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Sati [7]
3 years ago
10

On July 15, 2004, NASA launched the Aura spacecraft to study the earth's climate and atmosphere. This satellite was injected int

o an orbit 705 km above the earth's surface, and we shall assume a circular orbit.
(a) How many hours does it take this satellite to make one orbit?


(b) How fast (in km/s) is the Aura spacecraft moving?
Physics
1 answer:
Reika [66]3 years ago
7 0
On the earth surface g = GM/R^2
 And at the hieight h
 g" = GM/ [R+h] ^2
 g" = R^2/ [ R+h]^2 * g
 g" = {R/ [ R+h] }^2 * g
 g" = {6378/ 7083}^2 * 9.81 = 7.95 m/s^2
 --------------------------------------
T^2 = 4π^2 (R+h) /g"
T^2 = 4π^2 *7083 /7.95 
T = 187.54s = 0.052 hour.
 ========================
 Part B
 v = 2*π *[R+h] / T
 v = 2*π *[7.083] / 187.54s
 v = 0.24 km /s 
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An unlit match is held near (not touching) a Bunsen flame. The match doesn't get hot enough to light because
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The answer is option D)
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6 0
3 years ago
Determine the slit width that produces a diffraction pattern with the 2nd dark fringe at 6.2mm from the central fringe. The scre
Elanso [62]

Answer:

d= 0.242 mm

Explanation:

Slit width (d ) = ?

Screen distance ( D ) = 1.25 m

Wave length of light λ = 600 nm

Distance of n the dark fringe from centre

= n λ D / d

Here n = 2

so

6.2\times10^{-3}=\frac{2\times600\times10^{-9}}{d}

d=\frac{1500\times10^{-6}}{6.2}

d= 0.242 mm

4 0
3 years ago
A medical cyclotron used in the production of medical isotopes accelerates protons to 6.5 MeV. The magnetic field in the cyclotr
mart [117]

Answer:

diameter of largest orbit is 0.60 m

Explanation:

given data

isotopes accelerates KE = 6.5 MeV

magnetic field B = 1.2 T

to find out

diameter

solution

first we find velocity from kinetic energy equation

KE = 1/2 × m×v²   ........1

6.5 × 1.6 × 10^{-19} = 1/2 × 1.672 × 10^{-27} ×v²

v = 3.5 × 10^{7} m/s

so

radius will be

radius = \frac{m*v}{B*q}   ........2

radius =  \frac{1.672*10^{-27}*3.5*10^{7}}{1.2*1.6*10^{-19}}  

radius = 0.30

so diameter = 2 × 0.30

so diameter of largest orbit is 0.60 m

8 0
3 years ago
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the answer to your question!
8 0
3 years ago
A hiker begins a trip by first walking 25.0 km southeast from her car. She stops and sets up her tent for the night. On the seco
Sunny_sXe [5.5K]

Answer:

(a)

x_1=17.68km\\y_1=-17.68km\\x_2=20km\\y_2=34.64km

(b) r=41.32km

\alpha =24.23^o

Explanation:

Let us take the north direction to be the positive y-axis and the east to be positive x-axis.

First day:

25.0 km southeast, which implies 45^o south of east. The y-component will be negative and the x-component will be positive.

x_1=25cos45^o=17.68km

y_1=-25sin45^o=-17.68km

Second day:

She starts off at the stopping point of last day. This time, both the y- and x-components are positive.

x_2=40cos60^o=20km

y_2=40sin60^o=34.64km

Therefore, total displacements:

x=x_1+x_2=(17.68+20)km=37.68km

y=y_1+y_2=(-17.68+34.64)km=16.96km

Magnitude of displacements,

r=\sqrt{x^2+y^2}=41.32km

Direction,

\alpha =tan^{-1}(\frac{y}{x})=24.23^o

6 0
4 years ago
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