Answer:
The flux is 682.6 Wb.
Explanation:
Given that,
Vector field 
We need to calculate the flux
Using formula of flux

Put the value into the formula




Hence, The flux is 682.6 Wb.
Answer:
m = mass of the penny
r = distance of the penny from the center of the turntable or axis of rotation
w = angular speed of rotation of turntable
F = centripetal force experienced by the penny
centripetal force "F" experienced by the penny of "m" at distance "r" from axis of rotation is given as
F = m r w²
in the above equation , mass of penny "m" and angular speed "w" of the turntable is same at all places. hence the centripetal force directly depends on the radius .
hence greater the distance from center , greater will be the centripetal force to remain in place.
So at the edge of the turntable , the penny experiences largest centripetal force to remain in place.
Explanation:
Answer:
5 mg, 
Explanation:
First of all, let's rewrite the mass in grams using scientific notation.
we have:
m = 0.005 g
To rewrite it in scientific notation, we must count by how many digits we have to move the dot on the right - in this case three. So in scientific notation is

If we want to convert into milligrams, we must remind that
1 g = 1000 mg
So we can use the proportion

and we find

Answer:
Explanation:
Impulse = change in momentum
mv - mu , v and u are final and initial velocity during impact at surface
For downward motion of baseball
v² = u² + 2gh₁
= 2 x 9.8 x 2.25
v = 6.64 m / s
It becomes initial velocity during impact .
For body going upwards
v² = u² - 2gh₂
u² = 2 x 9.8 x 1.38
u = 5.2 m / s
This becomes final velocity after impact
change in momentum
m ( final velocity - initial velocity )
.49 ( 5.2 - 6.64 )
= .7056 N.s.
Impulse by floor in upward direction
= .7056 N.s
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:
