This would be a physical change because it can change back to its original form. This is like ripping paper. You can piece it back together and it still is paper.
The opposite of this is chemical change. Chemical change means the product has been changed completely like burning paper. The paper has now been turned to ash and it's impossible to change this back to its original form.
The magnitude of the source charge is 3 μC which generates 4286 N/C of the electric field. Option B is correct.
What does Gauss Law state?
It states that the electric flux across any closed surface is directly proportional to the net electric charge enclosed by the surface.
Where,
= electric force = 4286 N/C
= Coulomb constant = 
= charges = ?
= distance of separation = 2.5 m
Put the values in the formula,
Therefore, the magnitude of the source charge is 3 μC.
Learn more about Gauss's law:
brainly.com/question/1249602
Given the following in the problem:
Distances : 2.0 m and 4.0 m
Sound waves : 1700 hz
Speed of sound : 340 m/s
Get the wavelength of the sound by using the formula:
Lambda = speed of sound/sound waves
Lambda = 340 m/s / 1700 hz
Lambda = 0.2
Get the path length difference to the point from the two speakers
L1 = 4mL2 = sqrt (42+ 22) m
Delta = 4.47
x = delta / lambda
If the outcome is nearly an integer, the waves strengthen at the point. If it is nearly an integer +0.5 the waves interfere destructively at the point. If it is neither the point is somewhat in in the middle.
Solving x = (4.47 – 4) / (0.2) = 2.35 an integer +0.5 so it’s a point of destructive interference.
Answer:
Explanation:
As we know that tension force in the string will be equal to the centripetal force on the string
so we will have
now we have
now we have
now when string length is 0.896 m and its speed is 71.5 m/s then we will have
Answer:
Explanation:
Given that,
Mass of ball m = 2kg
Ball traveling a radius of r1= 1m.
Speed of ball is Vb = 2m/s
Attached cord pulled down at a speed of Vr = 0.5m/s
Final speed V = 4m/s
Let find the transverse component of the final speed using
V² = Vr²+ Vθ²
4² = 0.5² + Vθ²
Vθ² = 4²—0.5²
Vθ² = 15.75
Vθ =√15.75
Vθ = 3.97 m/s.
Using the conservation of angular momentum,
(HA)1 = (HA)2
Mb • Vb • r1 = Mb • Vθ • r2
Mb cancels out
Vb • r1 = Vθ • r2
2 × 1 = 3.97 × r2
r2 = 2/3.97
r2 = 0.504m
The distance r2 to the hole for the ball to reach a maximum speed of 4m/s is 0.504m
The required time,
Using equation of motion
V = ∆r/t
Then,
t = ∆r/Vr
t = (r1—r2) / Vr
t = (1—0.504) / 0.5
t = 0.496/0.5
t = 0.992 second
