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2 years ago

Plzzz Helppppppppppppppppppp

Physics

1 answer:

White raven [17]2 years ago

4 0

1gw

that's the answer tnx

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Determine W (or fuel energy) required to launch a satellite of mass m at rest from a launching pad placed at the surface earth,

jeka57 [31]

Answer:

5. 9GmM/(10R)

Explanation:

m is the mass of the satellite

M is the mass of the earth

W is the energy required to launch the satellite

Energy at earth surface = Potential energy (PE) + W

W = Energy at earth surface - Potential energy (PE)

But PE = $-\frac{GMm}{R}$

Therefore: W = Energy at earth surface - $\frac{GMm}{R}$

Energy at earth surface (E) at an altitude of 5R = $-\frac{GMm}{5r} +\frac{1}{2}mV^2$

But $V=\sqrt{\frac{GM}{5R} }$

Therefore: $E=-\frac{GMm}{5R}+\frac{1}{2}m(\sqrt{\frac{GM}{5R} } )^2= -\frac{GMm}{5R}+\frac{GMm}{10R} = -\frac{GMm}{10R}$

W = E - PE

$W=-\frac{GMm}{10R}-(-\frac{GMm}{R})=-\frac{GMm}{10R}+\frac{GMm}{R}=\frac{9GMm}{10R} \\W=\frac{9GMm}{10R}$

7 0

3 years ago

if a bowling ball hits a wall a force of 6 N, the wall exerts a force of how much back. on the bowling ball

grigory [225]

It would exert the same back right?

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3 years ago

The human activities in two locations are described below:

Bogdan [553]

Air pollution level is higher in Location B, because poisonous fumes are produced when coal burns.

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3 years ago

Read 2 more answers

A brand of earplugs reduces the sound intensity level by 27 dB.By what factor do these earplugs reduce the acoustic intensity?

Fofino [41]

Answer:

Explanation:

27dB = 2.7 B

So I / I₀ = 10⁻²°⁷ , I₀ is intensity of main sound and I is intensity after reduction.

= 1.99 X 10⁻³

So intensity will reduce by 1.99 X 10⁻³ .

3 0

2 years ago

A 1-kg rock is suspended from the tip of a horizontal meterstick at the 0-cm mark so that the meterstick barely balances like a

tigry1 [53]

Explanation:

Given that,

Mass if the rock, m = 1 kg

It is suspended from the tip of a horizontal meter stick at the 0-cm mark so that the meter stick barely balances like a seesaw when its fulcrum is at the 12.5-cm mark.

We need to find the mass of the meter stick. The force acting by the stone is

F = 1 × 9.8 = 9.8 N

Let W be the weight of the meter stick. If the net torque is zero on the stick then the stick does not move and it remains in equilibrium condition. So, taking torque about the pivot.

$9.8\times 12.5=W\times (50-12.5)\\\\W=\dfrac{9.8\times 12.5}{37.5}$

W = 3.266 N

The mass of the meters stick is :

$m=\dfrac{W}{g}\\\\m=\dfrac{3.266}{9.81}\\\\m=0.333\ kg$

So, the mass of the meter stick is 0.333 kg.

5 0

3 years ago