2 years ago

A rocket can fly into space because !

1 answer:

7 0

Newton 3rd law of motion which says action and reaction are equal and opposite. just like if you fill up a ballon with air it losses mommentum it loses the greater it flies around.

so rocket achieve thrust and moves when you exert a huge amount of mass at high speed

and also its able to move or fly to space due to rocket staging

You might be interested in

A charge of uniform volume density (40 nC/m3) fills a cube with 8.0-cm edges. What is the total electric flux through the surfac

GREYUIT [131]

Answer:

The flux through the surface of the cube is $2.314\ Nm^{2}/C$

Solution:

As per the question:

Edge of the cube, a = 8.0 cm = $8.0\times 10^{- 2}\ m$

Volume Charge density, $\rho_{v} = 40 nC/m^{3} = 40\times {- 9}\ C/m^{3}$

Now,

To calculate the electric flux:

$\phi = \frac{q}{\epsilon_{o}}$ (1)

where

$\phi$ = electric flux

$\epsilon_{o} = 8.85\times 10^{- 12}\ F/m$ = permittivity of free space

Volume Charge density for the given case is given by the formula:

$\rho_{v} = \frac{Total\ charge, q}{Volume of cube, V}$ (2)

Volume of cube, $V = a^{3}$

Thus

$V = (8.0\times 10^{- 2})^{3} = 5.12\times 10^{- 4}\ m^{3}$

Thus from eqn (2), the total charge is given by:

$q = \rho_{v}V = 40\times {- 9}\times 5.12\times 10^{- 4}$

$q = 2.048\times 10^{-11}\ F = 20.48\ pF$

Now, substitute the value of 'q' in eqn (1):

$\phi = \frac{2.048\times 10^{-11}}{8.85\times 10^{- 12}} = 2.314\ Nm^{2}/C$

5 0

2 years ago

Briana swings a ball on the end of a rope in a circle. The rope is 1.5 m long. The ball completes a full circle every 2.2 s. Wha

schepotkina [342]

The radius of the circular path is 1.5 m.

The circumference is then
$1.5\ m*2\pi=3\pi\ m$

The ball moves 3π m every 2.2 s, so the speed is
$\frac{3\pi\ m}{2.2\ s}\approx 4.3\ m/s$

9 0

2 years ago

Read 2 more answers

Learning Goal: To practice Problem-Solving Strategy 19.1 Work in Ideal-gas Processes. A cylinder with initial volume VVV contain

Leokris [45]

Answer:

The work done on the gas is equal to the area under the curve pv diagram w = area of triangle = 1/2 (base)(height) = 1/2 (BC)(Ac) = 1/2 (3v - v)(3p - p) = 1/2 (9 vp - 3 vp - 3vp + vp) = 4 vp/2 W = 2 vp

Check attachment for the diagrammatic representation