One of the scientists suggests that he can build a
cooling system for the theoretical photovoltaic cells
Experiment 2, which will keep the cells 1°C cooler
dan normal but decrease their efficiency by 1%. The
teoretical photovoltaic cells capturing which frequency
ranges
, if any, would benefit from this cooling system?

Answer:

Hope It Help

Explanation:

That's all I know

Answer:

the correct answers are a and c

Explanation:

In an electrical circuit there are two important quantities to measure, such as voltage and current.

Voltage is the potential difference between two points in a circuit

current is the number of electrons you pass through a given point per unit of time.

Now let's analyze each answer

a) true. The potential difference across an element

b) False. The potential difference is u field there is no physical entity that moves

c) True. The current is electrons in motion and these pass through the given element

d) False. There is a physical quantity that passes through the point

the correct answers are a and c

Answer:

the kinetic energy of the combined 1.5 kg block when it reaches the bottom of the plane is 10.62 J

Explanation:

Given that the data in the question;

angle of inclination with respect to the ground [tex]\theta[/tex] = 30°

length of plane d = 2m

m₁ = 1 kg

m₂ = 0.5 kg

now, velocity of the first block at midpoint;

[tex]\frac{1}{2}[/tex]mv² = mgsin[tex]\theta[/tex][tex]\frac{d}{2}[/tex]

[tex]\frac{1}{2}[/tex]v² = gsin[tex]\theta[/tex][tex]\frac{d}{2}[/tex]

v² = gsin[tex]\theta[/tex]d

v = √( gsin[tex]\theta[/tex]d)

g is 9.8 m/s

so we substitute

v = √( 9.8 × sin30° × 2)

v = √( 19.6 )

v =  3.13 m/s

Now, velocity just after collision of the blocks will be;

(m₁ + m₂)v₂ = m₁v

v₂ = m₁v / (m₁ + m₂)

we substitute

v₂ = (1 × 3.13) / (1 + 0.5)

v₂ = 3.13 / 1.5

v₂ = 2.0866 m/s

now, final kinetic energy will be;

[tex]KE_f[/tex] = (m₁ + m₂)gsin[tex]\theta[/tex][tex]\frac{d}{2}[/tex] + Initial Kinetic energy

[tex]KE_f[/tex] = (m₁ + m₂)gsin[tex]\theta[/tex][tex]\frac{d}{2}[/tex] + [tex]\frac{1}{2}[/tex]mv₂²

we substitute

[tex]KE_f[/tex] = [(1 + 0.5)9.8 × sin30 × [tex]\frac{2}{2}[/tex]] + [[tex]\frac{1}{2}[/tex] × 1.5 × 2.0866 ]

[tex]KE_f[/tex] = 7.35 + 3.2654

[tex]KE_f[/tex] = 10.62 J

Therefore, the kinetic energy of the combined 1.5 kg block when it reaches the bottom of the plane is 10.62 J