EC_1 + EP_1 = EC2 + EP_2
EC_2 = 0
EC_2 = EP_1 - EP_2
EC_2 = mg(H_1 - H_2) = 0.20 kg * 9.8 m/s^2 * (3.25 m - 1.5m) = 3.43 J
I think so it ans wold be d.because if the source approaches the observer more and more wavefront will pass and it get squeezed.so wavelength decreases and frequency increases.
Answer:
a) 200A
b) 10.2V
c) 2.04kW
d)
I=80A
V=4.08V
P=0.326kW
Explanation:
Here we have a circuit of one power source and two resistors in series, the first question is asking for the current, so according to Ohm's Law:
Where R is the equivalent resistance of the resistors in series
![R=0.0510+0.0090=0.0600[ohm]](https://tex.z-dn.net/?f=R%3D0.0510%2B0.0090%3D0.0600%5Bohm%5D)
To calculate the voltage dropped by the motor we have to apply the voltage divider rule:
The power dissipated supplied to the motor is given by:
now solving adding a 0.0900 ohm resistor:
Answer:
4.2 J
Explanation:
Specific heat capacity: This is defined as the amount of a heat required to rise a unit mass of a substance through a temperature of 1 K
From specific heat capacity,
Q = cmΔt.............................. Equation 1
Where Q = amount of energy absorbed or lost, c = specific heat capacity of water, m = mass of water, Δt = Temperature rise.
Given: m = 1 g = 0.001 kg, Δt = 1 °C
Constant : c = 4200 J/kg.°C
Substitute into equation 1
Q = 0.001×4200(1)
Q = 4.2 J.
Hence the energy absorbed or lost = 4.2 J
15.49 should be the answer if that is 12 watt battery.
