If a bicycle travels 1200m, and it takes 300seconds, what is its speed in m/s.

A 6.41 $\mu C$ particle moves through a region of space where an electric field of magnitude 1270 N/C points in the positive $x$

Marat540 [252]

Answer:

The particle's velocity is 212.15 m/s.

Explanation:

Given that,

Charge of particle, $q=6.41\ \mu C=6.41\times 10^{-6}\ C$

The magnitude of electric field, E = 1270 N/C

The magnitude of magnetic field, B = 1.28 T

Net force, $F=6.4\times 10^{-3}\ N$

We need to find the magnitude of the particle's velocity. the net force acting on the particle is given by Lorentz force as :

$F=qE+qvB\\\\v=\dfrac{F-qE}{qB}\\\\v=\dfrac{6.4\times 10^{-3}-6.41\times 10^{-6}\times 1270}{6.41\times 10^{-6}\times 1.28}\\\\v=-212.15\ m/s$

So, the particle's velocity is 212.15 m/s.

5 0

3 years ago

Niobium metal becomes a superconductor when cooled below 9K. Itssuperconductivity is destroyed when the surface magnetic fieldex

elena-14-01-66 [18.8K]

Answer:

the maximum current is 500 A

Explanation:

Given the data in the question;

the B field magnitude on the surface of the wire is;

B = μ₀i / 2πr

we are to determine the maximum current so we rearrange to find i

B2πr = μ₀i

i = B2πr / μ₀

given that;

diameter d = 2 mm = 0.002 m

radius = 0.002 / 2 = 0.001 m

B = 0.100 T

we know that permeability; μ₀ = 4π × 10⁻⁷ Tm/A

so we substitute

i = (0.100)(2π×0.001 ) / 4π × 10⁻⁷

i = 500 A

Therefore, the maximum current is 500 A

4 0

3 years ago

What is the maximum speed with which a 1200-kg car can round a turn of radius 94.0 m on a flat road if the coefficient of static

JulijaS [17]

Answer:

Explanation:

For the car to turn at the about the centripetal force must not be greater than the static friction between the tires and the road

we will use the expression relating centripetal force and static friction below

let U represent the coefficient of static friction

Given that

U= 0.50

mass m= 1200-kg

radius r= 94.0 m

Assuming g= 9.81 m/s^2

$U*m*g=\frac{mv^2}{r}$

$U*g=\frac{v^2}{r}$

substituting our given data in to expression we can solve for the speed V

$0.5*9.81=\frac{v^2}{94}$

making v the subject of formula we have

$0.5*9.81=\frac{v^2}{94}\\\v= \sqrt{0.5*9.81*94} \\\\v= \sqrt{461.07} \\\\v= 21.47$

v= 21.47m/s

<em><u>hence the maximum velocity of the car is 21.47m/s</u></em>

7 0

3 years ago

Which of the following graphs shows the relationship between two variables that obey the inverse square law?

blondinia [14]

Well, if we've been paying attention in class, we already KNOW that the electrostatic force changes as the inverse square of the distance, and the top graph is conveniently labeled "Electrostatic Force".

But if we didn't already know that, we'd have to examine the graphs, and find the one where 'y' changes like 1/x² .

The top graph does that. After 1 unit of time, the force is 350. Double the time to 2 units, and the force should drop to 1/4 of 350 ... sure enough, it's a little less than 90. Double the time again, to 4 units, and it should drop to 1/4 of a little less than 90 ... by golly, it's down below 30.

The first graph is what an inverse square looks like. Now that you've worked out this graph, you'll know an inverse square relationship whenever you see it.

8 0

3 years ago

HEYYYYYY HEYYYYYYYYY

weeeeeb [17]

Answer: i dont know what is happening but I'm answering with another HEEEYYYYYYY HEYYYYYYYY

Explanation:

8 0

2 years ago

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