(a) On the axes below, sketch the graphs of the horizontal and vertical components of the sphere’s velocity as a function of tim
e between , when the sphere is launched and , when the sphere hits the target. Label for the horizontal component of the sphere’s velocity and the vertical component of the sphere’s velocity.
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:
Answer:
c.=0 and
Explanation:
We are given that two particles collide and stick together.
If there is no external force act on the two particles then ,it is inelastic collision.
Inelastic collision: There is some loss of kinetic energy but the momentum is conserved.
According to law of conservation of momentum
Initial momentum=Final momentum
Change in momentum=Final momentum-Initial momentum=0
Change in momentum=
Initial kinetic energy is greater than final kinetic energy.
Change in kinetic energy=Final kinetic energy-kinetic energy=- negative
Hence, option c is true.
c.=0 and
Answer:
Charge,
Explanation:
It is given that,
Electric field strength, E = 180000 N/C
Distance from a small object, r = 2.8 cm = 0.028 m
Electric field at a point is given by :
Q is the charge on an object
So, the charge on the object is . Hence, this is the required solution.