The diagram shows Ned’s movement as he left his house and traveled to different places throughout the day. Which reference point
2 answers:
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m = mass of the ice added = ?
M = mass of water = 1.90 kg
= specific heat of the water = 4186 J/(kg ⁰C)
= specific heat of the ice = 2000 J/(kg ⁰C)
= latent heat of fusion of ice to water = 3.35 x 10⁵ J/kg
= initial temperature of ice = 0 ⁰C
= initial temperature of water = 79 ⁰C
T = final equilibrium temperature = 8 ⁰C
using conservation of heat
Heat gained by ice = Heat lost by water
m (T -
) + m
= M
(
- T)
inserting the values
m (4186) (8 - 0) + m (3.35 x 10⁵ ) = (1.90) (4186) (79 - 8)
m = 1.53 kg
A) The motion of the ball consists of two indipendent motions on the horizontal (x) and vertical (y) axis. The laws of motion in the two directions are:
where
- the horizontal motion is a uniform motion, with constant speed
, where 
and 
- the vertical motion is an uniformly accelerated motion, with constant acceleration
, initial position h (the height of the building) and initial vertical velocity 
(with a negative sign, since it points downwards)
The ball strikes the ground after a time t=3.00 s, so we can find the distance covered horizontally by the ball by substituting t=3.00 s into the equation of x(t):
b) To find the height from which the ball was thrown, h, we must substitute t=3.00 s into the equation of y(t), and requiring that y(3.00 s)=0 (in fact, after 3 seconds the ball reaches the ground, so its vertical position y(t) is zero). Therefore, we have:
which becomes
c) We want the ball to reach a point 10.0 meters below the level of launching, so we want to find the time t such that
If we substitute this into the equation of y(t), we have
whose solution is
(the other solution is negative, so it has no physical meaning). Therefore, the ball reaches a point 10 meters below the level of launching after 1.18 s.
where
- the horizontal motion is a uniform motion, with constant speed
- the vertical motion is an uniformly accelerated motion, with constant acceleration
The ball strikes the ground after a time t=3.00 s, so we can find the distance covered horizontally by the ball by substituting t=3.00 s into the equation of x(t):
b) To find the height from which the ball was thrown, h, we must substitute t=3.00 s into the equation of y(t), and requiring that y(3.00 s)=0 (in fact, after 3 seconds the ball reaches the ground, so its vertical position y(t) is zero). Therefore, we have:
which becomes
c) We want the ball to reach a point 10.0 meters below the level of launching, so we want to find the time t such that
If we substitute this into the equation of y(t), we have
whose solution is