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jeka94
4 years ago
9

What factors affect the amount of solar energy that reaches earth's surface

Physics
2 answers:
zalisa [80]4 years ago
8 0
Weather
Particles in air such as smoke smog
bazaltina [42]4 years ago
7 0
Latitude~ due to earth spherical shape the solar rays have more intensity around the equatorial regions

 cloud cover: clouds have a big impact on the amount of solar radition reaching earth surface
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Sasha lifts a couch 8.2 meters from the ground floor of her house to the attic. If the couch has a mass of 120 kg, what is the g
gavmur [86]

As we know that gravitational potential energy is given by

U = mgH

here we have

m = mass = 120 kg

g = 9.81 m/s^2

h = height = 8.2 m

now from above formula

U = 120kg (9.81 m/s^2) (8.2 m)

U = 9653.04 J

so above is the gravitational potential energy of the couch

4 0
3 years ago
Read 2 more answers
There is the more role of the moon than the sun to occur tides in the oceans,why?
Nana76 [90]

Answer:

The ocean tides on earth are caused by both the moon's gravity and the sun's gravity. ... Even though the sun is much more massive and therefore has stronger overall gravity than the moon, the moon is closer to the earth so that its gravitational gradient is stronger than that of the sun.

8 0
2 years ago
Suppose a log's mass is 5 kg. After burning, the mass of the ash is 1 kg. explain what May have happened to the other 4 kg.
ladessa [460]

The other 4 kg of mass may have departed the scene
of the fire, in the form of gases and smoke particles.

7 0
3 years ago
Answer pls will give 20 points per person
LuckyWell [14K]

Answer: question 1 is b I believe

Explanation:

every action has an opposite reaction

7 0
3 years ago
Read 2 more answers
A steel ball of mass 0.500 kg is fastened to a cord that is 70.0 cm long and fixed at the far end. The ball is then released whe
Liula [17]

Answer:

a) v₁fin = 3.7059 m/s   (→)

b) v₂fin = 1.0588 m/s     (→)

Explanation:

a) Given

m₁ = 0.5 Kg

L = 70 cm = 0.7 m

v₁in = 0 m/s   ⇒  Kin = 0 J

v₁fin = ?

h<em>in </em>= L = 0.7 m

h<em>fin </em>= 0 m   ⇒    U<em>fin</em> = 0 J

The speed of the ball before the collision can be obtained as follows

Einitial = Efinal

⇒ Kin + Uin = Kfin + Ufin

⇒ 0 + m*g*h<em>in</em> = 0.5*m*v₁fin² + 0

⇒ v₁fin = √(2*g*h<em>in</em>) = √(2*(9.81 m/s²)*(0.70 m))

⇒ v₁fin = 3.7059 m/s   (→)

b)  Given

m₁ = 0.5 Kg

m₂ = 3.0 Kg

v₁ = 3.7059 m/s    (→)

v₂ = 0 m/s

v₂fin = ?

The speed of the block just after the collision can be obtained using the equation

v₂fin = 2*m₁*v₁ / (m₁ + m₂)

⇒  v₂fin = (2*0.5 Kg*3.7059 m/s) / (0.5 Kg + 3.0 Kg)

⇒  v₂fin = 1.0588 m/s     (→)

7 0
4 years ago
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