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ella [17]
3 years ago
6

A satellite is in a circular orbit around the Earth at an altitude of 3.18x10 m. Find the period and th orbital speed of the sat

ellite? A. T= 2.94 h, B. T= 3.23 h, v 5610 m/s C. T= 1.75 h, v = 5920 m/s D. T 1.12 h, v 4980 m/s E. T 2.58 h, v 6460 m/s
Physics
1 answer:
Leona [35]3 years ago
8 0

Answer:

112.17 m/s

56.427 years

Explanation:

h = 3.18 x 10^10 m

R = 6.4 x 10^6 m

r = R + h = 3.18064 x 10^10 m

M = 6 x 10^24 kg

The formula for the orbital velocity is given by

v = \sqrt{\frac{G M }{r}}

v = \sqrt{\frac{6.67 \times 10^{-11}\times 6\times 10^{24}  }{3.18064\times 10^{10}}}

v = 112.17 m/s

Orbital period, T = 2 x 3.14 x 3.18064 x 10^10 / 112.17

T = 0.178 x 10^10 s

T = 56.427 years

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Eduardwww [97]

Answer:

Motion with constant velocity of magnitude 1 m/s (uniform motion) for 4 seconds in a positive direction and then for 2 seconds uniform motion with constant velocity of magnitude 3 m/s in reverse direction .

Explanation:

The graph shows a constant velocity of 1 m/s for 4 seconds in the positive direction. After that, between 4 seconds and 6 seconds, the object reverses its motion with constant velocity of magnitude 3m/s.

4 0
3 years ago
Momentum for a system can be conserved in one direction while not being conserved in another. What is the anglebetween the direc
katrin [286]

Answer:

90°

Explanation:

The angle will be 90° when momentum for a system can be conserved in one direction while not being conserved in another.

The example can be

If we apply force on an object horizontally in west direction, then as in other direction south or north we cannot apply the principal of momentum conservation.

5 0
3 years ago
(a) How many fringes appear between the first diffraction-envelope minima to either side of the central maximum in a double-slit
Ainat [17]

Answer:

a

The number of fringe is  z  = 3 fringes

b

The  ratio is I = 0.2545I_o

Explanation:

a

 From the question we are told that

        The wavelength is  \lambda = 600 nm

        The distance between the slit is  d = 0.117mm = 0.117 *10^{-3} m

        The width of the slit is  a = 35.7 \mu m = 35.7 *10^{-6}m

let  z be the number of fringes that appear between the first diffraction-envelope minima to either side of the central maximum in a double-slit pattern is  and this mathematically represented as

             z = \frac{d}{a}

Substituting values

             z = \frac{0.117*10^{-3}}{35.7 *10^{-6}}  

             z  = 3 fringes

b

   From the question  we are told that the order  of the bright fringe is  n = 3

   Generally the intensity of  a pattern  is mathematically represented as

                 I = I_o cos^2 [\frac{\pi d sin \theta}{\lambda} ][\frac{sin (\pi a sin \frac{\theta}{\lambda } )}{\pi a sin \frac{\theta}{\lambda} } ]

Where I_o is the intensity  of the  central fringe

 And  Generally  sin \theta = \frac{n \lambda }{d}

               I = I_o co^2 [ \frac{\pi (\frac{n \lambda}{d} )}{\lambda} ] [\frac{\frac{sin (\pi a (\frac{n \lambda}{d} ))}{\lambda} }{\frac{\pi a (\frac{n \lambda}{d} )}{\lambda} } ]

               I = I_o cos^2 (n \pi)[\frac{\frac{sin(\pi a (\frac{n \lambda}{d} ))}{\lambda} )}{ \frac{ \pi a (\frac{n \lambda }{d} )}{\lambda} } ]

               I = I_o cos^2 (3 \pi) [\frac{sin (\frac{3 \pi }{6} )}{\frac{3 \pi}{6} } ]

                I = I_o (1)(0.2545)

                  I = 0.2545I_o

6 0
3 years ago
Because of your success in physics class you are selected for an internship at a prestigious bicycle company in its research and
tiny-mole [99]

To develop the problem it is necessary to apply the equations related to the moment of inertia.

The given values can be defined as,

M = 1.0 kg

r = 0.5 m

m = 10 g

I = 0.280 kg.m^2

According to the definition of the moment of inertia applied to the exercise we can arrive at the equation that,

I = I_{rim} + n * I_{spoke}

Where n is the number of spokes necessary to construct the wheel.

I_{rim} = M*r^2 = 1.0 * 0.5^2

I_{spoke} = \frac{1}{3} * m * r^2 = \frac{1}{3}* 10 * 10^-3 * 0.5^2

Replacing the values at the general equation we have,

0.280 = 1.0 * 0.5^2 + n * (1/3 * 10 * 10^-3 * 0.5^2 )

Solving for n,

n = 36

Therefore the number of spokes necessary to construct the wheel is 36

PART B) The mass of the wheel is given by the sum of all masses and the total spokes, then

M_w= M + n*m

M_w = 1.0 + 36* 10 * 10^{-3} Kg

M_w = 1.36 Kg

Therefore the mass of the wheel must be of 1.36Kg

4 0
3 years ago
A capacitor is charged to a potential difference of 3 volt it delivers 30% store energy to lamp what is the final potential diff
nexus9112 [7]

Answer:

3.98V

Explanation:

Given

Pontential difference V as 3v

Energy delivered is 30%,

Recall that Enery E=1/2cv^2 from this E=V^2(since Current C is not provided we can assume a value 2)

So E=V^2

E=3^2=9

At full charge E=9,30%of 9,0.3*9=2.7 energy in capacitor is 9-2.7=6.3

But E=V^2

✓E=V

✓6.3=3.98V

4 0
3 years ago
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