The orbital with the lowest energy is 3s.
The independent variable is what type of powder is used, the dependent variable is how long itching lasts and the constant is the amount of powder used (50 grams)
The recoil velocity of cannon is (4) 5.0 m/s
Explanation:
We can find the recoil velocity from the law of conservation of momentum.
The recoil velocity is velocity of body 2 after release of body 1, i.e. velocity of cannon after release of clown.
Let v2 be cannon's velocity, v1 be clown's velocity given = 15 m/sec
m1 be clown's mass = 100kg and m2 be cannon's mass given = 500kg.
So recoil velocity of cannon v2 is given by,
v2 = -(m1÷m2)v1
v2 = -(100÷500)15
v2 = -5 m/s
where the minus sign refers to the direction of cannon's recoil velocity being opposite to that of clown.
Hence, option (4)5.0 m/s is the correct answer.
Answer:
The kinetic energy of the proton at the end of the motion is 1.425 x 10⁻¹⁶ J.
Explanation:
Given;
initial velocity of proton, 
= 3 x 10⁵ m/s
distance moved by the proton, d = 3.5 m
electric field strength, E = 120 N/C
The kinetic energy of the proton at the end of the motion is calculated as follows.
Consider work-energy theorem;
W = ΔK.E
where;
K.Ef is the final kinetic energy
W is work done in moving the proton = F x d = (EQ) x d = EQd
Therefore, the kinetic energy of the proton at the end of the motion is 1.425 x 10⁻¹⁶ J.
