Part a)
Magnitude of electric field is given by force per unit charge
Part b)
Electrostatic force on the proton is given as
F = qE
PART C)
Gravitational force is given by
PART d)
Ratio of electric force to weight
I NEED THIS ANSWER ASAP PLEASE!!! What happens when a light ray passes through the focal point, and then reflects off of a conve
1 answer:
You might be interested in
Answer:
The final translational seed at the bottom of the ramp is approximately 4.84 m/s
Explanation:
The given parameters are;
The radius of the sphere, R = 2.50 cm
The mass of the sphere, m = 4.75 kg
The translational speed at the top of the inclined plane, v = 3.00 m/s
The length of the inclined plane, l = 2.75 m
The angle at which the plane is tilted, θ = 22.0°
We have;
+
=
+
K = (1/2)×m×v²×(1 + I/(m·r²))
I = (2/5)·m·r²
K = (1/2)×m×v²×(1 + 2/5) = 7/10 × m×v²
U = m·g·h
h = l×sin(θ)
h = 2.75×sin(22.0°)
∴ 7/10×4.75×3.00² + 4.75×9.81×2.75×sin(22.0°) = 7/10 × 4.75ײ + 0
7/10×4.75×3.00² + 4.75×9.81×2.75×sin(22.0°) ≈ 77.93
∴ 77.93 ≈ 7/10 × 4.75ײ
² = 77.93/(7/10 × 4.75)
≈ √(77.93/(7/10 × 4.75)) ≈ 4.84
The final translational seed at the bottom of the ramp, ≈ 4.84 m/s.
Answer:
A. α = - 1.047 rad/s²
B. θ = 14.1 rad
C. θ = 2.24 rev
Explanation:
A.
We can use the first equation of motion to find the acceleration:
where,
ωf = final angular speed = 0 rad/s
ωi = initial angular speed = (30 rpm)(2π rad/1 rev)(1 min/60 s) = 3.14 rad/s
t = time = 3 s
α = angular acceleration = ?
Therefore,
<u>α = - 1.047 rad/s²</u>
B.
We can use the second equation of motion to find the angular distance:
<u>θ = 14.1 rad</u>
C.
θ = (14.1 rad)(1 rev/2π rad)
<u>θ = 2.24 rev</u>