1. 0.16 N
The weight of a man on the surface of asteroid is equal to the gravitational force exerted on the man:
where
G is the gravitational constant
is the mass of the asteroid
m = 100 kg is the mass of the man
r = 2.0 km = 2000 m is the distance of the man from the centre of the asteroid
Substituting, we find
2. 1.7 m/s
In order to stay in orbit just above the surface of the asteroid (so, at a distance r=2000 m from its centre), the gravitational force must be equal to the centripetal force
where v is the minimum speed required to stay in orbit.
Re-arranging the equation and solving for v, we find:
Answer:
option D
Explanation:
Sunspots are the spot that appears on the sun, this spot appears darker than the surrounding surface of the sun.
Sun magnetic field goes through a cycle and this cycle is called the Sunspot cycle. Every 11 years the magnetic field of the sun completely flips. This sunspot cycle affects activity on the surface of the sun.
Sunspot cycle is the pattern of solar activity where an average number of sunspot gradually increase and decrease.
Hence, the correct answer is option D
Answer:
d = 10.076 m
Explanation:
We need to obtain the velocity of the ball in the y direction
Vy = 24.5m/s * sin(35) = 14.053 m/s
To obtain the distance, we use the formula
vf^2 = v0^2 -2*g*d
but vf = 0
d = -vo^2/2g
d = (14.053)^2/2*(9.8) = 10.076 m
Answer:
Magnitude of the force between the charges is F = 1.92×10^20N
Explanation:
Given the magnitude of force according to coulombs law
F =K[(q1*q2)/r2]
Where q1 and q2 are the charges
r is the distance between the charges
K is the coulombs constant
Substituting the given values, we have;
F = 8.98×10^9 × 1.5×10^6 × 3.2×10^4/1.5²
F = 43.1×10^19/2.25
F = 19.16×10^19N
F = 1.92×10^20N
