Answer:
In space there is no air resistance. on earth there is
in space there is no opposite forces acting on stopping the ball, so if you throw it it will go on forever.
on earth there is air resistance and gravity, this will pull the ball towards the ground and slow it down.
Answer:
N₂ = 503.8 N
Explanation:
given,
mass of bottom block = 37 Kg
mass of middle block = 18 Kg
mass of the top block = 16 Kg
force acting on the top block = 170 N
force on the block at top
N₁ be the normal force from block at middle
now,
N₁ = 170 + m g
N₁ = 170 + 16 x 9.8
now, force on block at middle
N₂ be the normal force exerted by the bottom block
N₂ = N₁ + m₂ g
N₂ = 326.8 + 18 x 9.8
N₂ = 503.8 N
hence, normal force by bottom block is equal to N₂ = 503.8 N
Answer: The answer is A just answered it
Explanation:
Answer:
Yes, it can can be completed adiabatically
Explanation:
To solve the problem we will resort to the theory of thermodynamics,
It is necessary to develop this problem to resort to the A-11E tables in English Units for R134a (since the problem requires it, if it were SI just to change to that table)
State 1 indicates that the refrigerant is at 60 ° F,
In the first table (attached image of the value taken) the value of the entropy is
For State 2 the refrigerant is at 50% quality and at a pressure of
In table 2 of the refrigerant (for the pressure values) we perform the reading and we have to
We know that,
The change in enthalpy would be given by
<em>The change in enthalpy is positive, so the process can be completed adiabatically</em>
ΔVl = L di/dt
i = i₀e -t/T
di/dt = i₀ × (-1/T) e -t/T
ΔVl = L× (-I/T i₀e -t/T
ΔVl = -L/T i₀e -t/T
b. 15mm, i₀ = 36mA, T = 1.1m
t= Os
ΔVl = 0,491V
C. t = 1ms
ΔVl = 0.198V
t = 2ms
ΔVl = 0.08V
E. t = ms
ΔVl = 0.032V