Answer:

Explanation:
We are given that
Current in wire=40 A
Magnetic field=
T( vertically downward)
We have to find the resultant magnitude of the magnetic field 29 cm above the wire and 29 cm below the wire.
According to Bio-Savart law, the magnetic field exerted by the wire at distance R is given by

We have R=29 cm=
1 m=100 cm
Substitute the values in the given formula

The resultant magnetic field is given by

Substitute the values then we get


The resultant magnitude of magnetic field is same above and below the wire as it is at same distance.
The resultant magnitude of the magnetic field 29 cm below the wire=
Hence, the resultant magnitude of the magnetic field 29 cm above the wire=
Answer:
1800/300 = 6ropes
Explanation:
The engine weighs 1800N and the person exerts a force of 300N, so for him to lift the engine and exerting a force of 300N all through we divide the weight of the engine by the force exerted to know how many ropes are used. Which makes it 6 thereby each rope uses 300N to lift the engine.
Given Information:
Angular displacement = θ = 51 cm = 0.51 m
Radius = 1.8 cm = 0.018 m
Initial angular velocity = ω₁ = 0 m/s
Angular acceleration = α = 10 rad/s
²
Required Information:
Final angular velocity = ω₂ = ?
Answer:
Final angular velocity = ω₂ = 21.6 rad/s
Explanation:
We know from the equations of kinematics,
ω₂² = ω₁² + 2αθ
Where ω₁ is the initial angular velocity that is zero since the toy was initially at rest, α is angular acceleration and θ is angular displacement.
ω₂² = (0)² + 2αθ
ω₂² = 2αθ
ω₂ = √(2αθ)
We know that the relation between angular displacement and arc length is given by
s = rθ
θ = s/r
θ = 0.51/0.018
θ = 23.33 radians
finally, final angular velocity is
ω₂ = √(2αθ)
ω₂ = √(2*10*23.33)
ω₂ = 21.6 rad/s
Therefore, the top will be rotating at 21.6 rad/s when the string is completely unwound.
Huh??????????????????????????
Most of the problem depends on which object you observe. For the speed, take the absolute value of the derivative of the polynomial interpolation of position verses time.