Following a collision between a large spacecraft and an asteroid, a copper disk of radius 28.0 m and thickness 1.20 m, at a temp
erature of 850°C, is floating in space, rotating about its axis with an angular speed of 20.0 rad/s. As the disk radiates infrared light, its temperature falls to 20.0°C. No external torque acts on the disk.
A) Find the change in kinetic energy of the disk.
B) Find the change in internal energy of the disk.
C) Find the amount of energy it radiates.
A) Find the change in kinetic energy of the disk.
B) Find the change in internal energy of the disk.
C) Find the amount of energy it radiates.
1 answer:
You might be interested in
Answer:
The thrown rock strike 2.42 seconds earlier.
Explanation:
This is an uniformly accelerated motion problem, so in order to find the arrival time we will use the following formula:
So now we have an equation and unkown value.
for the thrown rock
for the dropped rock
solving both equation with the quadratic formula:
we have:
the thrown rock arrives on t=5.4 sec
the dropped rock arrives on t=7.82 sec
so the thrown rock arrives 2.42 seconds earlier (7.82-5.4=2.42)
Answer:
-0.79 J
Explanation:
We are given that
We have to find the work done by the electric force on the moving point charge.
Work done,
Where
Using the formula
The magnitude of electric field is produced by the electrons at a certain distance.
E = kQ/r²
where:
E = electric field produced
Q = charge
r = distance
k = Coulomb Law constant 9 x10^9<span> N. m</span>2<span> / C</span><span>2
Given are the following:
Q = </span><span>1.602 × 10^–19 C
</span><span>r = 38 x 10^-9 m
Substitue the given:
E = </span>
E = 998.476 kN/C
E = kQ/r²
where:
E = electric field produced
Q = charge
r = distance
k = Coulomb Law constant 9 x10^9<span> N. m</span>2<span> / C</span><span>2
Given are the following:
Q = </span><span>1.602 × 10^–19 C
</span><span>r = 38 x 10^-9 m
Substitue the given:
E = </span>
E = 998.476 kN/C