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Rina8888 [55]
2 years ago
15

The mass per unit length of a 14-gauge copper wire is 18.5 g/m. If the wire is placed running along the horizontal x-axis (east-

west) in a region in which there is a uniform 0.288-T magnetic field directed along the horizontal y-axis (northward). Calculate the magnitude of the minimum current in the wire that could allow the wire to levitate if the conventional current points to the east.
Physics
1 answer:
zmey [24]2 years ago
7 0

Answer:

0.6295 A

Explanation:

I=mg/BL put values in this formula.  

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Compare the components that make up the windsurfer, his board, and his surroundings.
frozen [14]

Answer:

Sample Response: The windsurfer, his board, and the air and water around him are all made of matter. That matter is made up of very small particles called atoms.

Explanation:

i just finished lesson on edgenuity :)

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2 years ago
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3) A defense football player on one team tackles the other team’s quarterback, who is running down the field. The quarterback is
alisha [4.7K]
Newton's first law says that an object at rest tends to stay at rest while an object in motion stays in motion at a constant velocity unless acted upon by an outside force so the amount of force behind the defensive football player (N) was greater than the quarterback's so he was able to over power him which is also called unbalanced forces
5 0
2 years ago
A sailboat starts from rest and accelerates at a rate of 0.21 m/s^2 over a distance of 280 m. find the magnitude of the boat's f
sasho [114]

We use the kinematic equations,

v=u+at                                          (A)

S= ut + \frac{1}{2} at^2                  (B)

Here, u is initial velocity, v is final velocity, a is acceleration and t is time.

Given,  u=0, a=0.21 \ m/s^2 and s= 280 m.

Substituting these values in equation (B), we get

280 \ m = 0 +\frac{1}{2} (0.21 m/s^2) t^2 \\\\ t^2 = \frac{280 \times 2}{0.21 } \\\\ t= 51.63 \ s.

Therefore from equation (A),

v = 0 + (0.21) \times (51.63 s)= 10.84 \ m/s

Thus, the magnitude of the boat's final velocity is 10.84 m/s and the time taken by boat to travel the distance 280 m is 51.63 s



8 0
2 years ago
A satellite, orbiting the earth at the equator at an altitude of 400 km, has an antenna that can be modeled as a 1.76-m-long rod
ivann1987 [24]

Answer:

The inducerd emf is 1.08 V

Solution:

As per the question:

Altitude of the satellite, H = 400 km

Length of the antenna, l = 1.76 m

Magnetic field, B = 8.0\times 10^{- 5}\ T

Now,

When a conducting rod moves in a uniform magnetic field linearly with velocity, v, then the potential difference due to its motion is given by:

e = - l(vec{v}\times \vec{B})

Here, velocity v is perpendicular to the rod

Thus

e = lvB           (1)

For the orbital velocity of the satellite at an altitude, H:

v = \sqrt{\frac{Gm_{E}}{R_{E}} + H}

where

G = Gravitational constant

m_{e} = 5.972\times 10^{24}\ kg = mass of earth

R_{E} = 6371\ km = radius of earth

v = \sqrt{\frac{6.67\times 10^{- 11}\times 5.972\times 10^{24}}{6371\times 1000 + 400\times 1000} = 7670.018\ m/s

Using this value value in eqn (1):

e = 1.76\times 7670.018\times 8.0\times 10^{- 5} = 1.08\ V

5 0
2 years ago
When a mass of 25 g is attached to a certain spring, it makes 20 complete vibrations in 4.0 s. what is the spring constant of th
earnstyle [38]

Answer: The spring  of the spring is 25 N/m.

Explanation:

Mass of the body = 25 g= 0.025 kg (1 kg = 1000 g)

Oscillation is 4 sec = 20

Oscillation in 1 sec =\frac{20}{4}=5

Frequency of the vibration of the spring = 5 s^{-1}=5 Hz

Force constant can be calculated bu using the relation between the frequency and, mass and spring constant 'k'

Frequency=\frac{1}{2\pi}\times \sqrt{\frac{k}{m}}

5 s^{-1}=\frac{1}{2\times 3.14}\times \sqrt{\frac{k}{0.025 kg}}

k=24.649 N/m\approx 25 N/m

The spring  of the spring is 25 N/m.

3 0
2 years ago
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