The direction of the magnetic force on the wire is west.
The magnetic force acting on the moving protons acts northward in the horizontal plane. If the thumb is up (current flows vertically up), the wrapped finger will be counterclockwise.
Therefore, the direction of the magnetic field is counterclockwise. Here, the magnetic field is pointing upwards (vertical magnetic field) and the electrons are moving east. Applying Fleming's left-hand rule here, we can see that the direction of force is along the south direction.
As the change in magnetic flux increases upwards, Lenz's law indicates that the induced magnetic field of the induced current must resist and the inside of the loop must be directed downwards. Using the right-hand rule, we can see that a clockwise current is induced.
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Answer:
129900
Explanation:
Given that
Mass of the particle, m = 1 g = 1*10^-3 kg
Speed of the particle, u = ½c
Speed of light, c = 3*10^8
To solve this, we will use the formula
p = ymu, where
y = √[1 - (u²/c²)]
Let's solve for y, first. We have
y = √[1 - (1.5*10^8²/3*10^8²)]
y = √(1 - ½²)
y = √(1 - ¼)
y = √0.75
y = 0.8660, using our newly gotten y, we use it to solve the final equation
p = ymu
p = 0.866 * 1*10^-3 * 1.5*10^8
p = 129900 kgm/s
thus, we have found that the momentum of the particle is 129900 kgm/s
Answer:
T = 676 N
Explanation:
Given that: f = 65 Hz, L = 2.0 m, and ρ = 5.0 g
= 0.005 kg
A stationary wave that is set up in the string has a frequency of;
f = 
⇒ T = 4
M
Where: t is the tension in the wire, L is the length of the wire, f is the frequency of the waves produced by the wire and M is the mass per unit length of the wire.
But M = L × ρ = (2 × 0.005) = 0.01 kg/m
T = 4 × 
×
× 0.01
= 4 × 4 ×4225 × 0.01
= 676 N
Tension of the wire is 676 N.
Answer:
-4,651.42N
Explanation:
Since we are not asked what to find, we can calculate the force applied by the track as shown;
F = ma
F = m(v-u)/t
Substitute the given value
F = 4400(7.6-52)/42
F = 4400(-44.4)/42
F = -4,651.43N
Hence the required force is -4,651.42N
