Answer: The temperature after another 5 minutes is 68.5°c
Explanation: Please see the attachments below
The given question is incomplete. The complete question is as follows.
In a nuclear physics experiment, a proton (mass kg, charge +e = C) is fired directly at a target nucleus of unknown charge. (You can treat both objects as point charges, and assume that the nucleus remains at rest.) When it is far from its target, the proton has speed m/s. The proton comes momentarily to rest at a distance m from the center of the target nucleus, then flies back in the direction from which it came. What is the electric potential energy of the proton and nucleus when they are m apart?
Explanation:
The given data is as follows.
Mass of proton = kg
Charge of proton =
Speed of proton =
Distance traveled =
We will calculate the electric potential energy of the proton and the nucleus by conservation of energy as follows.
=
where,
U =
Putting the given values into the above formula as follows.
U =
=
=
Therefore, we can conclude that the electric potential energy of the proton and nucleus is .
Use the Inverse square law, Intensity (I) of a light is inversely proportional to the square of the distance(d).
I=1/(d*d)
Let Intensity for lamp 1 is L1 distance be D1 so on, L2 D2 for Intensity for lamp 2 and its distance.
L1/L2=(D2*D2)/(D1*D1)
L1/15=(200*200)/(400*400)
L1=15*0.25
L1=3.75 <span>candela</span>
Answer:
The speed of the 270g cart after the collision is 0.68m/s
Explanation:
Mass of air track cart (m1) = 320g
Initial velocity (u1) = 1.25m/s
Mass of stationary cart (m2) = 270g
Velocity after collision (V) = m1u1/(m1+m2) = 320×1.25/(320+270) = 400/590 = 0.68m/s