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yan [13]
2 years ago
12

Plate tectonics is unique to which terrestrial planet?

Physics
2 answers:
const2013 [10]2 years ago
7 0
B. EARTH !!!!!!!!!!!!!!!!
Furkat [3]2 years ago
6 0
Answer:  [B]:  Earth .
_____________________________________________________
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PLEASE !! Help
n200080 [17]

Answer:

scalar,magnitude

Explanation:

scalar is an example of length/distance

magnitude is the length of a vector

8 0
2 years ago
An english scientist created the first periodic table of elements
olga nikolaevna [1]
Like a then it would be A something I guess it goes like that
8 0
3 years ago
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What are some examples of gravitational forces that keep an object in orbit? Pls give at least 2-3 example if you can.​
Lelu [443]

Answer:

Without any external forces a moving object will continue to move in a straight line. The gravitational force between the two objects will provide the centripetal force to keep the objects moving around one another.

1. satellite in orbit around the earth             (motion of earth is negligible)

2. moon in orbit around the earth     (center of motion several thousand miles        

                                                             from center of earth)

3. earth in orbit around sun    (center of rotation close to center of sun)

4. binary stars         (if masses of stars are equal center of rotation is in middle)

3 0
3 years ago
5. The condition of the road surface affects total stopping distance.<br>A. true<br>B. false​
Tom [10]

Answer:

A true

Explanation:

Can you please give me brainiest.

7 0
3 years ago
A projectile is fired at time t= 0.0 s from point 0 at the edge of a cliff, with initial velocity components of Vox = 30 m/s and
BARSIC [14]

Answer:

At t = 15.0 s the magnitude of the velocity is 58.31 m/s

Explanation:

The given parameters are;

V₀ₓ = 30 m/s

V_{0y} = 100 m/s

The time of flight of the projectile = 25 s

For projectile motion;

Vₓ = V₀ × cos(θ₀)

The magnitude of the velocity V = √(V₀ₓ² + V_{0y}²)

We have the magnitude of the initial velocity = √(30² + 100²) = 10·√109 m/s

cos(θ₀) = V₀ₓ/V₀ = 30/(10·√109) = 3/√109

θ₀ = cos⁻¹(3/√109) = 73.3°

The components of the velocity after time t is given by the relations;

Vₓ = V₀ × cos(θ₀) = 30 m/s

V_y =  V₀ × sin(θ₀) - g×t

When V_y = 0, we have;

0 =  V₀ × sin(θ₀) - g×t

g×t  =  V₀ × sin(θ₀)  = 10·√109×0.958 = 100 m/s

t = 100/g = 100/10 = 10 s

The time to reach maximum height = 10 s

At 15.0 seconds, we have;

V_y =  V₀ × sin(θ₀) - g×t = 10·√109×0.958  - 10×15 = -50 m/s

Therefore, the projectile is returning at 50 m/s

The magnitude of the velocity =√(30² + 50²) = 10·√34 m/s = 58.31 m/s.

5 0
3 years ago
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