Net Force = mass x acceleration
3500=1,000a
So a= 3500/1000
a=35/10
a=3.5 m/s^2
Question:
A wire 2.80 m in length carries a current of 5.20 A in a region where a uniform magnetic field has a magnitude of 0.430 T. Calculate the magnitude of the magnetic force on the wire assuming the following angles between the magnetic field and the current.
(a)60 (b)90 (c)120
Answer:
(a)5.42 N (b)6.26 N (c)5.42 N
Explanation:
From the question
Length of wire (L) = 2.80 m
Current in wire (I) = 5.20 A
Magnetic field (B) = 0.430 T
Angle are different in each part.
The magnetic force is given by

So from data

Now sub parts
(a)

(b)

(c)

Answer:
part A ⇒ u = 1.28 m
part B ⇒v = 0.43 m
Explanation:
for u is the distance to the object from the mirror and v is the distance from the mirror to the image.
Part A:
the mirror equation is given by:
1/f = 1/v + 1/u
but we told that, v = 1/3u:
1/f = 3/v + 1/u = 4/u
1/f = 4/u
f = u/4
u = 4f
= 4×(32×10^-2)
= 1.28 m
Therefore the distance from the mirror to the object is 1.28 m.
part B:
v = 1/3×u = 1/3×(1.28) = 0.43 m
The Ohm's law is I(the strength of the current flowing in a conductor)= V(the potential difference applied to the ends) divided by R(resistance)