The impulse was greater in the first experiment because the egg broke.
<h3>What is impulse?</h3>
The term impulse is defined a the product of the force and time. We know that the impulse is high when a large force acts for a short time.
From the experiment if the students, we can conclude that the impulse was greater in the first experiment because the egg broke.
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Answer:
<em>The distance covered by comet is </em>
Explanation:
Speed is defined as the rate of change of distance with time. It is given by the equation speed= 
Thus distance= 
In this problem it is given that speed of comet= 
time travelled by the comet= 4 hours
Thus distance= 
= 
= 
Answer:
The frequency is
Explanation:
From the question we are told that
The frequency of the tuning fork is 
The beat period is 
Generally the beat frequency is mathematically represented as


The beat frequency is also represented mathematically as

Where
is the frequency of the piano
So
Answer:

Explanation:
A 6.0-cm-diameter parallel-plate capacitor has a 0.46 mm gap.
What is the displacement current in the capacitor if the potential difference across the capacitor is increasing at 500,000V/s?
Let given is,
The diameter of a parallel plate capacitor is 6 cm or 0.06 m
Separation between plates, d = 0.046 mm
The potential difference across the capacitor is increasing at 500,000 V/s
We need to find the displacement current in the capacitor. Capacitance for parallel plate capacitor is given by :
, r is radius
Let I is the displacement current. It is given by :

Here,
is rate of increasing potential difference
So

So, the value of displacement current is
.
Answer:
The new distance is d = 0.447 d₀
Explanation:
The electric out is given by Coulomb's Law
F = k q₁ q₂ / r²
This electric force is in balance with tension.
We reduce the charge of sphere B to 1/5 of its initial value (
=q₂ = q₂ / 5) than new distance (d = n d₀)
dat
q₁ = 
q₂ = 
r = d₀
In order for the deviation to maintain the electric force it should not change, so we apply the Coulomb equation for the two points
F = k q₁ q₂ / d₀²
F = k q₁ (q₂ / 5) / (n d₀)²
.k q₁ q₂ / d₀² = q₁ q₂ / (5 n² d₀²)
5 n² = 1
n = √ 1/5
n = 0.447
The new distance is
d = 0.447 d₀