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kow [346]
3 years ago
13

Energy from the Sun travels to Earth as ______. a. mechanical energy a. mechanical energy b. chemical energy c. radiant energy d

. combustion
Physics
1 answer:
shepuryov [24]3 years ago
3 0

Answer:

Radiant energy

Explanation:

Radiant energy is energy that travels by waves or particles, particularly electromagnetic radiation such as heat or x-rays.

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A brick with a mass of 20 kg and a weight of 196 N is released on a frictionless plane at an angle of 30°. What is the accelerat
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Basically answer A

Explanation:

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Find the radius Rrigel of the star Rigel, the bright blue star in the constellation Orion that radiates energy at a rate of 2.7×
Zinaida [17]

Answer:

r = 1.61 x 10^{11} m

Explanation:

energy radiated (H) = 2.7 x 10^31 W

surface temperature (T)  = 11,000 k

assuming ε = 1 and taking σ = 5.67 x 10^{-8} W/m^{2}.K^{4}

we can find the radius of the star from the equation below

H = A x  ε x σ x T^{4}              

where area (A) = 4 x π x r^{2} (assuming it is a sphere)

therefore  the equation becomes

H = 4 x π x r^{2} x  ε x σ x T^{4}  

2.7 x 10^31  = 4 x π x r^{2} x  1 x 5.67 x 10^{-8}  x (11,000)^{4}

r = \sqrt{\frac{2.7 x 10^31}{4 x π x 1 x 5.67 x 10^{-8}  x (11,000)^{4}} }

r = 1.61 x 10^{11} m

       

4 0
3 years ago
The area of the piston to the master cylinder in a hydraulic braking system of a car is 0.4 square inches. If a force of 6.4 lb
Anit [1.1K]

Answer:

The force applied on one wheel during braking = 6.8 lb

Explanation:

Area of the piston (A) = 0.4 in^{2}

Force applied on the piston(F) = 6.4 lb

Pressure on the piston (P) = \frac{F}{A}

⇒ P = \frac{6.4}{0.4}

⇒ P = 16 \frac{lb}{in^{2} }

This is the pressure inside the cylinder.

Let force applied on the brake pad = F_{1}

Area of the brake pad (A_{1})= 1.7 in^{2}

Thus the pressure on the brake pad (P_{1}) =  \frac{F_{1} }{A_{1} }

When brake is applied on the vehicle the pressure on the piston is equal to pressure on the brake pad.

⇒ P = P_{1}

⇒ 16 = \frac{F_{1} }{A_{1} }

⇒ F_{1} = 16 × A_{1}

Put the value of A_{1} we get

⇒ F_{1} = 16 × 1.7

⇒ F_{1} = 27.2 lb

This the total force applied during braking.

The force applied on one wheel = \frac{F_{1} }{4} = \frac{27.2}{4} = 6.8 lb

⇒ The force applied on one wheel during braking.

7 0
3 years ago
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