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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.

Gekata [30.6K]

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 $1.6*10^5 N$

4 0

3 years ago

Read 2 more answers

A 15.0 kg fish swimming at 1.10 m/s suddenly gobbles up a 4.50 kg fish that is initially stationary. Ignore any drag effects of

Nana76 [90]

Answer

given.

Mass of big fish = 15 Kg

speed of big fish = 1.10 m/s

mass of the small fish = 4.50 Kg

speed of the fish after eating small fish =?

a) using conservation of momentum

m₁v₁ + m₂v₂ = (m₁+m₂) V

15 x 1.10 + 4.50 x 0 = (15 + 4.5)V

16.5 = 19.5 V

V = 0.846 m/s

b) Kinetic energy before collision

$KE_1 = \dfrac{1}{2}m_1v_1^2 + \dfrac{1}{2}m_2v_2^2$

$KE_1 = \dfrac{1}{2}\times 15 \times 1.1^2 + \dfrac{1}{2}m_2\times 0^2$

KE₁ = 9.075 J

Kinetic energy after collision

$KE_2= \dfrac{1}{2}(15+4.5)\times 0.846^2$

KE₂ = 6.98 J

Change in KE = 6.98 - 9.075 = -2.096 J

hence,

mechanical energy was dissipated during this meal = -2.096 J

5 0

3 years ago

Investigations allow for the control of variables

Sloan [31]

I believe that the statement is true. Investigations allow for the control of variables. Changing variables will lead you to observations that may prove your hypothesis. Hope this answers the question. Have a nice day. Please feel free to ask more questions.

8 0

3 years ago

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Consider a motor that exerts a constant torque of 25.0 nâ‹…m to a horizontal platform whose moment of inertia is 50.0 kgâ‹…m2. A

Crazy boy [7]

Answer:

W = 1884J

Explanation:

This question is incomplete. The original question was:

Consider a motor that exerts a constant torque of 25.0 N.m to a horizontal platform whose moment of inertia is 50.0kg.m^2 . Assume that the platform is initially at rest and the torque is applied for 12.0rotations . Neglect friction.

How much work W does the motor do on the platform during this process? Enter your answer in joules to four significant figures.

The amount of work done by the motor is given by:

$W=\Delta K$