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andriy [413]
2 years ago
13

If the mass of one of two particles is doubled and the distance between them is doubled, the force of attraction between the two

particles will
Physics
2 answers:
andre [41]2 years ago
7 0
The force of attraction between the two particles will remain the same, because when mass is doubled, force of attraction is doubled. However, when distance between their centers is doubled, then force of  attraction is halved. As such double and half cancel out each other and force of attraction remains the same.
iren2701 [21]2 years ago
5 0

Answer:

The force of attraction between the two particles will become half.

Explanation:

Consider force F between two objects with masses m and m' which are placed r distance apart from each other.

Mass of first object = m

Mass of second object = m'

Distance between the  two objects = r

F=G\frac{m\times m'}{r^2}...(1) (Gravitational force)

If the mass of one of two particles is doubled and the distance between them is doubled the force between them be F'.

Mass of first object = m

New mass of second object = M = 2m'

Distance between the  two objects = R = 2r

F'=G\times \frac{m\time M}{R^2}=\frac{m\times 2m'}{4r^2}=G\frac{m\times m'}{2r^2}...(2) (Gravitational force)

Dividing (1) from (2):

\frac{F}{F'}=\frac{G\frac{m\times m'}{r^2}}{G\frac{m\times m'}{2r^2}}

F=2F'

F'=\frac{F'}{2}

The force of attraction between the two particles will become half.

You might be interested in
Please help! will give brainlest!!!!!!!!!!!!
eimsori [14]

The force of friction is 19.1 N

Explanation:

According to Newton's second law, the net force acting on the bag is equal to the product between its mass and its acceleration:

\sum F = ma

where

\sum F is the net force

m is the mass

a is the acceleration

The bag is moving at constant speed, so its acceleration is zero:

a=0

Therefore the net force is zero as well:

\sum F = 0

Here we are interested only in the forces acting along the horizontal direction, therefore the net force is given by:

\sum F = F cos \theta - F_f = 0

where

F cos \theta is the horizontal component of the applied force, with

F = 22.5 N

\theta=32.0^{\circ}

F_f is the force of friction

And solving for F_f, we find

F_f =Fcos \theta=(22.5)(cos 32.0^{\circ})=19.1 N

Learn more about friction:

brainly.com/question/6217246

brainly.com/question/5884009

brainly.com/question/3017271

brainly.com/question/2235246

#LearnwithBrainly

7 0
2 years ago
Which one of the following represents an acceptable set of quantum numbers for an electron in an atom? (arranged as n, l, m l ,
Vitek1552 [10]

Answer:

The correct option that represents an acceptable set of quantum numbers for an electron in an atom is;

(b) 4, 3, -3, 1/2.

Explanation:

To solve the question, we note that the available options where the set of quantum numbers for an electron in an atom are arranged as n, l, m l , and ms are;

4, 4, 4, 1/2

4, 3, -3, 1/2

4, 3, 0, 0

4, 5, 7, -1/2

4, 4, -5, 1/2

Let us label them as a to as follows

(a) 4, 4, 4, 1/2

(b) 4, 3, -3, 1/2

(c) 4, 3, 0, 0

(d) 4, 5, 7, -1/2

(e) 4, 4, -5, 1/2

Next we note the rules for the assignment and arrangement of quantum numbers are as follows

Number                                   Symbol                Possible values

Principal Quantum Number  .......n........................1, 2, 3, ......n

Angular momentum quantum

number...............................................l.........................0, 1, 2, .......(n - 1)

Magnetic Quantum Number........m₁......................-l, ..., -1, 0, 1,.....,l  

Spin Quantum Number.................m_s.....................+1/2, -1/2

We are meant to analyze each of the arrangement for acceptability.

Therefore for (a),

we note that the angular momentum quantum number, l =4 , is equal to the principal quantum number n =4 which violates the rule as the maximum value of the angular momentum quantum number is (n-1) where the maximum value of the principal quantum number is n.

Therefore (a) is not acceptable.

