Answer:
-833.3 N/C
Explanation:
Kinetic energy, K, in terms of electric field, E, is given as:
K = qEr
q = charge = e = 1.6 × 10⁻¹⁹C
E = Electric field
r = distance = 0.3m
The electric field can be gotten by making E subject of formula:
E = K/(qr)
The electeic field needed to stop the electrons must be equal in magnitude to the electric field carried by these electrons:
E = (4.0 × 10⁻¹⁷)/(-1.6 × 10⁻¹⁹ * 0.3)
E = -833.3 N/C
This is the electric field needed to stop the electrons.
The negative sign means that the electric field must be in a direction opposite to the motion of the electrons.
Answer:
Explanation:
Moment of inertia of the metal rod pivoted in the middle
= M l² / 12
If the spring is compressed by small distance x twisting the rod by angle θ
restoring force by spring
= k x
moment of torque about axis
= k x l /2
= k θ( l /2 )² ( x / .5 l = θ )
=
moment of torque = moment of inertia of rod x angular acceleration
k θ( l /2 )² = M l² / 12 d²θ/dt²
d²θ/dt² = 3 k/M θ
acceleration = ω² θ
ω² = 3 k/M
ω = √ 3 k / M
Initially, the velocity vector is
. At the same height, the x-value of the vector will be the same, and the y-value will be opposite (assuming no air resistance). Assuming perfect reflection off the ground, the velocity vector is the same. After 0.2 seconds at 9.8 seconds, the y-value has decreased by
, so the velocity is
.
Converting back to direction and magnitude, we get ![\langle r,\theta \rangle=\langle \sqrt{29.11^2+24.23^2},tan^{-1}(\frac{29.11}{24.23}) \rangle = \langle 37.87,50.2^{\circ}\rangle](https://tex.z-dn.net/?f=%5Clangle%20r%2C%5Ctheta%20%5Crangle%3D%5Clangle%20%5Csqrt%7B29.11%5E2%2B24.23%5E2%7D%2Ctan%5E%7B-1%7D%28%5Cfrac%7B29.11%7D%7B24.23%7D%29%20%5Crangle%20%3D%20%5Clangle%2037.87%2C50.2%5E%7B%5Ccirc%7D%5Crangle)
Answer:
A machine in which work input equals work output. energy can be used to do work, work can be used to transfer energy. The change in the kinetic energy of an object is equal to the net work done on the object.
hope this helps
Answer:
8050 J
Explanation:
Given:
r = 4.6 m
I = 200 kg m²
F = 26.0 N
t = 15.0 s
First, find the angular acceleration.
∑τ = Iα
Fr = Iα
α = Fr / I
α = (26.0 N) (4.6 m) / (200 kg m²)
α = 0.598 rad/s²
Now you can find the final angular velocity, then use that to find the rotational energy:
ω = αt
ω = (0.598 rad/s²) (15.0 s)
ω = 8.97 rad/s
W = ½ I ω²
W = ½ (200 kg m²) (8.97 rad/s)²
W = 8050 J
Or you can find the angular displacement and find the work done that way:
θ = θ₀ + ω₀ t + ½ αt²
θ = ½ (0.598 rad/s²) (15.0 s)²
θ = 67.3 rad
W = τθ
W = Frθ
W = (26.0 N) (4.6 m) (67.3 rad)
W = 8050 J