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2 years ago

abin is doing work by lifting a bowling ball. Which statement could be made about the energy in this situation?

2 answers:

8 0

The statement that could be made about the energy in this situation would be :
It being transferred from his arms muscles to the ball.

The muscle contraction from his arms created a force that could be used to lift the ball up.

bekas [8.4K]2 years ago

3 0

Answer:

Explanation:

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A 60-μC charge is held fixed at the origin and a −20-μC charge is held fixed on the x axis at a point x = 1.0 m. If a 10-μC char

Aleksandr [31]

Answer:

Ek = 8,79 [J]

Explanation:

We are going to solve this problem, using the energy conservation principle

State 1 or initial state (charges at rest t=0)

E₁ = Ek + U₁

As charge are at rest Ek = 0

And U₁ has two components

U₁₂ = K * Q₁*Q₂ / 0,4 and U₃₂ = K*Q₃*Q₂ / 0,6

U₁₂ = 9*10⁹* 60*10⁻⁶*10*10⁻⁶/0,4 ⇒ U₁₂ = 9*60*10*10⁻³/0,4

U₃₂ = - 9*10⁹* 20*10⁻⁶*10*10⁻⁶/0,6 ⇒ U₃₂ = - 9*20*10*10⁻³/0,6

U₁₂ = 540*10⁻2/0,4 [J] ⇒13,5 [J]

U₃₂ = - 180*10⁻² /0,6 [J] ⇒ - 3 [J]

Then E₁ = E₁₂ + E₃₂

E₁ = 10,5 [J]

At the moment of Q₂ passing x = 40 cm or 0,4 m

E₂ = Ek + U₂

We can calculate the components of U₂ in this new configuration

U₂ = U₁₂ + U₃₂

U₁₂ = 9*10⁹* 60*10⁻⁶*10*10⁻⁶/0,7 ⇒ U₁₂ = 9*60*10*10⁻³/0,7

U₁₂ = 540*10⁻²/0,7 U₁₂ = 7,71 [J]

U₃₂ = - 9*10⁹* 20*10⁻⁶*10*10⁻⁶/0,3 ⇒ U₃₂ = - 9*20*10*10⁻³/0,3

U₃₂ = - 9*20*10⁻²/0,3

U₃₂ = - 6

U₂ = 7,71 -6

U₂ = 1,71 [J]

Then as

E₂ = Ek + U₂ and E₂ = E₁

Then

Ek + U₂ = E₁

Ek = 10,5 - U₂

Ek = 10,5 - 1,71

Ek = 8,79 [J]

5 0

1 year ago

the acceleration due to gravity on the moon is 1.6 m/s^2 what is the gravitational potential energy of a 1200 kg lander resting

Mumz [18]

Answer

672,000 Joules

Explanation

Gravitational potential energy (P.E) is the energy possessed by a body that is at a potential height from the ground.

IT is calculated by the formula;

P.E = mgh

Where m ⇒ mass

g ⇒ acceleration due to gravity

h ⇒ height from trhe ground.

P.E = 1200 × 1.6 × 350

= 672,000 Joules.

5 0

2 years ago

Let A be the last two digits and let B be the sum of the last three digits of your 8-digit student ID. (Example: For 20245347, A

zheka24 [161]

Answer:

The speed is 173 m/s.

Explanation:

Given that,

A = 47

B = 14

Length 1 urk = 58.0 m

An hour is divided into 125 time units named dorts.

3600 s = 125 dots

dorts = 28.8 s

Speed v= (25.0+A+B) urks/dort

We need to convert the speed into meters per second

Put the value of A and B into the speed

$v=25.0+47+14$

$v =86\ urk s/dort$

$v=86\times\dfrac{58.0}{28.8}$

$v=173.19\ m/s$

Hence, The speed is 173 m/s.

7 0

2 years ago

A string, 0.28 m long and vibrating in its third harmonic, excites an open pipe that is 0.82 m long into its second overtone res

S_A_V [24]

Answer: 98.

Explanation: it has been stated in the question that sound wave in the string and in the pipe resonated at a specific frequency, this simply implies that the frequency of sound wave in the string equals frequency of sound wave in pipe.

Fs = Fp.

The length (l) of the string is 0.28m and it is vibrating at it third harmonic.

The length of stationary wave on a string at third harmonic is given below as

l = 3λ/2

Where λ = wavelength of sound wave in pipe (λs)

By substituting l = 0.28m into the equation above, we have that

0. 28 = 3λs/2

3λs = 0.28 * 2

3λs = 0.56, λs= 0.56/ 3

λs = 0.187m

Thus the wavelength of wave in the string is 0.187m.

Sound from the string in the pipe is produced at the second overtone ( which is the third harmonic).

Therefore the length of air in the pipe at second overtone ( third harmonic) is given below as

l = 5λp/ 4, we need to get the wavelength of sound in the pipe.

Thus

λp = 4*l/5

λp = 4 * 0.82 / 5

λp = 0.656m.

The velocity of sound waves produced in the pipe is 345m/s thus the frequency of sound in the pipe is gotten using the formulae below

V = fpλp

V= velocity of sound in pipe, fp = frequency of sound in pipe, λp= wavelength of sound in pipe

345 = f / 0.656

fp = 525.92Hz.

As stated in the question, the frequency of sound in pipe is the same as that in string (fp = fs = f) , thus to get the velocity of sound wave in string we use the same formulae of

v = fλ

Where f = frequency of sound in pipe = frequency of sound in string = 525.92Hz.

λ = wavelength of sound in string = 0.187m

Thus v = 525.92 * 0.187 = 98.34 which is closest to 98.

3 0

1 year ago

if one sprinter runs the 400.0 m in 58 seconds and another can run the same distance in 60.0 seconds, by how much distance will

Sindrei [870]

They will win by approximately .2m
Divide 400/58=6.9m while the other person is 400/60=6.6 thus he will win by about .2m

6 0

2 years ago