Equations of the vertical launch:
Vf = Vo - gt
y = yo + Vo*t - gt^2 / 2
Here yo = 35.0m
Vo is unknown
y final = 0
t = 4.00 s
and I will approximate g to 10m/s^2
=> 0 = 35.0 + Vo * 4 - 5 * (4.00)^2 => Vo = [-35 + 5*16] / 4 = - 45 / 4 = -11.25 m/s
The negative sign is due to the fact that the initial velocity is upwards and we assumed that the direction downwards was positive when used g = 10m/s^2.
Answer: 11.25 m/s
Answer:
Being an elastic object, rubber ball will be an ideal choice as it will bounce off the bowling pit and will experience a large change in momentum in comparison with the beanbag which will either slow down or come to a halt upon hitting a bowling pit. That is why rubber ball will experience a greater impulse and the bowling pin will experience the negative impulse of the rubber ball.
For Rubber Ball
Upon elastic collision it will reverses the direction and move with velocity equal or less then original
change in momentum = P

For Beanbag
value of impulse will large if velocity is zero.

Explanation:
To solve this problem we will apply the theorem given in the conservation of energy, by which we have that it is conserved and that in terms of potential and kinetic energy, in their initial moment they must be equal to the final potential and kinetic energy. This is,


Replacing the 5100MJ for satellite as initial potential energy, 4200MJ for initial kinetic energy and 5700MJ for final potential energy we have that



Therefore the final kinetic energy is 3600MJ
Answer: 96N
Explanation:
To calculate the velocity of the impact On the persons head, we have
h = gt²/2
14 = 9.81t²/2
t² = 28/9.8
t² = 2.86
t = 1.69s
V = u + at
V = 0 + 9.81*1.69
V = 16.58m/s
a(average) = (v1² + v2²) /2Δy
a(average) = 16.58² + 0)/2 * 0.005
a(average) = 274.8964/0.01
a(average) = 27489.64m/s²
Using newton's second law of motion,
F(average) = m * a(average)
F(average) = 0.0035 * 27489.64
F(average) = 96.21N
Therefore the force needed by the acorn to do much damage starts from 96N
Answer:
194516 sheets
Explanation:
So the area of each sheet of paper is:
A = 0.216 * 0.279 = 0.060264 square meters
For the paper sheet to make the same effect as the atmospheric pressure P, then the gravity F from the paper sheet must be
F = AP = 0.060264 * 101325 = 6106 N
Let g = 9.81 m/s2, then the mass of paper needed to generate that gravity is
m = F/g = 6106 / 9.81 = 622.4 kg
If each sheet has a mass of 0.0032 kg, then the total number of sheets to have that much mass is
622.4 / 0.0032 = 194516 sheets