A car travels 50km in 2 hours.

A light ray traveling through a material with an index of refraction of 1.2 is incident on a material that has an index of refra

zzz [600]

Answer:

compared to the incident angle, the refracted angle is 45.56⁰

Explanation:

From Snell's law;

n₁sin(I) = n₂sin(r)

Where;

n₁ is the refractive index of light in medium 1 = 1.2

n₂ is the refractive index of light in medium 2 = 1.4

I is the incident angle

r is the refractive angle

$n = \frac{1}{sin(I)}\\\\sin(I) = \frac{1}{n}\\\\sin(I) =\frac{1}{1.2}\\\\sin(I) =0.8333\\\\I = sin^-{(0.8333)$

I = 56.439⁰

Applying snell's law

$n_1sin(I) = n_2sin(r)\\\\sin(r) = \frac{n_1sin(I) }{n_2}\\\\sin(r) = \frac{1.2*sin(56.439) }{1.4}\\\\sin(r) = 0.714\\\\r = sin^-(0.714)\\\\r = 45.56^o$

Therefore, compared to the incident angle, the refracted angle is 45.56⁰

3 0

2 years ago

A roller coaster travels down a 120 m track in 12.5 seconds how fast does the roller coaster go

KIM [24]

Answer:

9.6m/s

Explanation:

Using the equation S=d/t where s=speed, d=distance, and t=time

plug in the known variables

S=120m/12.5s

S=9.6m/s

4 0

3 years ago

A 0.877 mol sample of N2(g) initially at 298 K and 1.00 atm is held at constant volume while enough heat is applied to raise the

MissTica

Answer : The value of $q\text{ and }\Delta U$ is 286.2 J and 286.2 J respectively.

Explanation : Given,

Moles of sample = 0.877 mol

Change in temperature = 15.7 K

First we have to calculate the heat absorbed by the system.

Formula used :

$q=n\times c_v\times \Delta T$

where,

q = heat absorbed by the system = ?

n = moles of sample = 0.877 mol

$\Delta T$ = Change in temperature = 15.7 K

$c_v$ = heat capacity at constant volume of $N_2$ (diatomic molecule) = $\frac{5}{2}R$

R = gas constant = 8.314 J/mol.K

Now put all the given value in the above formula, we get:

$q=0.877mol\times \frac{5}{2}\times 8.314J/mol.K\times 15.7K$

$q=286.2J$

Now we have to calculate the change in internal energy of the system.

$\Delta U=q+w$

As we know that, work done is zero at constant volume. So,

$\Delta U=q=286.2J$

Therefore, the value of $q\text{ and }\Delta U$ is 286.2 J and 286.2 J respectively.

8 0

3 years ago

Read 2 more answers

6. An article, which costs Rs. 850 is sold at a profit of 6%.

geniusboy [140]

Answer:

b) Rs.901 is the selling price

Explanation:

selling price = cost price+profit percentage/100 x cost price

cost price = 850

profit percentage = 6%

selling price = 850+6/100 x 850 = 901

soln 2

if 100 % = 850

therefore what about 106% = ?

106 x 805/100 = 901

ans = 901

if you need any clarification or more explanation pls do mention on the comment section.i would like to help more

Hope this helps and if it does mark as branliest answer thx

5 0

2 years ago

A 0.200-m uniform bar has a mass of 0.795 kg and is released from rest in the vertical position, as the drawing indicates. The s

aleksklad [387]

Explanation:

Since, the rod is present in vertical position and the spring is unrestrained.

So, initial potential energy stored in the spring is $U_{s}$ = 0

And, initial potential gravitational potential energy of the rod is $U_{g} = \frac{mgL}{2}$ .

It is given that,

mass of the bar = 0.795 kg

g = 9.8 $m/s^{2}$

L = length of the rod = 0.2 m

Initial total energy T = $\frac{mgL}{2}$

Now, when the rod is in horizontal position then final total energy will be as follows.

T = $\frac{1}{2}kx^{2} + I \omega^{2}$

where, I = moment of inertia of the rod about the end = $\frac{mL^{2}}{3}$

Also, $\omega = \frac{\nu}{L}$

where, $\nu$ = speed of the tip of the rod

x = spring extension

The initial unstrained length is $x_{o}$ = 0.1 m

Therefore, final length will be calculated as follows.

x' = $\sqrt{(0.2)^{2} + (0.1)^{2}}$ m

Then, x = $x' - x_{o}$

x = $\sqrt{(0.2)^{2} + (0.1)^{2}}$ m - 0.1 m

= 0.1236 m

k = 25 N/m

So, according to the law of conservation of energy

$\frac{mgL}{2} = \frac{1}{2}kx^{2} + \frac{1 \times mL^{2}}{2 \times 3}(\frac{\nu}{L})^{2}$

$\frac{mgL}{2} = \frac{1}{2}kx^{2} + \frac{1}{6}mv^{2}$

Putting the given values into the above formula as follows.

$\frac{mgL}{2} = \frac{1}{2}kx^{2} + \frac{1}{6}mv^{2}$

$\frac{0.795 kg \times 9.8 \times 0.2 m}{2} = \frac{1}{2} \times 27 N/m \times (0.1236)^{2} + \frac{1}{6} \times 0.795 \times v^{2}$

v = 2.079 m/s

Thus, we can conclude that tangential speed with which end A strikes the horizontal surface is 2.079 m/s.

7 0

3 years ago