Answer:
24 m to the right/ to the east
Using the conservation of momentumwhere momentum = mass x velocity
so the let x be the initial velocity of the ball B
(30 kg) (-10 m/s) + (10 kg) (x) = (30 kg) (30 m/s) + ( 10 kg) (-10 m/s)
-300 + 10 x = 900 - 10010x = 900 - 100 + 30010x = 1100x = 1100 / 10x = 110 m/s
Acceleration = (change in speed) / (time for the change)
Change in speed = (speed at the end) - (speed at the beginning).
Change in speed = (132 m/s) - (zero m/s) = 132 m/s
Acceleration = (132 m/s) / (3.94 s)
Acceleration = (132/3.94) m/s²
<em>Acceleration = 33.5 m/s² </em>(about 3.4 G's)
Answer:
100°C
Explanation:
The heat gained by the ice equals the heat lost by the steam, so the total heat transfer equals 0.
Heat lost by the steam as it cools to 100°C:
q = mCΔT
q = (3 kg) (2.00 kJ/kg/K) (100°C − 120°C)
q = -120 kJ
Total heat so far is negative.
Heat lost by the steam as it condenses:
q = -mL
q = -(3 kg) (2256 kJ/kg)
q = -6768 kJ
Heat absorbed by the ice as it warms to 0°C:
q = mCΔT
q = (6 kg) (2.11 kJ/kg/K) (0°C − (-40°C))
q = 506.4 kJ
Heat absorbed by the ice as it melts:
q = mL
q = (6 kg) (335 kJ/kg)
q = 2010 kJ
Heat absorbed by the water as it warms to 100°C:
q = mCΔT
q = (6 kg) (4.18 kJ/kg/K) (100°C − 0°C)
q = 2508 kJ
The total heat absorbed by the ice by heating it to 100°C is 5024.4 kJ.
If the steam is fully condensed, it loses a total of -6888 kJ.
Therefore, the steam does not fully condense. The equilibrium temperature is therefore 100°C
Answer:
The charge carried by the droplet is ![1.330\times10^{-19}\ C](https://tex.z-dn.net/?f=1.330%5Ctimes10%5E%7B-19%7D%5C%20C)
Explanation:
Given that,
Distance =8.4 cm
Time = 0.250 s
Suppose tiny droplets of oil acquire a small negative charge while dropping through a vacuum in an experiment. An electric field of magnitude
points straight down and if the mass of the droplet is ![2.93\times10^{-15} kg](https://tex.z-dn.net/?f=2.93%5Ctimes10%5E%7B-15%7D%20kg)
We need to calculate the acceleration
Using equation of motion
![s=ut+\dfrac{1}{2}at^2](https://tex.z-dn.net/?f=s%3Dut%2B%5Cdfrac%7B1%7D%7B2%7Dat%5E2)
Put the value into the formula
![8.4\times10^{-2}=0+\dfrac{1}{2}\times a\times(0.250)^2](https://tex.z-dn.net/?f=8.4%5Ctimes10%5E%7B-2%7D%3D0%2B%5Cdfrac%7B1%7D%7B2%7D%5Ctimes%20a%5Ctimes%280.250%29%5E2)
![a=\dfrac{8.4\times10^{-2}\times2}{(0.250)^2}](https://tex.z-dn.net/?f=a%3D%5Cdfrac%7B8.4%5Ctimes10%5E%7B-2%7D%5Ctimes2%7D%7B%280.250%29%5E2%7D)
![a=2.688\ m/s^2](https://tex.z-dn.net/?f=a%3D2.688%5C%20m%2Fs%5E2)
We need to calculate the charge carried by the droplet
Using formula of electric filed
![E=\dfrac{F}{q}](https://tex.z-dn.net/?f=E%3D%5Cdfrac%7BF%7D%7Bq%7D)
![q=\dfrac{ma}{E}](https://tex.z-dn.net/?f=q%3D%5Cdfrac%7Bma%7D%7BE%7D)
Put the value into the formula
![q=\dfrac{2.93\times10^{-15}\times2.688}{5.92\times10^4}](https://tex.z-dn.net/?f=q%3D%5Cdfrac%7B2.93%5Ctimes10%5E%7B-15%7D%5Ctimes2.688%7D%7B5.92%5Ctimes10%5E4%7D)
![q=1.330\times10^{-19}\ C](https://tex.z-dn.net/?f=q%3D1.330%5Ctimes10%5E%7B-19%7D%5C%20C)
Hence, The charge carried by the droplet is ![1.330\times10^{-19}\ C](https://tex.z-dn.net/?f=1.330%5Ctimes10%5E%7B-19%7D%5C%20C)