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Tcecarenko [31]
3 years ago
14

What is a satellite

Physics
2 answers:
g100num [7]3 years ago
7 0

Answer:

an artificial body placed in orbit around the earth or moon or another planet in order to collect information or for communication.

Explanation:

Look it up on google

Anuta_ua [19.1K]3 years ago
3 0

Answer:

A satellite is an artificial body put in orbit around the earth or another planet in order to collect information or for communication.

A satellite could also be a celestial body orbiting the Earth or another planet.

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Which statement best describes perigee?
pantera1 [17]

Answer:

A. The closest point in the Moon's orbit to Earth

Explanation:

The perigee is defined as the closest point in the orbit of an object (such as a satellite) from the centre of the Earth. In this case, the Earth's satellite is the Moon, so the perigee is defined as the closest point in the Moon's orbit to Earth. so option A is the correct one.

Let's see instead the names of the other options:

B. The farthest point in the Moon's orbit to Earth  --> this point is called apogee

C. The closest point in Earth's orbit of the Sun  --> this point is called perihelion

D. The Sun's orbit that is closest to the Moon --> this point has no specific name

8 0
3 years ago
Read 2 more answers
If a trapezoidal channel has side slopes of 1:1, hydraulic depth is 5 feet, the bottom width is 8 feet, flow is 2,312 cubic feet
Evgesh-ka [11]

Answer:

S = \dfrac{1}{2.5}

Explanation:

given,

side slope = 1 : 1

hydraulic depth(y) = 5 ft

bottom width (b)= 8 ft

x = 1

Q = 2,312 ft³/s

n = 0.013

slope of channel = ?

R = \dfrac{A}{P}

R = \dfrac{y(b + xy)}{b+2y\sqrt{1+x^2}}

R = \dfrac{5(8+ 5)}{8+2\times 5\sqrt{1+1^2}}

R = 4.69 m

using manning's equation

Q = \dfrac{1}{n}AR^{2/3} S^{1/2}

2312= \dfrac{1}{0.013}\times (5(8+5))\times 4.69^{2/3} S^{1/2}

2312=14009.37\times S^{1/2}

S = 0.406

S = \dfrac{1}{2.5}

5 0
3 years ago
Is it possible for an object to be in motion without any external force applied? justify
Rudiy27
Newton’s first law is commonly stated as:
An object at rest stays at rest and an object in motion stays in motion.
However, this is missing an important element related to forces. We could expand it by stating:
An object at rest stays at rest and an object in motion stays in motion at a constant speed and direction unless acted upon by an unbalanced force.
By the time Newton came along, the prevailing theory of motion—formulated by Aristotle—was nearly two thousand years old. It stated that if an object is moving, some sort of force is required to keep it moving. Unless that moving thing is being pushed or pulled, it will simply slow down or stop. Right?
This, of course, is not true. In the absence of any forces, no force is required to keep an object moving. An object (such as a ball) tossed in the earth’s atmosphere slows down because of air resistance (a force). An object’s velocity will only remain constant in the absence of any forces or if the forces that act on it cancel each other out, i.e. the net force adds up to zero. This is often referred to as equilibrium. The falling ball will reach a terminal velocity (that stays constant) once the force of air resistance equals the force of gravity.

Hope this help
8 0
3 years ago
Why is the curve between 1950 and 1980 relatively flat and centered around zero degrees difference from the baseline? (Hint: how
zimovet [89]

Look at the title of the graph, in small print under it.

Each point is "compared to 1950-1980 baseline". So the set of data for those years is being compared to itself. No wonder it matches up pretty close !

3 0
3 years ago
Star A and Star B have measured stellar parallax of 1.0 arc second and 0.75 arc second, respectively. Which star is closer? How
zhuklara [117]

Answer:

Star A is closer than Star B

Explanation:

As we know that in parallax method of distance measurement the angle subtended by the star when it covers a distance of one Parsec arc length, it is known as parallax angle

Here we can say

angle = \frac{1 Parsec}{distance}

so we have

distance = \frac{1 Parsec}{angle}

so here we have

angle subtended by Star A = 1 arc sec

angle subtended by star B = 0.75 arc sec

now we have

distance for star A is given as

d_a = \frac{1 Parsec}{1} = 1 Parsec

distance of star B is given as

d_b = \frac{1 Parsec}{0.75} = 1.33 Parsec

So star A is closer than star B

7 0
3 years ago
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