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2 years ago

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A proton is moving at 105 m/s at a point where the potential is 10 V. Later, it is at a place where the potential is 5 V. What i

s its speed there, assuming energy is conserved?

1 answer:

Taya2010 [7]2 years ago

8 0

Answer:

The speed is $7.07\times10^{4}\ m/s$

Explanation:

Given that,

Speed of proton $v= 10^{5}\ m/s$

Final potential = 10 v

Initial potential = 5 V

We need to calculate the speed

Using formula of energy

$\dfrac{1}{2}mv^2=eV$

$v^2=\dfrac{2eV}{m}$

The speed of the particle is directly proportional to the potential.

$v^2\propto V$

Put the value into the formula

$(10^{5})\propto 10$ ....(I)

For 5 V,

$v^2\propto 5$ .....(II)

From equation (I) and (II)

$\dfrac{(10^{5})^2}{v^2}=\dfrac{10}{5}$

$v=70710.67\ m/s$

$v=7.07\times10^{4}\ m/s$

Hence, The speed is $7.07\times10^{4}\ m/s$

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A proton (mass=1.67x10^-27 kg, charge= 1.60x10^-19 C) moves from point A to point under the influence of an electrostatic force

Tom [10]

Answer:

$VB - VA = - 33.4$

Explanation:

Generally the workdone in moving the proton is mathematically represented as

$W = KE_f - KE_i$

Where $KE_i \ and \ KE_f \ are\ the\ initial \ and \ final \ kinetic \ energy$

So

$KE_i = \frac{1}{2} m v_a^2$

Here $v_a$ is the velocity at A with value 50 m/s

So

$KE_i = \frac{1}{2} (1.67*10^{-27}) * 50^2$

$KE_i = 2.09 *10^{-24} \ J$

Also

$KE_f = \frac{1}{2} m v_b^2$

Here $v_a$ is the velocity at A with value $80 km/s = 80000 m/s$

=> $KE_f = \frac{1}{2} (1.67*10^{-27}) * 80000^2$

=> $KE_f = 5.34 *10^{-18} \ J$

So

$W = 5.34 *10^{-18} - 2.09 *10^{-24}$

$W = 5.34 *10^{-18} m/s$

Now this workdone is also mathematically represented as

$W = q * V$

So

$q * V = 5.34 *10^{-18}$

Here $q = 1.60*10^{-19} C$

So

$V = \frac{5.34 *10^{-18} }{1.60*10^{-19}}$

$V = 33.4 \ V$

Generally proton movement is in the direction of the electric field it means that $VA>VB$

So

$VB - VA = - 33.4$

8 0

2 years ago

Help meh in this question plzzz <br>

iragen [17]

The Moment of Inertia of the Disc is represented by $I = \frac{15}{32}\cdot M\cdot R^{2}$ . (Correct answer: A)

Let suppose that the Disk is a Rigid Body whose mass is uniformly distributed. The Moment of Inertia of the element is equal to the Moment of Inertia of the entire Disk minus the Moment of Inertia of the Hole, that is to say:

$I = I_{D} - I_{H}$ (1)

Where:

$I_{D}$ - Moment of inertia of the Disk.

$I_{H}$ - Moment of inertia of the Hole.

Then, this formula is expanded as follows:

$I = \frac{1}{2}\cdot M\cdot R^{2} - \frac{1}{2}\cdot m\cdot \left(\frac{1}{2}\cdot R^{2} \right)$ (1b)

Dimensionally speaking, Mass is directly proportional to the square of the Radius, then we derive the following expression for the Mass removed by the Hole ( $m$ ):

$\frac{m}{M} = \frac{R^{2}}{4\cdot R^{2}}$

$m = \frac{1}{2}\cdot M$

And the resulting equation is:

$I = \frac{1}{2}\cdot M\cdot R^{2} -\frac{1}{2}\cdot \left(\frac{1}{4}\cdot M \right) \cdot \left(\frac{1}{4}\cdot R^{2} \right)$

$I = \frac{1}{2} \cdot M\cdot R^{2} - \frac{1}{32}\cdot M\cdot R^{2}$

$I = \frac{15}{32}\cdot M\cdot R^{2}$

The moment of inertia of the Disc is represented by $I = \frac{15}{32}\cdot M\cdot R^{2}$ . (Correct answer: A)

Please see this question related to Moments of Inertia: brainly.com/question/15246709

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2 years ago

Two Polaroids are aligned so that the initially unpolarized light passing through them is a maximum. At what angle should one of

steposvetlana [31]

To solve this problem it is necessary to apply the law of Malus which describes the change in the Intensity of Light when it crosses a polarized surface.

Mathematically the expression is given as

$I = I_0 cos^2\theta$

Where,

$I_0$ = Initial Intensity

I = Final Intensity after pass through the polarizer

$\theta$ = Angle between the polarizer and the light

Since it is sought to reduce the intensity by half the relationship between the two intensities will be given as

$\frac{I}{I_0} = \frac{1}{2}$

Using the Malus Law we have,

$I = I_0 cos^2\theta$

$cos^2\theta = \frac{I}{I_0}$

$cos^2\theta = \frac{1}{2}$

$\theta = cos^{-1}(\frac{1}{2})^2$

$\theta = 75.52\°$

Angle with respect to maximum is $90-75.52 = 14.48\°$

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escribe the differences between the terrestrial and jovian planets and classify each of the 8 planets to the proper type

Radda [10]

Answer:

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Explanation:

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san4es73 [151]

Answer:

a.

Explanation:

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