B
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As the box compresses the spring, the spring performs
-1/2 (85 N/m) (0.065 m)² ≈ -0.18 J
of work on the box. By the work energy theorem, the total work performed on the box (which is done only by the spring since there's no friction) is equal to the change in the box's kinetic energy. At full compression, the box has zero instantaneous speed, so
<em>W</em> = ∆<em>K</em> ==> -0.18 J = 0 - 1/2 (2.5 kg) <em>v</em> ²
where <em>v</em> is the box's speed when it first comes into contact with the spring. Solve for <em>v</em> :
<em>v</em> ² ≈ 0.14 m²/s² ==> <em>v</em> ≈ 0.38 m/s
Answer:
t=0.0625s
Explanation:
F=number of swings/time taken
DATA
Frequency=4.0Hz
number of swings from Q to R
=1/4
time taken=?
Frequency=number of swings/time taken
make t the subject of the formula
t=n/f
substitute the given date
t=0.25/4.0
t=0.0625s
option A is collect
Answer:
1753246.75325 V/m
Explanation:
d = Distance of separation = 1.54 cm
V = Potential difference = 27 kV
When the voltage is divided by the distance between the plates we get the electric field.
Electric field is given by
The magnitude of the electric field in the region between the plates is 1753246.75325 V/m
Answer:did you get an answer ?
Explanation:
