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DIA [1.3K]
3 years ago
7

A vertical wire carries a current vertically upward in a region where the magnetic field vector points toward the north. What is

the direction of the magnetic force on this current due to the field?
Physics
1 answer:
KengaRu [80]3 years ago
5 0

Answer:

West

Explanation:

We can solve the problem by using the right-hand rule. We can apply the rule as follows:

- Index finger: direction of the current

- middle finger: direction of the magnetic field

- thumb: direction of the force exerted on the wire

By applying the rule to this situation, we have:

- index finger: upward (current)

- middle finger: north (magnetic field)

- thumb: west (force)

So, the direction of the force exerted on the wire is to the west.

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3 years ago
Wha is the movement of the balloon when brought near to the can after removing your hand?​
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3 years ago
A cardinal (Richmondena cardinalis) of mass 4.00×10−2 kg and a baseball of mass 0.146 kg have the same kinetic energy. What is t
sp2606 [1]

Answer:

0.5232

Explanation:

the cardinal and the baseball has the same kinetic energy

\frac{1}{2} m_{c} v_{c} ^{2}  = \frac{1}{2} m_{b} v_{b} ^{2}

m_{c}v_{c}^{2} = m_{b}v_{b}^{2}

[4.0X 10^{-2} ]v_{c} ^{2} = [0.146]v_{b} ^{2}

\frac{v_{c} }{v_{b} } = \sqrt{\frac{0.146}{4.0 X 10^{-2} } }

\frac{v_{c} }{v_{b} } = 1.910

Ratio of momentum

\frac{p_{c}}{p_{b}}  = \frac{m_{c}v_{c}}{m_{b}v_{b}} = \frac{ (4.0 X 10^{-2} )(1.910v_{b} )}{0.146v_{b} }

\frac{p_{c}}{p_{b}} = \frac{4.0 X  10 ^{-2} X 1.910 }{0.146}  = 0.5232

6 0
4 years ago
After a big snowfall, you take your favorite rocket‑powered sled out to a wide field. The field is 223 m223 m across, and you kn
pogonyaev

Answer:

11.7 s

Explanation:

In this problem, the rocket is moving in a uniform accelerated motion. We have the following data:

d = 223 m, the distance that the sled has to cover

a=3.25 m/s^2, the acceleration of the rocket

We can use therefore the following SUVAT equation:

d=ut+\frac{1}{2}at^2

where

d is the distance

u = 0 is the initial velocity of the sled (it starts from rest)

t is the time

a is the acceleration

Re-arranging the equation and substituting the numbers, we find the time it takes for the rocket to cross the field:

d=\frac{1}{2}at^2\\t=\sqrt{\frac{2d}{a}}=\sqrt{\frac{2(223)}{3.25}}=11.7 s

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3 years ago
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