Because of the power of the light it is very strong so that is why light reaches all the way to earth
Answer:
Explanation:
Using Boyles law
Boyle's law states that, the volume of a given gas is inversely proportional to it's pressure, provided that temperature is constant
V ∝ 1 / P
V = k / P
VP = k
Then,
V_1 • P_1 = V_2 • P_2
So, if we want an increase in pressure that will decrease volume of mercury by 0.001%
Then, let initial volume be V_1 = V
New volume is V_2 = 0.001% of V
V_2 = 0.00001•V
Let initial pressure be P_1 = P
So,
Using the equation above
V_1•P_1 = V_2•P_2
V × P = 0.00001•V × P_2
Make P_2 subject of formula by dividing be 0.00001•V
P_2 = V × P / 0.00001 × V
Then,
P_2 = 100000 P
So, the new pressure has to be 10^5 times of the old pressure
Now, using bulk modulus
Bulk modulus of mercury=2.8x10¹⁰N/m²
bulk modulus = P/(-∆V/V)
-∆V = 0.001% of V
-∆V = 0.00001•V
-∆V = 10^-5•V
-∆V/V = 10^-5
Them,
Bulk modulus = P / (-∆V/V)
2.8 × 10^10 = P / 10^-5
P = 2.8 × 10^10 × 10^-5
P = 2.8 × 10^5 N/m²
We are asked to solve and determine the magnitude of the current flowing through the first device. In order for us to have a better understanding of the problem, we can refer to the attached picture which contains electric circuit diagram. Since it the problem we are already given with an electromotive source or the voltage supply and since the two resistance is in parallel, it would clearly mean that the voltage drop in each resistance is just the same. The resistance 1 uses the 40 volts at the same time the resistance 2 uses 40 volts also. Solving further for the current, we can apply Ohm's law which V = IR where "V" represents the voltage, the "I" represents the current and "R" represents the resistance.
Such as the solution in obtaining current is shown below:
I = V / R, substitute values we have it
I = 40 volts / 1208 ohms
I = 0.0331 Amperes
Therefore, the current flowing in the first device is
0.033 Amperes or 33 milliAmperes.
Answer:
d = 1700 meters
Explanation:
During a rainy day, as a result of colliding clouds an observer saw lighting and a heard thunder sound. The time between seeing the lighting and hearing the sound was 5 second, t = 5 seconds
Speed of sound, v = 340 m/s (say)
Let d is the distance of the colliding cloud from the observer. The distance covered by the object. It is given by :
![d=v\times t](https://tex.z-dn.net/?f=d%3Dv%5Ctimes%20t)
d = 1700 meters
So, the distance of the colliding cloud from the observer is 1700 meters. Hence, this is the required solution.