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2 years ago

PLEASE CAN PEOPLE HELP WITH MY STUFF I SUCK AT CONCENTRAITING IN SCHOOL

Physics

2 answers:

Musya8 [376]2 years ago

8 0

Answer:

I can help what subject and grade do you need help with?

Explanation:

Im in 8th so anything above 8th i cant help with sorry!

NeTakaya2 years ago

3 0

Answer:

maybe:)

Explanation:

it really just depends on the subject and grade for me, I'm good in history and writing!

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The work done by an external force to move a -6.70 μc charge from point a to point b is 1.20×10−3 j .

ASHA 777 [7]

Answer:

108.7 V

Explanation:

Two forces are acting on the particle:

- The external force, whose work is $W=1.20 \cdot 10^{-3}J$

- The force of the electric field, whose work is equal to the change in electric potential energy of the charge: $W_e=q\Delta V$

where

q is the charge

$\Delta V$ is the potential difference

The variation of kinetic energy of the charge is equal to the sum of the work done by the two forces:

$K_f - K_i = W + W_e = W+q\Delta V$

and since the charge starts from rest, $K_i = 0$ , so the formula becomes

$K_f = W+q\Delta V$

In this problem, we have

$W=1.20 \cdot 10^{-3}J$ is the work done by the external force

$q=-6.70 \mu C=-6.7\cdot 10^{-6}C$ is the charge

$K_f = 4.72\cdot 10^{-4}J$ is the final kinetic energy

Solving the formula for $\Delta V$ , we find

$\Delta V=\frac{K_f-W}{q}=\frac{4.72\cdot 10^{-4}J-1.2\cdot 10^{-3} J}{-6.7\cdot 10^{-6}C}=108.7 V$

4 0

3 years ago

Read 2 more answers

The 630-nm light from a helium-neon laser irradiates a grating. The light then falls on a screen where the first bright spot is

dimaraw [331]

Answer:

464.8 nm

Explanation:

The second wavelength of light can be calculated using the next equation:

$\lambda = \frac{x*d}{L}$

Where:

λ : is the wavelength of light

x: is the distance from the central maximum

d: is the distance between the spots

L: is the lenght from the screen to the bright spot

For the first wavelength of light we have:

$\lambda_{1} = \frac{x_{1}*d}{L}$

$630 \cdot 10^{-9} m = \frac{0.61 m*d}{L}$

$\frac{d}{L} = \frac{630 \cdot 10^{-9} m}{0.61 m} = 1.033 \cdot 10^{-6}$ (1)

For the second wavelength of light we have:

$\lambda_{2} = \frac{x_{2}*d}{L}$

$\lambda_{2} = 0.45 m*\frac{d}{L}$ (2)

By entering equation (1) into equation (2) we have:

$\lambda_{2} = 0.45 m* 1.033 \cdot 10^{-6} = 4.648 \cdot 10^{-7} m = 464.8 nm$

Therefore, the second wavelength is 464.8 nm

I hope it helps you!

3 0

3 years ago

If the velocity of a car changes from 0 meters per second (m/s) to 100 m/s in 10 seconds, what is the acceleration over that 10

neonofarm [45]

Answer:

10m/s²

Explanation:

Given parameters:

Initial velocity = 0m/s

Final velocity = 100m/s

Time taken = 10s

Unknown:

Acceleration = ?

Solution:

Acceleration is the rate of change of velocity with time.

A = $\frac{v - u}{t}$

v = final velocity

u = initial velocity

t = time taken

So, insert the parameters and solve;

A = $\frac{100 - 0}{10}$ = 10m/s²

3 0

3 years ago

Two lasers are shining on a double slit, with slit separationd.Laser 1 has a wavelength of d/20, whereas laser 2 has a wavelengt

Sophie [7]

Answer:

a) laser 1 has the maximum closest to the central maximum

b) y₂ –y₁ = L 1.66 10⁻²

Explanation:

a), B1, B2) The expression that describes the constructive interference for a double slit is

d sin θ = m λ

The pattern is observed on a screen

tan θ = y / L

Since the angles are very small

tan θ = sin θ / cos θ = sin θ = y/L

d y / L = m λ

In this case the laser has a wavelength

λ ₁ = d/20

We substitute

d y / L = m d / 20

m = 1

y₁ = L / 20

For the laser 2 λ ₂= d / 15

y₂ = L / 15

When examining the two expressions, laser 1 has the maximum closest to the central maximum

b) the difference between the two patterns is

y₂- y₁ = L (1/15 - 1/20)

y₂ –y₁ = L 1.66 10⁻²

C) laser 1 second maximum

y₁ ’= 2 L / 20

y₁ ’= L 0.1

Laser 2 third minimum

To have a minimum, the equation must be satisfied

d sin θ = (m + ½) λ

d y / L = (m + ½) λ

d y / L = (m + ½) d / 15

y = L (m +1/2) / 15

m = 3

y₂’= L (3 + ½) / 15

y₂’= L 0.2333

The difference is

y₁ ’- y₂’ = L (0.1 - 0.2333)

y₁ ’–y₂’ = L (-0.133)

3 0

3 years ago

A beam of light passes through a liquid into air. Angle 1, angle of

Ede4ka [16]

Answer: 1.57

Explanation:

This described situation is known as Refraction, a phenomenon in which light bends or changes its direction when passing through a medium with a index of refraction different from the other medium.

In this context, the index of refraction is a number that describes how fast light propagates through a medium or material.

According to Snell’s Law:

$n_{1}sin(\theta_{1})=n_{2}sin(\theta_{2})$ (1)

Where:

$n_{1}$ is the first medium index of refraction (the value we want to know)

$n_{2}=1$ is the second medium index of refraction (air)

$\theta_{1}=23\°$ is the angle of incidence

$\theta_{2}=38\°$ is the angle of refraction

Now, let's find $n_{1}$ from (1):

$n_{1}=n_{2}\frac{sin \theta_{2}}{sin \theta_{1}}$ (2)

Substituting the known values:

$n_{1}=1\frac{sin(38\°)}{sin(23\°)}$

Finally:

$n_{1}=1.57$

3 0

4 years ago