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1 year ago

For two objects in space with very different masses, the gravitational force causes:

1 answer:

7 0

The answer is B. The less massive object to orbit the more massive object.
Explanation: The more massive the object, the larger its gravitational pull.
Example: the moon orbits the earth because the moon’s mass is smaller than earth, earth orbits the sun because the earth’s mass is smaller than the sun..

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Suppose two comets, comet A and comet B, were orbiting the sun, having the same average orbital radii. If comet A had a higher o

zhannawk [14.2K]

Answer:

a. A

Explanation:

Kepler's First Law says that the orbits of planets are ellipses with the sun at one focus of the ellipse. Moreover, Kepler’s Second Law says that the line joining the planet to the sun sweeps out equal areas in equal times as it moves along its orbit. Finally, Kepler’s Third Law says that the ratio of the squares of the periods for two planets is equal to the ratio of the cubes of their semi-major axes.

By these laws, the comet A will have lower orbital speed.

3 0

2 years ago

A factory emits pollutants at a rate of 25 g/s. The factory is located between two mountain ranges resulting in an effective val

Lostsunrise [7]

Answer:

$1.25\ \mu\text{g/m}^3$

Explanation:

v = Velocity of the breeze = 4 m/s

w = Width of the valley = 5000 m

h = Height of the valley = 1000 m

Volumetric flow rate is given by

$\dot{V}=vwh\\\Rightarrow \dot{V}=4\times 5000\times 1000\\\Rightarrow \dot{V}=2\times10^{7}\ \text{m}^3/\text{s}$

$\dot{m}$ = Mass flow rate of pollutant = 25 g/s = $25\times 10^6\ \mu\text{g/s}$

Concentration is given by

$C=\dfrac{\dot{m}}{\dot{V}}\\\Rightarrow C=\dfrac{25\times 10^6}{2\times 10^7}\\\Rightarrow C=1.25\ \mu\text{g/m}^3$

The steady state concentration of pollutants in the valley, is $1.25\ \mu\text{g/m}^3$ .

6 0

2 years ago

With the water trap being used, all the possible sources of error would be eliminated.

Phoenix [80]

Answer:

True

Explanation:

7 0

2 years ago

a sensor light installed on the edge of a home can detect motion for a distance of 50 feet in front and with a range of motion o

Dvinal [7]

Answer:

4363.3231 feets²

Explanation:

Given that :

Distance, r = 50 ft

θ = 200°

The arc length of area covered :

Arc length = θ/360° * πr²

Arc length = (200/360) * 50 ft ^2 * π

Arc length = 0.5555555 * 2500 * π

Arc length = 4363.3231 feets²

7 0

2 years ago

Alkaline batteries have the advantage of putting out constant voltage until very nearly the end of their life. How long will an

Sidana [21]

Answer:

Explanation:

Voltage, V = 1.58 V

Power, P = 1 W

1 A.h

Charge, Q = 1 A.h = 1 x 3600 A.s = 3600 C

Power x time = Voltage x charge

1 x t = 1.58 x 3600

t = 1.58 x 3600 second

t = 1.58 hours

8 0

3 years ago