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worty [1.4K]
3 years ago
7

Explain why during a cold winter, the air temperatures are generally higher on snowy days than on clear days.

Physics
1 answer:
Aleks04 [339]3 years ago
4 0
I don’t know that is weird
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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

#SPJ4

6 0
2 years ago
When a vehicle travels at a steady speed the driving force and the
Ivanshal [37]

Answer:

.......

Must be equal

Explanation:

..

3 0
3 years ago
Date<br>Page<br>What are the advantages of alcohol<br>thermometric liquid?<br>​
Leya [2.2K]

Answer:

low freezing point. high vapour pressure.

<em>HOPE</em><em> </em><em>IT</em><em> </em><em>WILL</em><em> </em><em>HELP</em><em> </em><em>U</em><em>! </em><em>!</em><em>!</em><em>!</em><em>!</em><em>!</em>

3 0
3 years ago
An electric toy with a resistance of 2.50 Ω is operated by a 3.00-V battery. (a) What current does the toy draw? (b) Assuming th
Leno4ka [110]

Answer:

a) The current is i = 1.2 A

b) The charge is Q = 17280 C

c) The energy is E = 43200 J

Explanation:

a) The current is given by the ohm's law wich is:

i = V/R = 3/2.5 = 1.2 A

b) Since the charge is steady we can use the following equation to find the charge amount in that time:

i = Q/t

Q = t*i

Where t is in seconds, so we have 4h * 3600 = 14400 s

Q = 1.2*14400 = 17280 C

c)  The energy is the power delivered to the toy multiplied by the time:

P = 1.2*2.5 = 3 W

E = P*t = 3*14400 = 43200 J

7 0
3 years ago
How do resistors in parallel affect the total resistance?
4vir4ik [10]

Answer:

They're going to increase the total resistance as R_{T} = \sum\limits_{i=1}^N \left(\frac{1}{R_i} \right)^{-1}

Explanation:

If the resistors are in parallel, the potential difference is the same for each resistor. But the total current is the sum of the currents that pass through each of the resistors. Then

I = I_1 + I_2 + ... + I_N

where

I_i = \frac{V_i}{R_i}

but

V_i = V_j = V for i,j= 1, 2,..., N

so

I = \frac{V}{R_1}+ \frac{V}{R_2} + ... + \frac{V}{R_N} = \left(\frac{1}{R_1} +\frac{1}{R_2} + ... + \frac{1}{R_N}\right)V = \frac{V}{R_T}

where

R_T = \left(\frac{1}{R_1} +\frac{1}{R_2} + ... + \frac{1}{R_N}\right)^{-1} =\sum\limits_{i=1}^N \left(\frac{1}{R_i} \right)^{-1}

4 0
3 years ago
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