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3 years ago

How much potential energy foes a 5kg mass have when its 2 meters above the ground?(Hint :PE=mgh)

Physics

1 answer:

frez [133]3 years ago

5 0

Answer:

Explanation:

The potential energy of a body can be found by using the formula

PE = mgh

where

m is the mass

h is the height

g is the acceleration due to gravity which is 9.8 m/s²

From the question we have

PE = 5 × 9.8 × 2

We have the final answer as

Hope this helps you

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A string of length 100 cm is held fixed at both ends and vibrates in a standing wave pattern. The wavelengths of the constituent

azamat

The wavelengths of the constituent travelling waves CANNOT be 400 cm.

The given parameters:

<em>Length of the string, L = 100 cm</em>

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The wavelengths of the constituent travelling waves is calculated as follows;

$L = \frac{n \lambda}{2} \\\\n\lambda = 2L\\\\\lambda = \frac{2L}{n}$

for first mode: n = 1

$\lambda = \frac{2\times 100 \ cm}{1} \\\\\lambda = 200 \ cm$

for second mode: n = 2

$\lambda = \frac{2L}{2} = L = 100 \ cm$

For the third mode: n = 3

$\lambda = \frac{2L}{3} \\\\\lambda = \frac{2 \times 100}{3} = 67 \ cm$

For fourth mode: n = 4

$\lambda = \frac{2L}{4} \\\\\lambda = \frac{2 \times 100}{4} = 50 \ cm$

Thus, we can conclude that, the wavelengths of the constituent travelling waves CANNOT be 400 cm.

The complete question is below:

A string of length 100 cm is held fixed at both ends and vibrates in a standing wave pattern. The wavelengths of the constituent travelling waves CANNOT be:

A. 400 cm

B. 200 cm

C. 100 cm

D. 67 cm

E. 50 cm

Learn more about wavelengths of travelling waves here: brainly.com/question/19249186

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3 years ago

I'm walking 1.6m/s to 7-11 and it started to rain so I sped up to 2.7m/s in 1.2

olga nikolaevna [1]

Answer:

Explanation:

$a = \frac{v_f-v_0}{t}$ which is the final velocity minus the initial velocity in the numerator, and the change in time in the denominator. For us:

$a=\frac{2.7-1.6}{1.2}$ so

a = .92 m/s/s (NOT negative because you're speeding up)

5 0

3 years ago

LVULAN

3241004551 [841]

For this case we have that by definition, the kinetic energy is given by the following formula:

$k= \frac {1} {2} * m * v ^ 2$

Where:

m: It is the mass

v: It is the velocity

According to the data we have to:

$m = 100 \ kg\\v = 9 \frac {m} {s}$

Substituting the values we have:

$k = \frac {1} {2} * (100) * (9) ^ 2\\k = \frac {1} {2} * (100) * 81\\k = 50 * 81\\k = 4050$

finally, the kinetic energy is $4050 \ J$

Answer:

Option A

7 0

3 years ago

A 10.0 kg weather rocket generates a thrust of 230 NN . The rocket, pointing upward, is clamped to the top of a vertical spring.

blondinia [14]

Answer: 0.2m

Explanation: Firstly only the Rocket's Weight Compress the spring which can be found by

$F_r=M_r*g\\F_r=10*9.81\\F_r=98.1N$

According to Hooks Law

$F_r=k*x\\x=F_r/k\\x=98.1/480\\x=0.2m$

The part b and c of this question is done in the attachment

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3 years ago

Which one?? please someone quick!

LiRa [457]

Answer:

i think its second law of motion.

Explanation:

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3 years ago

Read 2 more answers

How much potential energy foes a 5kg mass have when its 2 meters above the ground?(Hint :PE=m*g*h)​

How much potential energy foes a 5kg mass have when its 2 meters above the ground?(Hint :PE=mgh)