vf^2 = 2ad
vf^2 = 2(9.81)(44m)
vf^2 = 863.28
vf = √863.28
vf = 29.4 - using equations of motion
ME = PE + KE
ME = mgh + 1/2mv^2
ME = (1)(9.81)(44) + 1/2(1)(3^2)
ME = 431.64 + 4.5
ME = 436.14 - using conservation of energy
hope this helps :)
Answer: i think that the third option is correct
the force of the ball goes to the floor making it go down and up reflecting a little lower force back to the ball and sending the ball back up
hope this helps
brainliest if possible
Answer:
Explanation:
THE GIVEN sheet can be taken as two horizontal force with surface charge density is
at one surface is ∈_1 =
at oher surface is ∈_2=
the magnitude of electric field due to surface charge is given as
So, electric field at P (2 CM below from surface is) = E_1 +E_2
Answer:
a) 6.26 m/s
b) 7.67 m/s
Explanation:
The potential energy at height h0 is initially ...
PE0 = mgh0
At height h1, the potential energy is ...
PE1 = mgh1
The difference in potential energy is converted to kinetic energy:
PE0 -PE1 = KE1 = (1/2)m(v1)^2
Solving for v1, we have ...
mg(h0 -h1) = (1/2)m(v1)^2
2g(h0 -h1) = (v1)^2
v1 = √(2g(h0 -h1))
__
a) When the body is 1 m high, its speed is ...
v = √(2(9.8)(3 -1)) ≈ 6.26 m/s . . . at 1 m high
__
b) When the body is 0 m high, its speed is ...
v = √(2(9.8)(3 -0)) ≈ 7.67 m/s . . . when it reaches the ground
Because the gravitational field strength on the moon (1.6N/kg)is smaller than the gravitational field strength on the earth (9.8N/kg). Weight= mass×gravitational field strength. Your mass is the same but as g is smaller, your weight decreases. weight is the gravitational force, the downwards force towards the centre of the planet, and as it will be smaller on the moon, the force of attraction between the moon and you is smalller so when you jumb you will reach a higher point above the ground than when you jumb on the earth. Basically the moon 'pulls' you less than the earth.