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3 years ago

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A wire carries a steady current of 2.60 A. A straight section of the wire is 0.750 m long and lies along the x axis within a uni

form magnetic field, vector B = 1.50 k T. If the current is in the positive x direction, what is the magnetic force on the section of wire?

1 answer:

ad-work [718]3 years ago

7 0

Answer:

The magnetic force on the section of wire is $-2.925\hat{j}\ N$ .

Explanation:

Given that,

Current $I = 2.60\hat{i}\ A$

Length = 0.750 m

Magnetic field $B = 1.50\hat{k}\ T$

We need to calculate the magnetic force on the section of wire

Using formula of magnetic force

$\vec{F}=l\vec{I}\times\vec{B}$

$\vec{F}=0.750\times2.60\hat{i}\times1.50\hat{k}$

Since, $\hat{i}\times\hat{k}=-\hat{j}$

$\vec{F}=-2.925\hat{j}\ N$

Hence, The magnetic force on the section of wire is $-2.925\hat{j}\ N$ .

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An object is given a very small amount of charge. Which of the following could

spayn [35]

5.4*10^-19 C

Explanation:

For the purposes of this question, charges essentially come in packages that are the size of an electron (or proton since they have the same magnitude of charge). The charge on an electron is -1.6*10^-19

Therefore, any object should have a charge that is a multiple of the charge of an electron - It would not make sense to have a charge equivalent to 1.5 electrons since you can't exactly split the electron in half. So the charge of any integer number of electrons can be transferred to another object.

Charge = q(electron)*n(#electrons)

Since 5.4/1.6 = 3.375, we know that it can not be the right answer because the answer is not an integer.

If you divide every other option listed by the charge of an electron, you will get an integer number.

(16*10^-19 C)/(1.6*10^-19C) = 10

(-6.4*10^-19 C)/(1.6*10^-19C) = -4

(4.8*10^-19 C)/(1.6*10^-19C) = 3

(5.4*10^-19 C)/(1.6*10^-19C) = 3.375

(3.2*10^-19C)/(1.6*10^-19C) = 2

etc.

I hope this helps!

3 0

2 years ago

Electric fields up to 2.00 × 10 5 N/C have been measured inside of clouds during electrical storms. Neglect the drag force due t

katrin2010 [14]

Answer:

$1.9161676647\times 10^{13}\ m/s^2$ or $1.9532799844\times 10^{12}g$

23.4843749996 m

Yes

Explanation:

E = Electric field = $2\times 10^5\ N/C$

c = Speed of light = $3\times 10^8\ m/s$

m = Mass of proton= $1.67\times 10^{-27}\ kg$

q = Charge of electron = $1.6\times 10^{-19}\ C$

Acceleration is given by

$a=\dfrac{Eq}{m}\\\Rightarrow a=\dfrac{2\times 10^5\times 1.6\times 10^{-19}}{1.67\times 10^{-27}}\\\Rightarrow a=1.9161676647\times 10^{13}\ m/s^2$

Dividing by g

$\dfrac{a}{g}=\dfrac{1.9161676647\times 10^{13}}{9.81}\\\Rightarrow a=1.9532799844\times 10^{12}g$

The acceleration is $1.9161676647\times 10^{13}\ m/s^2$ or $1.9532799844\times 10^{12}g$

$v^2-u^2=2as\\\Rightarrow s=\dfrac{v^2-u^2}{2a}\\\Rightarrow s=\dfrac{(0.1\times 3\times 10^8)^2-0^2}{2\times 1.9161676647\times 10^{13}}\\\Rightarrow s=23.4843749996\ m$

The distance is 23.4843749996 m

The gravitational field is very small compared to the electric field so the effects of gravity can be ignored.

5 0

3 years ago

A rocket car on a horizontal rail has an initial mass of 2500 kg and an additional fuel mass of 1000 kg. At time t0 the rocket m

slamgirl [31]

Answer: Acceleration of the car at time = 10 sec is 108 $m/s^{2}$ and velocity of the car at time t = 10 sec is 918.34 m/s.

Explanation:

The expression used will be as follows.

$M\frac{dv}{dt} = u\frac{dM}{dt}$

$\int_{t_{o}}^{t_{f}} \frac{dv}{dt} dt = u\int_{t_{o}}^{t_{f}} \frac{1}{M} \frac{dM}{dt} dt$

= $u\int_{M_{o}}^{M_{f}} \frac{dM}{M}$

$v_{f} - v_{o} = u ln \frac{M_{f}}{M_{o}}$

$v_{o} = 0$

As, $v_{f} = u ln (\frac{M_{f}}{M_{o}})$

u = -2900 m/s

$M_{f} = M_{o} - m \times t_{f}$

= $2500 kg + 1000 kg - 95 kg \times t_{f}s$

= $(3500 - 95t_{f})s$

$v_{f} = -2900 ln(\frac{3500 - 95 t_{f}}{3500}) m/s$

Also, we know that

a = $\frac{dv_{f}}{dt_{f}} = \frac{u}{M} \frac{dM}{dt}$

= $\frac{u}{3500 - 95 t} \times (-95) m/s^{2}$

= $\frac{95 \times 2900}{3500 - 95t} m/s^{2}$

At t = 10 sec,

$v_{f}$ = 918.34 m/s

and, a = 108 $m/s^{2}$

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2 years ago

Lenny loves physics and math. In which Energy career pathway would these interests be the most helpful?

vazorg [7]

Answer:

it is D

Explanation:

cause i took the test

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3 years ago

Read 2 more answers

what is the total distance traveled by a bike rider who rides for three hours at 40 km/hr and then two more hours at 50 km/hr?

marta [7]

Answer:

220 km

Explanation:

3 hours at 40 km/hr

40 km = 1 hour

40 km times 3 = 120 km

2 hours at 50 km/hr

50 km = 1 hour

50 times 2 = 100 km

Total distance

120 km + 100 km = 220 km

4 0

3 years ago