(b) Here we note that

The principal quantum number n = 4 ∈ (1, 2, 3, ......n) → acceptable

The angular momentum quantum number l = 3 ∈ (0, 1, 2, .......(n - 1)) → acceptable

The magnetic quantum number m₁ = -3 ∈ (-l, ..., -1, 0, 1,.....,l)  → acceptable

The spin quantum number m_s = 1/2 ∈ (+1/2, -1/2) → acceptable

Therefore (b) 4, 3, -3, 1/2 represents an acceptable set of quantum numbers for an electron in an atom.

(c) Here we have

The principal quantum number n = 4 ∈ (1, 2, 3, ......n) → acceptable

The angular momentum quantum number l = 3 ∈ (0, 1, 2, .......(n - 1)) → acceptable

The magnetic quantum number m₁ = 0 ∈ (-l, ..., -1, 0, 1,.....,l)  → acceptable

The spin quantum number m_s = 0 ∉ (+1/2, -1/2) → not acceptable

Therefore (c) 4, 3, 0, 0 does not represents an acceptable set of quantum numbers for an electron in an atom.

(d) Here we have;

The principal quantum number n = 4 ∈ (1, 2, 3, ......n) → acceptable

The angular momentum quantum number l = 5 ∉ (0, 1, 2, .......(n - 1)) → not acceptable

The magnetic quantum number m₁ = 7 ∉ (-l, ..., -1, 0, 1,.....,l)  → acceptable

The spin quantum number m_s = -1/2 ∈ (+1/2, -1/2) → acceptable

Therefore (d) 4, 5, 7, -1/2 does not represents an acceptable set of quantum numbers for an electron in an atom.

(e) Here we have;

The principal quantum number n = 4 ∈ (1, 2, 3, ......n) → acceptable

The angular momentum quantum number l = 4 ∉ (0, 1, 2, .......(n - 1)) → not acceptable

The magnetic quantum number m₁ = -5 ∉ (-l, ..., -1, 0, 1,.....,l)  → acceptable

The spin quantum number m_s = 1/2 ∈ (+1/2, -1/2) → acceptable

Therefore (e) 4, 4, -5, 1/2 does not represents an acceptable set of quantum numbers for an electron in an atom.

3 0
2 years ago
Батискаф витримує тиск 60 МПа. Чи можна провести дослідження
Elanso [62]

1) Yes

2) 6.34\cdot 10^9 N

Explanation:

1)

To solve this part, we have to calculate the pressure at the depth of the batyscaphe, and compare it with the maximum pressure that it can withstand.

The pressure exerted by a column of fluid of height h is:

p=p_0+\rho g h

where

p_0 = 101,300 Pa is the atmospheric pressure

\rho is the fluid density

g=10 m/s^2 is the acceleration due to gravity

h is the height of the column of fluid

Here we have:

\rho=1030 kg/m^3 is the sea water density

h = 5440 m is the depth at which the bathyscaphe is located

Therefore, the pressure on it is

p=101,300+(1030)(10)(5440)=56.1\cdot 10^6 Pa = 56.1 MPa

Since the maximum pressure it can withstand is 60 MPa, then yes, the bathyscaphe can withstand it.

2)

Here we want to find the force exerted on the bathyscaphe.

The relationship between force and pressure on a surface is:

p=\frac{F}{A}

where

p is hte pressure

F is the force

A is the area of the surface

Here we have:

p=56.1\cdot 10^6 Pa is the pressure exerted

The bathyscaphe has a spherical surface of radius

r = 3 m

So its surface is:

A=4\pi r^2

Therefore, we can find the force exerted on it by re-arranging the previous equation:

F=pA=4\pi pr^2 = 4\pi (56.1\cdot 10^6)(3)^2=6.34\cdot 10^9 N

6 0
2 years ago
A wave has a frequency of 46 Hz and a wavelength of 1.7 meters. What is the wave speed wave?
arlik [135]

Answer:

Explanation:

(1.7 m/cycle)(46 cycle/s) = 78.2 m/s

4 0
2 years ago
You're a safety engineer reviewing plans for a university's new high-rise dorm. The elevator motors draw 20 A and behave electri
Leya [2.2K]

Answer:

a) R `= 3.5 ohms

b) energy decipated =   560J

Explanation:

V = I . R

R = V / I

R `= 70 / 20

R `= 3.5 ohms

2)energy decipated = 1/2ij²

energy decipated = 1/2 x 2.8 x (20)²

energy decipated =   560J

3 0
2 years ago
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