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Do Planets with more mass have more gravity than planets with less mass

STatiana [176]

Here you go it was too long to type

8 0

2 years ago

What is the best description of the end result of chemical bonding for most

forsale [732]

A. Getting a full set of valence electrons

Explanation:

The best description of the end result of chemical bonding for most atoms is the getting of a full set of valence electrons.

Atoms reacts with one another in order to complete valence electronic shell.

The valence electron shell is the outermost energy level of an atom.
It is from this energy level that electrons are lost or gained to form bonds.
All atoms wants to be like the noble gases whose valence electronic shell is completely filled up
This is the crux of chemical bonding
The attraction that is produced from the interaction leads to bond formation

learn more:

Chemical bond brainly.com/question/10903097

#learnwithBrainly

7 0

2 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

2 years ago

A horse shoe magnet is placed on a mass balance such that a uniform magnetic field of magnitude B runs between it from North to

BARSIC [14]

Answer:

(a) Measured change in mass (Δm) = BVL/Rg

(b) Total measured mass M' = M - BVL/Rg

Explanation:

Current (I) across is coil is given by the formula;

I = V/R ------------------------1

The magnetic force is given by the formula;

Fb = B*I*L -------------------2

Putting equation 1 into equation 2, we have;

Fb = B*V*L/R -------------------3

Change in mass (Δm) is given as:

Δm = Fb/g -----------------------4

Putting equation 3 into equation 4, we have;

Δm = BVL/Rg

Therefore, change in mass (Δm) = BVL/Rg

2. Since B runs from North to South and current running from East to West, then the magnetic force is directed upward.

Therefore,

Total measure mass M' = M - BVL/Rg

4 0

2 years ago

A rod of length L and mass M has a nonuniform mass distribution. The linear mass density (mass per length) is λ=cx2, where x is

grigory [225]

Answer:

Explanation:

λ=c x²

c = λ / x²

λ is mass / length

so its dimensional formula is ML⁻¹

x is length so its dimensional formula is L

c = λ / x²

= ML⁻¹ / L²

= ML⁻³

B )

We shall find out the mass of the rod with the help of given expression of mass per unit length and equate it with given mass that is M

The mass in the rod is symmetrically distributed on both side of middle point.

we consider a small strip of rod of length dx at x distance away from middle point

its mass dm = λdx = cx² dx

By integrating it from -L to +L we can calculate mass of whole rod , that is

M = ∫cx² dx

= [c x³ / 3] from -L/2 to +L/2

= c/3 [ L³/8 + L³/8]

M = c L³/12

c = 12 M L⁻³

C ) Moment of inertia of rod

∫dmx²

= ∫λdxx²

= ∫cx²dxx²

= ∫cx⁴dx

= c x⁵ / 5 from - L/2 to L/2

= c / 5 ( L⁵/ 32 +L⁵/ 32)

= (2c / 160)L⁵

= (c / 80) L⁵

= (12 M L⁻³/80)L⁵

= 3/20 ML²

=

4 0

2 years ago