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Leni [432]
4 years ago
11

"A pendulum is pulled back from its equilibrium (center) position and then released. When the pendulum bob is halfway between th

e high point and the low point in its swing, is the total energy kinetic energy, potential energy, or both
Physics
1 answer:
kumpel [21]4 years ago
3 0

Answer:

Option C, The total energy consists of half the original potential energy and half of the original potential energy converted to kinetic energy.

Explanation:

Complete question

A pendulum is pulled back from its equilibrium (center) position and then released. When the pendulum bob is halfway between the high point and the low point in its swing, is the total energy kinetic energy, potential energy, or both? Explain.

The total energy is kinetic energy only.

The total energy is potential energy only.

The total energy consists of half the original potential energy and half of the original potential energy converted to kinetic energy.

The total energy consists of one-fourth the original potential energy and three-fourths of the original potential energy converted to kinetic energy.

Solution

Total energy is the sum of kinetic energy and potential energy and as a pendulum moves back and forth, there is continuous transformation of energy from one form to the other form. i.e from kinetic energy to potential energy and vice versa.  

When the pendulum is released from some position, the potential energy  in it start converting into kinetic energy with the increase in speed of motion of pendulum bob

Hence, option C is correct

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The displacement of a wave traveling in the positive x-direction is y(x, t)|= (3.5 cm)cos(2.7x − 92t), where x is in m and t is
zubka84 [21]

Answer

Given,

y(x, t) = (3.5 cm) cos(2.7 x − 92 t)

comparing the given equation with general equation

y(x,t) = A cos(k x - ω t)

 A = 3.5 cm  , k = 2.7 rad/m    , ω = 92 rad/s

we know,

a) ω =2πf

   f = 92/ 2π

   f = 14.64 Hz

b) Wavelength of the wave

 we now, k = 2π/λ

      2π/λ = 2.7

      λ = 2 π/2.7

      λ = 2.33 m

c) Speed of wave

     v = ν λ

     v = 14.64 x 2.33

     v = 34.11 m/s

5 0
4 years ago
A 75 kg astronaut floating in space throws a 5 kg rock at 5 m/sec. How fast does the astronaut move backwards?
JulijaS [17]

Answer:

The astronaut will get a velocity 0.064ms−1 opposite to the direction of the object.

4 0
3 years ago
An ocean thermal energy conversion system is being proposed for electric power generation. Such a system is based on the standar
defon

Answer:

Explanation:

Dear Student, this question is incomplete, and to attempt this question, we have attached the complete copy of the question in the image below. Please, Kindly refer to it when going through the solution to the question.

To objective is to find the:

(i) required heat exchanger area.

(ii) flow rate to be maintained in the evaporator.

Given that:

water temperature = 300 K

At a reasonable depth, the water is cold and its temperature = 280 K

The power output W = 2 MW

Efficiency \zeta = 3%

where;

\zeta = \dfrac{W_{out}}{Q_{supplied }}

Q_{supplied } = \dfrac{2}{0.03} \ MW

Q_{supplied } = 66.66 \ MW

However, from the evaporator, the heat transfer Q can be determined by using the formula:

Q = UA(L MTD)

where;

LMTD = \dfrac{\Delta T_1 - \Delta T_2}{In (\dfrac{\Delta T_1}{\Delta T_2} )}

Also;

\Delta T_1 = T_{h_{in}}- T_{c_{out}} \\ \\ \Delta T_1 = 300 -290 \\ \\ \Delta T_1 = 10 \ K

\Delta T_2 = T_{h_{in}}- T_{c_{out}} \\ \\ \Delta T_2 = 292 -290 \\ \\ \Delta T_2 = 2\ K

LMTD = \dfrac{10 -2}{In (\dfrac{10}{2} )}

LMTD = \dfrac{8}{In (5)}

LMTD = 4.97

Thus, the required heat exchanger area A is calculated by using the formula:

Q_H = UA (LMTD)

where;

U = overall heat coefficient given as 1200 W/m².K

66.667 \times 10^6 = 1200 \times A \times 4.97 \\ \\  A= \dfrac{66.667 \times 10^6}{1200 \times 4.97} \\ \\  \mathbf{A = 11178.236 \ m^2}

The mass flow rate:

Q_{H} = mC_p(T_{in} -T_{out} )  \\ \\  66.667 \times 10^6= m \times 4.18 (300 -292) \\ \\ m = \dfrac{  66.667 \times 10^6}{4.18 \times 8} \\ \\  \mathbf{m = 1993630.383 \ kg/s}

3 0
3 years ago
A force of 45 newtons is applied on an object, moving it 12 meters away in the same direction as the force. What is the magnitud
NARA [144]
<h2>Answer: 540 J</h2>

Explanation:

The Work W done by a Force F refers to the release of potential energy from a body that is moved by the application of that force to overcome a resistance along a path.  

Now, when the applied force is constant and the direction of the force and the direction of the movement are parallel, the equation to calculate it is:  

W=(F)(d) (1)  

In this case both (the force and the distance in the path) are parallel (this means they are in the same direction), so the work W performed is the product of the force exerted to push the box F=45N by the distance traveled d=12m.

Hence:  

W=(45N)(12m)   (2)

W=540Nm=540J

6 0
3 years ago
A 350-N child is in a swing that is attached to a pair of ropes 2.10 m long. Find the gravitational potential energy of the chil
o-na [289]

Answer:

a)  U = 735 J , b) U = 125.7 J , c)   U = 0 J

Explanation:

The gravitational power energy is

      U = mg y - mg y₀

The last value is a constant, for simplicity we can make it zero, if the lowest point is at the origin of the coordinate system, which in this case we will place in the lowest part

a) Rope is horizontal

The height in this case is the same length of the rope

     y = 2.10 m

    w = mg = 350 N

    U = 350 2.10

    U = 735 J

b) when the angle is 34º

     y = L - L cos 34

    y = L (1- cos34)

    y = 2.10 (1- cos 34)

    y = 0.359 m

    U = 350 0.359

    U = 125.7 J

c) in this case this point coincides with the reference system

     y = 0

     U = 0 J

4 0
4 years ago
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