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sammy [17]
3 years ago
6

If a liquid has a refractive index of 1.5323, than the speed of light through that liquid is _______ x 108 m/s. Type your numeri

cal answer rounded to the 2nd decimal place (i.e. if you calculate 4.56792 x 108 m/s, you would type 4.57).
Physics
1 answer:
wariber [46]3 years ago
5 0

Answer:

The speed of the light through that liquid is v=1.95\times 10^8\ m/s.

Explanation:

Given that,

The refractive index of the liquid, \mu=1.5323

We need to find the speed of light through that liquid. Let the speed of light in liquid is v. The relation between the refractive index and the speed of light in the medium. So,

\mu=\dfrac{c}{v}

v=\dfrac{c}{\mu}

v=\dfrac{3\times 10^8\ m/s}{1.5323}

v=1.95\times 10^8\ m/s

So, the speed of the light through that liquid is v=1.95\times 10^8\ m/s. Hence, this is the required solution.

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Practice entering numbers that include a power of 10 by entering the diameter of a hydrogen atom in its ground state, dH = 1.06
Slav-nsk [51]

Answer:

1.06085\times 10^{-10}\ m

Explanation:

h = Planck's constant = 6.626\times 10^{-34}\ m^2kg/s

m = Mass of electron = 9.11\times 10^{-31}\ kg

k = Coulomb constant = 8.99\times 10^{9}\ Nm^2/C^2

e = Charge of electron = 1.6\times 10^{-19}\ C

n = 1 (ground state)

Angular momentum is given by

L=mvr

From Bohr's atomic model we have

L=\dfrac{nh}{2\pi}

mvr=\dfrac{nh}{2\pi}\\\Rightarrow v=\dfrac{nh}{2\pi mr}

The centripetal force will balance the electrostatic force

\dfrac{ke^2}{r^2}=\dfrac{mv^2}{r}\\\Rightarrow \dfrac{ke^2}{r}=mv^2\\\Rightarrow \dfrac{ke^2}{r}=m(\dfrac{nh}{2\pi mr})^2\\\Rightarrow r=\dfrac{n^2h^2}{4\pi^2mke^2}\\\Rightarrow r=\dfrac{1^2\times (6.626\times 10^{-34})^2}{4\pi^2 \times 9.11\times 10^{-31}\times 8.99\times 10^{9}\times (1.6\times 10^{-19})^2}\\\Rightarrow r=5.30426\times 10^{-11}\ m

The diameter is 2\times 5.30426\times 10^{-11}=1.06085\times 10^{-10}\ m

7 0
3 years ago
The Steamboat Geyser in Yellowstone National Park shoots water into the air at 48.0 m/s. How
qwelly [4]

Answer:

The maximum height reached by the water is 117.55 m.

Explanation:

Given;

initial velocity of the water, u = 48 m/s

at maximum height the final velocity will be zero, v = 0

the water is going upwards, i.e in the negative direction of gravity, g = -9.8 m/s².

The maximum height reached by the water is calculated as follows;

v² = u² + 2gh

where;

h is the maximum height reached by the water

0 = u² + 2gh

0 = (48)² + ( 2 x -9.8 x h)

0 = 2304 - 19.6h

19.6h = 2304

h = 2304 / 19.6

h = 117.55 m

Therefore, the maximum height reached by the water is 117.55 m.

7 0
3 years ago
Read 2 more answers
After being struck by a bowling ball, a 1.3 kg bowling pin sliding to the right at 5.0 m/s collides head-on with another 1.3 kg
GuDViN [60]

Answer:

a) 4.2m/s

b) 5.0m/s

Explanation:

This problem is solved using the principle of conservation of linear momentum which states that in a closed system of colliding bodies, the sum of the total momenta before collision is equal to the sum of the total momenta after collision.

The problem is also an illustration of elastic collision where there is no loss in kinetic energy.

Equation (1) is a mathematical representation of the the principle of conservation of linear momentum for two colliding bodies of masses m_1 and m_2 whose respective velocities before collision are u_1 and u_2;

m_1u_1+m_2u_2=m_1v_1+m_2v_2..............(1)

where v_1 and v_2 are their respective velocities after collision.

Given;

m_1=1.3kg\\u_1=5m/s\\m_2=1.3kg\\u_2=0m/s

Note that u_2=0 because the second mass m_2 was at rest before the collision.

Also, since the two masses are equal, we can say that m_1=m_2=m so that equation (1) is reduced as follows;

mu_1+mu_2=mv_1+mv_2\\\\m(u_1+u_2)=m(v_1+v_2)..............(2)

m cancels out of both sides of equation (2), and we obtain the following;

u_1+u_2=v_1+v_2.............(3)

a) When v_1=0.8m/s, we obtain the following by equation(3)

5+0=0.8+v_2\\hence\\v_2=5-0.8\\v_2=4.2m/s

b) As m_1 stops moving v_1=0, therefore,

5+0=0+v_2\\v_2=5m/s

5 0
3 years ago
During the execution of a play, a football player carries the ball for a distance of 33 m in the direction 58° north of east. To
podryga [215]

Answer:28 m

Explanation:

Given

Direction is 58^{\circ} North of east i.e. 58 ^{\circ} with x axis

Also ball moved by 33 m

therefore its east component is 33cos58=17.48 m

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8 0
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What force is required to accelerated a body with a mass of 15 kilograms at a rate of 8 m/s2
rjkz [21]

Force  =  (mass) · (acceleration)

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           =  120 kg-m/s²  =  120 newtons
7 0
3 years ago
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