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3 years ago

most responsible for the formation of a star is most responsible for the formation of a star is blank Force

1 answer:

5 0

That would be gravitational force. Check out creation.com. Search up "star-formation" in their search bar. Hope this helped.

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What is the period of a wave traveling 5 m/s if its wavelength is 20 m/s

Ilia_Sergeevich [38]

Speed of wave is given as

$v = 5 m/s$

Wavelength of the wave is given as

$\lambda = 20 m$

now from the formula of wave time period we can say

$speed = \frac{wavelength}{time period}$

$5 = \frac{20}{T}$

$T = \frac{20}{5}$

$T = 4 s$

so it will have time period of T = 4 s

7 0

3 years ago

What is a non contact force that attracts all objects to the centre of the earth

beks73 [17]

Answer:

Gravity. The weight-force or weight of an object is the force because of Gravity, which acts on the object attracting it towards the centre of the earth.

Hope this helps, 

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4 0

2 years ago

Friction is generated when ______ interfere with each other on sliding surfaces. ...?

cricket20 [7]

According to its definition, friction is generated when atoms interfere with each other on sliding surfaces.

5 0

3 years ago

Please help

JulsSmile [24]

Answer:

Explanation:

because it is the only one that has something to do with heat keyword would be boiling

7 0

3 years ago

A long, thin rod parallel to the y-axis is located at x = - 1 cm and carries a uniform positive charge density λ = 1 nC/m . A se

zheka24 [161]

Answer:

The electric field at origin is 3600 N/C

Solution:

As per the question:

Charge density of rod 1, $\lambda = 1\ nC = 1\times 10^{- 9}\ C$

Charge density of rod 2, $\lambda = - 1\ nC = - 1\times 10^{- 9}\ C$

Now,

To calculate the electric field at origin:

We know that the electric field due to a long rod is given by:

$\vec{E} = \frac{\lambda }{2\pi \epsilon_{o}{R}$

Also,

$\vec{E} = \frac{2K\lambda }{R}$ (1)

where

K = electrostatic constant = $\frac{1}{4\pi \epsilon_{o} R}$

R = Distance

$\lambda$ = linear charge density

Now,

In case, the charge is positive, the electric field is away from the rod and towards it if the charge is negative.

At x = - 1 cm = - 0.01 m:

Using eqn (1):

$\vec{E} = \frac{2\times 9\times 10^{9}\times 1\times 10^{- 9}}{0.01} = 1800\ N/C$

$\vec{E} = 1800\ N/C$ (towards)

Now, at x = 1 cm = 0.01 m :

Using eqn (1):

$\vec{E'} = \frac{2\times 9\times 10^{9}\times - 1\times 10^{- 9}}{0.01} = - 1800\ N/C$

$\vec{E'} = 1800\ N/C$ (towards)

Now, the total field at the origin is the sum of both the fields:

$\vec{E_{net}} = 1800 + 1800 = 3600\ N/C$

7 0

3 years ago