Answer:
is the drop in the water temperature.
Explanation:
Given:
- mass of ice,
- mass of water,
Assuming the initial temperature of the ice to be 0° C.
<u>Apply the conservation of energy:</u>
- Heat absorbed by the ice for melting is equal to the heat lost from water to melt ice.
<u>Now from the heat equation:</u>
......................(1)
where:
latent heat of fusion of ice
specific heat of water
change in temperature
Putting values in eq. (1):
is the drop in the water temperature.
Answer:
The velocity of the bag will be 16.08 m/s
Explanation:
Given:
Mass of the astronaut, M = 124 kg
Mass of the bag, m = 10 lg
Initial speed of the astronaut with bag, = 1.20 m/s
Applying the concept of conservation of momentum
we have
(M + m) = Mv₁ + mv₂
here,
v₁ is the final velocity of the astronaut = 0
v₂ is final velocity of the bag
thus, on substituting the values, we have
(124 + 10)1.20 = 124 × 0 + 10v₂
or
v₂ = 16.08 m/s
Hence, <u>the velocity of the bag will be 16.08 m/s</u>
Answer: 12.0 m/s^2
Explanation:
Let be the angular acceleration of the end of the rod
Taking torque about the link, we have:
Torque is also given in terms of moment of inertia of the rod and its angular acceleration i.e.
From equations (i) and (ii) we have:
The acceleration of the end of the rod farthest from the link is given by:
Answer:
The time taken for the commercial Jet liner to reach the end of its runway is 10.18 s.
Explanation:
Given;
average acceleration of the commercial Jet liner, a = 3g = 3 x 9.8 m/s² = 29.4 m/s²
distance traveled by the commercial Jet liner, s = 1542 m
The time taken for the commercial Jet liner to reach the end of its runway is calculated as follows;
s = ut + ¹/₂at²
where;
u is the initial velocity of the commercial Jet liner = 0
s = 0 + ¹/₂at²
s = ¹/₂at²
2s = at²
Therefore, the time taken for the commercial Jet liner to reach the end of its runway is 10.18 s.