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3 years ago

What is the altitude of the Sun at noon on December 22, as seen from a place on the Tropic of Cancer?

Physics

1 answer:

scZoUnD [109]3 years ago

4 0

Answer:

108.217 °

Explanation:

Day of year = 356 = d (Considering year of 365 days)

Latitude of Tropic of Cancer = 23.5 °N

Declination angle

δ = 23.45×sin[(360/365)(d+284)]

⇒δ = 23.45×sin[(360/365)(356+284)]

⇒δ = 5.2832 °

Altitude angle at solar noon

90+Latitude-Declination angle

= 90+23.5-5.2832

= 108.217 °

∴ Altitude angle of the Sun as seen from the tropic of cancer on December 22 is 108.217 °

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A wire is stretched between two posts. Another wire is stretched between two posts that are twice as far apart. The tension in t

Margarita [4]

Answer: 996m/s

Explanation:

Formula for calculating velocity of wave in a stretched string is

V = √T/M where;

V is the velocity of wave

T is tension

M is the mass per unit length of the wire(m/L)

Since the second wire is twice as far apart as the first, it will be L2 = 2L1

Let V1 and V2 be the speed of the shorter and longer wire respectively

V1 = √T/M1... 1

V2 = √T/M2... 2

Since V1 = 249m/s, M1 = m/L1 M2 = m/L2 = m/2L1

The equations will now become

249 = √T/(m/L1) ... 3

V2 = √T/(m/2L1)... 4

From 3,

249² = TL1/m...5

From 4,

V2²= 2TL1/m... 6

Dividing equation 5 by 6 we have;

249²/V2² = TL1/m×m/2TL1

{249/V2}² = 1/2

249/V2 = (1/2)²

249/V2 = 1/4

V2 = 249×4

V2 = 996m/s

Therefore the speed of the wave on the longer wire is 996m/s

3 0

3 years ago

Consider a cyclotron in which a beam of particles of positive charge q and mass m is moving along a circular path restricted by

Ulleksa [173]

A) $v=\sqrt{\frac{2qV}{m}}$

B) $r=\frac{mv}{qB}$

C) $T=\frac{2\pi m}{qB}$

D) $\omega=\frac{qB}{m}$

E) $r=\frac{\sqrt{2mK}}{qB}$

Explanation:

When the particle is accelerated by a potential difference V, the change (decrease) in electric potential energy of the particle is given by:

$\Delta U = qV$

where

q is the charge of the particle (positive)

On the other hand, the change (increase) in the kinetic energy of the particle is (assuming it starts from rest):

$\Delta K=\frac{1}{2}mv^2$

where

m is the mass of the particle

v is its final speed

According to the law of conservation of energy, the change (decrease) in electric potential energy is equal to the increase in kinetic energy, so:

$qV=\frac{1}{2}mv^2$

And solving for v, we find the speed v at which the particle enters the cyclotron:

$v=\sqrt{\frac{2qV}{m}}$

When the particle enters the region of magnetic field in the cyclotron, the magnetic force acting on the particle (acting perpendicular to the motion of the particle) is

$F=qvB$

where B is the strength of the magnetic field.

This force acts as centripetal force, so we can write:

$F=m\frac{v^2}{r}$

where r is the radius of the orbit.

Since the two forces are equal, we can equate them:

$qvB=m\frac{v^2}{r}$

And solving for r, we find the radius of the orbit:

$r=\frac{mv}{qB}$ (1)

The period of revolution of a particle in circular motion is the time taken by the particle to complete one revolution.

It can be calculated as the ratio between the length of the circumference ( $2\pi r$ ) and the velocity of the particle (v):

$T=\frac{2\pi r}{v}$ (2)

From eq.(1), we can rewrite the velocity of the particle as

$v=\frac{qBr}{m}$

Substituting into(2), we can rewrite the period of revolution of the particle as:

$T=\frac{2\pi r}{(\frac{qBr}{m})}=\frac{2\pi m}{qB}$

And we see that this period is indepedent on the velocity.

The angular frequency of a particle in circular motion is related to the period by the formula

$\omega=\frac{2\pi}{T}$ (3)

where T is the period.

The period has been found in part C:

$T=\frac{2\pi m}{qB}$

Therefore, substituting into (3), we find an expression for the angular frequency of motion:

$\omega=\frac{2\pi}{(\frac{2\pi m}{qB})}=\frac{qB}{m}$

And we see that also the angular frequency does not depend on the velocity.

For this part, we use again the relationship found in part B:

$v=\frac{qBr}{m}$

which can be rewritten as

$r=\frac{mv}{qB}$ (4)

The kinetic energy of the particle is written as

$K=\frac{1}{2}mv^2$

So, from this we can find another expression for the velocity:

$v=\sqrt{\frac{2K}{m}}$

And substitutin into (4), we find:

$r=\frac{\sqrt{2mK}}{qB}$

So, this is the radius of the cyclotron that we must have in order to accelerate the particles at a kinetic energy of K.

Note that for a cyclotron, the acceleration of the particles is achevied in the gap between the dees, where an electric field is applied (in fact, the magnetic field does zero work on the particle, so it does not provide acceleration).

6 0

3 years ago

Physics Homework

katrin2010 [14]

Explanation:

a. Average speed = distance / time

= 100 m / 70 s

= 1.43 m/s

b. Average displacement = displacement / time

= 0 m / 70 s

= 0 m/s

Distance is the length of the path traveled. Displacement is the difference between the final position and initial position.

4 0

3 years ago

if i roll pieces of allium foil in a ball what property of aluminium foil changes as a result of my action

MArishka [77]

Answer:

Malleability

Explanation:

The ability to roll aluminum foil in a ball is the property of metal called malleability.

Metallic bonds in metals makes it possible for them to take different shape without losing internal cohesion.
Metals are malleable and they posses a wide range of physical properties.
A malleable metal can be beaten into sheets and molded into many shapes.

6 0

3 years ago

An electric pump rated at 1000W, takes 16 seconds to pump 100 kg of water out to a tank at a height of 10 m. what is the efficie

TEA [102]

Answer:

$\boxed{\sf Efficiency \ of \ the \ motor = 62.5 \ \%}$