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Tom [10]
3 years ago
10

Can a plane mirror produce a real image? Explain.

Physics
1 answer:
Komok [63]3 years ago
4 0
I don't think so, because in order to produce an image, you need a surface behind the mirror. The light will hit the mirror, then it will bounce it back in your eyes and you see the image.
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A car with a mass of 3 Kg and velocity of 40 m/s collided with a truck of a velocity of 60 m/s, if the momentum is conserved wha
Luden [163]

Answer:

the mass of the truck is 2 kg.

Explanation:

Given;

mass of the car, m₁ = 3 kg

initial velocity of the car, u₁ = 40 m/s

initial velocity of the truck, u₂ = 60 m/s

let the mass of the truck = m₂

Apply the principle of conservation of linear momemtum;

m₁u₁ = m₂u₂

m₂ = (m₁u₁) / u₂

m₂ = (3 x 40) / (60)

m₂ = 2 kg

Therefore, the mass of the truck is 2 kg.

4 0
3 years ago
An airplane accelerates down a runway at 3 m/s2 for 32 s until is finally lifts off the ground. Determine the distance traveled
Alex Ar [27]

Answer:

12 or 24

Explanation:

i think it is i hope it is right

4 0
3 years ago
Which statement describes steps involved in the production of hydroelectric power?
viktelen [127]

Answer:

Answer is D.......Falling water turns a turbine that helps generate electricity.

Explanation:

Hydropower plants capture the energy of falling water to generate electricity. A turbine converts the kinetic energy of falling water into mechanical energy. Then a generator converts the mechanical energy from the turbine into electrical energy.

4 0
3 years ago
Read 2 more answers
An arrow movirg 48.3 m/s has 5.22<br> kg•m/s of momentum. What is its<br> mass?
spin [16.1K]

Answer:

0.11 kg

Explanation:

Ft = MV

Ft = momentum 5.22kg m/s

M = mass

V = velocity 48.3m/s

Therefore

5.22 = M x 48.3

Divide both sides by 48.3

5.22/48.3 = M x 48.3/48.3

0.11 = M

M = 0.11kg

6 0
3 years ago
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

3 0
3 years ago
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