Two children push on opposite sides of a door during play. Both push horizontally and perpendicular to the door. One child pushe
1 answer:
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Answer:
d) 1.2 mT
Explanation:
Here we want to find the magnitude of the magnetic field at a distance of 2.5 mm from the axis of the coaxial cable.
First of all, we observe that:
- The internal cylindrical conductor of radius 2 mm can be treated as a conductive wire placed at the axis of the cable, since here we are analyzing the field outside the radius of the conductor. The current flowing in this conductor is
I = 15 A
- The external conductor, of radius between 3 mm and 3.5 mm, does not contribute to the field at r = 2.5 mm, since 2.5 mm is situated before the inner shell of the conductor (at 3 mm).
Therefore, the net magnetic field is just given by the internal conductor. The magnetic field produced by a wire is given by
where
is the vacuum permeability
I = 15 A is the current in the conductor
r = 2.5 mm = 0.0025 m is the distance from the axis at which we want to calculate the field
Substituting, we find:
The speed is 0.956 m / s.
<u>Explanation</u>:
The kinetic energy is equal to the product of half of an object's mass, and the square of the velocity.
K.E = 1/2 m
where K.E represents the kinetic energy,
m represents the mass,
v represents the velocity.
K.E = 1/2 m
1.10 10^42 = 1/2
3.26
10^31
= (1.10
10^42
2) / (3.26
10^31)
v = 0.956 m / s.
Answer:
d= 7.32 mm
Explanation:
Given that
E= 110 GPa
σ = 240 MPa
P= 6640 N
L= 370 mm
ΔL = 0.53
Area A= πr²
We know that elongation due to load given as
A= 42.14 mm²
πr² = 42.14 mm²
r=3.66 mm
diameter ,d= 2r
d= 7.32 mm