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Vesna [10]
3 years ago
15

During the first 6 years of its operation, the Hubble Space Telescope circled the Earth 37,000 times, for a total of 1,280,000,0

00 km. It takes the Hubble Space Telescope about 90 minutes to make one complete orbit. How fast is it traveling (in units of km/min)? Remember, distance = time x speed. Round your answer to the nearest integer and don't include the units for this answer.
Physics
1 answer:
oksian1 [2.3K]3 years ago
5 0

Answer:

v = 384km/min

Explanation:

In order to calculate the speed of the Hubble space telescope, you first calculate the distance that Hubble travels for one orbit.

You know that 37000 times the orbit of Hubble are 1,280,000,000 km. Then, for one orbit you have:

d=\frac{1,280,000,000km}{37,000}=34,594.59km

You know that one orbit is completed by Hubble on 90 min. You use the following formula to calculate the speed:

v=\frac{d}{t}=\frac{34,594.59km}{90min}=384.38\frac{km}{min}\approx384\frac{km}{min}

hence, the speed of the Hubble is approximately 384km/min

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Answer:

44.4cm

Explanation:

glass has an index of refraction .n = 1.54

radii of curvature of 40 cm R1 = 40 by

radii of curvature of 600 cm R2 = 60

Now, by lens maker formula

1/f = (n - 1) (1/R1 - 1/R2)

Putting in the given values for n = 1.54 , we get f = 22.2

\frac{1}{f} = (1.54 -1) (\frac{1}{40cm} -\frac{1}{(-60cm)} )

\frac{1}{f} = 0.0225

f = 1 / 0.0225

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so, focal length in air will be  = 44.4 cm

6 0
3 years ago
How long would it take a drag racer to increase her speed from 10m/s to 20 m/s if her car accelerates at a uniform rate of 15 m/
weeeeeb [17]

Answer:

t = 0.67 [s]

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To solve this problem we must use the following kinematics equation.

v_{f} =v_{i} +(a*t)\\where:

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a = aceleration = 15 [m/s^2]

Now replacing in the equation we have:

20 = 10 + (15*t)

t = (20-10)/15

t = 0.67 [s]

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In separate experiments, a large number of particles (all with the same charge but with a wide variety of masses, speeds, and sp
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The particles that move in orbits of the same radius have the same momentum.

<h3>Orbital angular momentum:</h3>

A point particle's three-dimensional angular momentum is traditionally represented by the pseudovector r p, which is the cross product of the particle's position vector r (relative to some origin) and momentum vector, which in Newtonian physics is denoted by p = mv.

L = mrV_{prep} = mr²w is the particle's orbital angular momentum in units of magnitude. The part of the particle's velocity that is here perpendicular to the axis of rotation is designated as V_{perp.} The right-hand rule indicates the direction of the angular momentum. In isolated systems, the angular momentum is conserved.

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4 0
2 years ago
A boat has a porthole window, of
jekas [21]

Answer:

F = 534.6[N]

Explanation:

We must find the pressure exerted by the water at the depth of the boat, by means of the following equation.

P=Ro*g*h

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Ro = density of sea water = 1027 [kg/m³]

g = gravity acceleration = 9.81 [m/s²]

h = wáter Depth =  6.25 [m]

Now replacing:

P=1027*9.81*6.25\\P=62967.93[Pa]

The net force is:

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7 0
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