1000 miles = 1610km = 1.61x10^6m
2 weeks = 14 days = 14x24x1440
V=d/t = 1.61x10^6/14x24x1440
= 3.33m/s
Answer:
1.5024
Explanation:
Draw a diagram. Put the two cells in series. Now draw 3 resistors. Two of them equal 0.26 ohms each. The third one is the lightbulb which is 12 ohms.
R = 0.26 + 0.26 + 12 = 12.52
The bulb has a voltage of 2.88 volts across it. You can get the current from that.
i = E / R
i = 2.88 / 12 =
i = 0.24 amps.
Now you can get the voltage drop across the two cells.
E = ?
R = 0.26
i = 0.24 amps
E = 0.26 * 0.24
E = 0. 0624
Finally divide the 2.88 by 2 to get 1.44
Each cell has an emf of 1.44 + 0.0624 = 1.5024
Velocity of submarine A is vs = 11.0m/s
frequency emitted by submarine A. F = 55.273 × 10∧3HZ
Velocity of submarine B = vO = 3.00m/s
The given equation is
f' = ((V + vO) ((v - vS)) × f
The observer on submarine detects the frequency f'.
The sign of vO should be positive as the observer of submarine B is moving away from the source of submarine A.
The speed of the sound used in seawater is 1533m/s
The frequency which is detected by submarine B is
fo = fs (V -vO/ v +vs)
= 53.273 × 10∧3hz) ((1533 m/s - 4.5 m/s)/ (1533 m/s +11 m/s)
fo = 5408 HZ
