Answer: The volume of the diver's at the same depth, at a temperature of 21.0°C and a pressure of 141.20 kPa is 1.66 L
Explanation:
The combined gas equation is,
where,
= initial pressure of gas = 101.30 kPa
= final pressure of gas = 141.20 kPa
= initial volume of gas = 2.40 L
= final volume of gas = ?
= initial temperature of gas =
= final temperature of gas =
Now put all the given values in the above equation, we get:
The volume of the diver's at the same depth, at a temperature of 21.0°C and a pressure of 141.20 kPa is 1.66 L
The Halogens and the Alkali Metals.
Answer:
Velocity of Afrom B=21m/s
Acceleration of A from B=1.68m/s°2
Explanation:
Given
Radius r=150m
Velocity of a Va= 54km/hr
Va=54*1000/3600=15m/s
Velocity of b Vb=82km/hr
VB=81*1000/3600=22.5mls
The velocity of Car A as observed from B is VBA
VB= VA+VBA
Resolving the vector into X and Y components
For X component= 15cos60=7.5m/s
Y component=22 5sin60=19.48m/s
VBA= √(X^2+Y^2)
VBA= ✓(7.5^2+19.48^2)=21m/s
For acceleration of A observed from B
A=VA^2/r= 15^2/150=1.5m/s
Resolving into Xcomponent=1.5cos60=0.75m/s
Y component=3cos60=1.5
Acceleration BA=√(0.75^2+1.5^2)
1.68m/s
Answer:
At the Launch point
Explanation:
We are told that after the explosion, one of the halves landed at the point Y, which is a distance of 2R to the right side of the launch point.
This means that the only way one half of the object would have gone double the original distance of R is if it's velocity was two times that of the initial velocity. Furthermore, the only way that will make sense is if the other half fell straight down. So In conclusion, if air resistance is neglected, the other half will land at the initial velocity point which is at launch point.
<u>Answer:</u>
The time taken for the ball to hit the floor as 1.02 seconds
<u>Explanation:</u>
As per the given question, the ball leaves at a speed from the table with an initial velocity of 10 m/s, we have the equation
where Vf represents the final velocity
Vi represents the initial velocity
a represents the acceleration and
t represents the time
after rearranging
= 1.022 seconds