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Andre45 [30]
2 years ago
15

Someone please help me!

Physics
1 answer:
DIA [1.3K]2 years ago
5 0

Answer:

1

Explanation:

I = 12/12 = 1

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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. <span>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.</span>
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:

<em>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. </em>

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

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

W=\Delta K

W= 1/2*I*\omega f^2-1/2*I*\omega o^2

Where I = 50kg.m^2 and ωo = rad/s. We need to calculate ωf.

By using kinematics:

\omega f^2=\omega o^2+2*\alpha*\theta

But we don't have the acceleration yet. So, we have to calculate it by making a sum of torque:

\tau=I*\alpha

\alpha=\tau/I     =>     \alpha = 0.5rad/s^2

Now we can calculate the final velocity:

\omega f = 8.68rad/s

Finally, we calculate the total work:

W= 1/2*I*\omega f^2 = 1883.56J

Since the question asked to "<em>Enter your answer in joules to four significant figures.</em>":

W = 1884J

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3 years ago
A block starting from rest slides down the length of an 18 plank with an acceleration of 4.0 meters per second. How long does th
Snowcat [4.5K]

Answer:

\boxed{\text{\sf \Large 3.0 s}}

Explanation:

Use distance formula

\displaystyle d=ut+\frac{1}{2} at^2

u= \text{\sf  initial velocity}\\d= \text{\sf  distance}\\a= \text{\sf  acceleration}\\t= \text{\sf  time taken}

\displaystyle 18=0 \times t+\frac{1}{2} \times 4 \times t^2

t=3

4 0
3 years ago
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