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Gnom [1K]
2 years ago
14

Time period of a simple pendulum is measured at Karachi. What change will occur in the time period, if it is measured on mount e

verest, explain?
Physics
1 answer:
Reil [10]2 years ago
5 0
The period of a pendulum is given by T=2 \pi  \sqrt{ \frac{l}{g} }
since Karachi is near sea level, g is larger than it is on Mt. Everest.  That means the pendulum will have a larger period on Mt. Everest than it would in Karachi.
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3. A ray of light consisting of blue light (wavelength 480 nm) and red light (wavelength 670 nm) is incident on a thick piece of
Alex Ar [27]

Answer:

The angular separation between the refracted red and refracted blue beams while they are in the glass is 42.555 - 42.283 = 0.272 degrees.

Explanation:

Given that,

The respective indices of refraction for the blue light and the red light are 1.4636 and 1.4561.

A ray of light consisting of blue light (wavelength 480 nm) and red light (wavelength 670 nm) is incident on a thick piece of glass at 80 degrees.

We need to find the angular separation between the refracted red and refracted blue beams while they are in the glass.

Using Snell's law for red light as :

n_1\sin\theta_1=n_2\sin\theta_2\\\\\theta_2=\sin^{-1}((\dfrac{n_2}{n_1})\sin\theta_1)\\\\\theta_2=\sin^{-1}((\dfrac{1}{1.4561})\sin(80))\\\\\theta_2=42.555

Again using Snell's law for blue light as :

n_1\sin\theta_1=n_2\sin\theta'_2\\\\\theta'_2=\sin^{-1}((\dfrac{n_2}{n_1})\sin\theta_1)\\\\\theta'_2=\sin^{-1}((\dfrac{1}{1.4636 })\sin(80))\\\\\theta'_2=42.283

The angular separation between the refracted red and refracted blue beams while they are in the glass is 42.555 - 42.283 = 0.272 degrees.

7 0
3 years ago
A cylinder of mass 250 kg and radius 2.60 m is rotating at 4.00 rad/s on a frictionless.
aleksandrvk [35]

Answer:

The angular momentum of a cylinder, when it is rotating with constant angular velocity is Lini =Iωi

. When two cylinders are added to the rotating cylinder, which are identical in their dimensions, the moment of inertia of the entire system increases (since mass increases). The final moment of inertia will be 3I

Since friction exist, all the cylinders start rotating with same angular velocity, the new angular velocity can be calculated using conservation of angular momentum

Thus, Iωi =3Iωf ⟹ωf =ωi/3 = 0.33ωi

8 0
3 years ago
If an atomic nucleus were the size of a dime how far away might one of its electrons be?
mihalych1998 [28]
The radius of a nucleus of hydrogen is approximately r_{n1}=1\cdot 10^{-15}m, while we can use the Borh radius as the distance of an electron from the nucleus in a hydrogen atom: r_{e1}=5.3 \cdot 10^{-11}m

The radius of a dime is approximately r_{n2} = 9\cdot 10^{-3}m: if we assume that the radius of the nucleus is exactly this value, then we  can find how far is the electron by using the proportion
r_{n1}:r_{e1}=r_{n2}:r_{e2}
from which we find
r_{e2}= \frac{r_{e1} r_{n2}}{r_{n1}}= \frac{(5.3 \cdot 10^{-11}m)(9\cdot 10^{-3}m)}{1 \cdot 10^{-15}m}=477 m

So, if the nucleus had the size of a dime, we would find the electron approximately 500 meters away.
6 0
3 years ago
13. Now we will examine the electric field of a dipole. The magnitude and direction of the electric field depends on the distanc
Gennadij [26K]

This question is incomplete, the complete question is;

Now we will examine the electric field of a dipole. The magnitude and direction of the electric field depends on the distance and the direction. We will investigate in detail just two directions. With charges available in the simulation (all the charges are either positive or negative 1 nC increments).

how do you create a dipole with dipole moment 1 x 10-9 Cm with a direction for the dipole moment pointing to the right. Make a table below that shows the amounts of  charge and the distance between the charges. There are many correct answers

Answer:

Given the data in question;

Dipole moment P = 1 × 10⁻⁹ C.m

now dipole pointing to the right;

               P→

_{-\theta } (-) ---------------->(+) _{+\theta }

               d

so let distance between the dipoles be d

∴ P = d\Theta

Let \Theta_{1} = 1 nC

so

P = d\Theta

1 × 10⁻⁹ =  1 × 10⁻⁹  × d

d = (1 × 10⁻⁹) / (1 × 10⁻⁹)

d = 1 m

Also Let \Theta_{2} = 2 nC

so

P = d\Theta

1 × 10⁻⁹ =  2 × 10⁻⁹  × d

d = (1 × 10⁻⁹) / (2 × 10⁻⁹)  

d = 0.5 m

Also Let \Theta_{3} = 3 nC

so

P = d\Theta

1 × 10⁻⁹ =  3 × 10⁻⁹  × d

d = (1 × 10⁻⁹) / (3 × 10⁻⁹)

d = 0.33 m

such that;

charge                 distance

1 nC                        1.00 m      

2 nC                       0.50 m

3 nc                        0.33 m

4 nC                       0.25 m

5 nC                       0.20 m      

5 0
3 years ago
A 0.2 kg block of ice at 0°C is placed into a Styrofoam calorimeter cup with an unknown mass of water at 20°C. When thermal equi
Zanzabum

Answer:

Mass of the metal, m = 0.20 kg = 200 g

Initial temperature of the metal, T

1

= 150

o

C

Final temperature of the metal, T

2

= 40

o

C

Calorimeter has water equivalent of mass, m = 0.025 kg = 25 g

Volume of water, V = 150 cm

3

Mass (M) of water at temperature T = 27

o

C:

150×1=150g

Fall in the temperature of the metal:

ΔT

m

=T

1

-T

2

=150−40=110

o

C

Specific heat of water, C

w

=4.186J/g/K

Specific heat of the metal =C

Heat lost by the metal, =mCT .... (i)

Rise in the temperature of the water and calorimeter system: T

1

−T=40−27=13

o

C

Heat gained by the water and calorimeter system: =m

1

C

w

T=(M+m)C

w

T ....(ii)

Heat lost by the metal = Heat gained by the water and colorimeter system

mCΔT

m

=(M+m)C

w

T

w

200×C×110=(150+25)×4.186×13

C=(175×4.186×13)/(110×200)=0.43Jg

−1

k

−1

If some heat is lost to the surroundings, then the value of C will be smaller than the actual value.

4 0
3 years ago
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