Lauren is comparing rocks from two different locations. She has the following

A 3.00-kg ball swings rapidly in a complete vertical circle of radius 2.00 m by a light string that is fixed at one end. The bal

Brut [27]

Answer:

The gravity does a work of - 117.6 Joules.

The tension does not do work as the force is perpendicular to the direction of motion at any point in the trajectory.

Explanation:

The work done by the gravity simply is the difference in gravitational potential energy multiplied by -1:

$W_g = - \Delta E_p = - (mgh_f - m g h_i)$

where m is the mass of the ball, g is the acceleration due to gravity, $h_f$ is the final height and $h_i$ is the initial height.

So, if the radius is 2.00 m, then the difference of height will be 4 meters:

$W_g = - mg (h_f - h_i)$

$W_g = - 3.00 \ kg \ 9.8 \frac{m}{s^2} \ 4 \m$

$W_g = - 117.6 Joules$

As the tension is perpendicular to the velocity of the ball, the force is always perpendicular to the direction of motion. So, the differential of work will be:

$dW = \vec{F} d\vec{r} = 0$

6 0

2 years ago

How does the configuration of the electric field occur between a "parallel plate" setup in a lab?What is the effect of conductor

sleet_krkn [62]

Explanation:

The magnitude of the electric field between the plates is given by

E = -ΔV/d

minus sign indicates Potential decreases in the direction of electric field

where

ΔV is the potential difference between the plates

D is the distance between the plates.

The work done when carrying an electrical charge on an equipotential surface between one position to the other is zero W= q(V-V)=0 The electric field lines of force are always perpendicular to an equipotential surface. That conductor in an equipotential surface as direction E is at right angles to an eauipotential surface The intensity of the electric field along an equipotential surface is always zero. Equipotential surfaces never collide with each other as this would mean that at that point, there are two alternative values that are not true.

8 0

2 years ago

Light propagate faster through medium “a” than medium “b”

dangina [55]

1) Medium "b" has more optical density

2) Light must hit the interface between the two mediums perpendicularly

Explanation:

1)

Refraction occurs when light propagates from a medium into a second medium.

The optical density of a medium is given by its index of refraction, which is defined as:

$n=\frac{c}{v}$

where

c is the speed of light in a vacuum

v is the speed of light in a medium

Higher index of refraction means higher optical density, and light propagater slower into a medium with higher optical density.

In this problem, light propagates faster through medium "a" than medium "b": this means that medium "a" has lower refractive index of medium "b", and so "b" has more optical density.

2)

We can answer this part by referring to Snell's law, which gives the relationship between the direction of the incident ray and of the refracted ray when light passes through the interface between two media:

$n_1 sin \theta_1 = n_2 sin \theta_2$

where

$n_1, n_2$ are the index of refraction of the two mediums

$\theta_1, \theta_2$ are the angle of incidence and of refraction (the angle that light makes with the normal to the surface in medium 1 and medium 2)

Here we want the direction of propagation of the light ray not to change: this means that it must be

$sin \theta_1 = sin \theta_2$ (1)

However, here we have two mediums "a" and "b" with different index of refraction, so

$n_1\neq n_2$

Therefore the only angle that can satisfy eq.(1) is

$\theta_1 = \theta_2 = 0$

So, the light must hit the surface perpendicular to the interface between the two mediums.

Learn more about refraction:

brainly.com/question/3183125

brainly.com/question/12370040

#LearnwithBrainly

3 0

2 years ago

A parallel-plate capacitor has a plate area of 0.2m^2 and a plate separation of 0.1mm. To obtain an electric field of 2.0 × 10^6

Oduvanchick [21]

Answer:

3.536*10^-6 C

Explanation:

The magnitude of the charge is expresses as Q = CV

C is the capacitance of the capacitor

V is the voltage across the capacitor

Get the capacitance

C = ε0A/d

ε0 is the permittivity of the dielectric = 8.84 x 10-12 F/m

A is the area = 0.2m²

d is the plate separation = 0.1mm = 0.0001m

Substitute

C = 8.84 x 10-12 * 0.2/0.0001

C = 1.768 x 10-8 F

Get the potential difference V

Using the formula for Electric field intensity

E = V/d

2.0 × 10^6 = V/0.0001

V = 2.0 × 10^6 * 0.0001

V = 2.0 × 10^2V

Get the charge on each plate.

Q = CV

Q = 1.768 x 10-8 * 2.0 × 10^2

Q = 3.536*10^-6 C

Hence the magnitude of the charge on each plate should be 3.536*10^-6 C

5 0

2 years ago

____ [38]

Answer:

Explanation:

D = 8.27 m ⇒ R = D / 2 = 8.27 m / 2 = 4.135 m

ω = 0.66 rev/sec = (0.66 rev/sec)*(2π rad/1 rev) = 4.1469 rad/s

We can apply the equation

Ff = W ⇒ μ*N = m*g (I)

then we have

N = Fc = m*ac = m*(ω²*R)

Returning to the equation I

μ*N = m*g ⇒ μ*m*ω²*R = m*g ⇒ μ = g / (ω²*R)

Finally

μ = (9.81 m/s²) / ((4.1469 rad/s)²*4.135 m) = 0.1379

3 0

3 years ago