Answer:
7.6 g
Explanation:
"Well lagged" means insulated, so there's no heat transfer between the calorimeter and the surroundings.
The heat gained by the copper, water, and ice = the heat lost by the steam
Heat gained by the copper:
q = mCΔT
q = (120 g) (0.40 J/g/K) (40°C − 0°C)
q = 1920 J
Heat gained by the water:
q = mCΔT
q = (70 g) (4.2 J/g/K) (40°C − 0°C)
q = 11760 J
Heat gained by the ice:
q = mL + mCΔT
q = (10 g) (320 J/g) + (10 g) (4.2 J/g/K) (40°C − 0°C)
q = 4880 J
Heat lost by the steam:
q = mL + mCΔT
q = m (2200 J/g) + m (4.2 J/g/K) (100°C − 40°C)
q = 2452 J/g m
Plugging the values into the equation:
1920 J + 11760 J + 4880 J = 2452 J/g m
18560 J = 2452 J/g m
m = 7.6 g
734 is the answer for the chronic blood exchange service of new france
Answer:
vₓ = xg/2y
Explanation:
In this question, let us find the time it takes for the ball on the right that has zero initial velocity to reach the ground.
By newton equation of motion we know that
y = v₀ t - ½ g t²
t = 2y / g
This is the time it takes for the ball on the right to reach the ground; at this time the ball on the left travels a distance
vₓ = x/t
vₓ = xg/2y
vₓ = xg/2y
Where we assume that x and y are known.
Explanation:
Echoes occur due to the reflection of sound from any obstacle, but not all the reflected sound waves lead to the phenomenon of echo. For the echo to be heard it actually depends upon the human perception as well, human ears can encounter the difference between the sound wave directly form the source and the reflected sound waves only if there is a minimum time gap of one-tenth of a second. For this time gap in the atmosphere at normal temperature and pressure the obstacle must be at least 7 meters away from the sound source.
Answer:
a)Q=71.4 μ C
b)ΔV' = 10.2 V
Explanation:
Given that
C ₁= 8.7 μF
C₂ = 8.2 μF
C₃ = 4.1 μF
The potential difference of the battery, ΔV= 34 V
When connected in series
1/C = 1/C ₁ + 1/C₂ + 1/C₃
1/ C= 1/8.4 +1 / 8.4 + 1/4.2
C=2.1 μF
As we know that when capacitor are connected in series then they have same charge,Q
Q= C ΔV
Q= 2.1 x 34 μ C
Q=71.4 μ C
b)
As we know that when capacitor are connected in parallel then they have same voltage difference.
Q'= C' ΔV'
C'= C ₁+C₂+C₃ (For parallel connection)
C'= 8.4 + 8.4 + 4.2 μF
C'=21 μF
Q'= C' ΔV'
Q'=3 Q
3 x 71.4= 21 ΔV'
ΔV' = 10.2 V
