Answer:
The pressure of the remaining gas in the tank is 6.4 atm.
Explanation:
Given that,
Temperature T = 13+273=286 K
Pressure = 10.0 atm
We need to calculate the pressure of the remaining gas
Using equation of ideal gas
For a gas
Where, P = pressure
V = volume
T = temperature
Put the value in the equation
....(I)
When the temperature of the gas is increased
Then,
....(II)
Divided equation (I) by equation (II)
Hence, The pressure of the remaining gas in the tank is 6.4 atm.
The acceleration is the principal subordinate of the speed if the speed is steady the subsidiary is invalid if the speed is diminishing the subsidiary is negative. When discussing so much stuff we consider the momentary esteem.
<span>Note that when you back off, you back off by and large yet can locally in time quicken a tiny bit, suppose amid 1/tenth of a sec since you achieved a segment of the street which was slanting. In any case, this does not change the way that when the speed diminishes, the quickening is negative.</span>
I think the gravity doesn't affect the mass of an object. Only it's weight can be compared
Answer:
24.8m/s
Explanation:
Given data
m1= 10kg
u1=25m/s
m2=17kg
u2=16m/s
v1=10m/s
v2=??
Applying the conservation of linear momentum
m1u1+m2u2=m1v1+m2v2
substitute
10*25+17*16=10*10+17*v2
250+272=100+17v2
522=100+17v2
522-100=17v2
422=17v2
Divide both sides by 17
v2= 422/17
v2= 24.8 m/s
Hence the velocity of the red cart is 24.8m/s in the opposite direction of the blue cart
