Answer:
720 J
Explanation:
The gravitational potential energy that Essam loses for every metre is given by:
where
m=72 kg is Essam's mass
is the gravitational field strength
is the difference in height
By substituting the numbers into the formula, we find
Answer:
<h2><em>V(water)= 237 mL=237×10^-6 m^3</em></h2><h2><em>ρ(water)=1000 kg/m^3</em></h2><h2><em>
m=</em><em>ρ×V=(1000)×(237×10^-6)</em></h2><h2><em>
m= 237×10^-3 = 0.237 kg</em></h2><h2><em>
m= 237 gram.</em></h2>
consider the right direction as positive and left direction as negative.
M = mass of the ball = 5 kg
m = mass of stone = 1.50 kg
= initial velocity of the ball before collision = 0 m/s
= initial velocity of the stone before collision = 12 m/s
= final velocity of the ball after collision = ?
= final velocity of the stone after collision = - 8.50 m/s
using conservation of momentum
M + m = M + m
(5) (0) + (1.5) (12) = 5 + (1.50) (- 8.50)
= 6.15 m/s
h = height gained by the ball
using conservation of energy
Potential energy gained by ball at Top = kinetic energy at the bottom
Mgh = (0.5) M
(9.8) h = (0.5) (6.15)²
h = 1.93 m
Answer:
x = 1,185 m
, t = 4/3 s
, F = - 4 N
Explanation:
For this exercise we use Newton's second law
F = m a = m dv /dt
β - α t = m dv / dt
dv = (β – α t) dt
We integrate
v = β t - ½ α t²
We evaluate between the lower limits v = v₀ for t = 0 and the upper limit v = v for t = t
v-v₀ = β t - ½ α t²
the farthest point of the body is when v = v₀ = 0
0 = β t - ½ α t²
t = 2 β / α
t = 2 4/6
t = 4/3 s
Let's find the distance at this time
v = dx / dt
dx / dt = v₀ + β t - ½ α t2
dx = (v₀ + β t - ½ α t2) dt
We integrate
x = v₀ t + ½ β t - ½ 1/3 α t³
x = v₀ 4/3 + ½ 4 (4/3)² - 1/6 6 (4/3)³
The body comes out of rest
x = 3.5556 - 2.37
x = 1,185 m
The value of force is
F = β - α t
F = 4 - 6 4/3
F = - 4 N
Models show how the atoms in a compound are connected.
