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yawa3891 [41]
2 years ago
9

Why type of energy is present after a solar panel changes forms?

Physics
2 answers:
NNADVOKAT [17]2 years ago
8 0

Explanation:

Solar technologies convert sunlight into electrical energy either through photovoltaic (PV) panels or through mirrors that concentrate solar radiation. This energy can be used to generate electricity or be stored in batteries or thermal storage.

Svetllana [295]2 years ago
5 0
Light and heat should be the type of energy


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The transfer of energy when particles of a fluid move from one place to another is called
siniylev [52]

Answer:

Answer below!!!!

Explanation:

Convection is the transfer of thermal energy by particles moving through a fluid.

Hope I Helped!!!

;)

5 0
2 years ago
Human reaction times are worsened by alcohol. How much further (in feet) would a drunk driver's car travel before he hits the br
sweet-ann [11.9K]

Answer:

A drunk driver's car travel 49.13 ft further than a sober driver's car, before it hits the brakes

Explanation:

Distance covered by the car after application of brakes, until it stops can be found by using 3rd equation of motion:

2as = Vf² - Vi²

s = (Vf² - Vi²)/2a

where,  

Vf = Final Velocity of Car = 0 mi/h

Vi = Initial Velocity of Car = 50 mi/h

a = deceleration of car  

s = distance covered

Vf, Vi and a for both drivers is same as per the question. Therefore, distance covered by both car after application of brakes will also be same.

So, the difference in distance covered occurs before application of brakes during response time. Since, the car is in uniform speed before applying brakes. Therefore, following equation shall be used:

s = vt

FOR SOBER DRIVER:

v = (50 mi/h)(1 h/ 3600 s)(5280 ft/mi) = 73.33 ft/s

t = 0.33 s

s = s₁

Therefore,

s₁ = (73.33 ft/s)(0.33 s)

s₁ = 24.2 ft

FOR DRUNK DRIVER:

v = (50 mi/h)(1 h/ 3600 s)(5280 ft/mi) = 73.33 ft/s

t = 1 s

s = s₂

Therefore,

s₂ = (73.33 ft/s)(1 s)

s₂ = 73.33 ft

Now, the distance traveled by drunk driver's car further than sober driver's car is given by:

ΔS = s₂ - s₁

ΔS = 73.33 ft - 24.2 ft

<u>ΔS = 49.13 ft</u>

6 0
3 years ago
What is the current in a 160V circuit if the resistance is 20Ω
Mkey [24]

Answer:

I =V/R

I= 160/20 = 8 Amperes

3 0
2 years ago
The current in a single-loop circuit with one resistance R is 6.3 A. When an additional resistance of 3.4 Ω is inserted in serie
Dmitry [639]

Answer:

10.15Ω

Explanation:

From ohm's law,

V = IR...................... Equation 1

Where V = Voltage, I = current, R = resistance.

Assume the voltage across the resistance = V,

Given: I = 6.3 A

Substitute into equation 1

V = 6.3R.................. Equation 2

When an additional resistance of 3.4 Ω is inserted in series with R,

The voltage remain the same, but the current changes

Total Resistance(Rt) = (R+3.4)Ω, I' = 4.72 A

Also from ohm' law,

V = I'Rt............... Equation 3

Substitute the value of I'  and Rt into equation 3

V = 4.72(R+3.4)............... Equation 5.

Divide equation 2 by equation 5

V/V = 6.3R/4.72(R+3.4)

1 = 1.335R/(R+3.4)

1 = 1.335R/(R+3.4)

R+3.4 = 1.335R

3.4 = 1.335R-R

3.4 = 0.335R

R = 3.4/0.335

R = 10.15Ω

6 0
2 years ago
Read 2 more answers
A satellite of mass m is in a circular orbit of radius R2 around a spherical planet of radius R1 made of a material with density
Amiraneli [1.4K]

Answer:

a)      K = 2/3 π G m ρ R₁³ / R₂ ,  b) U = - G m M / r

Explanation:

The law of universal gravitation is

     F = G m M / r²

Part A

Let's use Newton's second law

     F = m a

The acceleration is centripetal

     a = v² / R₂

     

      G m M / R₂² = m v² / R₂

      v² = G M / R₂

They give us the density of the planet

    ρ = M / V

    V = 4/3 π R₁³

    M =   ρ V

    M =   ρ 4/3 π R₁³

    v² = 4/3 π G  ρ R₁³ / R₂

    K = ½ m v²

    K = ½ m (4/3 π G ρ R₁³ / R₂)

    K = 2/3 π G m ρ R₁³ / R₂

Part B

Potential energy and strength are related

     F = - dU / dr

     ∫ dU = - ∫ F. dr

The force was directed towards the center and the vector r outwards therefore there is an angle of 180º between the two cos 180 = -1

    U- U₀ = G m M ∫ dr / r²

    U - U₀ = G m M (- r⁻¹)

We evaluate for

    U - U₀ = -G m M (1 / r_{f} -  1 /r_{i})

They indicate that for ri = ∞     U₀ = 0

    U = - G m M / r

6 0
3 years ago
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