Which is the correct unit for electric power

Which planet formed near the sun where the solar system temperatures were very hugh

Gnom [1K]

mercury is the closest planet from the sun

8 0

3 years ago

Read 2 more answers

Calculate the magnitude of the force between two 3.60 mC point challenges 9.3 cm apart

vladimir1956 [14]

Q = magnitude of charge on each of the two point charge = 3.60 mC = 3.60 x 10⁻³ C

r = distance between the two point charges = 9.3 cm = 0.093 m

k = constant = 9 x 10⁹ Nm²/C²

F = magnitude of the force between the two point charges = ?

according to coulomb's law , force between two charges is given as

F = k Q²/r²

inserting the values

F = (9 x 10⁹) (3.60 x 10⁻³)²/(0.093)²

F = 1.35 x 10⁷ N

3 0

3 years ago

A moon orbits a planet every 42 hours with a mean orbital radius of .002819 AU. The mass of the moon is 8.932 x 1022 kg. Using N

Pepsi [2]

Answer:

The mass of the planet is $1.9407\times10^{27}\ kg$

Explanation:

Given that,

Time period = 42 hours = 151200 sec

Orbital radius = 0.002819 AU = 421716397.5 m

Mass of moon $m=8.932\times10^{22}\ kg$

We need to calculate the mass of the planet

Using Kepler’s third law

$T^2\propto a^3$

$T^2=\dfrac{4\pi^2}{G(M+m)}\times a^3$

Where, a = orbital radius

T = time period

G = gravitational constant

M = mass of moon

m = mass of planet

Put the value into the formula

$(151200)^2=\dfrac{4\pi^2}{6.673\times10^{-11}(8.932\times10^{22}+m)}\times(421716397.5)^3$

$(8.932\times10^{22}+m)=\dfrac{4\pi^2}{6.673\times10^{-11}}\times\dfrac{(421716397.5)^3}{(151200)^2}$

$(8.932\times10^{22}+m)=1.94087\times10^{27}$

$m=1.94087\times10^{27}-8.932\times10^{22}$

$m=1.9407\times10^{27}\ kg$

Hence, The mass of the planet is $1.9407\times10^{27}\ kg$

8 0

4 years ago

How do the meetings of the words exercise and fitness differ?

ICE Princess25 [194]

Exercise is the activity and fitness is a lifestyle and done with time

7 0

3 years ago

A 10.0g piece of copper wire, sitting in the sun reaches a temperature of 80.0 C. how many Joules are released when the copper c

Zolol [24]

Answer:

150.8 J

Explanation:

The heat released by the copper wire is given by:

$Q=mC_s \Delta T$

where:

m = 10.0 g is the mass of the wire

Cs = 0.377 j/(g.C) is the specific heat capacity of copper

$\Delta T=40.0 C - 80.0 C=-40.0 C$ is the change in temperature of the wire

Substituting into the equation, we find

$Q=(10.0 g)(0.377 J/gC)(-40.0^{\circ})=-150.8 J$

And the sign is negative because the heat is released by the wire.

6 0

3 years ago