During the first stage of a multistage rocket, a satellite is:

PLEASE HELP I NEED HELP PERIODTT

SVETLANKA909090 [29]

Answer:

Explanation:

Liluzivert

5 0

2 years ago

Read 2 more answers

How did the theory of relativity change the law of conservation of energy?

Vinil7 [7]

I believe that the first one is correct because if you have a ball that Weighs 5 pounds and you push it down a ramp it will hit the end with more force than a ball that weighs 2 pounds

3 0

3 years ago

A thin, metallic spherical shell of radius 0.227 m has a total charge of 6.03 × 10 − 6 C placed on it.

KATRIN_1 [288]

Answer:

Explanation:

Given

radius $r=0.227 m$

Charge on surface $Q=6.03\times 10^{-6} C$

Point Charge inside sphere $q=1.15\times 10^{-6} C$

Electric Field at $r=0.735 m$

Treating Surface charge as Point charge and applying Gauss law

$E_{total}A=\frac{q_{enclosed}}{\epsilon _0}$

where A=surface area up to distance r

$E_{total}=\frac{Q+q}{4\pi r^2}$

$E_{total}=\frac{6.03\times 10^{-6}+1.15\times 10^{-6}}{4\pi (0.735)^2\times 8.85\times 10^{-12}}$

$E_{total}=1.194\times 10^{5} N/C$

3 0

3 years ago

Prove that..<br>please help<br>

GaryK [48]

$\large{ \boxed{ \bf{ \color{red}{Universal \: law \: of \: gravitation}}}}$

Every object in the universe attracts every other object with a force which is proportional to the product of their masses and inversely proportional to the square of the distance between them. The forces along the line joining the centre of the two objects.

❍ Let us consider two masses m1 and m2 line at a separation distance d. Let the force of attraction between the two objects be F.

According to universal law of gravitation,

$\large{ \longrightarrow{ \rm{F \propto m_1 m_2}}}$

Also,

$\large{ \longrightarrow{ \rm{ F \propto \dfrac{1}{ {d}^{2} } }}}$

Combining both, We will get

$\large{ \longrightarrow{ \rm{F \propto \dfrac{ m_1 m_2}{ {d}^{2}}}}}$

Or, We can write it as,

$\large{ \longrightarrow{ \rm{F \propto \: G \dfrac{ m_1 m_2}{ {d}^{2} }}}}$

Where, G is the constant of proportionality and it is called 'Universal Gravitational constant'.

☯️ Hence, derived !!

<u>━━━━━━━━━━━━━━━━━━━━</u>

8 0

2 years ago

A (20*20) cm² loop has a resistance of 0.10 Ω. A magnetic field perpendicular to the loop is B = 4t - 2t², where B is in tesla a

Ilya [14]

Answer with Explanation:

We are given that

Area of loop= $(20\times 20) cm^2=400\times 10^{-4} m^2$

$1 cm^2=10^{-4} m^2$

Resistance, R= $0.1\Omega$

B= $4t-2t^2$

We know that magnetic flux

$\phi=BA$

Emf , $E=\mid \frac{d\phi}{dt}\mid =\mid\frac{d(BA}{dt}\mid =\mid A\frac{dB}{dt}=400\times 10^{-4}\times \frac{4t-2t^2}{dt}\mid =\mid400\times 10^{-4}\times(4-4t)\mid$