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valentina_108 [34]
2 years ago
15

A thin copper rod 1.0 m long has a mass of 0.050 kg and is in a magnetic field of 0.10 t. What minimum current in the rod is nee

ded in order for the magnetic force to balance the weight of the rod?
Physics
1 answer:
slamgirl [31]2 years ago
8 0

Answer:

i = 4.9 A

Explanation:

Force on a current carrying rod due to magnetic field is given as

F = iLB

here we know that

i =current in the rod

B = 0.10 T

L = 1.0 m

now magnetic force is balanced by the weight of the rod

so we will have

iLB = mg

i(1.0)(0.10) = 0.05 \times 9.8

i = 4.9 A

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Explanation:

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2 years ago
An electron starts from rest 3.00 cm from the center of a uniformly charged sphere of radius 2.00 cm. if the sphere carries a to
Sidana [21]
Answer:
velocity = 7.26 * 10^6 m/sec

Explanation:
The rule that is used to solve this problem is shown in the attached image.
The variables are as follows:
k = 8.99 * 10^9 Nm^2 / C^2
e is the electron charge = -1.6 * 10^-19 C
q is the charge given = 1 * 10^-9 C
m is the mass of the electron = 9.11 * 10^-31
r1 is the radius of starting point = 3 cm = 0.03 m
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Hope this helps :)

7 0
2 years ago
A slender rod is 80.0 cm long and has mass 0.370 kg . A small 0.0200-kg sphere is welded to one end of the rod, and a small 0.05
nataly862011 [7]

Answer:

1.10 m/s

Explanation:

Linear speed is given by

v=r\omega

Kinetic energy is given by

KE=0.5I\omega^{2}

Potential energy

PE= mgh

From the law of conservation of energy, KE=PE hence

0.5I\omega^{2}=mgh where m is mass, I is moment of inertia, \omega is angular velocity, g is acceleration due to gravity and h is height

Substituting m2-m1 for m and 0.5l for h, \frac {2v}{L} for \omega we obtain

0.5I(\frac {2v}{L})^{2}=0.5Lg(m2-m1)

(\frac {2v}{L})^{2}=\frac {gl(m2-m1)}{I} and making v the subject

v^{2}=\frac {gl^{3}(m2-m1)}{4I}

v=\sqrt {\frac {gl^{3}(m2-m1)}{4I}}

For the rod, moment of inertia I=\frac {ML^{2}}{12} and for sphere I=MR^{2} hence substituting 0.5L for R then I=M(0.5L)^{2}

For the sphere on the left hand side, moment of inertia I

I=m1(0.5L)^{2} while for the sphere on right hand side, I=m2(0.5L)^{2}

The total moment of inertia is therefore given by adding

I=\frac {ML^{2}}{12}+ m1(0.5L)^{2}+ m2(0.5L)^{2}=\frac {L^{2}(M+3m1+3m2)}{12}

Substituting \frac {L^{2}(M+3m1+3m2)}{12} for I in the equation v=\sqrt {\frac {gL^{3}(m2-m1)}{4I}}

Then we obtain

v=\sqrt {\frac {gL^{3}(m2-m1)}{4(\frac {L^{2}(M+3m1+3m2)}{12})}}=\sqrt {\frac {3gL^{3}(m2-m1)}{L^{2}(M+3m1+3m2)}}

This is the expression of linear speed. Substituting values given we get

v=\sqrt {\frac {3*9.81*0.8^{3}(0.05-0.02)}{0.8^{2}(0.39+3(0.02)+3(0.05))}} \approx 1.08 m/s

8 0
2 years ago
A particle at 9 AM is moving towards the east at 4 ms At 12 noon, it changes its velocity and starts moving towards the north un
nexus9112 [7]

Answer:

Explanation:

Acceleration is the time rate of change of velocity.

Acceleration and velocity are vectors

If east and north are the positive directions, the east moving vector is reduced to zero and the north moving vector increases from zero to 4 m/s.

There are 3 hours or 10800 seconds between 10 AM and 1 PM

a1 = √((-4)² + 4²) / 10800 = (√32) / 10800 m/s² ≈ 4.2 x 10⁻⁴ m/s²

There are 14400 seconds between 10 AM and 2 PM

The velocity changes are still the same

a2 = √((-4)² + 4²) / 10800 = (√32) / 14400 m/s² ≈ 3.9 x 10⁻⁴ m/s²

7 0
2 years ago
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Artyom0805 [142]

Answer:

75 m

Explanation:

The horizontal motion of the projectile is a uniform motion with constant speed, since there are no forces acting along the horizontal direction (if we neglect air resistance), so the horizontal acceleration is zero.

The horizontal component of the velocity of the projectile is

v_x = 15 m/s

and it is constant during the motion;

the total time of flight is

t = 5 s

Therefore, we can apply the formula of the uniform motion to find the horizontal displacement of the projectile:

d= v_x t =(15 m/s)(5 s)=75 m

5 0
2 years ago
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