A 10.0 kg weather rocket generates a thrust of 230 NN . The rocket, pointing upward, is clamped to the top of a vertical spring.
The bottom of the spring, whose spring constant is 480 N/mN/m , is anchored to the ground. Part A Initially, before the engine is ignited, the rocket sits at rest on top of the spring. How much is the spring compressed
1 answer:
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Answer:
145 m
Explanation:
Given:
Wavelength (λ) = 2.9 m
we know,
c = f × λ
where,
c = speed of light ; 3.0 x 10⁸ m/s
f = frequency
thus,
substituting the values in the equation we get,
f = 1.03 x 10⁸Hz
Now,
The time period (T) =
or
T = = 9.6 x 10⁻⁹ seconds
thus,
the time interval of one pulse = 100T = 9.6 x 10⁻⁷ s
Time between pulses = (100T×10) = 9.6 x 10⁻⁶ s
Now,
For radar to detect the object the pulse must hit the object and come back to the detector.
Hence, the shortest distance will be half the distance travelled by the pulse back and forth.
Distance = speed × time = 3 x 10^8 m/s × 9.6 x 10⁻⁷ s) = 290 m {Back and forth}
Thus, the minimum distance to target = = 145 m
Answer:
EP = 49.05Joules (J)
Explanation:
The equation for Potential energy (EP) is
EP = m g h
We are given the values below (do convert them into SI units)
m = 0.0025kg
h = 2000m
g = 9.81m/
Substitute the values into the equation and solve for EP
EP = 0.0025 * 2000 * 9.81
EP = 49.05Joules (J)