Answer:
the horizontal distance covered by the cannonball before it hits the ground is 327.5 m
Explanation:
Given;
height of the cliff, h = 210 m
initial horizontal velocity of the cannonball, Ux = 50 m/s
initial vertical velocity of the cannonball, Uy = 0
The time for the cannonball to reach the ground is calculated as;
The horizontal distance covered by the cannonball before it hits the ground is calculated as;
Therefore, the horizontal distance covered by the cannonball before it hits the ground is 327.5 m
Answer:
there is friction between the two things
Explanation:
Answer:
a) 4.485 kg b) 3.94 kg
Explanation:
since the maximum tension the line can stand is 44 N and for question a the speed is constant (acceleration must be zero since the velocity or speed is not changing), F(tension) = mass * acceleration due to gravity (g) .
44 = m * 9.81m/s^2
m = 44/9.81 = 4.485kg
b) F(tension) = ma + mg ( where a is the acceleration of the body and g is the acceleration of the gravity)
44 = m (a +g)
44 = m (1.37 + 9.81)
44/11.18 = m
m = 3.94 kg
Answer:
The kinetic energy K of the moving charge is K = 2kQ²/3d = 2Q²/(4πε)3d = Q²/6πεd
Explanation:
The potential energy due to two charges q₁ and q₂ at a distance d from each other is given by U = kq₁q₂/r.
Now, for the two charges q₁ = q₂ = Q separated by a distance d, the initial potential energy is U₁ = kQ²/d. The initial kinetic energy of the system K₁ = 0 since there is no motion of the charges initially. When the moving charge is at a distance of r = 3d, the potential energy of the system is U₂ = kQ²/3d and the kinetic energy is K₂.
From the law of conservation of energy, U₁ + K₁ = U₂ + K₂
So, kQ²/d + 0 = kQ²/3d + K
K₂ = kQ²/d - kQ²/3d = 2kQ²/3d
So, the kinetic energy K₂ of the moving charge is K₂ = 2kQ²/3d = 2Q²/(4πε)3d = Q²/6πεd
