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2 years ago

Refractive index measures the speed of light in a vacuum to its speed in

1 answer:

8 0

Refractive index of a medium = speed of light in vacuum
-------------------------------------
speed of light in that medium

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While standing at the edge of the roof of a building, a man throws a stone upward with an initial speed of 6.95 6.95 m/s. The st

qwelly [4]

Answer:

18.1347 m/s

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

g = Acceleration due to gravity = 9.81 m/s² = a

$v^2-u^2=2as\\\Rightarrow s=\dfrac{v^2-u^2}{2a}\\\Rightarrow s=\dfrac{0^2-6.95^2}{2\times -9.81}\\\Rightarrow s=2.4619\ m$

Total height the ball falls is 2.4619+14.3 = 16.7619 m

$v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times 9.81\times 16.7619+0^2}\\\Rightarrow v=18.1347\ m/s$

The speed at which the stone reaches the ground is 18.1347 m/s

8 0

3 years ago

What is the kinetic energy of a golf ball with a mass of 0.046 kg traveling at 15.5<br> m/s?

Serjik [45]

Answer:

Explanation:

The kinetic energy of an object can be found by using the formula

$k = \frac{1}{2} m {v}^{2} \\$

m is the mass

v is the velocity

From the question we have

$k = \frac{1}{2} \times 0.046 \times {15.5}^{2} \\ = 0.023 \times 240.25 \\ = 5.52575$

We have the final answer as

Hope this helps you

3 0

2 years ago

Show that the electric potential along the axis of a uniformly charged disk of radius R and charge density sigma is given by by

OlgaM077 [116]

Explanation:

Area of ring $\ 2{\pi} a d a$

Charge of on ring $d q=-(\ 2{\pi} a d a)$

Charge on disk

$Q=-\left(\pi R^{2}\right)$

$\begin{aligned}d v &=\frac{k d q}{\sqrt{x^{2}+a^{2}}} \\&=2 \pi-k \frac{a d a}{\sqrt{x^{2}+a^{2}}} \\v(1) &=2 \pi c k \int_{0}^{R} \frac{a d a}{\sqrt{x^{2}+a^{2}}} \cdot_{2 \varepsilon_{0}}^{2} R \\&=2 \pi \sigma k[\sqrt{x^{2}+a^{2}}]_{0}^{2} \\&=\frac{2 \pi \sigma}{4 \pi \varepsilon_{0}}[\sqrt{z^{2}+R^{2}}-(21)] \\&=\frac{\sigma}{2}(\sqrt{2^{2}+R^{2}}-2)\end{aligned}$

Note: Refer the image attached

8 0

3 years ago

A red water balloon is thrown horizontally from the top of a bridge. At the same instant, a yellow water balloon is dropped off

Advocard [28]

The time of fall for the two balloons is equal.

Here the conditions given resembles a free fall. Since the air resistance is negligible, gravitational force is the only force acting on both the balloons which means the two balloons are falling in a vacuum. In a vacuum, if two objects of different mass are thrown from the same height, both objects will reach the ground at the same time.

Free fall is nothing but a condition in which an object falls solely under the influence of gravity.

Therefore, the time of fall for both yellow and red water balloons is equal no matter the weight where air resistance is negligible.

To know more about free fall

brainly.com/question/13299152

#SPJ4

7 0

1 year ago

A hollow non-conducting spherical shell has inner radius R1 = 5 cm and outer radius R2 = 19 cm. A charge Q = -35 nC lies at the

Gnom [1K]

Answer:

a. E = -13.8 kN/C

b. E = +8.51 kN/C

Explanation:

We will apply Gauss' Law to the regions where the electric field is asked.

Gauss' Law states that if you draw an imaginary surface enclosing a charge distribution, then the electric field through the imaginary surface is equal to the total charge enclosed by this surface divided by electric permittivity.

$\int\vec{E}d\vec{a} = \frac{Q_{\rm enc}}{\epsilon_0}$

a. For this case, we will draw the imaginary surface between the inner and outer shell of the sphere. The total charge enclosed by this surface will be equal to the sum of the charge Q at the center and charge of the shell within the volume from R1 and r.

Here, r = 0.5(R1+R2) = 12 cm.

$E4\pi r^2 = \frac{Q_{\rm enc}}{\epsilon_0}\\Q_{\rm enc} = Q + \rho V_{\rm enc} = Q + (Ar) (\frac{4}{3}\pi (r^3 - R_1^3)) = (-35\times 10^{-9}) + (16\times 10^{-6})(12\times 10^{-2})(\frac{4}{3}\pi((12\times 10^{-2})^3 - (5\times 10^{-2})^3)) = -2.21\times 10^{-8}~C$

$E = \frac{-2.21\times 10^{-8}}{4\pi (12\times10^{-2})^2 \epsilon_0} = -1.38\times 10^4~N/C\\E = -13.8~kN/C$

b. For this case, we will draw the imaginary surface on the outside of the shell.

The total charge enclosed by this surface will be equal to the sum of the charge at the center and the total charge of the shell.

$Q_{\rm enc} = Q + \rho V = Q + (Ar)[\frac{4}{3}\pi (R_2^3 - R_1^3)]\\Q_{\rm enc} = (-35\times 10^{-9}) + [(16\times 10^{-6})(38\times 10^{-2})][\frac{4}{3}\pi((19\times 10^{-2})^3 - (5\times 10^{-2})^3)]\\Q_{\rm enc} = 1.36\times 10^{-7}~C$

$E = \frac{1.36\times 10^{-7}}{4\pi (38\times10^{-2})^2 \epsilon_0} = 8.51\times 10^3~N/C\\E = 8.51~kN/C$

7 0

3 years ago