Increasing the what of two objects will cause the gravitational force between the objects to decrease

How is this formula ( Rt= r1 x r2 / r1 + r2) which is used to calculate total resistance of two parallel resistors found from th

Maru [420]

Explanation:

The total resistance used to calculate the total resistance of the two parallel resistor is given by :

$R_t=\dfrac{r_1\times r_2}{r_1+r_2}$

Taking reciprocal of the above equation.

$\dfrac{1}{R_t}=\dfrac{r_1+r_2}{r_1\times r_2}$

We can also write the above equation as :

$\dfrac{1}{R_t}=\dfrac{r_1}{r_1\times r_2}+\dfrac{r_2}{r_1\times r_2}$

On simplification of above equation,

$\dfrac{1}{R_t}=\dfrac{1}{r_2}+\dfrac{1}{r_1}$

or

$\dfrac{1}{R_t}=\dfrac{1}{r_1}+\dfrac{1}{r_2}$

Hence, proved.

4 0

3 years ago

According to the concept of length contraction, what happens to the length of an object as it approaches the speed of light and

lord [1]

When an object moves its length contracts in the direction of motion. The faster it moves the shorter it gets in the direction of motion.
The object in this question moves and then stops moving. So it's length first contracts and then expands to its original length when the motion stops.
The speed doesn't have to be anywhere near the speed of light. When the object moves its length contracts no matter how fast or slow it's moving.

8 0

4 years ago

You measure an electric field of 1.36×106 N/C at a distance of 0.158 m from a point charge. There is no other source of electric

marysya [2.9K]

Answer:

The Electric flux will be $0.42\times10^6\ \rm N.m^2/C$

Explanation:

Given

Strength of the Electric Field at a distance of 0.158 m from the point charge is

$E=1.36\times10^6\ \rm N/C$

We know that the flux of the Electric Field can be calculated by using Gauss Law which is given by

$\int E.dA=\dfrac{q_{in}}{\epsilon_0}\\$

Let consider a sphere of radius 0.158 m as Gaussian Surface at a distance of 0.158 m from the point charge and Let $\phi$ be the flux of the Electric Field coming out\passing through it which is given by

$\phi=\int E.dA=1.36\times10^6 \times4\pi \times 0.158^2\\\\=0.42\times10^6\ \rm N.m^2/C$

It can be observed that same amount of flux which is passing through the Gaussian sphere of radius 0.158 is also passing through the Gaussian sphere of radius 0.142 m at a distance of 0.142 m from its centre.

Also it can be observed that the charge inside the two Gaussian Sphere mentioned have same value so the Flux of electric field through them will also be same.

So the electric flux through the surface of sphere that has given charge at its centre and that has radius 0.142 m is $0.42\times10^6\ \rm N.m^2/C$

8 0

3 years ago

If the length of a building is 2 1 2 times the width and each dimension is increased by 7 ft, then the perimeter is 266 ft. Find

N76 [4]

Answer:

The original dimensions of the building is 95 ft × 38 ft.

Explanation:

Let the original length be 'l' and original width be 'w'.

Given:

Original length (l) = $2\frac{1}{2}\times original\ width$