Answer:
q = 2,95 10-6 C
Explanation:
The magnetic force on a particle is described by the equation
F = q v x B
Where bold indicate vectors
Let's make the vector product
vxB =
v x B = 1.20 106 [i ^ (4 0.130) - j ^ (3 0.130)]
vx B = 1.20 106 [0.52 i ^ - 0.39j ^]
As they give us the force module, let's use Pythagoras' theorem,
|v xB | =1.20 10⁶ √( 0.52² + 0.39²)
|v x B| = 1.20 10⁶ 0.65
v xB = 0.78 10⁶
Let's replace and calculate
2.30 = q 0.78 10⁶
q = 2.3 / 0.78 106
q = 2,95 10-6 C
Answer:
Weight required = 194.51 N
Explanation:
The elongation is given by
Length , L= 1.6 m
Diameter, d = 1.1 mm
Area
Change in length, ΔL = 2.8 mm = 0.0028 m
Young's modulus of copper, E = 117 GPa = 117 x 10⁹ Pa
Substituting,
Weight required = 194.51 N
The orbital radius is:
Explanation:
The problem is asking to find the radius of the orbit of a satellite around a planet, given the orbital speed of the satellite.
For a satellite in orbit around a planet, the gravitational force provides the required centripetal force to keep it in circular motion, therefore we can write:
where
G is the gravitational constant
M is the mass of the planet
m is the mass of the satellite
r is the radius of the orbit
v is the speed of the satellite
Re-arranging the equation, we find:
Learn more about circular motion:
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Answer:
0.0389 cm
Explanation:
The current density in a conductive wire is given by
where
I is the current
A is the cross-sectional area of the wire
In this problem, we know that:
- The fuse melts when the current density reaches a value of
- The maximum limit of the current in the wire must be
I = 0.62 A
Therefore, we can find the cross-sectional area that the wire should have:
We know that the cross-sectional area can be written as
where d is the diameter of the wire.
Re-arranging the equation, we find the diameter of the wire: