A system absorbed 44 joules of heat from its surroundings. After doing work, the increase in the internal energy of the system i
2 answers:
We can solve the problem by using the first law of thermodynamics:
where
is the variation of internal energy of the system
Q is the heat absorbed by the system
W is the work done by the system
In this problem, the heat absorbed by the system is Q=+44 J, and the increase in internal energy is
. So we can rearrange the equation to calculate the work done:
and the work is positive, which means that it is done by the system on the surrounding.
where
Q is the heat absorbed by the system
W is the work done by the system
In this problem, the heat absorbed by the system is Q=+44 J, and the increase in internal energy is
and the work is positive, which means that it is done by the system on the surrounding.
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The period of a pendulum is given by
where L is the pendulum length and g is the gravitational acceleration.
We can write down the ratio between the period of the pendulum on the Moon and on Earth by using this formula, and we find:
where the labels m and e refer to "Moon" and "Earth".
Since the gravitational acceleration on Earth is
while on the Moon is 
, the ratio between the period on the Moon and on Earth is
where L is the pendulum length and g is the gravitational acceleration.
We can write down the ratio between the period of the pendulum on the Moon and on Earth by using this formula, and we find:
where the labels m and e refer to "Moon" and "Earth".
Since the gravitational acceleration on Earth is
Answer:
a) t=1s
y = 10.1m
v=5.2m/s
b) t=1.5s
y =11.475 m
v=0.3m/s
c) t=2s
y =10.4 m
v=-4.6m/s (The minus sign (-) indicates that the ball is already going down)
Explanation:
Conceptual analysis
We apply the free fall formula for position (y) and speed (v) at any time (t).
As gravity opposes movement the sign in the equations is negative.:
y = vi*t - ½ g*t2 Equation 1
v=vit-g*t Equation 2
y: The vertical distance the ball moves at time t
vi: Initial speed
g= acceleration due to gravity
v= Speed the ball moves at time t
Known information
We know the following data:
Vi=15 m / s
t=1s ,1.5s,2s
Development of problem
We replace t in the equations (1) and (2)
a) t=1s
=15-4.9=10.1m
v=15-9.8*1 =15-9.8 =5.2m/s
b) t=1.5s
=22.5-11.025=11.475 m
v=15-9.8*1.5 =15-14.7=0.3m/s
c) t=2s
= 30-19.6=10.4 m
v=15-9.8*2 =15-19.6=-4.6m/s (The minus sign (-) indicates that the ball is already going down)