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Masteriza [31]
2 years ago
5

A point charge of 6.8 C moves at 6.5 × 104 m/s at an angle of 15° to a magnetic field that has a field strength of 1.4 T.

Physics
2 answers:
Alexxx [7]2 years ago
5 0

The correct answer is A) 1.6 x 10-1 N

Gekata [30.6K]2 years ago
4 0

As per the question, the point charge is given as [q] = 6.8 C

The velocity of the charged particle [v] = 6.5*10^{4}\ m/s

The magnetic field [B] = 1.4 T

The angle made between magnetic field and velocity [\theta] = 15 degree.

We are asked to calculate the magnetic force experienced by the charged  particle.

The magnetic force experienced by the charged particle is calculated as -

Magnetic force \vec F=q(\ \vec V \times \vec B)

                        i.e F = =qVBsin\theta

                                   =6.8*6.5*10^{4}*1.4*sin15

                                   =61.88*10^{4}sin15

                                   =61.88*10^4*0.2588\ N

                                   =16.016*10^4\ N

                                   =1.6*10^5\ N

Hence, the force experienced by the charged particle is C i.e 1.6*10^5 N

                                   

             

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A 62 kg skier is moving at 6.5 m/s on frictionless horizontal snow-covered plateau when she encounters a rough patch 3.50 m long
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(B). The speed of skier is 10.57 m/s.

Explanation:

Given that,

Mass of skier = 62 kg

Speed = 6.5 m/s

Length = 3.50 m

Coefficient kinetic friction = 0.30

Height = 2.5 m

(A) we need to calculate the work done by friction in crossing the patch

Using formula of work done

W=-\mu mg\times l

Put the value into the formula

W=-0.30\times62\times9.8\times3.50

W=-637.98\ J

The work done by friction in crossing the patch is -637.98 J.

(B) we need to calculate the speed of skier

Using conservation of energy

K.E_{i}+U_{i}-W_{friction}=K.E_{f}+U_{f}

\dfrac{1}{2}mv_{1}^2+mgh-\mu mgl=\dfrac{1}{2}mv_{2}^2+U_{f}

Final potential energy is zero

So, \dfrac{1}{2}mv_{1}^2+mgh-\mu mgl=\dfrac{1}{2}mv_{2}^2

\dfrac{1}{2}v_{2}^2=\dfrac{1}{2}v_{1}^2+gh-\mu gl

Put the value into the formula

\dfrac{1}{2}v_{2}^2=\dfrac{1}{2}\times6.5^2+9.8\times2.5+0.30\times9.8\times3.50

v_{2}=\sqrt{2\times55.915}

v_{2}=10.57\ m/s

The speed of skier is 10.57 m/s.

Hence,  (A).The work done by friction in crossing the patch is -637.98 J.

(B).The speed of skier is 10.57 m/s.

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