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storchak [24]
3 years ago
11

In a certain region of space, a uniform electric field is in the x direction. A particle with negative charge is carried from x

5 20.0 cm to x 5 60.0 cm. (i) Does the electric po- tential energy of the charge-field system (a) increase, (b) re- main constant, (c) decrease, or (d) change unpredictably
Physics
1 answer:
Ostrovityanka [42]3 years ago
5 0

Answer:

(a) increase

Explanation:

On a line graph; we have the x-axis to the positive side and the negative side .In a positive x-axis direction, the force is usually positive, vice versa the negative side as well.

The change in the potential energy of a charge field-system can be given as:

\delta U= -q(EdsCos \theta)

where;

q = positive test charge

E = Electric field

ds = displacement between thee charge positions

θ  = Angle between the electric field and the displacement.

Given that:

Charge of the particle = -q

displacement = (60.0 -20.0)cm = 40.0 cm

θ = 0

Replacing our values in the above equation, we have:

\delta U = -(-q)(60Cos 0)

\delta U = qEds

Since the potential energy of the system is positive, therefor the electric potential energy also increases.

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What is the momentum of a 200 kg truck travelling at 20 m/s?
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A photovoltaic panel of dimension 2 m × 4 m is installed on the roof of a home. The panel is irradiated with a solar flux of GS
Flura [38]

Answer:

(a) the electrical power generated for still summer day is 1013.032 W

(b)the electrical power generated for a breezy winter day is 1270.763 W

Explanation:

Given;

Area of panel = 2 m × 4 m, = 8m²

solar flux  GS = 700 W/m²

absorptivity of the panel, αS = 0.83

efficiency of conversion, η = P/αSGSA = 0.553 − 0.001 K⁻¹ Tp

panel emissivity , ε = 0.90

Apply energy balance equation to determine he electrical power generated;  

transferred energy + generated energy = 0

(radiation + convection) +  generated energy = 0

[\alpha_sG_s-\epsilon \alpha(T_p^4-T_s^4)]-h(T_p-T_\infty) - \eta \alpha_s G_s = 0

[\alpha_sG_s-\epsilon \alpha(T_p^4-T_s^4)]-h(T_p-T_\infty) - (0.553-0.001T_p)\alpha_s G_s

(a) the electrical power generated for still summer day

T_s = T_{\infty} = 35 ^oC = 308 \ k

[0.83*700-0.9*5.67*10^{-8}(T_p_1^4-308^4)]-10(T_p_1-308) - (0.553-0.001T_p_1)0.83*700 = 0\\\\3798.94-5.103*10^{-8}T_p_1^4 - 9.419T_p_1 = 0\\\\Apply \  \ iteration \ method \ to \ solve \ for \ T_p_1\\\\T_p_1 = 335.05 \ k

P = \eta \alpha_s G_s A = (0.553-0.001 T_p_1)\alpha_s G_s A \\\\P = (0.553-0.001 *335.05)0.83*700*8 \\\\P = 1013.032 \ W

(b)the electrical power generated for a breezy winter day

T_s = T_{\infty} = -15 ^oC = 258 \ k

[0.83*700-0.9*5.67*10^{-8}(T_p_2^4-258^4)]-10(T_p_2-258) - (0.553-0.001T_p_2)0.83*700 = 0\\\\8225.81-5.103*10^{-8}T_p_2^4 - 29.419T_p_2 = 0\\\\Apply \  \ iteration \ method \ to \ solve \ for \ T_p_2\\\\T_p_2 = 279.6 \ k

P = \eta \alpha_s G_s A = (0.553-0.001 T_p_2)\alpha_s G_s A \\\\P = (0.553-0.001 *279.6)0.83*700*8 \\\\P = 1270.763 \ W

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