Answer:
The time taken by missile's clock is 
Solution:
As per the question:
Speed of the missile, 
Now,
If 'T' be the time of the frame at rest then the dilated time as per the question is given as:
T' = T + 1
Now, using the time dilation eqn:




(1)
Using binomial theorem in the above eqn:
We know that:

Thus eqn (1) becomes:


Now, putting appropriate values in the above eqn:


Answer:
remains the same, but the apparent brightness is decreased by a factor of four.
Explanation:
A star is a giant astronomical or celestial object that is comprised of a luminous sphere of plasma, binded together by its own gravitational force.
It is typically made up of two (2) main hot gas, Hydrogen (H) and Helium (He).
The luminosity of a star refers to the total amount of light radiated by the star per second and it is measured in watts (w).
The apparent brightness of a star is a measure of the rate at which radiated energy from a star reaches an observer on Earth per square meter per second.
The apparent brightness of a star is measured in watts per square meter.
If the distance between us (humans) and a star is doubled, with everything else remaining the same, the luminosity remains the same, but the apparent brightness is decreased by a factor of four (4).
Some of the examples of stars are;
- Canopus.
- Sun (closest to the Earth)
- Betelgeuse.
- Antares.
- Vega.
Just remember
Voltage = current times resistance
current = voltage over resistance
Current = 9/3 = 3
Explanation:
Let magnitude of the two forces be x and y.
Resultant at right angle R1= √15N) and at
60 degrees be R2= √18N.
Now, R1 = √(x² + y²) = √15,
R2= √(x² + y² +2xycos50) = √18.
So x² + y² = 15,
and x² + y² + 1.29xy = 18,
therefore 1.29xy = 3,
y = 3/1.29x.
y = 2.33/x
Now, x2 + (2.33/x)2 = 15,
x² + 5.45/x² = 15
multiply through by x²
x⁴ + 5.45 = 15x²
x⁴ - 15x2 + 5.45 = 0
Now find the roots of the equation, and later y. The two values of x will correspond to the
magnitudes of the two vectors.
Good luck
Answer: A
Explanation: We know that f=p*n
f=50*300=15000 Hz = 15kHz.
Have a great day! <3