Answer:
Heat energy required (Q) = 3,000 J
Explanation:
Find:
Mass of water (M) = 200 g
Change in temperature (ΔT) = 15°C
Specific heat of water (C) = 1 cal/g°C
Find:
Heat energy required (Q) = ?
Computation:
Q = M × ΔT × C
Heat energy required (Q) = Mass of water (M) × Change in temperature (ΔT) × Specific heat of water (C)
Heat energy required (Q) = 200 g × 15°C × 1 cal/g°C
Heat energy required (Q) = 3,000 J
According to the conservation of mechanical energy, the kinetic energy just before the ball strikes the ground is equal to the potential energy just before it fell.
Therefore, we can say KE = PE
We know that PE = m·g·h
Which means KE = m·g·h
We can solve for h:
h = KE / m·g
= 20 / (0.15 · 9.8)
= 13.6m
The correct answer is: the ball has fallen from a height of 13.6m.
Answer:
The initial velocity was U=22.14m/s
Explanation:
Step one :
Applying the third equation of motion
v² = u²+ 2as
Where v= Final velocity
U =initial velocity
a= acceleration due to gravity
S= distance or displacement
Step two :
V= 0
a= 9.81m/s²
S=25m
U=?
Step three :
Substituting into the equation we have
0²=U²+2*9.81*25
0=U²+490.5
U²=-490.5
U=√490.5
U=22.14m/s
