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3 years ago

An equipotential surface must be

Physics

1 answer:

devlian [24]3 years ago

8 0

Answer: An equipotential surface must be perpendicular to the electric field at every point. The correct option is B.

Explanation: Equipotential surface is any surface over which the potential difference is constant. Examples include:

-The surface of a conductor in electrostatic field.

- surface of a liquid in equilibrium.

Equipotential surface are perpendicular to the electric field at every point because components of electric field intensity along the surface is zero ( this is because no work is done while moving a charge from one point to another on an equipotential surface).

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Red light of wavelength 651 nm produces photoelectrons from a certain photoemissive material. Green light of wavelength 521 nm p

Mnenie [13.5K]

Answer:

material work function is 0.956 eV

Explanation:

given data

red wavelength 651 nm

green wavelength 521 nm

photo electrons = 1.50 × maximum kinetic energy

to find out

material work function

solution

we know by Einstein photo electric equation that is

for red light

h ( c / λr ) = Ф + kinetic energy

for green light

h ( c / λg ) = Ф + 1.50 × kinetic energy

now from both equation put kinetic energy from red to green

h ( c / λg ) = Ф + 1.50 × (h ( c / λr ) - Ф)

Ф =( hc / 0.50) × ( 1.50/ λr - 1/ λg)

put all value

Ф =( 6.63 × $10^{-34}$ (3 × $10^{8}$ ) / 0.50) × ( 1.50/ λr - 1/ λg)

Ф =( 6.63 × $10^{-34}$ (3 × $10^{8}$ ) / 0.50 ) × ( 1.50/ 651× $10^{-9}$ - 1/ 521 × $10^{-9}$ )

Ф = 1.5305 × $10^{-19}$ J × ( 1ev / 1.6 × $10^{-19}$ J )

Ф = 0.956 eV

material work function is 0.956 eV

4 0

3 years ago

Read 2 more answers

How fast can a 4000 kg truck travel around a 70 m radius turn without skidding if its tires share a 0.6 friction coefficient wit

aleksley [76]

Answer:

Velocity of truck will be 20.287 m /sec

Explanation:

We have given mass of the truck m = 4000 kg

Radius of the turn r = 70 m

Coefficient of friction $\mu =0.6$

Centripetal force is given $F=\frac{mv^2}{r}$

And frictional force is equal to $F_{frictional}=\mu mg$

For body to be move these two forces must be equal

So $\frac{mv^2}{r}=\mu mg$

$v=\sqrt{\mu rg}=\sqrt{0.6\times 70\times 9.8}=20.287m/sec$

7 0

3 years ago

A particle moving along the x-axis has its velocity described by the function vx =2t2m/s, where t is in s. its initial position

Nesterboy [21]

The position of the object at time t =2.0 s is <u>6.4 m.</u>

Velocity vₓ of a body is the rate at which the position x of the object changes with time.

Therefore,

$v_x= \frac{dx}{dt}$

Write an equation for x.

$dx=v_xdt\\ x=\int v_xdt$

Substitute the equation for vₓ =2t² in the integral.

$x=\int v_xdt\\ =\int2t^2dt\\ =\frac{2t^3}{3} +C$

Here, the constant of integration is C and it is determined by applying initial conditions.

When t =0, x = 1. 1m

$x= \frac{2t^3}{3} +C\\ x_0=1.1\\ x= (\frac{2t^3}{3} +1.1)m$

Substitute 2.0s for t.

$x= (\frac{2t^3}{3} +1.1)m\\ =\frac{2(2.0)^3}{3} +1.1\\ =6.43 m$

The position of the particle at t =2.0 s is <u>6.4m</u>

5 0

3 years ago

The acceleration vector of a particle in projectile motion ________.

Alex73 [517]

Answer:

Points downward, and its magnitude is 9.8 m/s^2

Explanation:

The motion of a projectile consists of two independent motions:

- A uniform horizontal motion, with constant velocity and zero acceleration. In fact, there are no forces acting on the projectile along the horizontal direction (if we neglect air resistance), so the acceleration along this direction is zero.

- A vertical motion, with constant acceleration g = 9.8 m/s^2 towards the ground (downward), due to the presence of gravity wich "pulls" the projectile downward.

The total acceleration of the projectile is given by the resultant of the horizontal and vertical components of the acceleration. But we said that the horizontal component is zero, therefore the total acceleration corresponds just to its vertical component, therefore it is a vector with magnitude 9.8 m/s^2 which points downward.

4 0

3 years ago

The field between two charged parallel plates is kept constant. If the two plates are brought closer together, the potential dif

Dovator [93]

The answer is B, because it will lose potential energy.

3 0

3 years ago