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Ainat [17]
1 year ago
14

A conductor carrying a current I = 15.0 A is directed along the positive x axis and perpendicular to a uniform magnetic field. A

magnetic force per unit length of 0.120 N/m acts on the conductor in the negative y direction. Deter- mine (a) the magnitude
Physics
1 answer:
slavikrds [6]1 year ago
3 0

The magnitude of the magnetic field in the region through the current passes is 8×10⁻³  T.

<h3>What is a magnetic field?</h3>

The magnetic field is defined as the field the magnetic materials generate or when an electric charge moves in a field region that generates the magnetic field.

As we know:

F = ilB

F is the magnetic force

i is the current

l is the length

B is the magnitude of the magnetic field

Here F/l = 0.120 N/m

i = 15 A

B = (F/l)(1\i)

B = (0.120)(1/15)

B = 0.120/15

B = 8×10⁻³  T

Thus, the magnitude of the magnetic field in the region through the current passes is 8×10⁻³  T.

Learn more about the magnetic field here:

brainly.com/question/14848188

#SPJ6

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Answer:

A screw.

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What is the acceleration of a proton moving with a speed of 7.0 m/s at right angles to a magnetic field of 1.7 t ?
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We know that a charge moving in a magnetic field is subject to the force:
F = q · v · B

But we also know that:
F = m · a

Therefore, it must be:
m · a = <span>q · v · B

And solving for a:
</span>a = <span>q · v · B / m

Recall that for a proton:
q = 1.6</span>×10⁻¹⁹ C
m = 1.673×10⁻²⁷ kg

Now, you can find:
a = 1.6×10⁻¹⁹ · 7.0 · 1.7 / <span>1.673×10⁻²⁷
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8 0
3 years ago
A fighter plane flying at constant speed 420 m/s and constant altitude 3300 m makes a turn of curvature radius 11000 m. On the g
Arada [10]

Answer:

"Apparent weight during the "plan's turn" is  519.4 N

Explanation:

The "plane’s altitude" is not so important, but the fact that it is constant tells us that the plane moves in a "horizontal plane" and its "normal acceleration" is \mathrm{a}_{\mathrm{n}}=\frac{v^{2}}{R}

Given that,

v = 420 m/s

R = 11000 m

Substitute the values in the above equation,

a_{n}=\frac{420^{2}}{11000}

a_{n}=\frac{176400}{11000}

a_{n}=16.03 \mathrm{m} / \mathrm{s}^{2}

It has a horizontal direction. Furthermore, constant speed implies zero tangential acceleration, hence vector a = vector a N. The "apparent weight" of the pilot adds his "true weight" "m" "vector" "g" and the "inertial force""-m" vector a due to plane’s acceleration, vectorW_{\mathrm{app}}=m(\text { vector } g \text { -vector a })

In magnitude,

| \text { vector } g-\text { vector } a |=\sqrt{\left(g^{2}+a^{2}\right)}

| \text { vector } \mathrm{g}-\text { vector } \mathrm{a} |=\sqrt{\left(9.8^{2}+16.03^{2}\right)}

| \text { vector } \mathrm{g}-\text { vector } \mathrm{a} |=\sqrt{(96.04+256.96)}

| \text { vector } \mathrm{g}-\text { vector } \mathrm{a} |=\sqrt{353}

| \text { vector } \mathrm{g}-\text { vector } \mathrm{a} |=18.78 \mathrm{m} / \mathrm{s}^{2}

Because vector “a” is horizontal while vector g is vertical. Consequently, the pilot’s apparent weight is vector

\mathrm{W}_{\mathrm{app}}=(18.78 \mathrm{m} / \mathrm{s}^ 2)(53 \mathrm{kg})=995.77 \mathrm{N}

Which is quite heavier than his/her true weigh of 519.4 N

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Calculate the force between two objects that have masses of 70 kilograms and 2,000 kilograms separated by a distance of 1 meter.
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▪▪▪▪▪▪▪▪▪▪▪▪▪  {\huge\mathfrak{Answer}}▪▪▪▪▪▪▪▪▪▪▪▪▪▪

The Gravitational Force between given objects will be ~

  • 9.34 \times  {10}^{ - 6}  \:  \: N

\large \boxed{ \mathfrak{Step\:\: By\:\:Step\:\:Explanation}}

We know that ~

\huge\boxed{\mathrm{F = \dfrac{ Gm_1m_2}{ r²}}}

where ~

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Let's calculate the force ~

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