Answer:
F = 85696.5 N = 85.69 KN
Explanation:
In this scenario, we apply Newton's Second Law:
where,
F = Upthrust = ?
m = mass of space craft = 5000 kg
g = acceleration due to gravity on surface of Kepler-10b = (1.53)(9.81 m/s²)
g = 15.0093 m/s²
a = acceleration required = 2.13 m/s²
Therefore,
<u>F = 85696.5 N = 85.69 KN</u>
Answer:
Option 3: -48 cm
Explanation:
We are given:
refractive index; n = 1.5
radius of curvature; r2 = 24 cm
Formula for the focal length is given as;
1/f = (n - 1) × [(1/r1) - (1/r2)]
As r1 tends to infinity, 1/r1 = 0
Thus,we now have;
1/f = (n - 1) × (-1/r2)
Plugging in the relevant values;
1/f = (1.5 - 1) × (-1/24)
1/f = -0.02083333333
f = -1/0.02083333333
f = -48 cm
Electric field = potential difference
-----------------------------
distance between plates
Distance between plates = 45
----------
500
= 0.09 meters.
Let's call the constant acceleration a.
At a time t, its speed will thus be v(t)=a*t+v0 where v0 is its initial speed, here 10 m/s. Hence v(t)=a*t+10.
From there we can deduce the position P(t)=a*t^2/2+10t+p0 where p0 is the initial position, here 0.
Hence P(t)=a*t^2/2+10t
Let's call T the time at which it's at 50 m/s, we know that P(T)=225m and that v(T)=50 m/s hence a*T+10=50 thus a=40/T and P(T)=(40/2+10)T=30T
Hence T=225/30=7.5
It took 7.5 seconds
