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Delicious77 [7]

3 years ago

14

And are facing each other of the two and the friend are on the separate more you when the moving. Your friend weighs a lot on yo

u is push off of each other, the force exerts you exert on him. Less than

1 answer:

puteri [66]3 years ago

7 0

Buddy, I think you need to evaluate your question and fix it. Because it's not making any sense, whatsoever.

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CaHeK987 [17]

Potential energy is the answer

6 0

3 years ago

A rope has one end tied to a vertical support. You hold the other end so that the rope is horizontal. If you move the end of the

Strike441 [17]

Answer:

c. 2 m/s

Explanation:

The relationship between speed, frequency and wavelength of a wave is given by:

$v=f \lambda$

where

v is the speed of the wave

f is its frequency

$\lambda$ is the wavelength

For the transverse wave in this problem, we have:

$f = 4 Hz$ is the frequency

$\lambda=0.5 m$ is the wavelength

Substituting these numbers into the equation, we find the speed of the wave:

$v=(4 Hz)(0.5 m)=2 m/s$

7 0

3 years ago

John pushes a box with a constant force as shown in the graph below.

andrew11 [14]

From the graph, it can be seen that the constant force that John exerted in order to move the object is 14N. Work is calculated by multiplying the force with the distance to which the object moves in parallel with the direction of the force.
Work = Force x displacement
Work = (14 N) x (8 m)
Work = 112 J
The closest value is 110J. Thus, the answer to this item is the second choice.

4 0

3 years ago

Argon gas enters steadily an adiabatic turbine at 900 kPa and 450C with a velocity of 80 m/s and leaves at 150 kPa with a veloc

Crazy boy [7]

Answer:

Temperature at the exit = $267.3 C$

Explanation:

For the steady energy flow through a control volume, the power output is given as

$W_{out}= -m_{f}(h_{2}-h_{1} + \frac{v_{2}^{2}}{2} - \frac{v_{1}^{2}}{2})$

Inlet area of the turbine = $60cm^{2}= 0.006m^{2}$

To find the mass flow rate, we can apply the ideal gas laws to estimate the specific volume, from there we can get the mass flow rate.

Assuming Argon behaves as an Ideal gas, we have the specific volume $v_{1}$

as

$v_{1}=\frac{RT_{1}}{P_{1}}=\frac{0.2081\times723}{900}=0.1672m^{3}/kg$

$m_{f}=\frac{1}{v_{1}}\times A_{1}V_{1} = \frac{1}{0.1672}\times(0.006)(80)=2.871kg/sec$

for Ideal gasses, the enthalpy change can be calculated using the formula

$h_{2}-h_{1}=C_{p}(T_{2}-T_{1})$

hence we have

$W_{out}= -m_{f}((C_{p}(T_{2}-T_{1}) + \frac{v_{2}^{2}}{2} - \frac{v_{1}^{2}}{2})$

$250= -2.871((0.5203(T_{2}-450) + \frac{150^{2}}{2\times 1000} - \frac{80^{2}}{2\times 1000})$

<em>Note: to convert the Kinetic energy term to kilojoules, it was multiplied by 1000</em>

evaluating the above equation, we have $T_{2}=267.3C$

Hence, the temperature at the exit = $267.3 C$

5 0

3 years ago

2. What is the power rating of an engine capable of lifting a 100 kg object 5 m vertically

Leya [2.2K]

Work=f.d
Work=100*50 = 500
Power = work/time = 500/4
=125 watt

7 0

3 years ago