a. t=0.553 s
b. vox(horizontal speed) = 3.62 m/s
<h3>Further explanation</h3>
Given
h = 1.5 m
x = 2 m
Required
a. time
b. vo=initial speed
Solution
Free fall motion
a. h = 1/2 gt²(vertical motion=h=voyt+1/2gt²⇒voy = 0)
t = √2h/g
t = √2.1.5/9.8
t=0.553 s
b. x=vox.t(horizontal motion)
vox=x/t
vox=2/0.553
vox=3.62 m/s
Answer:
41.9 g
Explanation:
We can calculate the heat released by the water and the heat absorbed by the steel rod using the following expression.
Q = c × m × ΔT
where,
c: specific heat capacity
m: mass
ΔT: change in temperature
If we consider the density of water is 1.00 g/mol, the mass of water is 125 g.
According to the law of conservation of energy, the sum of the heat released by the water (Qw) and the heat absorbed by the steel (Qs) is zero.
Qw + Qs = 0
Qw = -Qs
cw × mw × ΔTw = -cs × ms × ΔTs
(4.18 J/g.°C) × 125 g × (21.30°C-22.00°C) = -(0.452J/g.°C) × ms × (21.30°C-2.00°C)
ms = 41.9 g
