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rewona [7]
4 years ago
11

Two objects attract each other gravitationally with a force of 2.5 x 10^-10N when they are 0.25 m apart. Their total mass is 4.0

0 kg. Find their individual masses.
Physics
1 answer:
In-s [12.5K]4 years ago
6 0

Answer:

M = 3.9406 kg and m = 0.0594 kg

Explanation:

The gravitational force between two bodies is directly proportional to the product of their masses and inversely proportional to the square of the distance that separates them. Mathematically it is expressed as follows:

Fg = (G×M×m)/r²    Formula (1)

Where:

Fg is the gravitational force (N)

G is the universal gravitation constant, G = 6.67 × 10⁻¹¹ (N×m²)/kg²

M and m are the masses of the bodies that interact  (kg).

r is the distance that separates them (m).

Known Data

Fg = 2.5 × 10⁻¹⁰ N

r = 0.25 m

G = 6.67 × 10⁻¹¹ (N×m²)/kg²

Problem development

We propose 2 equations

M + m = 4kg

M = 4 - m equation (1)

We replace in formula (1)

2.5 × 10⁻¹⁰ = (6.67 × 10⁻¹¹ × M × m)/(0.25)²

2.5 × 10⁻¹⁰ × (0.25)² = (6.67 × 10⁻¹¹ × M × m)

(2.5 × 10⁻¹⁰ × (0.25)²)/(6.67 × 10⁻¹¹) = M × m

M × m = 0.234 equation (2)

We replace M = 4 - m in equation (2)

(4 - m) × m = 0.234

4m - m² = 0.234

m² - 4m + 0.234 = 0 (quadratic equation)

We apply the formula for the quadratic equation and obtain 2 values for m that meet the conditions:

m = 3.9406 kg or m =  0.0594 kg

We replace m in equation (1)

M = 4 - 3.9406 = 0.0594 kg or M = 4 - 0.0594 = 3.9406

To meet the condition that M + m must give 4 kg, one mass must be equal 3.9406 and the other must equal 0.0594, then:

M = 3.9406 kg and m = 0.0594 kg

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If the coefficient of kinetic friction between a 22 kg kg crate and the floor is 0.27, what horizontal force is required to move
alina1380 [7]

Answer:

58.27 N

Explanation:

the data we have is:

mass: m=22kg

coefficient of friction: \mu =0.27

and we also know the acceleration of gravity is g=9.81m/s^2

We need to do an analysis of horizontal and vertical forces acting on the object:

-------

Vertically the forces acting on the object:

  • Normal force N (acting up from the object)
  • weight: w=mg (acting down from)

so the sum of forces in the vertical axis "y" are:

F_{y}=N-w\\F_{y}=N-mg

from Newton's second Law we know that F=ma, so:

ma_{y}=N-mg

and since the object is not accelerating in the vertical direction (the movement is only horizontal) a_{y}=0, and:

0=N-mg\\N=mg

-----------

now let's analyze the horizontal forces

  • frictional force: f= \mu N and since N=mg  --> f=\mu mg
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and the two forces just mentioned must be opposite, thus the sum of forces in the "x" axis is:

F=ma_{x}=F-f\\ma_{x}=F-\mu mg

and we are told that the crate moves at a steady speed, thus there is no acceleration: a_{x}=0

and we get:

0=F-\mu mg\\F=\mu mg

substituting known values:

F=(0.27)(22kg)(9.81m/s^2)\\F=58.27N

3 0
3 years ago
an airplane is flying through a thundercloud at a height of 2000m (this is very dangerous thing to do because of updrafts, turbu
Vedmedyk [2.9K]

Answer

In this question we have given,

Height of plane, h1=2000m

Height at which charge concentration is 40C, h2=3000m

Height at which charge concentration is -40C, h3=1000m

charge concentaration, q1=40C

charge concentaration, q2=-40C

let the charge concentrations at height h2 and h3 as point charges

Now we will first find the electric feild on plane due to positive charge q1=40

E1= k*q1/(h1-h2)..............(1)

Here k=8.98755*10^9N.m^2/C^2

q1=40C

put values of k, q1 , h1 and h2 in equation 1


[tex]E1=(8.98755*10^9)*(40)/(2000-3000)^2\\

E1=[tex]E= 359502+359502\\E=719004 V/mV/m[/tex]

similarly electric feild due to negative charge q2=-40

[tex]E2=(8.98755*10^9)*(-40)/(2000-1000)^2\\

E2=359502V/m

Total electric feild E at the aircraft is given as

E= E1+ E2\\...............(2)

Put values of  E1 and E2 in equation2

\\E=359502+359502\\E= 719004V/m\\

therefore s Total electric feild E at the aircraft is E= 719004V/m

3 0
3 years ago
When the plutonium bomb was tested in New Mexico in 1945, approximately 1 gram of matter was converted into energy. Suppose anot
gayaneshka [121]

Answer:

The value is E =  1.35 *10^{14} \ J

Explanation:

From the question we are told that

    The mass of matter converted to energy on first test is  m  =  1 \  g  = 0.001 \  kg

    The mass of matter converted to energy on second test m_1 =  1.5 \  g = 1.5 *10^{-3} \ kg

    Generally the amount of energy that was released by  the explosion is  mathematically  represented as  

         E =  m * c^2

=>       E =  1.5 *10^{-3}  * [ 3.0 *10^{8}]^2

=>       E =  1.35 *10^{14} \ J

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A charged particle enters into a uniform magnetic field such that its velocity vector is perpendicular to the magnetic field vec
Ugo [173]

Answer:

a circular path

Explanation:

In a magnetism field if a charged particle having a charge of magnitude '' enters  such that its velocity vector V is 90° to the direction of the magnetic field "B'', then it will experience a force, called Lorentz force F

F = V\times B

According to the property of cross-product, the Lorentz force (F) acting on the particle will be perpendicular to the instantaneous position of the particle, making the path of the particle to be a circular path.

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Two separate but nearby coils are mounted along the same axis. A power supply controls the flow of current in the first coil, an
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The ammeter will indicate that a current is flowing in the second coil only when current changes in the first coil.

<h3>Induced emf</h3>

An emf is induced in a coil placed in a magnetic field when a current carrying conductor moves in the field.

emf = NdФ/dt

where;

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Thus, the ammeter will indicate that a current is flowing in the second coil only when current changes in the first coil.

Learn more about induced emf here: brainly.com/question/13744192

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3 0
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