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3 years ago

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The two forces in each pair can either both act on the same body or they can act on different bodies. The two forces in each pai

r can either both act on the same body or they can act on different bodies.
a.true
b.false

1 answer:

Marrrta [24]3 years ago

6 0

Answer:

The correct answer in which peer forces each act in a different body

Explanation:

Newton's law of action and reaction states that the two forces have equal magnitude, but in the opposite direction, each acting on a body. Which creates acceleration in these, if they act in the same body it would be canceled so there would be no movement.

The correct answer in which peer forces each act in a different body

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1. Two wires - A and B - with circular cross-sections have identical lengths and are made of the same material. Yet, wire A has

PolarNik [594]

Answer:

Diameter of wire B is 2 times the diameter of wire A

Explanation:

We have given two wire A and B

They are made up of same material and length of both the wire is sane

So $l_A=l_B$

Let the resistivity of both the wire is $\rho$

It is given that wire A has 4 times the resistance as wire B

So $R_A=4R_B$

So $\frac{\rho l_A}{a_A}=4\frac{\rho l_B}{a_B}$ ( As $l_A=l_B$ )

$\frac{a_A}{a_B}=\frac{1}{4}$

$\frac{d_A^2}{d_B^2}=\frac{1}{4}$

$\frac{d_A}{d_B}=\frac{1}{2}$

$d_B=2d_A$

So diameter of wire B is 2 times the diameter of wire A

6 0

2 years ago

What is the strength of the electric field ep 0.90 mm from a proton?

dem82 [27]

The electric field strength is inversely related to the square of the distance.so the strength of the electric field is given by

$E=\frac{1}{4\pi \epsilon _{0} } \frac{q}{r^{2} } = \frac{k q}{r^{2} }$

Here, $\frac{1}{4\pi \epsilon _{0} } = k$ is constant depend upon medium and its value is $9.0 \times10^{9} \ N m^2/C^2$ and q is charge and r is the distance.

Given $r = 0.90 mm = 9.0 \times 10^{-4} m$ and we know the charge of proton, $q = 1.6\times 10^{-19} \ C$ .

Therefore,

$E=\frac{9.0 \times10^{9} \ N m^2/C^2 \times1.6\times 10^{-19} \ C }{(9.0 \times 10^{-4} m)^2} = 0.177 \times 10^{-2} \ N/C \\\\ E= 1.77 \times 10^{-3} N/C$

4 0

3 years ago

Which of the following shows kinetic energy being converted into potential energy?

Komok [63]

B) a rock being tossed high into the air

3 0

3 years ago

Given three different locations on Earth's surface, where will the weight of a person be greatest?

Feliz [49]

Answer:

Explanation:

In order to answer this question, we simply have to refer to the laws of the equations of gravitational mechanics.

The equation given by Newton tells us that

$F = \frac{Gm_{1} m_{2} }{r^{2} }$

In the case where we compare a specific place where the Force of Gravity is greater or lesser, we focus on the term assigned to the Planet's Radius.

In the case of $G, m_{1} ,m_{2}$ , we understand that they are constant.

We can easily notice that the more the Radius (Height seen from a viewer on the ground), the lower the force will be.

In other words, the smaller the radius in which the measurement is made with respect to the center of the earth, the greater the gravitational force.

In that order of ideas the smallest radio has South Pole, which is about 6356 km from the center of the Earth on the Equator line

4 0

2 years ago

An electric dipole is formed from ±1.00nC charges spaced 3.00 mm apart. The dipole is at the origin, oriented along the x-axis.

Ronch [10]

Answer:

Value of electric field along the axis and equitorial axis $E=31.25\ N/c$ and $E = 15.625\ N/c$ respectively.

Explanation:

Given :

Distance between charges , $d = 3 \ mm =\dfrac{3}{1000}\ m=3\times 10^{-3}\ m.$

Magnitude of charges , $q=1\ nC = 10^{-9}\ C.$

Dipole moment , $p=qL=10^{-9}\times 3\times 10^{-3}=3\times 10^{-12} \ C\ m.$

Case A) (x,y) = (12.0 cm, 0 cm) :

Electric field of dipole in its axis ,

$E=\dfrac{2kp}{r^3}$

Putting all values and $r=12\times 10^{-2}\ m.$

We get , $E=31.25\ N/c.$

Case B) (x,y) = (0 cm, 12.0 cm) :

Electric field of dipole on equitorial axis ,

$E = \dfrac{kp}{r^3}$

Putting all values and $r=12\times 10^{-2}\ m.$

We get , $E = 15.625\ N/c.$

Hence , this is the required solution.

7 0

3 years ago