Answer:
4.98 m
Explanation:
Given that
Width of the mirror, d = 0.6 m
Organist distance to the mirror, s = 0.78 m
Distance between the singer and the organist, S = 5.7 + 0.78 = 6.48 m
Width of north wall, D?
Using the simple relationship
D/S = d/s, on rearranging
D = dS /s
D = (0.6 * 6.48) / 0.78
D = 3.888 / 0.78
D = 4.98 m
Therefore, we can conclude that the Width of north wall is 4.98 m
Answer:
i can not read that sorry
Answer:
C) upward
Explanation:
The problem can be solved by using the right-hand rule.
First of all, we notice at the location of the negatively charged particle (above the wire), the magnetic field produced by the wire points out of the page (because the current is to the right, so by using the right hand, putting the thumb to the right (as the current) and wrapping the other fingers around it, we see that the direction of the field above the wire is out of the page).
Now we can apply the right hand rule to the charged particle:
- index finger: velocity of the particle, to the right
- middle finger: direction of the magnetic field, out of the page
- thumb: direction of the force, downward --> however, the charge is negative, so we must reverse the direction --> upward
Therefore, the direction of the magnetic force is upward.
Answer:
7.344 s
Explanation:
A = 0.15 x 0.3 m^2 = 0.045 m^2
N = 240
e = - 2.5 v
B1 = 0.1 T
B2 = 1.8 T
ΔB = B2 - B1 = 1.8 - 0.1 = 1.7 T
Δt = ?
e = - dФ/dt
e = - N x A x ΔB/Δt
- 2.5 = - 240 x 0.045 x 1.7 / Δt
2.5 = 18.36 / Δt
Δt = 7.344 s
Answer:
Resistance of the iron rod, R = 0.000077 ohms
Explanation:
It is given that,
Area of iron rod, 
Length of the rod, L = 35 cm = 0.35 m
Resistivity of Iron, 
We need to find the resistance of the iron rod. It is given by :



So, the resistance of the rod is 0.000077 ohms. Hence, this is the required solution.