According to the Law of Universal Gravitation, the gravitational force is directly proportional to the mass, and inversely proportional to the distance. In this problem, let's assume the celestial bodies to be restricted to the planets and the Sun. Since the distance is specified, the other factor would be the mass. Among all the celestial bodies, the Sun is the most massive. So, the Sun would cause the strongest gravitational pull to the satellite.
Answer:
The resistance that will provide this potential drop is 388.89 ohms.
Explanation:
Given;
Voltage source, E = 12 V
Voltage rating of the lamp, V = 5 V
Current through the lamp, I = 18 mA
Extra voltage or potential drop = 12 V - 5 V = 7 V
The resistance that will provide this potential drop (7 V) is calculated as follows:
Therefore, the resistance that will provide this potential drop is 388.89 ohms.
Answer:
Explanation:
Given that,
The mutual inductance of the two coils is
M = 300mH = 300 × 10^-3 H
M = 0.3 H
Current increase in the coil from 2.8A to 10A
∆I = I_2 - I_1 = 10 - 2.8
∆I = 7.2 A
Within the time 300ms
t = 300ms = 300 × 10^-3
t = 0.3s
Second Coil resistance
R_2 = 0.4 ohms
We want to find the current in the second coil,
The same induced EMF is in both coils, so let find the EMF,
From faradays law
ε = Mdi/dt
ε = M•∆I / ∆t
ε = 0.3 × 7.2 / 0.3
ε = 7.2 Volts
Now, this is the voltage across both coils,
Applying ohms law to the second coil, V=IR
ε = I_2•R_2
0.72 = I_2 • 0.4
I_2 = 0.72 / 0.4
I_2 = 1.8 Amps
The current in the second coil is 1.8A
Answer:
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Explanation:
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Answer:
The sound level of the 26 geese is 
Explanation:
From the question we are told that
The sound level is 
The number of geese is 
Generally the intensity level of sound is mathematically represented as
The intensity of sound level in dB for one goose is mathematically represented as
![Z_1 = 10 log [\frac{I}{I_O} ]](https://tex.z-dn.net/?f=Z_1%20%3D%2010%20log%20%5B%5Cfrac%7BI%7D%7BI_O%7D%20%5D)
Where I_o is the threshold level of intensity with value 
is the intensity for one goose in 
For 26 geese the intensity would be
Then the intensity of 26 geese in dB is
![Z_{26} = 10 log[\frac{26 I }{I_o} ]](https://tex.z-dn.net/?f=Z_%7B26%7D%20%3D%2010%20log%5B%5Cfrac%7B26%20I%20%7D%7BI_o%7D%20%5D)
![Z_{26} = 10 log (\ \ 26 * [\frac{ I }{I_o} ]\ \ )](https://tex.z-dn.net/?f=Z_%7B26%7D%20%3D%2010%20log%20%28%5C%20%5C%2026%20%2A%20%20%5B%5Cfrac%7B%20I%20%7D%7BI_o%7D%20%5D%5C%20%5C%20%29)
![Z_{26} = 10 log (\ \ 26 \ \ ) * (\ \ 10 log [\frac{ I }{I_o} ]\ \ )](https://tex.z-dn.net/?f=Z_%7B26%7D%20%3D%2010%20log%20%28%5C%20%5C%2026%20%20%5C%20%5C%20%29%20%2A%20%20%20%28%5C%20%5C%20%2010%20log%20%5B%5Cfrac%7B%20I%20%7D%7BI_o%7D%20%5D%5C%20%5C%20%29)
From the law of logarithm we have that
![Z_{26} = 10 log 26 + 10 log [\frac{I}{I_0} ]](https://tex.z-dn.net/?f=Z_%7B26%7D%20%3D%2010%20log%2026%20%2B%20%2010%20log%20%5B%5Cfrac%7BI%7D%7BI_0%7D%20%5D)
