What is the speed of a wave with 3Hz frequency and a wavelength of 9 m

A beam of light traveling through a liquid (of index of refraction n1 = 1.47) is incident on a surface at an angle of θ1 = 59° w

frosja888 [35]

Answer:

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

(b) $n_{2} = 1.349$

(c) $v_{1} = 2.04\times 10^{8}\ m/s$

(d) $v_{2} = 2.22\times 10^{8}\ m/s$

Solution:

As per the question:

Refractive index of medium 1, $n_{1} = 1.47$

Angle of refraction for medium 1, $\theta_{1} = 59^{\circ}$

Angle of refraction for medium 2, $\theta_{1} = 69^{\circ}$

Now,

(a) The expression for the refractive index of medium 2 is given by using Snell's law:

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

where

$n_{2}$ = Refractive Index of medium 2

Now,

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

(b) The refractive index of medium 2 can be calculated by using the expression in part (a) as:

$n_{2} = \frac{1.47\times sin59^{\circ}}{sin69^{\circ}}$

$n_{2} = 1.349$

(c) To calculate the velocity of light in medium 1:

We know that:

$Refractive\ index,\ n = \frac{Speed\ of\ light\ in vacuum,\ c}{Speed\ of\ light\ in\ medium,\ v}$

Thus for medium 1

$n_{1} = \frac{c}{v_{1}$

$v_{1} = \frac{c}{n_{1} = \frac{3\times 10^{8}}{1.47} = 2.04\times 10^{8}\ m/s$

(d) To calculate the velocity of light in medium 2:

For medium 2:

$n_{2} = \frac{c}{v_{2}$

$v_{2} = \frac{c}{n_{1} = \frac{3\times 10^{8}}{1.349} = 2.22\times 10^{8}\ m/s$

5 0

2 years ago

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5.0 kg, with a bullet of mass 0.1 kg. The target was mounted on

kari74 [83]

Answer:

306 m/s

Explanation:

Law of conservation of momentum

m1v1 + m2v2 = (m1+m2)vf

m1 is the bullet's mass so it is 0.1 kg

v1 is what we're trying to solve

m2 is the target's mass so it is 5.0 kg

v2 is the targets velocity, and since it was stationary, its velocity is zero

vf is the velocity after the target is struck by the bullet, so it is 6.0 m/s

plugging in, we get

(0.1 kg)(v1) + (5.0 kg)(0 m/s) = (0.1 kg + 5.0 kg)(6.0 m/s)

(0.1)(v1) + 0 = 30.6

(0.1)(v1) = 30.6

v1 = 306 m/s

8 0

3 years ago

An automobile of mass 1.46 cross times to the power of blank 10 cubed kg rounds a curve of radius 25.0 m with a velocity of 15.0

Vsevolod [243]

The centripetal force exerted on the automobile while rounding the curve is $1.31\times10^4 N$

<u>Explanation:</u>

given that

$Mass\ of\ the\ automobile\ m =1.46\times 10^3 kg\\radius\ of\ the\ curve\ r =25 m\\velocity\ of\ the\ automobile\ v=15m/s\\$

Objects moving around a circular track will experience centripetal force towards the center of the circular track.

$centripetal\ force=mv^2/r\\=1.46\times10^3\times15^2/25\\=1.46\times 10^3\times 225/15\\=1.31\times 10^4 N$

4 0

3 years ago

If the volume of a container of gas is reduced, what will happen to the pressure inside the container?.

AfilCa [17]

Answer:

The pressure of the gas will increase

Explanation:

When gas is put into a container, for example, a balloon, the gas expands to fill the space it can occupy. Since gas is not a solid or a liquid, its particles are all over the place - they are constantly moving and vibrating. As such, when too much gas is blown into a balloon, it will pop. So, when the volume of the container decreases, the pressure of the gas will increase the smaller it gets. Vice versa, the greater the space, the less pressure that will be present in the container.

7 0

2 years ago

Every chemical element goes through natural exponential decay, which means that over time its atoms fall apart. The speed of eac

Naddik [55]

Answer:

t = (ti)ln(Ai/At)/ln(2)

t = 14ln(16)/ln(2)

Solving for t

t = 14×4 = 56 seconds

Explanation:

Let Ai represent the initial amount and At represent the final amount of beryllium-11 remaining after time t

At = Ai/2^n ..... 1

Where n is the number of half-life that have passed.

n = t/half-life

Half life = 14

n = t/14

At = Ai/2^(t/14)

From equation 1.

2^n = Ai/At

Taking the natural logarithm of both sides;

nln(2) = ln(Ai/At)

n = ln(Ai/At)/ln(2)

Since n = t/14

t/14 = ln(Ai/At)/ln(2)

t = 14ln(Ai/At)/ln(2)

Ai = 800

At = 50

t = 14ln(800/50)/ln(2)

t = 14ln(16)/ln(2)

Solving for t

t = 14×4 = 56 seconds

Let half life = ti

t = (ti)ln(Ai/At)/ln(2)

4 0

3 years ago

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