Answer:
The best estimate of the depth of the well is 2.3 sec.
Explanation:
Given that,
Record time,





We need to find the best estimate of the depth of the well
According to record time,
We can write of the record time





Here, all time is nearest 2.3 sec.
So, we can say that the best estimate of the depth of the well is 2.3 sec.
Hence, The best estimate of the depth of the well is 2.3 sec.
Answer: I am pretty sure that you should pick radio waves.
Explanation: The scientist should use radio waves. I think this because you can use the radio waves to analyze the signals from outer space. This will work much better than anything there, to analyze it the best possible.
The best I could do.
Energy of a wave:
E = nhc/λ
3000 = (n x 6.63 x 10⁻³⁴ x 3 x 10⁸)/(510 x 10⁻⁹)
n = 7.69 x 10 ²¹ photons per second per meter²
2.70 cm² = 2.70/10,000 m²
= 2.7 x 10⁻⁴
Photons per second = 7.69 x 10 ²¹ x 2.7 x 10⁻⁴
= 2.08 x 10¹⁸ photons per second
The way to do this is very easy so do 4125 x 2 = ? then the ? will be times by 2 again after the answer to both of those is your answer!!!
Answer:
Explanation:
Given that,
Mass of ball m = 2kg
Ball traveling a radius of r1= 1m.
Speed of ball is Vb = 2m/s
Attached cord pulled down at a speed of Vr = 0.5m/s
Final speed V = 4m/s
Let find the transverse component of the final speed using
V² = Vr²+ Vθ²
4² = 0.5² + Vθ²
Vθ² = 4²—0.5²
Vθ² = 15.75
Vθ =√15.75
Vθ = 3.97 m/s.
Using the conservation of angular momentum,
(HA)1 = (HA)2
Mb • Vb • r1 = Mb • Vθ • r2
Mb cancels out
Vb • r1 = Vθ • r2
2 × 1 = 3.97 × r2
r2 = 2/3.97
r2 = 0.504m
The distance r2 to the hole for the ball to reach a maximum speed of 4m/s is 0.504m
The required time,
Using equation of motion
V = ∆r/t
Then,
t = ∆r/Vr
t = (r1—r2) / Vr
t = (1—0.504) / 0.5
t = 0.496/0.5
t = 0.992 second