In a quarter of an hour it will travel 1/4 x 90 = 22.5Km
Answer:
The heat energy required, Q = 6193.8 J
Explanation:
Given,
The mass of ice cube, m = 18.6 g
The heat of fusion of ice, ΔHₓ = 333 J/g
The heat energy of a substance is equal to the product of the mass and heat of fusion of that substance. It is given by the equation,
<em> Q = m · ΔHₓ joules</em>
Substituting the given values in the above equation
Q = 18.6 g x 333 J/g
= 6193.8 J
Hence, the heat required to melt the ice cube is, Q = 6193.8 J
Answer:
h> 2R
Explanation:
For this exercise let's use the conservation of energy relations
starting point. Before releasing the ball
Em₀ = U = m g h
Final point. In the highest part of the loop
Em_f = K + U = ½ m v² + ½ I w² + m g (2R)
where R is the radius of the curl, we are considering the ball as a point body.
I = m R²
v = w R
we substitute
Em_f = ½ m v² + ½ m R² (v/R) ² + 2 m g R
em_f = m v² + 2 m g R
Energy is conserved
Emo = Em_f
mgh = m v² + 2m g R
h = v² / g + 2R
The lowest velocity that the ball can have at the top of the loop is v> 0
h> 2R
Answer:
2.47 m/s
Explanation:
Momentum = Mass X Velocity
If they were locked together, it means its a perfectly inelastic collision. Therefore,
Total momentum before = Total momentum after
Total momentum before = (20 X 20) - (18 X 17)
= 94
Total momentum after = 94
Y = Object speed after collision
94 = (20+18)Y
Y = 2.47368421 m/s
Answer:
At low pressure- 
At high pressure- 
Explanation:
Initial speed, 
Final speed, 
Net horizontal force due to rolling friction 
mg where m is mass, g is acceleration due to gravity, 
is coefficient of rolling friction
From kinematic relation, 
For each tire,
Making the subject
Under low pressure of 40 Psi, d=18 m
Therefore,
At a pressure of 105 Psi, d=93.7
Therefore,
