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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.
You might be interested in
1) A uniform wooden beam, with mass of 120 and length L = 4 m, is supported as illustrated in the figure. If the static friction
Kobotan [32]

Answer:

1(a) 55.0°

1(b) 58.3°

2(a) 10.2 N

2(b) 2.61 m/s²

3(a) 76.7°

3(b) 12.8 m/s

3(c) 3.41 s

3(d) 21.8 m/s

3(e) 18.5 m

4(a) 7.35 m/s²

4(b) 31.3 m/s²

4(c) 12.8 m/s²

Explanation:

1) Draw a free body diagram on the beam.  There are five forces:

Weight force mg pulling down at the center of the beam,

Normal force Na pushing up at point A,

Friction force Na μa pushing left at point A,

Normal force Nb pushing perpendicular to the incline at point B,

Friction force Nb μb pushing up the incline at point B.

There are 3 unknown variables: Na, Nb, and θ.  So we're going to need 3 equations.

Sum of forces in the x direction:

∑F = ma

-Na μa + Nb sin φ − Nb μb cos φ = 0

Nb (sin φ − μb cos φ) = Na μa

Nb / Na = μa / (sin φ − μb cos φ)

Sum of forces in the y direction:

∑F = ma

Na + Nb cos φ + Nb μb sin φ − mg = 0

Na = mg − Nb (cos φ + μb sin φ)

Sum of torques about point B:

∑τ = Iα

-mg (L/2) cos θ + Na L cos θ − Na μa L sin θ = 0

mg (L/2) cos θ = Na L cos θ − Na μa L sin θ

mg cos θ = 2 Na cos θ − 2 Na μa sin θ

mg = 2 Na − 2 Na μa tan θ

Substitute:

Na = 2 Na − 2 Na μa tan θ − Nb (cos φ + μb sin φ)

0 = Na − 2 Na μa tan θ − Nb (cos φ + μb sin φ)

Na (1 − 2 μa tan θ) = Nb (cos φ + μb sin φ)

1 − 2 μa tan θ = (Nb / Na) (cos φ + μb sin φ)

2 μa tan θ = 1 − (Nb / Na) (cos φ + μb sin φ)

Substitute again:

2 μa tan θ = 1 − [μa / (sin φ − μb cos φ)] (cos φ + μb sin φ)

tan θ = 1/(2 μa) − (cos φ + μb sin φ) / (2 sin φ − 2 μb cos φ)

a) If φ = 70°, then θ = 55.0°.

b) If φ = 90°, then θ = 58.3°.

2) Draw a free body diagram of each mass.  For each mass, there are four forces.  For mass A:

Weight force Ma g pulling down,

Normal force Na pushing perpendicular to the incline,

Friction force Na μa pushing parallel down the incline,

Tension force T pulling parallel up the incline.

For mass B:

Weight force Mb g pulling down,

Normal force Nb pushing perpendicular to the incline,

Friction force Nb μb pushing parallel up the incline,

Tension force T pulling up the incline.

There are four unknown variables: Na, Nb, T, and a.  So we'll need four equations.

Sum of forces on A in the perpendicular direction:

∑F = ma

Na − Ma g cos θ = 0

Na = Ma g cos θ

Sum of forces on A up the incline:

∑F = ma

T − Na μa − Ma g sin θ = Ma a

T − Ma g cos θ μa − Ma g sin θ = Ma a

Sum of forces on B in the perpendicular direction:

∑F = ma

Nb − Mb g cos φ = 0

Nb = Mb g cos φ

Sum of forces on B down the incline:

∑F = ma

-T − Nb μb + Mb g sin φ = Mb a

-T − Mb g cos φ μb + Mb g sin φ = Mb a

Add together to eliminate T:

-Ma g cos θ μa − Ma g sin θ − Mb g cos φ μb + Mb g sin φ = Ma a + Mb a

g (-Ma (cos θ μa + sin θ) − Mb (cos φ μb − sin φ)) = (Ma + Mb) a

a = -g (Ma (cos θ μa + sin θ) + Mb (cos φ μb − sin φ)) / (Ma + Mb)

a = 2.61 m/s²

Plug into either equation to find T.

T = 10.2 N

3i) Given:

Δx = 3.7 m

vᵧ = 0 m/s

aₓ = 0 m/s²

aᵧ = -10 m/s²

t = 1.25 s

Find: v₀ₓ, v₀ᵧ

Δx = v₀ₓ t + ½ aₓ t²

3.7 m = v₀ₓ (1.25 s) + ½ (0 m/s²) (1.25 s)²

v₀ₓ = 2.96 m/s

vᵧ = aᵧt + v₀ᵧ

0 m/s = (-10 m/s²) (1.25 s) + v₀ᵧ

v₀ᵧ = 12.5 m/s

a) tan θ = v₀ᵧ / v₀ₓ

θ = 76.7°

b) v₀² = v₀ₓ² + v₀ᵧ²

v₀ = 12.8 m/s

3ii) Given:

Δx = D cos 57°

Δy = -D sin 57°

v₀ₓ = 2.96 m/s

v₀ᵧ = 12.5 m/s

aₓ = 0 m/s²

aᵧ = -10 m/s²

c) Find t

Δx = v₀ₓ t + ½ aₓ t²

D cos 57° = (2.96 m/s) t + ½ (0 m/s²) t²

D cos 57° = 2.96t

Δy = v₀ᵧ t + ½ aᵧ t²

-D sin 57° = (12.5 m/s) t + ½ (-10 m/s²) t²

-D sin 57° = 12.5t − 5t²

Divide:

-tan 57° = (12.5t − 5t²) / 2.96t

-4.558t = 12.5t − 5t²

0 = 17.058t  − 5t²

t = 3.41 s

d) Find v

vₓ = aₓt + v₀ₓ

vₓ = (0 m/s²) (3.41 s) + 2.96 m/s

vₓ = 2.96 m/s

vᵧ = aᵧt + v₀ᵧ

vᵧ = (-10 m/s²) (3.41 s) + 12.5 m/s

vᵧ = -21.6 m/s

v² = vₓ² + vᵧ²

v = 21.8 m/s

e) Find D.

