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galben [10]
2 years ago
12

A proton is accelerated from rest through a potential difference V0 and gains a speed v0. If it were accelerated instead through

a potential difference of 2V0, what speed would it gain
Physics
1 answer:
Neporo4naja [7]2 years ago
6 0

Answer:

Explanation:

Let the charge on proton be q .

energy gain by proton in a field having potential difference of V₀

= V₀ q

Due to gain of energy , its kinetic energy becomes 1/2 m v₀²

where m is mass and v₀ is velocity of proton

V₀ q = 1/2 m v₀²

In the second case , gain of energy in electrical field

= 2 V₀q , if v be the velocity gained in the second case

2 V₀q = 1/2 m v²

1/2 m v² = 2 V₀q = 2 x 1/2 m v₀²

mv² = 2  m v₀²

v = √2 v₀

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A flat disk of radius 0.50 m is oriented so that the plane of the disk makes an angle of 30 degrees with a uniform electric fiel
nexus9112 [7]

Answer:

The electric flux is 280\ \rm N.m^2/C

Explanation:

Given:

  • Radius of the disc R=0.50 m
  • Angle made by disk with the horizontal \theta=30^\circ
  • Magnitude of the electric Field E=713.0\ \rm N/C

The flux of the Electric Field E due to the are dA in space can be found out by using Gauss Law which is as follows

\phi=\int E.dA

where

  • \phi is the total Electric Flux
  • E is the Electric Field
  • dA is the Area through which the electric flux is to be calculated.

Now according to question we have

=EA\cos\theta \\=713\times 3.14\times 0.5^2 \times \cos60^\circ\\=280\ \rm N.m^2/C

Hence the electric flux is calculated.

8 0
3 years ago
An automobile moves on a level horizontal road in a circle of radius 30 m. The coefficient of
blondinia [14]

Answer:

v = 12.12 m/s      

Explanation:

It is given that,

Radius of circle, r = 30 m

The coefficient friction between tires and road is 0.5,

The centripetal force is balanced by the force of friction such that,

v = 12.12 m/s

So, the maximum speed with which this car can round this curve is 12.12 m/s. Hence, this is the required solution.

5 0
3 years ago
A soccer player takes a corner kick, lofting a stationary ball 33.0° above the horizon at 15.0 m/s. If the soccer ball has a mas
Alexxandr [17]

Explanation:

It is given that,

Mass of the soccer ball, m = 0.425 kg

Speed of the ball, u = 15 m/s

Angle with horizontal, \theta=33^{\circ}

Time for which the player's foot is in contact with it, \Delta t = 5.1\times 10^{-2}\ s

Part A,

The x component of the soccer ball's change in momentum is given by :

\Delta p_x=mv\ cos\theta

\Delta p_x=0.425\times 15\ cos(33)

p_x=5.34\ kg-m/s

The y component of the soccer ball's change in momentum is given by :

\Delta p_y=mv\ sin\theta

\Delta p_y=0.425\times 15\ sin(33)

p_y=3.47\ kg-m/s

Hence, this is the required solution.

3 0
3 years ago
A coil of 40 turns is wrapped around a long solenoid of cross-sectional area 7.5×10−3m2. The solenoid is 0.50 m long and has 500
defon

To solve this problem it is necessary to apply the concepts related to mutual inductance in a solenoid.

This definition is described in the following equation as,

M = \frac{\mu_0 N_1 N_2A_1}{l_1}

Where,

\mu =permeability of free space

N_1 = Number of turns in solenoid 1

N_2 = Number of turns in solenoid 2

A_1= Cross sectional area of solenoid

l = Length of the solenoid

Part A )

Our values are given as,

\mu_0 = 4\pi *10^{-7}H/m

N_1 = 500

N_2 = 40

A = 7.5*10^{-4}m^2

l = 0.5m

Substituting,

M = \frac{\mu_0 N_1 N_2A_2}{l_1}

M = \frac{(4\pi *10^{-7})(500)(40)(7.5*10^{-4})}{0.5}

M = 3.77*10^{-4}H

PART B) Considering that many of the variables remain unchanged in the second solenoid, such as the increase in the radius or magnetic field, we can conclude that mutual inducantia will appear the same.

8 0
3 years ago
If the half-life of niobium-91 is 680 years, how long will it take for the decay described in the previous question?
Svet_ta [14]
2040

15.4+2.2/2 until it equals 2.2
( divide by 3)

680(years)*3 devisions = 2040
3 0
3 years ago
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