Answer:
The new radius is three times the initial radius
Explanation:
The magnetic force is given by the equation
F = q v x B
Where the blacks indicate vectors, q is the electric charge, v the velocity of the particle and B the magnetic field. The scalar magnitude of this force is
F = q v B sin θ
Where θ is the angle between the velocity and the magnetic field.
As the force is perpendicular to the velocity, the particle describes a circular motion, using Newton's second law we can find the acceleration that is centripetal.
F = ma
a = v² / r
q v B = mv² / r
R = mv / qB
Let's calculate for the new speed (v₂ = 3v)
R₂ = (m / qB) 3v
R₂ = 3 R
R₂ / R = 3
The new radius is three times the initial radius
The answer is C. 15 meters North
Displacement is the distance and direction from the origin. If you move 20 meters North, your displacement is 20 meters North. If you move 5 meters South, you are basically moving back the way you came by 5 meters, which means you are now 15 meters North.
I hope this helped! :)
The potential energy of the block is given by:
V = m*g*h
m mass
g = 9.81m/s²
h height
The potential energy of a spring is given by:
V = 0.5 * k * x²
k spring constant
x compression of the spring
If the block starts from rest it has potential energy, but no kinetic energy. As it slides down the incline potential energy is converted into kinetic energy. When the block hits the spring the kinetic energy is converted into spring's potential energy. If the spring is fully compressed and the block is at rest again, the block has transferred all its energy into the spring. No energy is lost. So we can write:
m * g * h = 0.5 * k * x²
m = 0.5 kg
g = 9.81 m/s²
h = 2.5m * sin 37° = 1,5 m
x = 0,6 m
Solve for k.
k = 2 * m * g * h / x² = 40.8 N/m
Answer:
D). 
Explanation:
As we know that temperature scale is linear so we will have
now we have
so the relation between two scales is given as
now we know that in kelvin scale the absolute temperature is 600 K
so now we have
so correct answer is
D).
Complete Question
An aluminum "12 gauge" wire has a diameter d of 0.205 centimeters. The resistivity ρ of aluminum is 2.75×10−8 ohm-meters. The electric field in the wire changes with time as E(t)=0.0004t2−0.0001t+0.0004 newtons per coulomb, where time is measured in seconds.
I = 1.2 A at time 5 secs.
Find the charge Q passing through a cross-section of the conductor between time 0 seconds and time 5 seconds.
Answer:
The charge is 
Explanation:
From the question we are told that
The diameter of the wire is 
The radius of the wire is 
The resistivity of aluminum is 
The electric field change is mathematically defied as
Generally the charge is mathematically represented as
Where A is the area which is mathematically represented as
So
Therefore
substituting values
![Q = 120 \int\limits^{t}_{0} { [ 0.0004t^2 - 0.0001t +0.0004] } \, dt](https://tex.z-dn.net/?f=Q%20%3D%20120%20%5Cint%5Climits%5E%7Bt%7D_%7B0%7D%20%7B%20%5B%200.0004t%5E2%20-%200.0001t%20%2B0.0004%5D%20%7D%20%5C%2C%20dt)
![Q = 120 [ \frac{0.0004t^3 }{3} - \frac{0.0001 t^2}{2} +0.0004t] } \left | t} \atop {0}} \right.](https://tex.z-dn.net/?f=Q%20%3D%20120%20%5B%20%5Cfrac%7B0.0004t%5E3%20%7D%7B3%7D%20-%20%5Cfrac%7B0.0001%20t%5E2%7D%7B2%7D%20%2B0.0004t%5D%20%7D%20%20%5Cleft%20%7C%20t%7D%20%5Catop%20%7B0%7D%7D%20%5Cright.)
From the question we are told that t = 5 sec
![Q = 120 [ \frac{0.0004t^3 }{3} - \frac{0.0001 t^2}{2} +0.0004t] } \left | 5} \atop {0}} \right.](https://tex.z-dn.net/?f=Q%20%3D%20120%20%5B%20%5Cfrac%7B0.0004t%5E3%20%7D%7B3%7D%20-%20%5Cfrac%7B0.0001%20t%5E2%7D%7B2%7D%20%2B0.0004t%5D%20%7D%20%20%5Cleft%20%7C%205%7D%20%5Catop%20%7B0%7D%7D%20%5Cright.)
![Q = 120 [ \frac{0.0004(5)^3 }{3} - \frac{0.0001 (5)^2}{2} +0.0004(5)] }](https://tex.z-dn.net/?f=Q%20%3D%20120%20%5B%20%5Cfrac%7B0.0004%285%29%5E3%20%7D%7B3%7D%20-%20%5Cfrac%7B0.0001%20%285%29%5E2%7D%7B2%7D%20%2B0.0004%285%29%5D%20%7D)
