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stira [4]
3 years ago
12

How many minutes of daylight do we gain after winter solstice

Physics
1 answer:
serg [7]3 years ago
4 0

Answer:

Depends on which hemisphere you are belong to and how much distance you are away from Ecuador.

Explanation:

Minutes of daylight is equal on everywhere only on the equinox days (21 March and 23 September). On other days it depends on the place that you are belong to. On winter solstice, places on Ecuador have 12 hours daylight. North side of Ecuador have less, south side of Ecuador have more hour of daylight.

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A planar loop consisting of seven turns of wire, each of which encloses 200 cm2, is oriented perpendicularly to a magnetic field
PtichkaEL [24]

Answer:

The induced current is 0.084 A

Explanation:

the area given by the exercise is

A = 200 cm^2 = 200x10^-4 m^2

R = 5 Ω

N = 7 turns

The formula of the emf induced according to Faraday's law is equal to:

ε = (-N * dφ)/dt = (N*(b2-b1)*A)/dt

Replacing values:

ε = (7*(38 - 14) * (200x10^-4))/8x10^-3 = 0.42 V

the induced current is equal to:

I = ε /R = 0.42/5 = 0.084 A

3 0
3 years ago
Read 2 more answers
Red light of wavelength 651 nm produces photoelectrons from a certain photoemissive material. Green light of wavelength 521 nm p
Mnenie [13.5K]

Answer:

material work function is 0.956 eV

Explanation:

given data

red wavelength 651 nm

green wavelength 521 nm

photo electrons = 1.50 × maximum kinetic energy

to find out

material work function

solution

we know by Einstein photo electric equation  that is

for red light

h ( c / λr ) = Ф +  kinetic energy

for green light

h ( c / λg ) = Ф +  1.50 × kinetic energy

now from both equation put kinetic energy from red to green

h ( c / λg ) = Ф +  1.50 × (h ( c / λr ) - Ф)

Ф =( hc / 0.50) × ( 1.50/ λr  - 1/ λg)

put all value

Ф =( 6.63 ×10^{-34} (3 ×10^{8} )  / 0.50) × ( 1.50/ λr  - 1/ λg)

Ф =( 6.63 ×10^{-34} (3 ×10^{8} ) / 0.50 ) × ( 1.50/ 651×10^{-9}   - 1/ 521 ×10^{-9})

Ф = 1.5305  ×10^{-19} J  × ( 1ev / 1.6 ×10^{-19} J )

Ф = 0.956 eV

material work function is 0.956 eV

4 0
3 years ago
Read 2 more answers
A parallel plate capacitor creates a uniform electric field of and its plates are separated by . A proton is placed at rest next
zalisa [80]

Complete Question

A parallel plate capacitor creates a uniform electric field of 5 x 10^4 N/C and its plates are separated by 2 x 10^{-3}'m. A proton is placed at rest next to the positive plate and then released and moves toward the negative plate. When the proton arrives at the negative plate, what is its speed?

Answer:

V=1.4*10^5m/s

Explanation:

From the question we are told that:

Electric field B=1.5*10N/C

Distance d=2 x 10^{-3}

At negative plate

Generally the equation for Velocity is mathematically given by

V^2=2as

Therefore

V^2=\frac{2*e_0E*d}{m}

V^2=\frac{2*1.6*10^{-19}(5*10^4)*2 * 10^{-3}}{1.67*10^{-28}}

V=\sqrt{19.2*10^9}

V=1.4*10^5m/s

5 0
3 years ago
Which competition is the “ most important soccer competition in the world “ ?
Ganezh [65]
It would be B. The World Cup

Hope this helps

Have a great day/night
8 0
3 years ago
A charge moves a distance of 1.8 cm in the direction of a uniform electric field having a magnitude of 214 N/C. The electrical p
Andrei [34K]

Answer:

 13.4 x 10 raise to power -19  C

Explanation:

. The distance moved by a charge in the direction of a uniform electric field is d= 1.8 cm =0.018 m

. The uniform electric field is  E = 214 N/M

, The decrease in electrical potential energy is   d(P.E) = 51.63 x 10 raise to power -19 J

Let the magnitude of the charge of the moving particle be q

which is given by the equation

d(P.E) =qEd

51.63 x 10 power -19 = q(214)(0.018)

51.63 x 10 power -19 =3.852q

by making q the formular,

q = 13.4 x 10 power -19 C  

5 0
3 years ago
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