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snow_tiger [21]
2 years ago
14

Why is it so hot along the equator?

Physics
2 answers:
elena55 [62]2 years ago
7 0
Its hot because it's closer to the sun and the sun is pretty hot. the rays of the sun project more energy on the equator
iren2701 [21]2 years ago
3 0
Y no me digas nada que te lo digo yo no te quiero ni no su what y el de la selección nacional de la selección de
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Hahahahha ok it’s B or C or it B
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2 years ago
A satellite s is moving in an elliptical orbit around the earth . the mass of satellite is very small compared to mass of earth.
Contact [7]
The best and most correct answer among the choices provided by your question is the second choice or letter B.

<span>A satellite (s) is moving in an elliptical orbit around the earth has its angular momentum towards the earth changing in direction, but not in magnitude.</span>

I hope my answer has come to your help. Thank you for posting your question here in Brainly. We hope to answer more of your questions and inquiries soon. Have a nice day ahead!
8 0
3 years ago
What are the disadvantages and advantages for electromagnetic spectrum​
Basile [38]

Answer:

Advantages: In small quantities they help your body produce vitamin D.

Disadvantages: They can cause sunburn or even skin cancer if used in large quantities.

Explanation:

3 0
3 years ago
12
Svet_ta [14]

Answer:

See below ~

Explanation:

An object will sink in water when its density is greater than that of water, which is 1 g/cm³.

Volume of the box is <u>1331 cm³</u>. (11³)

Maximum mass of sand will be 1331 g. [because 1331/1331 = 1 g/cm³]

  • Volume of sand = Mass of sand / Density of sand
  • Volume (sand) = 1331/3.5
  • Volume (sand) = 380.29 cm³

If the volume of sand is <u>greater than 380.29 cm³</u>, the box will sink in water.

6 0
2 years ago
A projectile is launched at an angle of 36.7 degrees above the horizontal with an initial speed of 175 m/s and lands at the same
Softa [21]

Answer:

a) The maximum height reached by the projectile is 558 m.

b) The projectile was 21.3 s in the air.

Explanation:

The position and velocity of the projectile at any time "t" is given by the following vectors:

r = (x0 + v0 · t · cos α, y0 + v0 · t · sin α + 1/2 · g · t²)

v = (v0 · cos α, v0 · sin α + g · t)

Where:

r = position vector at time "t"

x0 = initial horizontal position

v0 = initial velocity

t = time

α = launching angle

y0 = initial vertical position

g = acceleration due to gravity (-9.80 m/s² considering the upward direction as positive).

v = velocity vector at time t

a) Notice in the figure that at maximum height the velocity vector is horizontal. That means that the y-component of the velocity (vy) at that time is 0. Using this, we can find the time at which the projectile is at maximum height:

vy = v0 · sin α + g · t

0 = 175 m/s · sin 36.7° - 9.80 m/s² · t

-  175 m/s · sin 36.7° /  - 9.80 m/s² = t

t = 10.7 s

Now, we have to find the magnitude of the y-component of the vector position at that time to obtain the maximum height (In the figure, the vector position at t = 10.7 s is r1 and its y-component is r1y).

Notice in the figure that the frame of reference is located at the launching point, so that y0 = 0.

y = y0 + v0 · t · sin α + 1/2 · g · t²

y = 175 m/s · 10.7 s · sin 36.7° - 1/2 · 9.8 m/s² · (10.7 s)²

y = 558 m

The maximum height reached by the projectile is 558 m

b) Since the motion of the projectile is parabolic and the acceleration is the same during all the trajectory, the time of flight will be twice the time it takes the projectile to reach the maximum height. Then, the time of flight of the projectile will be (2 · 10.7 s) 21.4 s. However, let´s calculate it using the equation for the position of the projectile.

We know that at final time the y-component of the vector position (r final in the figure) is 0 (because the vector is horizontal, see figure). Then:

y = y0 + v0 · t · sin α + 1/2 · g · t²

0 = 175 m/s · t · sin 36.7° - 1/2 · 9.8 m/s² · t²

0 = t (175 m/s ·  sin 36.7 - 1/2 · 9.8 m/s² · t)

0 = 175 m/s ·  sin 36.7 - 1/2 · 9.8 m/s² · t

-  175 m/s ·  sin 36.7 / -(1/2 · 9.8 m/s²) = t

t = 21.3 s

The projectile was 21.3 s in the air.

7 0
3 years ago
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