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Diano4ka-milaya [45]
2 years ago
15

the electrical energy used to pump the water up to a mountain lake is 1.2 x 10 to the power 12 J only 6.2 x 10 to the power 11 J

of electrical energy is generated when the water is released calculate the efficiency of this energy storage scheme​
Physics
1 answer:
Andreyy892 years ago
4 0

Answer:

Explanation:

ake is 1.2 x 10^12 J. Only 6.2 x 10^11 J of electrical energy is generated when the water is released. Calculate the efficiency of this energy

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Water flows without friction vertically downward through a pipe and enters a section where the cross sectional area is larger. T
djverab [1.8K]

Answer:

v_{2} will be less than v_{1} and P_{2} will be greater than P_{1}.

Explanation:

As we know from the conservation of mass, the rate at which any amount of fluid mass (m_{1}) is entering in a system is equal to the rate at which the same amount of fluid mass (m_{2}) is leaving the system.

Rate of mass flow can be written as,

m = \rho A v

where \rho is the density of the fluid, A is the area through which the fluid is flowing and v is the velocity of the fluid.

Now, according to the problem, as the density of the fluid does not change, we can write

&& m_{1} = m_{2}\\&or,& \rho A_{1} v_{1} = \rho A_{2} v_{2}\\&or,& \dfrac{v_{2}}{v_{1}} = \dfrac{A_{1}}{A_{2}}

where A_{1} and A_{2} are the cross-sectional areas through which the fluid is passing and v_{1} and v_{2} are the velocities of the fluid through the respective cross-sectional areas.

As according to the problem, A_{2} > A_{1}, so from the above formula v_{2} < v_{1}.

Also we know that fluid pressure is created by the motion of the fluid through any area. When the fluid gains speed, some of its energy is used to move faster in the fluid’s direction of motion. It causes in a lower pressure.

So, as in this case v_{2} < v_{1} the pressure in the large cross-sectional area P_{2} will be greater than the pressure  P_{1} in the small cross sectional area, i.e.,

P_{2} > P_{1}.

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3 years ago
A person who has the condition CIPA is unable to feel pain or temperature. Which sensory receptors are affected by this conditio
xenn [34]

B........................

7 0
2 years ago
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An Atwood machine consists of two masses, mA = 6.8 kg and mB = 8.0 kg , connected by a cord that passes over a pulley free to ro
Lisa [10]

To solve this problem it is necessary to apply the concewptos related to Torque, kinetic movement and Newton's second Law.

By definition Newton's second law is described as

F= ma

Where,

m= mass

a = Acceleration

Part A) According to the information (and as can be seen in the attached graph) a sum of forces is carried out in mass B, it is obtained that,

\sum F = m_b a

m_Bg-T_B = m_Ba

T_B = m_Bg-m_Ba

In the case of mass A,

\sum F = m_A a

T_A = m_Ag-m_Aa

Making summation of Torques in the Pulley we have to

\sum\tau = I\alpha

T_BR_0-T_AR_0=I\alpha

T_B-T_A=I\frac{a}{R^2_0}

Replacing the values previously found,

(m_Bg-m_Ba )-(m_Ag-m_Aa )=I\frac{a}{R^2_0}

(m_B-m_A)g-(m_B+m_A)a=I\frac{a}{R^2_0}

a = \frac{(m_B-m_A)g}{\frac{I}{R_0^2}+(m_B+m_A)}

a = \frac{(m_B-m_A)g}{\frac{MR^2_0^2/2}{R_0^2}+(m_B+m_A)}

a =\frac{(m_B-m_A)g}{\frac{M}{2}+(m_B+m_A)}

Replacing with our values

a =\frac{(8-6.8)(9.8)}{\frac{0.8}{2}+(8+6.8)}

a=0.7736m/s^2

PART B) Ignoring the moment of inertia the acceleration would be given by

a' =\frac{(m_B-m_A)g}{(m_B+m_A)}

a' =\frac{(8-6.8)(9.8)}{(8+6.8)}

a' = 0.7945

Therefore the error would be,

\%error = \frac{a'-a}{a}*100

\%error = \frac{0.7945-0.7736}{0.7736}*100

\%error = 2.7%

8 0
2 years ago
What is the de Broglie - equation​
fomenos

Answer:

is this it?

Explanation:

λ = h/mv, where λ is wavelength, h is Planck's constant, m is the mass of a particle, moving at a velocity v. de Broglie suggested that particles can exhibit properties of waves.

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2 years ago
A sound wave can be considered as a displacement wave or a pressure wave? What phase difference exists between the displacement
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Where the displacement is a maximum the pressure is a minimum.
Where the displacement is zero, the pressure is a maximum.
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