What resistance would produce a current of 120A from a 6-volt battery?

The Moon requires about 1 month (0.08 year) to orbit Earth. Its distance from us is about 400,000 km (0.0027 AU). Use Kepler’s t

dem82 [27]

Answer:

$\frac{M_e}{M_s} = 3.07 \times 10^{-6}$

Explanation:

As per Kepler's III law we know that time period of revolution of satellite or planet is given by the formula

$T = 2\pi \sqrt{\frac{r^3}{GM}}$

now for the time period of moon around the earth we can say

$T_1 = 2\pi\sqrt{\frac{r_1^3}{GM_e}}$

here we know that

$T_1 = 0.08 year$

$r_1 = 0.0027 AU$

$M_e$ = mass of earth

Now if the same formula is used for revolution of Earth around the sun

$T_2 = 2\pi\sqrt{\frac{r_2^3}{GM_s}}$

here we know that

$r_2 = 1 AU$

$T_2 = 1 year$

$M_s$ = mass of Sun

now we have

$\frac{T_2}{T_1} = \sqrt{\frac{r_2^3 M_e}{r_1^3 M_s}}$

$\frac{1}{0.08} = \sqrt{\frac{1 M_e}{(0.0027)^3M_s}}$

$12.5 = \sqrt{(5.08 \times 10^7)\frac{M_e}{M_s}}$

$\frac{M_e}{M_s} = 3.07 \times 10^{-6}$

4 0

3 years ago

A gas sample is confined within a chamber that has a movable piston. A small load is placed on the piston; and the system is all

timurjin [86]

Explanation:

It is given that,

Total weight of the piston, W = F = 70 N

Area of the piston, $a=5\times 10^{-4}\ m^2$

Let P is the pressure exerted on the piston by the gas. The force per unit area is called the pressure exerted pressure of the gas. Mathematically, it is given by :

$P=\dfrac{F}{A}$

$P=\dfrac{70\ N}{5\times 10^{-4}\ m^2}$

$P=1.4\times 10^5\ Pa$

We know that the atmospheric pressure is given by :

$P_o=1.013\times 10^5\ Pa$

So, the pressure is given by :

$p=P+P_o$

$p=1.4\times 10^5+1.013\times 10^5$

$p=2.41\times 10^5\ Pa$

Hence, this is the required solution.

7 0

3 years ago

What does strong gravity mean?<br> PLZZZZZ HELP MEEEEEE

grigory [225]

<span>The force of attraction by which terrestrial bodies tend to fall toward the center of the earth strongly. If i'm wrong, correct me. otherwise i hope this helped.</span>

7 0

3 years ago

Read 2 more answers

A horizontal rectangular surface has dimensions 2.80 cm by 3.20 cm and is in a uniform magnetic field that is directed at an ang

koban [17]

Answer:

magnitude of the magnetic field 0.692 T

Explanation:

given data

rectangular dimensions = 2.80 cm by 3.20 cm

angle of 30.0°

produce a flux Ф = 3.10 × $10^{-4}$ Wb

solution

we take here rectangular side a and b as a = 2.80 cm and b = 3.20 cm

and here angle between magnitude field and area will be ∅ = 90 - 30

∅ = 60°

and flux is express as

flux Ф = $\int \vec{B}.d\vec{A}$ .................1

and Ф = BA cos∅ ............2

so B = $\frac{\phi }{Acos\theta }$

and we know

A = ab

so

B = $\frac{\phi }{abcos\theta }$ ..............3

put here value

B = $\frac{3.10\times 10^{-4} }{2.80 \times 10^{-2}\times 3.20 \times 10^{-2}\times cos60}$

solve we get

B = 0.692 T

8 0

3 years ago

The maximum height a typical person can jump is about 60cm (0.6m). By how much does the gravitational potential energy increase

sasho [114]

The gravitational potential energy will increase by 423.36 J

<h3>How to determine the potential energy at ground level</h3>

Mass (m) = 72 kg
Acceleration due to gravity (g) = 9.8 m/s²
Height (h) = 0 m
Potential energy at ground level (PE₁) =?

PE = mgh

PE₁ = 72 × 9.8 × 0

PE₁ = 0 J

<h3>How to determine the potential energy at 60 cm (0.6 m)</h3>

Mass (m) = 72 kg
Acceleration due to gravity (g) = 9.8 m/s²
Height (h) = 0.6 m
Potential energy at 60 cm (0.6 m) (PE₂) =?

PE = mgh

PE₂ = 72 × 9.8 × 0.6

PE₂= 423.36 J

<h3>How to determine the change in potential energy </h3>

Potential energy at ground level (PE₁) = 0 J
Potential energy at 60 cm (0.6 m) (PE₂) = 423.36 J
Change in potential energy =?

Change in potential energy = PE₂ - PE₁

Change in potential energy = 423.36 - 0

Change in potential energy = 423.36 J

Learn more about energy:

brainly.com/question/10703928

#SPJ1

8 0

1 year ago