D cos 57° = 2.96t

D = 18.5 m

4) Given:

R = 90 m

d = 140 m

v₀ = 0 m/s

at = 0.7t m/s²

The distance to the opposite side of the curve is:

140 m + (90 m) (π/2) = 281 m

a) Find Δx and v if t = 10.5 s.

at = 0.7t

Integrate:

vt = 0.35t² + v₀

vt = 0.35 (10.5)²

vt = 38.6 m/s

Integrate again:

Δx = 0.1167 t³ + v₀ t + x₀

Δx = 0.1167 (10.5)³

Δx = 135 m

The car has not yet reached the curve, so the acceleration is purely tangential.

at = 0.7 (10.5)

at = 7.35 m/s²

b) Find Δx and v if t = 12.2 s.

at = 0.7t

Integrate:

vt = 0.35t² + v₀

vt = 0.35 (12.2)²

vt = 52.1 m/s

Integrate again:

Δx = 0.1167 t³ + v₀ t + x₀

Δx = 0.1167 (12.2)³

Δx = 212 m

The car is in the curve, so it has both tangential and centripetal accelerations.

at = 0.7 (12.2)

at = 8.54 m/s²

ac = v² / r

ac = (52.1 m/s)² / (90 m)

ac = 30.2 m/s²

a² = at² + ac²

a = 31.3 m/s²

c) Given:

Δx = 187 m

v₀ = 0 m/s

at = 3 m/s²

Find: v

v² = v₀² + 2aΔx

v² = (0 m/s)² + 2 (3 m/s²) (187 m)

v = 33.5 m/s

ac = v² / r

ac = (33.5 m/s)² / 90 m

ac = 12.5 m/s²

a² = at² + ac²

a = 12.8 m/s²

5 0
3 years ago
I WILL GIVE BRAINLIEST IF SOMEONE GETS THIS......
pav-90 [236]

Answer:

Explanation:

a)

Firstly to calculate the total mass of the can before the metal was lowered we need to add the mass of the eureka can and the mass of the water in the can. We don't know the mass of the water but we can easily find if we know the volume of the can. In order to calculate the volume we would have to multiply the area of the cross section by the height. So we do the following.

100cm^{2} x 10cm = 1000cm^{3}

Now in order to find the mass that water has in this case we have to multiply the water's density by the volume, and so we get....

\frac{1g}{cm^{3} } x 1000cm^{3} = 1000g or 1kg

Knowing this, we now can calculate the total mass of the can before the metal was lowered, by adding the mass of the water to the mass of the can. So we get....

1000g + 100g = 1100g or 1.1kg

b)

The volume of the water that over flowed will be equal to the volume of the metal piece (since when we add the metal piece, the metal piece will force out the same volume of water as itself, to understand this more deeply you can read the about "Archimedes principle"). Knowing this we just have to calculate the volume of the metal piece an that will be the answer. So this time in order to find volume we will have to divide the total mass of the metal piece by its density. So we get....

20g ÷ \frac{8g}{cm^{3} } = 2.5 cm^{3}

c)

Now to find out the total mass of the can after the metal piece was lowered we would have to add the mass of the can itself, mass of the water inside the can, and the mass of the metal piece. We know the mass of the can, and the metal piece but we don't know the mass of the water because when we lowered the metal piece some of the water overflowed, and as a result the mass of the water changed. So now we just have to find the mass of the water in the can keeping in mind the fact that 2.5cm^{3} overflowed. So now we the same process as in number a) just with a few adjustments.

\frac{1g}{cm^{3} } x (1000cm^{3} - 2.5cm^{3}) = 997.5g

So now that we know the mass of the water in the can after we added the metal piece we can add all the three masses together (the mass of the can. the mass of the water, and the mass of the metal piece) and get the answer.

100g + 997.5g + 20g = 1117.5g or 1.1175kg

5 0
3 years ago
If the Sun were scaled down to the size of a grapefruit, about how far would you have to walk from our classroom to reach Alpha
Burka [1]

Answer:

2000 miles.

Explanation:

It's the Colorado scale model, in which sun is taken as the grapefruit and the distances are measured for different planets with respect to sun just for understanding.

6 0
3 years ago
A hair dryer uses 1200 watts of power. Current flow through
KonstantinChe [14]
The answer is: 120V

Power is the rate at which energy is supplied/transformed in time:
we can write:

V ddp in Volts represents Energy/Charge i.e. energy carried by each coulomb;

I current in Amperes represents Charge/time or coulombs passing each seconds.

combining them we have:

Power = energy/time = V • 1

or

1200 = V ⋅ 10
V = 1200/10 = 120V
6 0
3 years ago
Read 2 more answers
Help<br> How much current will flow when a 120 V power supply is connected to a 30<br> resistor ?
AVprozaik [17]
Current= voltage divided by resistance
120/30=4
7 0
2 years ago
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