Due at 11:59pm please help

A satellite orbits the Earth every 2 hours at an average distance from the Earth's centre of 8000km. (i) What is the average ang

AlexFokin [52]

Answer:

i)ω=3600 rad/s

ii)V=7059.44 m/s

iii)F=1245.8 N

Explanation:

i)

We know that angular speed given as

$\omega =\dfrac{d\theta}{dt}$

We know that for one revolution

θ=2π

Given that time t= 2 hr

So

ω=θ/t

ω=2π/2 = π rad/hr

ω=3600 rad/s

ii)

Average speed V

$V=\sqrt{\dfrac{GM}{R}}$

Where M is the mass of earth.

R is the distance

G is the constant.

Now by putting the values

$V=\sqrt{\dfrac{GM}{R}}$

$V=\sqrt{\dfrac{6.667\times 10^{-11}\times 5.98\times 10^{24}}{8000\times 10^3}}$

V=7059.44 m/s

iii)

We know that centripetal fore given as

$F=\dfrac{mV^2}{R}$

Here given that m= 200 kg

R= 8000 km

so now by putting the values

$F=\dfrac{mV^2}{R}$

$F=\dfrac{200\times 7059.44^2}{8000\times 10^3}$

F=1245.8 N

3 0

3 years ago

Compare a series circuit powered by six 1.5-volt batteries to a series circuit powered by a single 9-volt battery. Make sure the

lana [24]

Answer:

Both series circuits provide a total voltage of 9 volts to the two bulbs connected in series and the voltage will be equally divided among two bulbs and they will have same brightness. Therefore, both circuits will have same characteristics.

Explanation:

We are asked to compare two series circuits having equal number of light bulbs.

1st circuit is powered by 6 batteries each having a voltage of 1.5V

2nd circuit is powered by a single battery having a voltage of 9V.

The six batteries in the 1st circuit can be connected together in series or in parallel.

When the batteries are connected in series (positive terminal of one battery connected to negative terminal of another battery) their voltage gets added which means

Voltage of pack = number of batteries*voltage of each battery

Voltage of pack = 6*1.5

Voltage of pack = 9 volts

But the current remains same in the series connection since there is only path for the current to flow.

On the other hand, when the batteries are connected in parallel, the voltage remains same but the current increases.

Circuit 1:

In this circuit, we have 6 batteries each of 1.5 volts connected in series to provide a voltage of 9 volts.

We have connected 2 bulbs in this series circuit.

The voltage will be equally divided between two bulbs if both bulbs are identical in construction.

So there will be 4.5 volts across each bulb and both bulbs will have same brightness.

Circuit 2:

In this circuit, we have 1 battery which provide a voltage of 9 volts.

We have connected 2 bulbs in this series circuit just like in circuit 1.

The voltage will be equally divided between two bulbs if both bulbs are identical in construction.

So there will be 4.5 volts across each bulb and both bulbs will have same brightness.

Conclusion:

Both series circuits provide a total voltage of 9 volts to the two bulbs connected in series and the voltage will be equally divided among two bulbs and they will have same brightness. Therefore, both circuits will have same characteristics.

3 0

3 years ago

Burn rate can be affected by: A. Variations in chamber pressure B. Variations in initial grain temperature C. Gas flow velocity

Digiron [165]

Answer: D) All of the above

Explanation:

Burn rate can be affected by all of the above reasons as, variation in chamber pressure because the pressure are dependence on the burn rate and temperature variation in initial gain can affect the rate of the chemical reactions and initial gain in the temperature increased the burning rate. As, gas flow velocity also influenced to increasing the burn rate as it flowing parallel to the surface burning. Burn rate is also known as erosive burning because of the variation in flow velocity and chamber pressure.

4 0

3 years ago

The dam cross section is an equilateral triangle, with a side length, L, of 50 m. Its width into the paper, b, is 100 m. The dam

lisabon 2012 [21]

Answer:

Explanation:

In an equilateral trinagle the center of mass is at 1/3 of the height and horizontally centered.

We can consider that the weigth applies a torque of T = W*b/2 on the right corner, being W the weight and b the base of the triangle.

The weigth depends on the size and specific gravity.

W = 1/2 * b * h * L * SG

Then

Teq = 1/2 * b * h * L * SG * b / 2