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Ghella [55]
2 years ago
11

An object attached to one end of a spring makes 20 vibrations in 10 seconds. Its angular frequency is: 0. 79 rad/s 1. 57 rad/s 2

. 0 rad/s 6. 3 rad/s 12. 6 rad/s
Physics
1 answer:
morpeh [17]2 years ago
5 0

Angular frequency in radian per second for 20 vibrations in 10 seconds is 12.6 rad/s

<h3>What is Angular frequency?</h3>

Angular frequency is the number of vibrations in radian per second.

The total number of vibrations n is 20 and the time taken for these vibrations is 10 s

The frequency of the vibrations will be

f = 20 / 10 = 2 Hz

Angular frequency is related to the frequency as

ω = 2πf

ω=2π × 2

ω = 12.6 rad/s

Thus, the angular frequency is 12.6 rad/s.

Learn more about Angular frequency.

brainly.com/question/14244057

#SPJ

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Using the balanced chemical equation below, calculate how many moles of ammonia would be produced when 9.1 moles of hydrogen gas
strojnjashka [21]

6.07 moles

Explanation:

Given parameters:

Number of moles of reacting hydrogen gas = 9.1 moles

Unknown:

Number of moles of ammonia produced = ?

Solution:

  Balanced chemical equation:

           N₂   +     3H₂    →      2NH₃

We should work from the known to the unknown specie. We known the number of moles of hydrogen gas reacting. We simply relate this to that of the ammonia.

 From the reaction:

   3 mole of hydrogen gas produced 2 mole of ammonia

     9.1 moles of the hydrogen gas will produce:  \frac{2 x 9.1}{3}

                                                                            = 6.07 moles

Learn more:

Number of moles brainly.com/question/1841136

#learnwithbrainly

3 0
3 years ago
A force is applied to an ideal spring (initially in its equilibrium position) and does 1.9 JJ of work stretching it 2.2 cmcm . H
Verdich [7]

Answer:

W=16.58J

Explanation:

initial information we have

work: W=1.9J

stretched distance: x=2.2cm=0.022m

from this, we can find the value of the constant of the spring k, with the equation for work in a spring:

W=\frac{1}{2} kx^2

substituting known values:

1.9J=\frac{1}{2}k(0.022)^2\\

and clearing for k:

k=\frac{2(1.9J)}{0.022^2} \\k=7,851.24

and now we want to know how much work is done when we stretch the spring a distance of 6.5cm from equilibrium, so now x is:

x=6.5cm=0.065m

and using the same formula for work, with the value of k that we just found:

W=\frac{1}{2} kx^2

W=\frac{1}{2}(7851.24)(0.065)^2\\W=16.58J

5 0
3 years ago
Read 2 more answers
A chemical change means a new substance with new properties was<br> formed.<br> True<br> False
Contact [7]

Answer:

False

Explanation:

4 0
3 years ago
How does the elbow medical and lateral epicondylitis?
dlinn [17]
<span>Lateral epicondylitis, or “tennis elbow,” is an inflammation of the tendons that join the forearm muscles on the outside of the elbow. </span>The bony bump on the outside (lateral<span> side) of the </span>elbow<span> is called the </span>lateral epicondyle<span>. The ECRB muscle and tendon is usually involved in </span>tennis elbow<span>. </span><span>
Medial epicondylitis, or “golfer’s elbow,” is an inflammation of the tendons that attach your forearm muscles to the inside of the bone at your elbow. </span>It's identified by pain from the elbow to the wrist on the inside (medial<span> side) of the elbow. The pain is caused by damage to the tendons that bend the wrist toward the palm.</span>
4 0
3 years ago
Can someone please help
ki77a [65]

Answer:

Acceleration of that planet is 30 \frac{m}{s^{2} }.

Given:

initial speed of hammer = 0 \frac{m}{s}

time = 1 s

distance = 15 m

To find:

Acceleration due to gravity = ?

Formula used:

Distance covered by hammer is given by,

s = ut + \frac{1}{2} a t^{2}

s = distance

u = initial speed of hammer

t = time taken by hammer to reach ground

a = acceleration

Solution:

Distance covered by hammer is given by,

s = ut + \frac{1}{2} a t^{2}

s = distance

u = initial speed of hammer

t = time taken by hammer to reach ground

a = acceleration

u = 0

t = 1 s

s = 15 m

a = g

Thus substituting these value in above equation.

15 = 0 + \frac{1}{2} g 1^{2}

g = 15 × 2

g = 30 \frac{m}{s^{2} }

Thus, acceleration of that planet is 30 \frac{m}{s^{2} }.

8 0
3 years ago
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