Answer:
Pretty sure magnetic waves thank me later bro bro
Answer:
a = 3.27 m/s²
T = 275 N
Explanation:
Given that:
Mass m₁ = 42.p0 kg
Mass m₂ = 21.0 kg
Consider both masses to be in a whole system, then:
The acceleration can be determined as:

Making acceleration the subject in the above formula;




a = 3.27 m/s²
in the string, the tension is calculated using the formula:



T = 274.68 N
T ≅ 275 N
Answer:
Magnetic field at point having a distance of 2 cm from wire is 6.99 x 10⁻⁶ T
Explanation:
Magnetic field due to finite straight wire at a point perpendicular to the wire is given by the relation :
......(1)
Here I is current in the wire, L is the length of the wire, R is the distance of the point from the wire and μ₀ is vacuum permeability constant.
In this problem,
Current, I = 0.7 A
Length of wire, L = 0.62 m
Distance of point from wire, R = 2 cm = 2 x 10⁻² m = 0.02 m
Vacuum permeability, μ₀ = 4π x 10⁻⁷ H/m
Substitute these values in equation (1).

B = 6.99 x 10⁻⁶ T
Answer:
a)
v = 14.1028 m/s
∅ = 83.0765° north of east
b)
the required distance is 40.98 m
Explanation:
Given that;
velocity of the river u = 1.70 m/s
velocity of boat v = 14.0 m/s
Now to get the velocity of the boat relative to shore;
( north of east), we say
a² + b² = c²
(1.70)² + (14.0)² = c²
2.89 + 196 = c²
198.89 = c²
c = √198.89
c = 14.1028 m/s
tan∅ = v/u = 14 / 1.7 = 8.23529
∅ = tan⁻¹ ( 8.23529 ) = 83.0765° north of east
Therefore, the velocity of the boat relative to shore is;
v = 14.1028 m/s
∅ = 83.0765° north of east
b)
width of river = 340 m,
ow far downstream has the boat moved by the time it reaches the north shore in meters = ?
we say;
340sin( 90° - 83.0765°)
⇒ 340sin( 6.9235°)
= 40.98 m
Therefore, the required distance is 40.98 m