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sammy [17]
3 years ago
8

A hot water heater is operated by using solar power. if the solar collector has an area of 5.3 m2 , and the power delivered by s

unlight is 995 w/m2 , how long will it take to increase the temperature of 1 m3 of water from 20 ◦c to 65 ◦c? the specific heat of water is 4186 j/kg · ◦ c and the density of water is 1000 kg/m3 . answer in units of h.
Physics
1 answer:
OLEGan [10]3 years ago
4 0
Heat absorbed by the solar collector = Area*Irradiance = 5.3*995 = 5273.5 W

Heat Q in joules absorbed in t hours = Heat used to heat water. That is,

5273.5*t = mCΔT; where mass = volume*density = 1*1000 = 1000 kg

Therefore;
5273.5t = 1000*4186*(65-20) = 188370000
t = 188370000/5273.5 = 35720.11 seconds = 35720.11/(60*60) hours ≈ 9.92 hours.

It will take approximately 9.92 hours.
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An ideal gas is enclosed in a cylinder that has a movable piston on top. The piston has a mass m and an area A and is free to sl
elena-14-01-66 [18.8K]

Answer:

Explanation:

Given mass of piston \left ( m\right )

no. of moles =n

Given Pressure remains same

Temperature changes from T_1 to T_2

Work done\left ( W\right ) is given by=\int_{V_1}^{V_2}PdV

W=P\left ( V_2-V_1\right )

also PV_1=nRT_1

PV_2=nRT_2

W=nR\left ( T_2-T_1\right )

4 0
3 years ago
The moon has a diameter of 3.48 x 106 m and is a distance of 3.85 x 108 m from the earth. The sun has a diameter of 1.39 x 109 m
Mrrafil [7]

Answer:

0.00903 rad

0.00926 rad

6.268\times 10^{-6}

Explanation:

s = Diameter of the object

r = Distance between the Earth and the object

Angle subtended is given by

\theta=\frac{s}{r}

For the Moon

\theta_m=\dfrac{3.48\times 10^6}{3.85\times 10^8}\\\Rightarrow \theta_m=0.00903\ rad

The angle subtended by the Moon is 0.00903 rad

For the Sun

\theta_s=\dfrac{1.39\times 10^9}{1.5\times 10^{11}}\\\Rightarrow \theta_s=0.00926\ rad

The angle subtended by the Sun is 0.00926 rad

Area ratio is given by

\frac{A_m}{A_s}=\dfrac{\pi r_m^2}{\pi r_s^2}\\\Rightarrow \frac{A_m}{A_s}=\dfrac{d_m^2}{d_s^2}\\\Rightarrow \frac{A_m}{A_s}=\dfrac{(3.48\times 10^{6})^2}{(1.39\times 10^9)^2}\\\Rightarrow \frac{A_m}{A_s}=6.268\times 10^{-6}

The area ratio is 6.268\times 10^{-6}

3 0
3 years ago
Which of the following is the best example of a primary circular reaction?
olga55 [171]

Primary Circular Reactions (1-4 months): This substage involves coordinating sensation and new schemas. For example, a child may suck his or her thumb by accident and then later intentionally repeat the action. These actions are repeated because the infant finds them pleasurable.

4 0
3 years ago
a runner covers the last straigjt stretch of a race in 4 s. during that time, he speeds up from 5m/s to 9m/s.
PtichkaEL [24]

During that final period of time,
his acceleration is
                                (9 m/s - 5 m/s) / (4 sec) = 1 m/s² .

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8 0
3 years ago
A 21 N block rests on a horizontal surface. The coefficients of static and kinetic friction between the surface and the block ar
Georgia [21]

Answer:

a)15 N

b)12.6 N

Explanation:

Given that

Weight of block (wt)= 21 N

μs = 0.80 and μk = 0.60

We know that

Maximum value of static friction given as

Frs = μs m g = μs .wt

by putting the values

Frs= 0.8 x 21 = 16.8 N

Value of kinetic friction

Frk= μk m g = μk .wt

By putting the values

Frk= 0.6 x 21 = 12.6 N

a)

When T = 15 N

Static friction Frs= 16.8 N

Here the value of static friction is more than tension T .It means that block will not move and the value of friction force will be equal to the tension force.

Friction force = 15 N

b)

When T= 35 N

Here value of tension force is more than maximum value of static friction that is why block will move .We know that when body is in motion then kinetic friction will act on the body.so the value of friction force in this case will be 12.6 N

Friction force = 12.6 N

8 0
3 years ago
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