The acceleration due to gravity serves as the centripetal acceleration of the objects that orbits the Earth. The centripetal acceleration due to gravity is calculated through the equation,
a = v²/r
where v is the speed and r is the radius. Substituting the known values to the equation,
9.8 m/s² = (420 m/s)² / r
The value of r from the equation is 18000 m or equal to 18 km.
<em>Answer: 18 km</em>
Answer:
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That will be
<em>=</em><em> </em><em>1</em><em>5</em><em>0</em><em>0</em><em> </em><em>x</em><em> </em><em>1</em><em>5</em><em> </em><em>x</em><em> </em><em>4</em><em>5</em><em>0</em><em>0</em>
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Answer:
3.6 seconds
Explanation:
Given:
y₀ = y = 0 m
v₀ = 31 sin 35° m/s
a = -9.8 m/s²
Find: t
y = y₀ + v₀ t + ½ at²
0 = 0 + (31 sin 35°) t + ½ (-9.8 m/s²) t²
0 = 17.78t − 4.9t²
0 = t (17.78 − 4.9t)
t = 0 or 3.63
Rounded to the nearest tenth, the ball lands after 3.6 seconds.
Answer:
1.125m/s^2
Explanation:
Since acceleration is defined as the rate of change in velocity with respect to time. Mathematically
v^2= u^2+2as
Where a,v,u and s are the acceleration, final velocity, initial velocity and distance respectively.
a = ?
u = 0m/s
v = 15m/s
s = 100m
Substituting the values into the formula above
v^2= u^2+2as
15^2=0^2+2×a×100
225= 0+200a
225= 200a
Divide both sides by 200
225/200 = 200a/200
a= 1.125m/s^2
Hence the acceleration of the car is 1.125m/s^2.
Note that the car accelerated uniformly from rest, that was why the initial velocity was 0m/s
