Answer:
a) m=20000Kg
b) v=0.214m/s
Explanation:
We will separate the problem in 3 parts, part A when there were no coals on the car, part B when there is 1 coal on the car and part C when there are 2 coals on the car. Inertia is the mass in this case.
For each part, and since the coals are thrown vertically, the horizontal linear momentum p=mv must be conserved, that is, 
, were each velocity refers to the one of the car (with the eventual coals on it) for each part, and each mass the mass of the car (with the eventual coals on it) also for each part. We will write the mass of the hopper car as 
, and the mass of the first and second coals as 
and 
respectively
We start with the transition between parts A and B, so we have:
Which means
And since we want the mass of the first coal thrown () we do:
Substituting values we obtain
For the transition between parts B and C, we can write:
Which means
Since we want the new final speed of the car (
) we do:
Substituting values we obtain
a= v (1) - v (2)/ t = 14 - 10/20 = 4/20 = 0.2 m/s^2 south
Answer:
A bowling ball sitting on the rack: Potential
Sitting in the top of a tree: Potential
The chemical bonds in sugar: Potential
An archer with his bow drawn: Potential
A baseball thrown to second base: Kinetic
The wind blowing through your hair: Kinetic
A volleyball player spiking a ball: Kinetic
A bicyclist pedaling up a hill
: Kinetic
A bowling ball rolling down the alley: Kinetic
Walking down the street: Kinetic
Answer:
Drums, harps, recorders, and bagpipes.
Explanation:
Answer:
238.75⁰C .
Explanation:
coefficient of linear thermal expansion of aluminum and steel is 23 x 10⁻⁶ K⁻¹ and 12 x 10⁻⁶ K⁻¹ respectively .
Rise in temperature be Δ t .
Formula for linear expansion due to heat is as follows
l = l₀ ( 1 + α x Δt )
l is expanded length , l₀ is initial length , α is coefficient of linear expansion and Δt is increase in temperature .
For aluminum
l = 2.5 ( 1 + 23 x 10⁻⁶ Δt )
For steel
l = 2.506 ( 1 + 12 x 10⁻⁶ Δt )
Given ,
2.5 ( 1 + 23 x 10⁻⁶ Δt ) = 2.506 ( 1 + 12 x 10⁻⁶ Δt )
1 + 23 x 10⁻⁶ Δt = 1.0024 ( 1 + 12 x 10⁻⁶ Δt )
1 + 23 x 10⁻⁶ Δt = 1.0024 + 12.0288 x 10⁻⁶ Δt
10.9712 x 10⁻⁶ Δt = .0024
Δt = 218.75
Initial temperature = 20⁰C
final temperature = 218.75 + 20 = 238.75⁰C .
