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Zolol [24]
3 years ago
10

What will be the ratio of distances between the two charges of each pair of charges (1µC, 2µC) and (2µC, -3uC) so that force her

ween them is equal Use Gauss's law to derive an expression for the electric field due to an infinitely long straight wire of linear
charge density cm-1


20points!!!!!!!
​
Physics
1 answer:
Alenkinab [10]3 years ago
4 0

Answer: Anurag Mishra - Problems in Physics - Electricity and Magnetism ... Between two infinitely long wires having linear charge densities}. and -].·there are two points A ... the ratio of the electric force between them to t:-:c grav:tadonal force between them? (a) 10 8 ... d between the first two charges on the straight line at a distance

Explanation:

MORE POWER

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Name TWO WEAKNESSES of the model pictured below
Nikolay [14]

Answer:

Here are a few:

1) The orbital radius of these planets is ridiculously small an in no way representative of their actual radii.

2) The planets will only line up like that once every 5200 years, making this very unrepresentative of their usual relations - although this does make their order in distance from the sun.

3) The nebulae, comet, lens flare,  and other junk in the background is incorrect.

4) If this is meant as a representation of the planets, then Pluto should not be there as it is now considered a planetoid.

5) The planets are incorrectly scaled both to each other and to the sun.

7 0
2 years ago
Your teacher (175kg) and the lab (15kg) are 2m apart. What is the gravitational force between them? Draw a FBD, and label
Pachacha [2.7K]
The gravitational force between two objects is given by
F=G \frac{m_1 m_2 }{r^2}
where
G is the gravitational constant
m1 and m2 are the masses of the two objects
r is the separation between the two objects

In this problem, m_1 = 175 kg, m_2 =15 kg and r=2 m, therefore the gravitational force between the two objects is
F=(6.67 \cdot 10^{-11} m^3 kg^{-1} s^{-2}) \frac{(175 kg)(15 kg)}{(2m)^2}=4.38 \cdot10^{-8} N
7 0
2 years ago
A player kicks a soccer ball from ground level and sons at flying at an angle of 30° at a speed of 26MS how far did the ball tra
RSB [31]

Answer:

Explanation:

60 meters is he answer for this question

7 0
3 years ago
An electric dipole is formed from ±1.00nC charges spaced 3.00 mm apart. The dipole is at the origin, oriented along the x-axis.
Ronch [10]

Answer:

Value of electric field along the axis and equitorial axis  E=31.25\ N/c and E = 15.625\ N/c respectively.

Explanation:

Given :

Distance between charges , d = 3 \ mm =\dfrac{3}{1000}\ m=3\times 10^{-3}\ m.

Magnitude of charges , q=1\ nC = 10^{-9}\ C.

Dipole moment , p=qL=10^{-9}\times 3\times 10^{-3}=3\times 10^{-12} \ C\ m.

Case A) (x,y) = (12.0 cm, 0 cm) :

Electric field of dipole in its axis ,

E=\dfrac{2kp}{r^3}

Putting all values and r=12\times 10^{-2}\ m.

We get , E=31.25\ N/c.

Case B) (x,y) = (0 cm, 12.0 cm) :

Electric field of dipole on equitorial axis ,

E = \dfrac{kp}{r^3}

Putting all values and r=12\times 10^{-2}\ m.

We get , E = 15.625\ N/c.

Hence , this is the required solution.

7 0
3 years ago
What is the electric force acting between two charges of -0.0045 C and -0.0025 C that are 0.0060 m apart? Use Fe=kq1q2/r^2 and k
oksano4ka [1.4K]

Answer:

D. 2.8 × 10⁹ N

Explanation:

The force between two charges is directly proportional to the amount of charges at the two points and inversely proportional to the square of distance between the two points.

Fe= k Q₁Q₂/r²

Q₁= -0.0045 C

Q₂= -0.0025 C

r= 0.0060 m

k= 9.00 × 10 ⁹ Nm²/C²

Fe= (9.00 × 10 ⁹ Nm²/C²×-0.0045 C×-0.0025 C)/0.0060²

=2.8 × 10⁹ N

4 0
3 years ago
Read 2 more answers
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