Answer: its speed upon release is 26.05 m/s
Explanation:
Given that;
mass m = 0.244 kg
force F = 30.3 N
V1 = 14.7 m/s
r = 59.3 cm = 0.593 m
Vf = ?
we know that;
1/2mV1² + FπR = 1/2mVf²
so we substitute
[1/2×0.244×(14.7)²] + [30.3×π×0.593 = 1/2×0.244×Vf²
26.3629 + 56.4478 = 0.122Vf²
82.8107 = 0.122Vf²
Vf² = 82.8107 / 0.122Vf
Vf² = 678.7762
Vf = √678.7762
Vf = 26.05 m/s
Therefore its speed upon release is 26.05 m/s
Proxima Centauri is approximately 4x1013 kilometers away from the sun. Which multiplies up to 40,000,000,000,000 km. Hope I helped! Thanks for asking us here at Brainly! Feel free to ask more questions at anytime!
Answer:
Its not A..
Explanation:
I chose A - was incorrect
An employee who has integrity is ____. D. trustworthy and honesty. Hope this helps!
Hello,
<span>A police car parked on the side of the highway emits a 1200 Hz sound that bounces off a vehicle farther down the highway and returns with a frequency of 1250 Hz.
How fast is the vehicle going?
Doppler equation formula: </span>ƒL = ƒS(v - vL)/(v - vS)
The wave returns with a frequency of 1250 Hz, the <span>echo frequency is higher; the car must be traveling towards the police car.
</span><span>The wave echo is coming back towards the police car at the same speed as the sound wave travels towards the moving car so t</span><span>he relative speed between the cars is half of the speed of the echo.
* </span><span>speed of sound equals about 337 m/s </span>
2v / 337 = (1250/1200) - 1
<span>2v = 14.04 m/s </span>
<span>v = 7.02 m/s
</span>
Thus, the vehicle is going 7.02 m/s.
Faith xoxo