn 5cm rồi buông tay
Delvig [45]

Answer:

The maximum velocity is 1.58 m/s.

Explanation:

A spring pendulum with stiffness k = 100N/m is attached to an object of mass m = 0.1kg, pulls the object out of the equilibrium position by a distance of 5cm, and then lets go of the hand for the oscillating object. Calculate the achievable vmax.

Spring constant, K = 100 N/m

mass, m = 0.1 kg

Amplitude, A = 5 cm = 0.05 m

Let the angular frequency is w.

w = \sqrt{K}{m}\\\\w = \sqrt{100}{0.1}\\\\w = 31.6 rad/s

The maximum velocity is

v_{max} = w A\\\\v_{max} = 31.6\times 0.05 = 1.58 m/s

8 0
3 years ago
A cricket can travel approximately 8 m/s. How many meters could a cricket travel in 75<br> s?
nika2105 [10]

Answer:

Your answer will be 600meters

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2 years ago
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