Answer:
The amount of heat transfer is 21,000J .
Explanation:
The equation form of thermodynamics is,
ΔQ=ΔU+W
Here, ΔQ is the heat transferred, ΔU is the change in internal energy, and W is the work done.
Substitute 0 J for W and 0 J for ΔU
ΔQ = 0J+0J
ΔQ = 0J
The change in internal energy is equal to zero because the temperature changes of the house didn’t change. The work done is zero because the volume did not change
The heat transfer is,
ΔQ=Q (in
) −Q (out
)
Substitute 19000 J + 2000 J for Q(in) and 0 J for Q(out)
ΔQ=(19000J+2000J)−(0J)
=21,000J
Thus, the amount of heat transfer is 21,000J .
-- Multiply each side of the formula by 2
-- Then divide each side by t
-- Then subtract V(i) from each side.
Density=mass/volume therefore volume=mass/density; 55g/11.4g/cm^3= 4.82cm^3
Answer: vl = 2.75 m/s vt = 1.5 m/s
Explanation:
If we assume that no external forces act during the collision, total momentum must be conserved.
If both cars are identical and also the drivers have the same mass, we can write the following:
m (vi1 + vi2) = m (vf1 + vf2) (1)
The sum of the initial speeds must be equal to the sum of the final ones.
If we are told that kinetic energy must be conserved also, simplifying, we can write:
vi1² + vi2² = vf1² + vf2² (2)
The only condition that satisfies (1) and (2) simultaneously is the one in which both masses exchange speeds, so we can write:
vf1 = vi2 and vf2 = vi1
If we call v1 to the speed of the leading car, and v2 to the trailing one, we can finally put the following:
vf1 = 2.75 m/s vf2 = 1.5 m/s
Answer:
Explanation:
When the box is on the ramp , component of its weight along the ramp
= mg sinθ
Friction force acting on it in upward direction
=μ mg cosθ
For sliding
μ mg cosθ < mg sinθ
μ cosθ < sinθ
.5 x cos35 < sin35
.41 < .57
So the box will slide
When sliding starts , kinetic friction acts
Net force in downward direction
mgsinθ - μ mg cosθ
acceleration
= gsinθ - μ g cosθ
= 5.62 - .3 x 9.8 x cos35
= 5.62 - 2.4
= 3.22 m /s²
