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WITCHER [35]
2 years ago
13

A 4500-kg spaceship is in a circular orbit 190 km above the surface of Earth. It needs to be moved into a higher circular orbit

of 380 km to link up with the space station at that altitude. In this problem you can take the mass of the Earth to be 5.97 × 1024 kg.
I am confused how to solve this problem, please show how this answer is obtained.
Physics
1 answer:
Maslowich2 years ago
8 0

Answer:

Explanation:

Total energy of a satellite in an orbit , h height away

= -  GMm /2 ( R + h )

When h = 380 km

Total energy of a satellite = \frac{6.67\times10^{-11}\times5.97\times10^{24}\times 4500}{2\times(6378+380)\times10^3}

=  - 13.25 x 10¹⁰ J

When h = 190 km

Total energy of a satellite =

\frac{6.67\times10^{-11}\times5.97\times10^{24}\times 4500}{2\times(6378+190)\times10^3}

=   - 13.63 x 10¹⁰ J

Diff

= 38 x 10⁸ J Energy will be required.

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A certain alarm clock ticks four times each second, with each tick representing half a period. The balance wheel consists of a t
Semenov [28]

Answer:

a. I=2.77x10^{-8} kg*m^2

b. K=4.37 x10^{-6} N*m

Explanation:

The inertia can be find using

a.

I = m*r^2

m = 0.95 g * \frac{1 kg}{1000g}=9.5x10^{-4} kg

r=0.54 cm * \frac{1m}{100cm} =5.4x10^{-3}m

I = 9.5x10^{-4}kg*(5.4x10^{-3}m)^2

I=2.77x10^{-8} kg*m^2

now to find the torsion constant can use knowing the period of the balance

b.

T=0.5 s

T=2\pi *\sqrt{\frac{I}{K}}

Solve to K'

K = \frac{4\pi^2* I}{T^2}=\frac{4\pi^2*2.7702 kg*m^2}{(0.5s)^2}

K=4.37 x10^{-6} N*m

3 0
2 years ago
Little Tammy lines up to tackle Jackson to (unsuccessfully) prove the law of conservation of momentum. Tammy’s mass is 34.0 kg a
Naily [24]

Answer:

So Tammy must move with speed 4.76 m/s in opposite direction of Jackson

Explanation:

As per law of conservation of momentum we know that there is no external force on it

So here we can say that initial momentum of the system must be equal to the final momentum of the system

now we have

m_1v_1 + m_2v_2 = 0

final they both comes to rest so here we can say that final momentum must be zero

now we have

34 v + 54 (3 m/s) = 0

v = -4.76 m/s

8 0
2 years ago
ou purchase a rectangular piece of metal that has dimen- sions 5.0 * 15.0 * 30.0 mm and mass 0.0158 kg. The seller tells you tha
Natalija [7]

Answer: 7022.2kg/m³, yes, I was cheated

Explanation:

Density of an object is defined as the ratio of the mass of the object to its volume. Mathematically;

Density = Mass/Volume

Note that the unit of both mass and volume must be standard unit.

Given mass = 0.0158kg

Dimension of the metal = 5mm×15mm×30mm

Note that 1mm = 0.001m

The volume of the metal will be

0.005×0.015×0.03

= 0.00000225m³

Density = 0.0158/0.00000225

Average density of the metal = 7022.2kg/m³

Since the standard density of Gold is 19,320kg/m³ and is higher than the density prescribed for me, it shows the I was cheated.

4 0
2 years ago
The basic function of an automobile carburetor is to atomize the gasoline and mix it with air to promote rapid combustion. Assum
tatyana61 [14]

Answer:

A = 7.5 \times 10^6 m^2

Explanation:

Since volume of all the liquid is always conserved

so here we know that

V_{total} = N V

25 cm^3 = N(\frac{4}{3}\pir^3)

25 = N (\frac{4}{3} \pi (1.0 \times 10^{-5})^3)

N = 5.97 \times 10^{15}

now we know that total surface area is given as

A = N(4\pi r^2)

A = (5.97 \times 10^{15})(4\pi (1 \times 10^{-5})^2

A = 7.5 \times 10^6 m^2

5 0
2 years ago
An electron in a vacuum chamber is fired with a speed of 9800 km/s toward a large, uniformly charged plate 75 cm away. The elect
melisa1 [442]

Answer:

The plate's surface charge density is -8.056\times10^{-9}\ C/m^2

Explanation:

Given that,

Speed = 9800 km/s

Distance d= 75 cm

Distance d' =15 cm

Suppose we determine the plate's surface charge density?

We need to calculate the surface charge density

Using work energy theorem

W=\Delta K.E

W=\dfrac{1}{2}mv_{f}^2-\dfrac{1}{2}mv_{i}^2

Here, final velocity is zero

W=0-\dfrac{1}{2}mv_{i}^2...(I)

We know that,

W=-Fd

W=-E\times e\times d

W=-\dfrac{\lambda}{2\epsilon_{0}}\times e\times d...(II)

From equation (I) and (II)

-\dfrac{1}{2}mv_{i}^2=-\dfrac{\lambda}{2\epsilon_{0}}\times e\times d

Charge is negative for electron

\lambda=\dfrac{mv^2\epsilon_{0}}{(-e)d}

Put the value into the formula

\lambda=-\dfrac{9.1\times10^{-31}\times(9800\times10^{3})^2\times8.85\times10^{-12}}{1.6\times10^{-19}\times(75-15)\times10^{-2}}

\lambda=-8.056\times10^{-9}\ C/m^2

Hence, The plate's surface charge density is -8.056\times10^{-9}\ C/m^2

3 0
2 years ago
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