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lyudmila [28]
3 years ago
12

A 5.00 kg pendulum swings back and forth. At the top of its arc it reaches a height of 0.36 m. What is the velocity of the pendu

lum at the top of the arc?
0 m/s
7.2 m/s
2.7 m/s
Physics
1 answer:
azamat3 years ago
8 0

Answer:

Gravitational potential energy = (mass) x (gravity) x (height)

Kinetic energy (of a moving object) = (1/2) (mass) x (speed)²

When the pendulum is at the top of its swing, its potential energy is

(mass) x (gravity) x (height)

= (5 kg) x (9.8 m/s²) x (0.36 m)

= (5 x 9.8 x 0.36) joules

17.64 joules .

Energy is conserved ... it doesn't appear or disappear ... so that number is exactly the kinetic energy the pendulum has at the bottom of the swing, only now, it's kinetic energy:

17.64 joules = (1/2) x (mass) x (speed)²

17.64 joules = (1/2) x (5 kg) x (speed)²

Divide each side by 2.5 kg:

17.64 joules / 2.5 kg = speed²

Write out the units of joules:

17.64 kg-m²/s² / 2.5 kg = speed²

(17.64 / 2.5) (m²/s²) = speed²

7.056 m²/s² = speed²

Take the square root of each side: Speed = √(7.056 m²/s²) = 2.656 m/s.

Looking through the choices, we're overjoyed to see that one if them is ' 2.7 m/s '. Surely that's IT !

Hope this will help you ...

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Water is flowing in a pipe with a circular cross section but with varying cross-sectional area, and at all points the water comp
slamgirl [31]

(a) 5.66 m/s

The flow rate of the water in the pipe is given by

Q=Av

where

Q is the flow rate

A is the cross-sectional area of the pipe

v is the speed of the water

Here we have

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the radius of the pipe is

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(b) 0.326 m

The flow rate along the pipe is conserved, so we can write:

Q_1 = Q_2\\A_1 v_1 = A_2 v_2

where we have

A_1 = 0.212 m^2\\v_1 = 5.66 m/s\\v_2 = 3.60 m/s

and where A_2 is the cross-sectional area of the pipe at the second point.

Solving for A2,

A_2 = \frac{A_1 v_1}{v_2}=\frac{(0.212 m^2)(5.66 m/s)}{3.60 m/s}=0.333 m^2

And finally we can find the radius of the pipe at that point:

A_2 = \pi r_2^2\\r_2 = \sqrt{\frac{A_2}{\pi}}=\sqrt{\frac{0.333 m^2}{\pi}}=0.326 m

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ver shines light up to the surface of a flat glass-bottomed boat at an angle of 30 relative to the normal. If the index of refr
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Answer:

\beta = 41.68°

Explanation:

according to snell's law

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sin\alpha = \frac{n_w}{n_g}* sin30

sin\alpha = 0.443

now applying snell's law between air and glass, so we have

\frac{n_g}{n_a} = \frac{sin\alpha}{sin\beta}

sin\beta = \frac{n_g}{n_a} sin\alpha

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we know that sin\alpha = 0.443

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Answer:

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