When in use, a lamp is designed to draw

A charge of 8.4 × 10–4 C moves at an angle of 35° to a magnetic field that has a field strength of 6.7 × 10–3 T. If the magnetic

larisa86 [58]

Answer:

The charge is moving with the velocity of $1.1\times10^{4}\ m/s$ .

Explanation:

Given that,

Charge $q =8.4\times10^{-4}\ C$

Angle = 35°

Magnetic field strength $B=6.7\times10^{-3}\ T$

Magnetic force $F=3.5\times10^{-2}\ N$

We need to calculate the velocity.

The Lorentz force exerted by the magnetic field on a moving charge.

The magnetic force is defined as:

$F = qvB\sin\theta$

$v = \dfrac{F}{qB\sin\theta}$

Where,

F = Magnetic force

q = charge

B = Magnetic field strength

v = velocity

Put the value into the formula

$v =\dfrac{3.5\times10^{-2}}{8.4\times10^{-4}\times6.7\times10^{-3}\times\sin35^{\circ}}$

$v =\dfrac{3.5\times10^{-2}}{8.4\times10^{-4}\times6.7\times10^{-3}\times0.57}$

$v = 10910.36\ m/s$

$v = 1.1\times10^{4}\ m/s$

Hence, The charge is moving with the velocity of $1.1\times10^{4}\ m/s$ .

4 0

3 years ago

Help me please

agasfer [191]

Answer:

For a velocity versus time graph how do you know what the velocity is at a certain time?

Ans: By drawing a line parallel to the y axis (Velocity axis) and perpendicular to the co-ordinate of the Time on the x axis (Time Axis). The point on the slope of the graph where this line intersects, will be the desired velocity at the certain time.

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How do you know the acceleration at a certain time?

$Ans: We\ know\ that\ acceleration = \frac{Final\ velocity-Initial\ Velocity}{Time\ taken}$

Hence,

By dividing the difference of the Final and Initial Velocity by the Time Taken, we could find the acceleration.

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How do you know the Displacement at a certain time?

Ans: As Displacement equals to the area enclosed by the slope of the Velocity-Time Graph, By finding the area under the slope till the perpendicular at the desired time, we find the Displacement.

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4 0

2 years ago

Show how the alternative definition of power, found in your book, can be derived by substituting the definitions of work and spe

Harman [31]

Let us consider body moves a distance S due to the force F.

Hence the work by the body W = FS

If the force is not along the direction of displacement,then the work by a body for travelling a distance S will be -

$W=[ Fcos\theta]*S$ where $Fcos\theta$ is the component of the force along the direction of displacement.