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lyudmila [28]
2 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:
azamat2 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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5 0
3 years ago
How should the student use the data collected and the known quantities from the experiment to determine the initial total mechan
fiasKO [112]

It should be noted that the student should use K = 1/2mv² with the initial speed of the block for one trial.

<h3>Method of using the data collected.</h3>

From the complete information, the student wants to use the data collected and the known quantities from the experiment to determine the initial total mechanical energy of the block-ramp-Earth system for all trials in the experiment.

In this case, it's important to use K = 1/2mv² with the block's initial speed for one trial due to the fact that the initial speed is the same in all the trials.

Learn more about experiments on:

brainly.com/question/17274244

7 0
2 years ago
Two facing surfaces of two large parallel conducting plates separated by 8.5 cm have uniform surface charge densities such that
elena-s [515]

Answer:

positive plate

E = 5.764 KV / m

W = 490eV or 7.85 * 10^-17 J

E_p = 4.74 *10^(-12) eV

E_k = 490 eV

Explanation:

part a

The potential difference between two plates = 490 V

Distance between two plates = 8.5 cm

Answer: The positive plate is at higher potential because of convention.

part b

Electric Field between the plates

E = V / d

E = 490 / 0.085 = 5.764 KV / m

Answer: Electric Field between the plates E = 5.764 KV / m

part c

Work done by electric field

W = V*q

W = 490 * 1.602*10^-19

W = 7.85 * 10^-17 J

or W = 490 eV

Answer: Work done by electric field W = 490eV or 7.85 * 10^-17 J

part d

Potential Energy of an electron gained:

E_p = m_e * g * d / (1.602*10^-19)

E_p =  9.109*10^-31* 9.81 * 0.085 / (1.602*10^-19)

E_p = 4.74 *10^(-12) eV

Very very small E_p approximately 0

Answer: Potential Energy of an electron gained E_p = 4.74 *10^(-12) eV or 0.

part e

Kinetic Energy of an electron gained:

W - E_p = E_k

E_k = 490eV - 4.74*10^(-12)eV

E_k = 490 eV

Answer: Kinetic Energy of an electron gained E_k = 490 eV

7 0
3 years ago
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Wood isn’t a medium. Pls give brainliest
8 0
2 years ago
A pitcher throws a baseball that reaches the catcher in 0.75 s. The ball curves because it is spinning at an average angular vel
7nadin3 [17]

The change in angular displacement as a function of time is the definition given for angular velocity, this is mathematically described as

\omega = \frac{\theta}{t}

Here,

\theta = Angular displacement

t = time

The angular velocity is given as

\omega = 230rev/min

PART A) The angular velocity in SI Units will be,

\omega = 230rev/min (\frac{1min}{60s})(\frac{2\pi rad}{1rev})

\omega = \frac{23}{3}\pi rad/s \approx 24.08rad/s

PART B) From our first equation we can rearrange to find the angular displacement then

\theta = \omega t

Replacing,

\theta = (24.08)(0.75)

\theta = 18.06 rad

4 0
3 years ago
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