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Orlov [11]
3 years ago
8

If the new moon happens on January 15th, what shape will it be on February 6th?

Physics
1 answer:
Jobisdone [24]3 years ago
3 0

-- From January 15 to February 6 is a period of 22 days.

-- The period of the full cycle of moon phases is 29.53 days.

-- So those dates represent (22/29.53) = 74.5% of a full cycle of phases.

-- That's almost exactly 3/4 of a full cycle, so on February 6, the moon would be almost exactly at <em>Third Quarter</em>.  That's the <em>left half of a disk </em>(viewed from the northern hemisphere).

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hope this helps it's F

8 0
3 years ago
2. the dipole moment of a dipole in a 300-n/c electric field is initially perpendicular to the field, but it rotates so it is in
aleksandr82 [10.1K]

Work Done (W) by the field is-6x 107 J,

<h3>What is Electric dipole?</h3>

A pair of opposite, non-coplanar, equally powerful electric charges that are in opposition to one another. An atom is said to have a "induced electric dipole" if the center of the negative cloud of electrons has moved a little bit away from the nucleus due to an external electric field. When the external field is taken away, dipolarity is lost.

Electric field (E) = 300 N/C

Dipole moment (p) = 2 x 10° Cm

Solution:

From the formula we know.

U = -pE cosФ

Here,

p Denotes Dipole moment.

E Denotes Electric field.

Ф Denotes angle b/w them

Now, as given, firstly the dipole is perpendicular to the electric field, so

angle (Ф1) will be 90° and now the dipole is rotated such that they are in same

direction so the angle (Ф2) will be 0°

So, let's find Change in Potential energy which will be equal to the work done

by the electric field.

ΔU = Uf - Ui

ΔU = [-pE*cos Ф2] - [-pE *cos Ф1]

ΔU = [-pE*cos Ф2] + pE *cos Ф1

ΔU = pE * [cos Ф2+ cos Ф1]

Substituting the values,

ΔU = pE * [cos 0° + cos 90°]

ΔU = pE * (-1 +0)

ΔU = -pE

ΔU = -2x 10^-9 × 300

ΔU = 6 x 10^(-9+2)

ΔU = 6 x 10^-7

W = ΔU = -6 x 10^-7

W = - 6 x 10 7 J

Work Done (W) by the field is - 6 x 10-7 J.

To learn more about Work Done visit:

https://brainly.in/question/48222628

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6 0
1 year ago
8. A sprinter on a school track team is running north at a velocity of 6.0 m/s. After 5.0 s, she
Marysya12 [62]

Answer:

acc. = 4-(-6) /5= 10/5=2 m/s^2

6 0
2 years ago
​A piston–cylinder assembly contains 5.0 kg of air, initially at 2.0 bar, 30 oC. The air undergoes a process to a state where th
vlada-n [284]

Answer:

Explanation:

The process is isothermic,  as P V = constant .

work done = 2.303 n RT log P₁ / P₂

= 2.303 x 5 / 29 x 8.3 x 303  log 2 / 1 kJ

= 300.5k J

This energy in work done by the gas will come fro heat supplied as internal energy is constant due to constant temperature.

heat supplied  = 300.5k J

specific volume is volume per unit mass

v / m

pv = n RT

pv  = m / M  RT

v / m = RT / p M

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option B is correct.

5 0
3 years ago
A long, thin straight wire with linear charge density λ runs down the center of a thin, hollow metal cylinder of radius R. The c
Delvig [45]

Answer:

E=\dfrac{\lambda }{2\pi \varepsilon _or}

Explanation:

Given that

For straight wire

Charge density= λ

For hollow metal cylinder

Charge density=2 λ

We know that electric filed for wire given as

E_w=\dfrac{\lambda_{wire} }{2\pi \varepsilon _or}

E_w=\dfrac{\lambda }{2\pi \varepsilon _or}

Now the electric filed due to hollow metal cylinder

E_c=\dfrac{\lambda_{cylinder} }{2\pi \varepsilon _or}

E_c=\dfrac{2\lambda }{2\pi \varepsilon _or}

Now  by considering the Gaussian surface r<R then only electric fild due to wire will present.So

At r<R

E=\dfrac{\lambda }{2\pi \varepsilon _or}

5 0
3 years ago
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