The work done when a spring is stretched from 0 to 40cm is 4J.
What is work done?
Work done is the magnitude of force multiplied by displacement of an object. It is also the amount of energy transferred to an object when work is done on that.
The work done on the spring to stretch to 40cm is,
F = kx
where F is force, k is force constant.
k = F / x = 10 N / 20 * 10^-2 m = 50 N/m
W = 0.5 * k * (x)^2
where W = work done, k = force constant.
W = 0.5 x 50 x (40 x 10^-2)^2 = 4 J.
Therefore, the work done on the spring when it is stretched to 40cm is 4J.
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To solve the problem, use Kepler's 3rd law :
T² = 4π²r³ / GM
Solved for r :
r = [GMT² / 4π²]⅓
but first covert 6.00 years to seconds :
6.00years = 6.00years(365days/year)(24.0hours/day)(6...
= 1.89 x 10^8s
The radius of the orbit then is :
r = [(6.67 x 10^-11N∙m²/kg²)(1.99 x 10^30kg)(1.89 x 10^8s)² / 4π²]⅓
= 6.23 x 10^11m
Answer:
c
Explanation:
the loud noise can reduce the quality of the analog signal
Answer:
5. Solenoid B’s magnetic field is 2.7 times greater than solenoid A's magnetic field.
Explanation:
The magnetic field produced by a solenoid depends on number of coils, the length of the solenoid and the amount of current flowing through the coil.

Current in solenoid B is 2 times greater than the current in solenoid A.

Number of coils in B is 4 times greater than coils in A

Length of the solenoid B is 3 times greater than solenoid A


Thus, solenoid B's magnetic field is 2.7 times greater than magnetic field of solenoid A.
<span>The correct answer is: Towards
Explanation:
Doppler Effect is the apparent change in the frequency of a wave caused by the relative motion between the source of the wave and the observer. As the observer gets closer to the source, that observer will observe or sense the higher frequency of the wave (emitting by the source). Frequency in the case of sound waves is called the "sound's pitch." Therefore, the sound's pitch will increase if the Falcon moves towards the source of the sound. Hence, the correct option is "towards."</span>