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soldier1979 [14.2K]

3 years ago

11

Name the four major forces in the universe that act over long distance as I greater than the nucleus of an atom?

1 answer:

GalinKa [24]3 years ago

4 0

Forces in the universe that act over long distance, meaning the distance is greater than the diameter of the nucleus of the atom are:

1. Electrostatic force or Coulomb force: Fc=(k*Q₁*Q₂)/r²,

2. Gravitational force: Fg=(G*m₁*m₂)/r²,

3. Magnetic force: Fm=qvB,

4. London dispersion force, also known as one of the van der Waals forces.

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Andrew [12]

The last one, the soil will become weak & unable to support plant growth

4 0

3 years ago

A loader sack of total mass

vampirchik [111]

Question: A loader sack of total mass

is l000 grams falls down from

the floor of a lorry 200 cm high

Calculate the workdone by the

gravity of the load.

Answer:

19.6 Joules

Explanation:

Applying

W = mgh........................ Equation 1

Where W = Workdone by gravity on the load, m = mass of the loader sack, h = height, g = acceleration due to gravity

From the question,

Given: m = 1000 grams = (1000/1000) kilogram = 1 kg, h = 200 cm = 2 m

Constant: g = 9.8 m/s²

Substitute these values into equation 1

W = (1×2×9.8)

W = 19.6 Joules

Hence the work done by gravity on the load is 19.6 Joules

8 0

3 years ago

What is the relationship of an object's acceleration, its mass and the force applied to it?

Anon25 [30]

Answer:

F = ma

Explanation:

Newtons second law

8 0

2 years ago

If earth stops rotating then what will be the value of 'g' at poles

Marianna [84]

The value of 'g' is not affected by rotation at any place on Earth.

7 0

3 years ago

The reflective surface of a CD consists of spirals of equally spaced grooves. If you shine a laser pointer on a CD, each groove

Ipatiy [6.2K]

Answer:

d = 1.55 * 10⁻⁶ m

Explanation:

To calculate the distance between the adjacent grooves of the CD, use the formula, $d = \frac{m \lambda}{sin(A_{m}) }$ ..........(1)

The fringe number, m = 1 since it is a first order maximum

The wavelength of the green laser pointer, $\lambda$ = 532 nm = 532 * 10⁻⁹ m

Distance between the central maximum and the first order maximum = 1.1 m

Distance between the screen and the CD = 3 m

$A_{m}$ = Angle between the incident light and the diffracted light

From the setup shown in the attachment, it is a right angled triangle in which

$sin(A_{m}) = \frac{opposite}{Hypotenuse} \\sin(A_{m}) =\frac{1.1}{\sqrt{1.1^{2}+3^{2}}}$

$sin(A_{m} ) = 0.344\\A_{m} = sin^{-1} 0.344\\A_{m} = 20.14^{0}$

Putting all appropriate values into equation (1)

$d = \frac{1* 532*10^{-9} }{0.344 }\\d = 0.00000155 m\\d = 1.55 * 10^{-6} m$

3 0

3 years ago