Answer:
5.44×10⁶ m
Explanation:
For a satellite with period t and orbital radius r, the velocity is:
v = 2πr/t
So the centripetal acceleration is:
a = v² / r
a = (2πr/t)² / r
a = (2π/t)² r
This is equal to the acceleration due to gravity at that elevation:
g = MG / r²
(2π/t)² r = MG / r²
M = (2π/t)² r³ / G
At the surface of the planet, the acceleration due to gravity is:
g = MG / R²
Substituting our expression for the mass of the planet M:
g = [(2π/t)² r³ / G] G / R²
g = (2π/t)² r³ / R²
R² = (2π/t)² r³ / g
R = (2π/t) √(r³ / g)
Given that t = 1.30 h = 4680 s, r = 7.90×10⁶ m, and g = 30.0 m/s²:
R = (2π / 4680 s) √((7.90×10⁶ m)³ / 30.0 m/s²)
R = 5.44×10⁶ m
Notice we didn't need to know the mass of the satellite.
Answer:
The shortest distance in which you can stop the automobile by locking the brakes is 53.64 m
Explanation:
Given;
coefficient of kinetic friction, μ = 0.84
speed of the automobile, u = 29.0 m/s
To determine the the shortest distance in which you can stop an automobile by locking the brakes, we apply the following equation;
v² = u² + 2ax
where;
v is the final velocity
u is the initial velocity
a is the acceleration
x is the shortest distance
First we determine a;
From Newton's second law of motion
∑F = ma
F is the kinetic friction that opposes the motion of the car
-Fk = ma
but, -Fk = -μN
-μN = ma
-μmg = ma
-μg = a
- 0.8 x 9.8 = a
-7.84 m/s² = a
Now, substitute in the value of a in the equation above
v² = u² + 2ax
when the automobile stops, the final velocity, v = 0
0 = 29² + 2(-7.84)x
0 = 841 - 15.68x
15.68x = 841
x = 841 / 15.68
x = 53.64 m
Thus, the shortest distance in which you can stop the automobile by locking the brakes is 53.64 m
I think it is D because gas particles move the fastest but if not then it is A
Answer:
bounced back at oncoming waves
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If the incoming wave strikes a barrier at a 90 degree angle, the wave energy is, in fact bouncing back at oncoming waves.
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Hope this helps!