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4 years ago

Graph the linear equation y=2x-6

1 answer:

4 0

I can't draw you the graph right now because i'm out but look up graphing calculator and click on the first one, type in your equation and it will show you the graph

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Charlie is investigating friction. She will use the same amount of force to push two wooden balls across two level surfaces. The

Zanzabum

D the distance for trial 3 will be greater than trial 4

4 0

3 years ago

Read 2 more answers

A ray of light incident in air strikes a rectangular glass block of refractive index 1.50, at an angle of incidence of 45°. Calc

balandron [24]

Answer:

Approximately $28^{\circ}$ .

Explanation:

The refractive index of the air $n_{\text{air}}$ is approximately $1.00$ .

Let $n_\text{glass}$ denote the refractive index of the glass block, and let $\theta _{\text{glass}}$ denote the angle of refraction in the glass. Let $\theta_\text{air}$ denote the angle at which the light enters the glass block from the air.

By Snell's Law:

$n_{\text{glass}} \, \sin(\theta_{\text{glass}}) = n_{\text{air}} \, \sin(\theta_{\text{air}})$ .

Rearrange the Snell's Law equation to obtain:

$\begin{aligned} \sin(\theta_{\text{glass}}) &= \frac{n_{\text{air}} \, \sin(\theta_{\text{air}})}{n_{\text{glass}}} \\ &= \frac{(1.00)\, (\sin(45^{\circ}))}{1.50} \\ &\approx 0.471\end{aligned}$ .

Hence:

$\begin{aligned} \theta_{\text{glass}} &= \arcsin (0.471) \approx 28^{\circ}\end{aligned}$ .

In other words, the angle of refraction in the glass would be approximately $28^{\circ}$ .

7 0

2 years ago

Check all that apply. The magnetic force on the current-carrying wire is strongest when the current is parallel to the magnetic

dedylja [7]

Answer:

The direction of the magnetic force acting on a current-carrying wire in a uniform magnetic field is perpendicular to the direction of the field.

The direction of the magnetic force acting on a current-carrying wire in a uniform magnetic field is perpendicular to the direction of the current.

The magnetic force on the current-carrying wire is strongest when the current is perpendicular to the magnetic field lines.

Explanation:

The magnitude of the magnetic force exerted on a current-carrying wire due to a magnetic field is given by

$F=ILB sin \theta$ (1)

where I is the current, L the length of the wire, B the strength of the magnetic field, $\theta$ the angle between the direction of the field and the direction of the current.

Also, B, I and F in the formula are all perpendicular to each other. (2)

According to eq.(1), we see that the statement:

"The magnetic force on the current-carrying wire is strongest when the current is perpendicular to the magnetic field lines."

is correct, because when the current is perpendicular to the magnetic field, $\theta=90^{\circ}, sin \theta = 1$ and the force is maximum.

Moreover, according to (2), we also see that the statements

"The direction of the magnetic force acting on a current-carrying wire in a uniform magnetic field is perpendicular to the direction of the field. "

"The direction of the magnetic force acting on a current-carrying wire in a uniform magnetic field is perpendicular to the direction of the current. "

because F (the force) is perpendicular to both the magnetic field and the current.

5 0

3 years ago

A simple model of a hydrogen atom is a positive point charge +e (representing the proton) at the center of a ring of radius a wi

Norma-Jean [14]

Answer:

Now e is due to the ring at a

We say

1/4πEo(ea/ a²+a²)^3/2

= 1/4πEo ea/2√2a³

So here E is faced towards the ring

Next is E due to a point at the centre

E² = 1/4πEo ( e/a²)

Finally we get the total

Et= E²-E

= e/4πEo(2√2-1/2√2)

So the direction here is away from the ring

8 0

3 years ago

When an ideal gas does positive work in its surroundings, which of the ga's quantities increases?

Burka [1]

I think it's B hope it helps

4 0

3 years ago