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Anestetic [448]
2 years ago
12

A 0.40-kg mass attached to the end of a string swings in a vertical circle having a radius of 1.8 m. At an instant when the stri

ng makes an angle of 40 degrees below the horizontal, the speed of the mass is 5.0 m/s. What is the magnitude of the tension in the string at this instant
Physics
1 answer:
enyata [817]2 years ago
7 0

Answer:

T=8.1N

Explanation:

From the question we are told that:

Mass m=0.40

Radius r=1.8m

Angle Beneath the Horizontal \theta =40 \textdegree

Speed v=5.0m/s

The Tension Angle

 \alpha=90-\theta\\\\\alpha=90-40

 \alpha=50 \textdegree

Generally the equation for Tension is is mathematically given by

 T=\frac{mv^2}{r}+mgcos \alpha

 T=\frac{0.40*5^2}{1.8}+0.40*5cos50

 T=8.1N

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Answer with Explanation:

We are given that

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\phi=BA

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Substitute t=0 s

Then, I=1.6\mid (1-0)\mid=1.6 A

Substitute t=1 s

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t=2 s

Current, I=1.6\mid(1-2)\mid=1.6 A

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2 years ago
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video for A door 1 m wide, of mass 15 kg, is hinged at one side so that it can rotate without friction about a vertical axis. It
natka813 [3]

Answer:

\omega_f = 0.4\ rad/s

Explanation:

given,

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\omega_f= \dfrac{mv\dfrac{W}{2}}{\dfrac{1}{3}MW^2 + m(\dfrac{W}{2})^2}

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\omega_f= \dfrac{\dfrac{0.01\times 400}{2}}{\dfrac{15\times 1 }{3}+(\dfrac{0.01\times 1}{4})}

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3 years ago
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Answer:

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Equipotential surface are those surfaces that have the same potential at any point on the surface. Thus the potential difference at any point on the surface is zero due to same potential.

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Hope this helps you!

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