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2 years ago

6

A fish is 80 cm below the surface of a pond. What is the apparent depth (in cm) when viewed from a position almost directly abov

e the fish

1 answer:

Rudik [331]2 years ago

5 0

Answer:

Apparent depth (Da) = 60.15 cm (Approx)

Explanation:

Given:

Distance from fish (D) = 80 cm

Find:

Apparent depth (Da)

Computation:

We know that,

Refractive index of water (n2) = 1.33

So,

Apparent depth (Da) = D(n1/n2)

Apparent depth (Da) = 80 (1/1.33)

Apparent depth (Da) = 60.15 cm (Approx)

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A long, thin straight wire with linear charge density λ runs down the center of a thin, hollow metal cylinder of radius R. The c

netineya [11]

Answer:

$E=\frac{\lambda}{2\pi r\epsilon_0}$

Explanation:

We are given that

Linear charge density of wire= $\lambda$

Radius of hollow cylinder=R

Net linear charge density of cylinder= $2\lambda$

We have to find the expression for the magnitude of the electric field strength inside the cylinder r<R

By Gauss theorem

$\oint E.dS=\frac{q}{\epsilon_0}$

$q=\lambda L$

$E(2\pi rL)=\frac{L\lambda}{\epsilon_0}$

Where surface area of cylinder= $2\pi rL$

$E=\frac{\lambda}{2\pi r\epsilon_0}$

8 0

1 year ago

Two large parallel conducting plates carrying opposite charges of equal magnitude are separated by 2.20 cm. Part A If the surfac

alukav5142 [94]

Answer:

5308.34 N/C

Explanation:

Given:

Surface density of each plate (σ) = 47.0 nC/m² = $47\times 10^{-9}\ C/m^2$

Separation between the plates (d) = 2.20 cm

We know, from Gauss law for a thin sheet of plate that, the electric field at a point near the sheet of surface density 'σ' is given as:

$E=\dfrac{\sigma}{2\epsilon_0}$

Now, as the plates are oppositely charged, so the electric field in the region between the plates will be in same direction and thus their magnitudes gets added up. Therefore,

$E_{between}=E+E=2E=\frac{2\sigma}{2\epsilon_0}=\frac{\sigma}{\epsilon_0}$

Now, plug in $47\times 10^{-9}\ C/m^2$ for 'σ' and $8.85\times 10^{-12}\ F/m$ for $\epsilon_0$ and solve for the electric field. This gives,

$E_{between}=\frac{47\times 10^{-9}\ C/m^2}{8.854\times 10^{-12}\ F/m}\\\\E_{between}= 5308.34\ N/C$

Therefore, the electric field between the plates has a magnitude of 5308.34 N/C

5 0

1 year ago

3. A 4.1 x 10-15 C charge is able to pick up a bit of paper when it is initially 1.0 cm above the paper. Assume an induced charg

Anni [7]

Answer:

$\mathbf{1.51\times10^{-15}N}$

Explanation:

The computation of the weight of the paper in newtons is shown below:

On the paper, the induced charge is of the same magnitude as on the initial charges and in sign opposite.

Therefore the paper charge is

$q_{paper}=-4.1\times10^{-15}C$

Now the distance from the charge is

$r=1cm=0.01m$

Now, to raise the paper, the weight of the paper acting downwards needs to be managed by the electrostatic force of attraction between both the paper and the charge, i.e.

$mg=\frac{k_{e}q_{1}q_{2}}{r^{2}}$

$\Rightarrow W=mg$

$=\frac{9\times10^{9}\times(4.1\times10^{-15})^{2}}{0.01^{2}}$

$=\mathbf{1.51\times10^{-15}N}$

6 0

1 year ago

A 200 g hockey puck is launched up a metal ramp that is inclined at a 30° angle. The coefficients of static and kinetic friction

Vaselesa [24]

Answer:

H=1020.12m

Explanation:

From a balance of energy:

$\frac{m*Vo^2}{2} -mg*H=-Ff*d$ where H is the height it reached, d is the distance it traveled along the ramp and Ff = μk*N.

The relation between H and d is given by:

H = d*sin(30) Replace this into our previous equation:

$\frac{m*Vo^2}{2} -mg*d*sin(30)=-\mu_k*N*d$

From a sum of forces:

N -mg*cos(30) = 0 => N = mg*cos(30) Replacing this:

$\frac{m*Vo^2}{2} -mg*d*sin(30)=-\mu_k*mg*cos(30)*d$ Now we can solve for d:

d = 2040.23m

Thus H = 1020.12m

6 0

1 year ago

Read 2 more answers

While on a camping trip, a man would like to carry water from the lake to his campsite. He fills two, non‑identical buckets with

xenn [34]

Answer:

The bucket located in the direction opposite of displacement of the rod for balancing will be heavier.

Explanation:

Moment of a force is directly proportional to it's distance from the point about which rotation takes place. In the given case the man increases the lever arm(distance from point of rotation of the force) to balance the moment of the lighter bucket. Thus in the direction of the increase the lighter bucket will be placed.

6 0

2 years ago