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il63 [147K]
3 years ago
12

Which of these things poses the greatest hazard to communications satellites? a) photons from the Sun b) solar magnetic fields c

) particles from the Sun.
Physics
2 answers:
Sphinxa [80]3 years ago
7 0

Answer:

Your answer is B because the solar magnetic fields mess with the waves of energy and damages the connection with the satellites and what's their sending information to...

Explanation:

leva [86]3 years ago
3 0

Answer:

a

Explanation:

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Look at the graph
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Here, Carefully look at the graph.
When it is on x=10, it is approximately 10, (slightly less than 10)
Closest value would be 90, so y/x = 90/10 = 9 

So, the density of the graph would be 9 g/cm³

In short, Your Answer would be Option D

Hope this helps!
4 0
3 years ago
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Light moves at a speed of around 1 million miles per hour<br>O<br>True<br>False​
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Answer:

False

Explanation:

In miles per hour, light speed is about 670,616,629 mph

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3 years ago
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A baseball is batted from a height of 1.09 m with a speed of
kobusy [5.1K]

(a) The horizontal and vertical components of the ball’s initial velocity is 37.8 m/s and 12.14 m/s respectively.

(b) The maximum height above the ground reached by the ball is 8.6 m.

(c) The distance off course the ball would be carried is 0.38 m.

(d) The ball's velocity after 2.0 seconds if there is no crosswind is 38.53 m/s.

<h3>Horizontal and vertical components of the ball's velocity</h3>

Vx = Vcosθ

Vx = 39.7 x cos(17.8)

Vx = 37.8 m/s

Vy = Vsin(θ)

Vy = 39.7 x sin(17.8)

Vy = 12.14 m/s

<h3>Maximum height reached by the ball</h3>

H = \frac{v^2 sin^2(\theta)}{2g} \\\\H = \frac{(39.7)^2 \times (sin17.8)^2}{2(9.8)} \\\\H = 7.51 \ m

Maximum height above ground = 7.51 + 1.09 = 8.6 m

<h3>Distance off course after 2 second </h3>

Upward speed of the ball after 2 seconds, V = V₀y - gt

Vy = 12.14 - (2x 9.8)

Vy = - 7.46 m/s

Horizontal velocity will be constant = 37.8 m/s

Resultant speed of the ball after 2 seconds = √(Vy² + Vx²)

V = \sqrt{(-7.46)^2 + (37.8)^2} \\\\V = 38.53 \ m/s

<h3>Resultant speed of the ball and crosswind</h3>

V = \sqrt{38.52^2 + 4^2} \\\\V = 38.72 \ m/s

<h3>Distance off course the ball would be carried</h3>

d = Δvt = (38.72 - 38.53) x 2

d = 0.38 m

The ball's velocity after 2.0 seconds if there is no crosswind is 38.53 m/s.

Learn more about projectiles here: brainly.com/question/11049671

5 0
2 years ago
If the velocity of an object is doubled, its kinetic energy is ______
swat32
Increased by a factor of 4
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3 years ago
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