ASAP NEED HELP PLEASE

Suppose the glass paperweight has index of refraction n=1.38. a) find the value of θ for which the reflection on the vertical su

zimovet [89]

Answer:

a)θ=71.89°

b)NO

Explanation:

Given that

For glass n= 1.38

We know that for air n'=1

The angle for total internal reflection θc given as

sin θc=n'/n

By putting the values

sin θc=n'/n

sin θc=1/1.38

θc=46.43°

n'sinθ = n sinθref

sinθref = cosθc

n'sinθ = n cosθc

1 x sinθ =1.38 x cos 46.43°

θ=71.89°

b)

NO

8 0

3 years ago

Farmer Tamar airlifted her sheep from 1000 meters to 1,200 meters

attashe74 [19]

Question

What was the change in potential energy of the sheep?

Answer:

1962(m) J

Explanation:

Potential energy is given by

PE=mgh

Change in potential energy is given by

∆PE=mg(h2-h1)

Where ∆PE is the change in potential eneegy, g is acceleration due to gravity, h2 is final height and h1 is the initial height.

Substituting 9.81 m/s² for g, 1200 m for h2 and 1000 m for h1 then

∆PE=m*9.81(1200-1000)=1962m Joules

Since we are not given the mass of rhe sheep, the answer will be 1962(m) Joules

4 0

3 years ago

Metal sphere having an excess of +5 elementary charges has a net electric charge of

Natasha2012 [34]

Answer:

$q=+8.01\cdot 10^{-19}\ coulombs$

Explanation:

<u>Elementary charge</u>

The elementary charge, denoted by the symbol e is the electric charge carried by a proton or, equivalently, the magnitude of a negative electric charge carried by an electron, which has charge −e.

The value of the elementary charge is a fundamental constant in physics:

$\mathbf{e}=1.60217662 \cdot 10^{-19}\ coulombs$

If a metal sphere has an excess of +5 elementary charge, then it has a net charge of:

$q=5*\mathbf{e}=+5*1.60217662 \cdot 10^{-19}\ coulombs$

$\boxed{q=+8.01\cdot 10^{-19}\ coulombs}$

5 0

3 years ago

Air is compressed in a cylinder such that the volume changes from 100.0 to 10.0 in^3. The initial pressure is 50.0 psia and the

EleoNora [17]

Answer:

5953.42 J

Explanation:

Given:

Initial volume, $V_i$ = 100 in³

Final Volume, $V_f$ = 10 in³

Initial pressure = 50 psia

Temperature = 100° F = 310.93 K

For isothermal reversible process, work done is given as:

Work done = $-2.303RTlog_{10}\frac{V_f}{V_i}$

Where,

R is the ideal gas constant = 8.314 J/mol.K

or

Work done = $-2.303\times8.314\times310.93log_{10}\frac{10}{100}$

or

Work done = 5953.42 J

7 0

3 years ago

A pole-vaulter just clears the bar at 5.31 m and falls back to the ground. The change in the vaulter's potential energy during t

irina [24]

Answer:

Weight = 734.46 N

Explanation:

Given:

Initial height of the pole-vaulter is, $h_i=5.31\ m$

Final height of the pole-vaulter is, $h_f=0\ m$

Change in the vaulter's potential energy is, $\Delta U=-3900\ J$

We know that, change in potential energy is given as:

$\Delta U=mg(h_f-h_i)$

Where, 'm' is the mass of the object, 'g' is acceleration due to gravity and has a value of 9.8 m/s².

Now, weight of the object is given as the product of its mass and acceleration due to gravity. So, replace 'mg' by weight 'w'. So, the equation becomes,

$\Delta U = w(h_f-h_i)$

Now, rewriting in terms of 'w', we get:

$w=\dfrac{\Delta U}{(h_f-h_i)}$

Now, plug in all the given values and solve for 'w'. This gives,

$w=\dfrac{-3900\ J}{(0-5.31)\ m}\\\\\\w=\dfrac{3900}{5.31}\ N\\\\\\w=734.46\ N$

Therefore, weight of the vaulter is 734.46 N.

8 0

3 years ago