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zhannawk [14.2K]

4 years ago

6

If a home uses a large supply of solar panels to generate electricity, but has no battery system, surplus electricity that is pr

oduced is usually __________.

1 answer:

Alinara [238K]4 years ago

4 0

Answer:

Passed into the power grid for others to use the electricity

Explanation:

If a home uses a large supply of solar panels to generate electricity, but has no battery system, surplus electricity that is produced is usually passed into the power grid for others to use the electricity, generating a income to the homeowner

You might be interested in

(d) If η = 40% and TH = 427°C, what is TC, in °C?

Brrunno [24]

Answer:

$T_C=256.2^{\circ}C$

Explanation:

Given that,

Efficiency of heat engine, $\eta=40\%=0.4$

Temperature of hot source, $T_H=427^{\circ}C$

We need to find the temperature of cold sink i.e. $T_C$ . The efficiency of heat engine is given by :

$\eta=1-\dfrac{T_C}{T_H}$

$T_C=(1-\eta)T_H$

$T_C=(1-0.4)\times 427$

$T_C=256.2^{\circ}C$

So, the temperature of the cold sink is 256.2°C. Hence, this is the required solution.

3 0

3 years ago

State which of the three materials would allow the thermometer to measure the

nignag [31]

Answer:

i dont know

Explanation:im sorry to do this to you but you dont have to watch ads if you answer questions

6 0

2 years ago

When is a secondary source more helpful than a primary source?

UNO [17]

Answer:

I think the answer is C.

Explanation:

A primary source is a first hand account of an event while a secondary source is a retelling or second hand account meaning as many details will be prevalent.

8 0

2 years ago

As a result, the total energy in a ____ system (in other words, a system with no external forces) will remain constant

Alika [10]

Answer:

Isolated or Closed system, both are correct

4 0

3 years ago

The volume of water in the Pacific Ocean is about 7.0 × 10 8 km 3 . The density of seawater is about 1030 kg/m3. (a) Determine t

Novay_Z [31]

To solve the problem it is necessary to consider the concepts related to Potential Energy and Kinetic Energy.

Potential Energy because of a planet would be given by the equation,

$PE=\frac{GMm}{r}$

Where,

G = Gravitational Universal Constant

M = Mass of Ocean

M = Mass of Moon

r = Radius

From the data given we can calculate the mass of the ocean water through the relationship of density and volume, then,

$m = \rho V$

$m = (1030Kg/m^3)(7*10^8m^3)$

$m = 7.210*10^{11}Kg$

It is necessary to define the two radii, when the ocean is far from the moon and when it is facing.

When it is far away, it will be the total diameter from the center of the earth to the center of the moon.

$r_1 = 3.84*10^8 + 6.4*10^6 = 3.904*10^8m$

When it's near, it will be the distance from the center of the earth to the center of the moon minus the radius,

$r_2 = 3.84*10^8-6.4*10^6 - 3.776*10^8m$

PART A) Potential energy when the ocean is at its furthest point to the moon,

$PE_1 = \frac{GMm}{r_1}$

$PE_1 = \frac{(6.61*10^{-11})*(7.21*10^{11})*(7.35*10^{22})}{3.904*10^8}$

$PE_1 = 9.05*10^{15}J$

PART B) Potential energy when the ocean is at its closest point to the moon

$PE_2 = \frac{GMm}{r_2}$

$PE_2 = \frac{(6.61*10^{-11})*(7.21*10^{11})*(7.35*10^{22})}{3.776*10^8}$

$PE_2 = 9.361*10^{15}J$

PART C) The maximum speed. This can be calculated through the conservation of energy, where,

$\Delta KE = \Delta PE$

$\frac{1}{2}mv^2 = PE_2-PE_1$

$v=\sqrt{2(PE_2-PE_1)/m}$

$v = \sqrt{\frac{2*(9.361*10^{15}-9.05*10^{15})}{7.210*10^{11}}}$

$v = 29.4m/s$

8 0

4 years ago