What variable(s) determine the amount of potential energy an object has?

Calculate the force between two small charged spheres having charges of 3×10−7 C and 4×10−7 C placed 20 cm apart.

xxMikexx [17]

Answer:

27 . 10^-7 or 27/1000

Explanation:

We use the Coulomb Law

k = Coloumb Constant

q1 and q2 are the charges

d is the distance between the spheres

3 0

2 years ago

The diagram below shows the heating of an unknown substance (it is not water). Its melting point is? The diagram below shows the

Anettt [7]

Answer:

the answer is B

Melting point, temperature at which the solid and liquid forms of a pure substance can exist in equilibrium. As heat is applied to a solid, its temperature will increase until the melting point is reached. More heat then will convert the solid into a liquid with no temperature change.

4 0

2 years ago

"One of the main projects being carried out by the Hubble Space Telescope is to measure the distances of galaxies located in gro

pickupchik [31]

Answer:

finding Cepheid variable and measuring their periods.

Explanation:

This method is called finding Cepheid variable and measuring their periods.

Cepheid variable is actually a type of star that has a radial pulsation having a varying brightness and diameter. This change in brightness is very well defined having a period and amplitude.

A potent clear link between the luminosity and pulsation period of a Cepheid variable developed Cepheids as an important determinants of cosmic criteria for scaling galactic and extra galactic distances. Henrietta Swan Leavitt revealed this robust feature of conventional Cepheid in 1908 after observing thousands of variable stars in the Magellanic Clouds. This in fact turn, by making comparisons its established luminosity to its measured brightness, allows one to evaluate the distance to the star.

5 0

2 years ago

Returning once again to our table top example of a horizontal mass on a low-friction surface with m = 0.254 kg and k = 10.0 N/m

Julli [10]

Explanation:

Given that,

Mass = 0.254 kg

Spring constant [tex[\omega_{0}= 10.0\ N/m[/tex]

Force = 0.5 N

y = 0.628

We need to calculate the A and d

Using formula of A and d

$A=\dfrac{\dfrac{F_{0}}{m}}{\sqrt{(\omega_{0}^2-\omega^{2})^2+y^2\omega^2}}$ .....(I)

$tan d=\dfrac{y\omega}{(\omega^2-\omega^2)}$ ....(II)

Put the value of $\omega=0.628\ rad/s$ in equation (I) and (II)

$A=\dfrac{\dfrac{0.5}{0.254}}{\sqrt{(10.0^2-0.628)^2+0.628^2\times0.628^2}}$

$A=0.0198$

From equation (II)

$tan d=\dfrac{0.628\times0.628}{((10.0^2-0.628)^2)}$

$d=0.0023$

Put the value of $\omega=3.14\ rad/s$ in equation (I) and (II)

$A=\dfrac{\dfrac{0.5}{0.254}}{\sqrt{(10.0^2-3.14)^2+0.628^2\times3.14^2}}$

$A=0.0203$

From equation (II)

$tan d=\dfrac{0.628\times3.14}{((10.0^2-3.14)^2)}$

$d=0.0120$

Put the value of $\omega=6.28\ rad/s$ in equation (I) and (II)

$A=\dfrac{\dfrac{0.5}{0.254}}{\sqrt{(10.0^2-6.28)^2+0.628^2\times6.28^2}}$

$A=0.0209$

From equation (II)

$tan d=\dfrac{0.628\times6.28}{((10.0^2-6.28)^2)}$

$d=0.0257$

Put the value of $\omega=9.42\ rad/s$ in equation (I) and (II)

$A=\dfrac{\dfrac{0.5}{0.254}}{\sqrt{(10.0^2-9.42)^2+0.628^2\times9.42^2}}$

$A=0.0217$

From equation (II)

$tan d=\dfrac{0.628\times9.42}{((10.0^2-9.42)^2)}$

$d=0.0413$

Hence, This is the required solution.

5 0

2 years ago

A runner accelerated to 4 m/s^2 for 20 seconds before winning the race.How far did he/she run?

creativ13 [48]

Answer:

s=800 m

Explanation:

Given that,

Acceleration of a runner, a = 4 m/s²

Time, t = 20 seconds

We need to find the distance covered by her. Initially, she was at rest. It means its initial velocity is equal to 0. So, using second equation of motion as follows :

$s=ut+\dfrac{1}{2}at^2$

Herre, u = 0

$s=\dfrac{1}{2}at^2\\\\s=\dfrac{1}{2}\times 4\times (20)^2\\\\s=800\ m$

So, she will cover a distance of 800 m.

8 0

3 years ago