Answer:
Explanation:
Person A's velocity relative to the train is 0. Therefore, the pitch of the horn will not change.
Answer:
a) frequency = 0.1724 Hz
b) Period = 5.8 sec
c) speed = 7.04 m/s
d) acceleration = 7.62 m/s²
Explanation:
Given that;
radius = 6.5m
time period = 5.8 sec every circle
a) the frequency
frequency is the number of rotation in unit time
frequency = 1 / time period = 1/5.8
frequency = 0.1724 Hz
b) the period
period is time taken in one rotation
period = total time / rotation = 5.8 / 1
Period = 5.8 sec
c) the speed
speed = distance/time = circumference/time period = 2πr / t = (2π×6.5) / 5.8
speed = 7.04 m/s
d) acceleration
To find the acceleration we take the linear velocity squared divided by the radius of the circle.
so
acceleration = v² / r = (7.04)² / 6.5 = 49.5616 / 6.5
acceleration = 7.62 m/s²
The magnitude of work done by the gas is 279 J and the sign is negative so W = -279 J as work is done by the system.
<u>Explanation:</u>
According to first law of thermodynamics, the change in internal energy of the system is equal to the sum of the heat energy added or released from the system with the work done on or by the system. If the heat energy is added to the system to perform a certain work, then the heat energy is taken as positive, while it will be negative when the heat energy is released from the system.
Similarly, in this case, the heat energy of 597 J is added to the system. So the heat energy will be positive, while the gas expansion occurs means work is done by the system.
ΔU = Q+W
Since ΔU is the change in internal energy which is given as 318 J and the heat energy added to the system is Q = 597 J.
Then the work done by the gas = ΔU - Q = 318 J - 597 J = - 279 J.
As the work is done by the system, so it will be denoted in negative sign and the magnitude of work done by the gas is 279 J.
The component that’s dissolved is called the solvent
Answer:
Potential at B would be 100V
Explanation:
The electric potential is defined as the work done to bring a unit positive charge from infinity to some point in the field.
We always determine the potential with respect to some reference point. Let the potential at A be zero. If the potential at B is V, then work done to bring charge q from A to B = qV
which is the electric potential energy.
If instead we use some charge Q, the electric potential <em>energy</em> will be QV, but the electric potential will always be V.
