Given
m1 = 15kg
vi1 = +20 m/s
vf1 = +5 m/s
m2 = 10 kg
vi2= -15 m/s
vf2 = ?
Procedure
Conservation of momentum, general law of physics according to which the quantity called momentum that characterizes motion never changes in an isolated collection of objects; that is, the total momentum of a system remains constant.
The speed of the 10kg Ball is 7.5m/s
Hey!
Given ,
Acceleration, a = 5 m/s^2
time , t = 10 seconds
Initial velocity,u = 0
Final velocity , v = ?
We have ,
v = u + at
=> v = (0)+(5)(10)
=> v = 50 m/s
The answer should be D. hypothesis
Answer: The total resistance in a parallel circuit is always less than any of the branch resistances. Adding more parallel resistances to the paths causes the total resistance in the circuit to decrease. As you add more and more branches to the circuit the total current will increase because Ohm's Law states that the lower the resistance, the higher the current.
Explanation:
(a) The proton’s potential energy change is 3.6 x 10⁻¹⁸ J.
(b) The potential difference between the negative plate and a point midway between the plates is 11.25 V.
(c) The speed of the proton just before it hits the negative plate is 6.57 x 10⁴ m/s.
<h3>
Potential energy of the proton</h3>
U = qΔV
where;
- q is charge of the proton
- ΔV is potential difference
U = q(Ed)
U = (1.6 x 10⁻¹⁹)(1500 x 1.5 x 10⁻²)
U = 3.6 x 10⁻¹⁸ J
<h3>Potential difference between the negative plate and a point midway</h3>
ΔV = E(0.5d)
ΔV = 0.5Ed
ΔV = 0.5 (1500)(1.5 x 10⁻²)
ΔV = 11.25 V
<h3>Speed of the proton </h3>
U = ¹/₂mv²
U = mv²
v² = 2U/m
where;
- m is mass of proton = 1.67 x 10⁻²⁷ kg
v² = (2 x 3.6 x 10⁻¹⁸) / ( 1.67 x 10⁻²⁷)
v² = 4.311 x 10⁹
v = √(4.311 x 10⁹)
v = 6.57 x 10⁴ m/s
Thus, the proton’s potential energy change is 3.6 x 10⁻¹⁸ J.
The potential difference between the negative plate and a point midway between the plates is 11.25 V.
The speed of the proton just before it hits the negative plate is 6.57 x 10⁴ m/s.
Learn more about potential difference here: brainly.com/question/24142403
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