All categories

4 years ago

How can you make the moon go around in a bigger circle

Physics

2 answers:

zlopas [31]4 years ago

5 0

You would have to give it more mechanical energy.

Like, strap a bunch of powerful rockets to one side of the moon, with all of them pointing in the direction that the moon is already moving in its orbit. Then blast away.

NOTE: There aren't enough rockets or rocket fuel on Earth to make a difference, even if you used ALL of them. The mass of the moon is about

<em>73,476,730,900,000,000,000,000 kilograms</em>

(rounded to the nearest hundred trillion kilograms.)

That's a lot.

Alex_Xolod [135]4 years ago

5 0

I doubt it if you can try

You might be interested in

Any clue on this one

aniked [119]

3rd one I believe so

6 0

4 years ago

Gold is the most ductile of all metals. For example, one gram of gold can be drawn into a wire 2.05 km long. The density of gold

arsen [322]

Answer:

Resistance of gold wire, $R=1977 \times 10^3 ohm$

Explanation:

In this question we have given

Density of gold, $d=19.3\times 10^3 \frac{kg}{m^3}$

resistivity of gold, $r=2.44\times 10^{-8} ohm.m$

Length of wire, $L= 2.05 km$

Temperature, $T= 20^oC$

We know that relation between volume and density is given as

$Density= \frac{mass}{Volume}$

Therefore, volume occupied by one gram gold is given as,

$V=\frac{.001 kg}{19.3\times 10^3 Kg m^{-3}} = 5.181\times 10^{-8} m^3$ .........(1)

We Know that Volume of gold wire which is cylindrical in shape is given by following formula

$V=\pi \times r^2 \times L$ ......(2)

Here,

$A= \pi \times r^2$ ...........(3)

here A is the cross sectional area of cylendrical gold wire

From equation 2 and 3

we got

$V=A \times L$ ...............(4)

on comparing equation 1 and equation 4, we got,

$A \times L=5.181\times 10^{-8} m^3$

$A=\frac{5.181\times 10^{-8} m^3}{2050 m}$

$A=2.53\times 10^{-11}m^2$

we know that resistance and resistivity are related by following formula,

$Resistance = resistivity\times \frac{L}{A}$ ................(5)

Put values of resistivity, A and L in equation 5, we got

$R = \frac{2.44 \times 10^{-8} ohm.m \times 2050 m}{2.53\times 10^{-11} m^2}$

$R=1977 \times 10^3 ohm$

Therefore resistance of gold wire, $R=1977 \times 10^3 ohm$

7 0

3 years ago

A 0.30 kg ball of putty is thrown at a stationary 1.2 kg cart. The ball of putty is initially moving with a speed of 6.0 m/s. Th

Lemur [1.5K]

<span>In order to determine the speed of the entire assembly, we employ conservation of momentum. Momentum p = mv where m is the object's mass and v is the velocity.
The putty ball's initial momentum p1 = 0.3kg*6m/s = 1.8 kg*m/s
That momentum is conserved, so the momentum of the new system having mass 0.3 kg + 1.2 kg = 1.5 kg is:
1.8 kg*m/s = 1.5kg*v. Solving for v, we find that the velocity is 1.2 meters/second.</span>

3 0

3 years ago

A house is maintained at 1 atm and 24°C. Outdoor air at 5.5°C infiltrates into the house through the cracks and warm air inside

Lunna [17]

Answer:

5454gjb

Explanation:

512212njun][[]

;;

3 0

3 years ago

A block of wood of mass 300g and density 0.75 g/cm^3 is floating on the surace of a liquid of density 1.1 g/cm^3. What mass of l

Travka [436]

Answer:

The minimum mass of the lead for the combination to submerge is 155 g.

Explanation:

let M be the mass of the wood.

let m be the minimum mass of lead to be added for the combination to submerge.

let ρ1 be the density of the liquid.

let ρ2 be the density of the wood.

let ρ3 be the density of lead.

let g be the gravitational acceleration.

For the combination to submerge, the weight of the wood combined with the weight of lead should at least be equal to the buyant force, that is:

weight of wood and lead = buyant force

g×(M+m) = g×(ρ1)×(M/ρ2 + m/ρ3)

M+m = (ρ1)×(M/ρ2 + m/ρ3)

m - ρ1×m/ρ3 = (ρ1)×(M/ρ2) - M

m(1 - ρ1/ ρ3) = M(ρ1/ρ2 -1)

m = [M(ρ1/ρ2 -1)]/[(1 - ρ1/ ρ3)]

= [(300)(1.1/0.75 -1)]/[(1 - 1.1/ 11.3)]

= 155 g

Therefore, the minimum mass of the lead for the combination to submerge is 155 g.

4 0

4 years ago