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Gemiola [76]
3 years ago
12

Which force is always pulling but never pushes and is directly influenced by mass

Physics
2 answers:
Ne4ueva [31]3 years ago
8 0

Answer:

i believe it is gravity

Explanation:

the gravitational force is a force that attracts any two objects with mass. it pulls things together does not pushes them.

Lapatulllka [165]3 years ago
7 0
<h2>Answer: I know when it comes to magnetic objects the magnet always pulls not push.</h2>
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Two tiny conducting spheres are identical and carry charges of -19.8μC and +40.7μC. They are separated by a distance of 3.59 cm.
romanna [79]

Answer:

(a): \rm -5.627\times 10^3\ N.

(b):  \rm 7.626\times 10^2\ N.

Explanation:

<u>Given:</u>

  • Charge on one sphere, \rm q_1 = -19.8\ \mu C = -19.8\times 10^{-6}\ C.
  • Charge on second sphere, \rm q_2 = +40.7\ \mu C = +40.7\times 10^{-6}\ C.
  • Separation between the spheres, \rm r=3.59\ cm = 3.59\times 10^{-2}\ m.

Part (a):

According to Coulomb's law, the magnitude of the electrostatic force of interaction between two static point charges is given by

\rm F=k\cdot\dfrac{q_1q_2}{r^2}

where,

k is called the Coulomb's constant, whose value is \rm 9\times 10^9\ Nm^2/C^2.

From Newton's third law of motion, both the spheres experience same force.

Therefore, the magnitude of the force that each sphere experiences is given by

\rm F=k\cdot\dfrac{q_1q_2}{r^2}\\=9\times 10^9\times \dfrac{(-19.8\times 10^{-6})\times (+40.7\times 10^{-6})}{(3.59\times 10^{-2})^2}\\=-5.627\times 10^3\ N.

The negative sign shows that the force is attractive in nature.

Part (b):

The spheres are identical in size. When the spheres are brought in contact with each other then the charge on both the spheres redistributes in such a way that the net charge on both the spheres distributed equally on both.

Total charge on both the spheres, \rm Q=q_1+q_2=-19.8\ \mu C+40.7\ \mu C = 20.9\ \mu C.

The new charges on both the spheres are equal and given by

\rm q_1'=q_2'=\dfrac Q2 = \dfrac{20.9}{2}\ \mu C=10.45\ \mu C = 10.45\times 10^{-6}\ C.

The magnitude of the force that each sphere now experiences is given by

\rm F'=k\cdot \dfrac{q_1'q_2'}{r^2}'\\=9\times 10^9\times \dfrac{10.45\times 10^{-6}\times 10.45\times 10^{-6}}{(3.59\times 10^{-2})^2}\\=7.626\times 10^2\ N.

7 0
3 years ago
Consider two soap bubbles with radius r1 and r2 (r1 &lt;r2) connected via a valve. What happens if we open the valve​
Sedaia [141]

Complete Question

The  complete question is shown on the first uploaded image

Answer:

The pressure difference of the first bubble is   \Delta  P _1 =10  J/m^3

The pressure difference of the second bubble is  \Delta  P _2 =20  J/m^3

The pressure difference on the second bubble is higher than that of the first bubble so when the valve is opened pressure from second bubble will cause air to flow toward the first bubble making is bigger

Explanation:

From the question we are told that

    The  radius of the first bubble is  r_1 =  10 \ mm=0.01 \ m

      The radius of the second bubble is  r_2  =  5 \ mm  =  0.005 \ m

      The surface tension of the soap solution is  s =  25 \ mJ/m^2 = 25*10^{-3} J/m^2

Generally according to the Laplace's Law for a spherical membrane the pressure difference is mathematically represented as

         \Delta  P  =  \frac{4 s}{R}

Now the pressure difference for the first bubble is  mathematically evaluated as

        \Delta  P _1 =  \frac{4 s}{r_1}

substituting values  

       \Delta  P _1 =  \frac{4 *25 *10^{-3}}{0.01}

       \Delta  P _1 =10  J/m^3

Now the pressure difference for the second bubble is  mathematically evaluated as

        \Delta  P _2 =  \frac{4 s}{r_1}

       \Delta  P _2 =  \frac{4 *25 *10^{-3}}{0.005}

       \Delta  P _2 =20  J/m^3

3 0
3 years ago
A man walks 400 m in the direction 45° north of east. Represent this vector graphically by
Marrrta [24]

Answer:

Explanation:

Coordinate system is one that describe the location of an object in a given plane. It implies the use of axes (coordinates) and points.

Given that the man in the question walks 400 m due north of east. The cardinal points can be used in this case, with the north and east cardinals as the required axis.

scale = \frac{length on drawing}{original length}

         = \frac{10}{400}

         = \frac{1}{40}

scale = 1:40

This is a reduced scale which implies that 1 cm on the drawing is equal to 40 m on the original length.

The man's direction is 45^{o} north of east.

The graphical drawing of the vector is herewith attached to this answer.

5 0
3 years ago
The cross sectional area of Sphere 2 is increased to 3 times the cross sectional area of Sphere 1. The masses remain 1.0 kg and
ira [324]

Complete question is;

a. Two equal sized and shaped spheres are dropped from a tall building. Sphere 1 is hollow and has a mass of 1.0 kg. Sphere 2 is filled with lead and has a mass of 9.0 kg. If the terminal speed of Sphere 1 is 6.0 m/s, the terminal speed of Sphere 2 will be?

b. The cross sectional area of Sphere 2 is increased to 3 times the cross sectional area of Sphere 1. The masses remain 1.0 kg and 9.0 kg, The terminal speed (in m/s) of Sphere 2 will now be

Answer:

A) V_t = 18 m/s

B) V_t = 10.39 m/s

Explanation:

Formula for terminal speed is given by;

V_t = √(2mg/(DρA))

Where;

m is mass

g is acceleration due to gravity

D is drag coefficient

ρ is density

A is Area of object

A) Now, for sphere 1,we have;

m = 1 kg

V_t = 6 m/s

g = 9.81 m/s²

Now, making D the subject, we have;

D = 2mg/((V_t)²ρA))

D = (2 × 1 × 9.81)/(6² × ρA)

D = 0.545/(ρA)

For sphere 2, we have mass = 9 kg

Thus;

V_t = √[2 × 9 × 9.81/(0.545/(ρA) × ρA))]

V_t = 18 m/s

B) We are told that The cross sectional area of Sphere 2 is increased to 3 times the cross sectional area of Sphere 1.

Thus;

Area of sphere 2 = 3A

Thus;

V_t = √[2 × 9 × 9.81/(0.545/(ρA) × ρ × 3A))]

V_t = 10.39 m/s

5 0
4 years ago
Law of motion that says for every action there is an equal and opposite
hodyreva [135]

Answer: Newton's third law

Formally stated, Newton's third law is: For every action, there is an equal and opposite reaction. The statement means that in every interaction, there is a pair of forces acting on the two interacting objects. The size of the forces on the first object equals the size of the force on the second object.

Explanation:

4 0
3 years ago
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