Answer:
0.0979 N/c
Explanation:
Electric field, E is given as a product of resistivity and current density
E=jP where P is resistivity and j is current density
But the current density is given as
where I is current and A is area and
Substituting this into the first equation then
Given diameter of 0.259 cm= 0.00259 m and the radius will be half of it which is 0.001295 m
Answer:
40 N/m
Explanation:
The diagram attached is used to answer the question
We know from Hooke's law that extension is directly proportional to the applied force hence
F=kx where x is extension, F is applied force and k is the spring constant. Making k the subject of the formula then
From the attached diagram extension is given by subtracting unstretched spring from stretched spring hence extension, x=1-0.5=0.5m
Substituting 20 N for F and 0.5 m for x then
Answer:
At t = 15.0 s the magnitude of the velocity is 58.31 m/s
Explanation:
The given parameters are;
V₀ₓ = 30 m/s
= 100 m/s
The time of flight of the projectile = 25 s
For projectile motion;
Vₓ = V₀ × cos(θ₀)
The magnitude of the velocity V = √(V₀ₓ² + ²)
We have the magnitude of the initial velocity = √(30² + 100²) = 10·√109 m/s
cos(θ₀) = V₀ₓ/V₀ = 30/(10·√109) = 3/√109
θ₀ = cos⁻¹(3/√109) = 73.3°
The components of the velocity after time t is given by the relations;
Vₓ = V₀ × cos(θ₀) = 30 m/s
= V₀ × sin(θ₀) - g×t
When = 0, we have;
0 = V₀ × sin(θ₀) - g×t
g×t = V₀ × sin(θ₀) = 10·√109×0.958 = 100 m/s
t = 100/g = 100/10 = 10 s
The time to reach maximum height = 10 s
At 15.0 seconds, we have;
= V₀ × sin(θ₀) - g×t = 10·√109×0.958 - 10×15 = -50 m/s
Therefore, the projectile is returning at 50 m/s
The magnitude of the velocity =√(30² + 50²) = 10·√34 m/s = 58.31 m/s.
A) Upward
In order to find the direction of the magnetic force on the wire, we can use the right-hand rule: the index finger, the middle finger and the thumb of the right hand must be placed all of them perpendicular to each other.
So we have:
- Index finger: direction of current in the wire (from west to east)
- Middle finger: direction of magnetic field (from south to north)
- Thumb: direction of the force --> so it will be upward
So, the force will point upward.
B)
The magnitude of the force exerted by the magnetic field on the wire is given by
where
I = 1.50 A is the current in the wire
L = 2.20 m is the length of the wire
Substituting into the equation, we find