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jasenka [17]
1 year ago
6

A 1-kg mass at the earth's surface weighs about? a. 1 n. b. 5 n. c. 10 n. d. 12 n. e. none of these

Physics
1 answer:
ValentinkaMS [17]1 year ago
3 0

A 1-kg mass at the earth's surface weighs about C. 10N

The third planet from the Sun is the Earth. It is the seventh largest in terms of size and weighs roughly 5.98 1024 kg. The inherent quality of mass is unaffected by the environment of the object or the technique employed to quantify it.

Newton's law of gravitation can be used to estimate the mass of the Earth. This is set to the fundamental formula, which reads: force (F) = mass (m) times acceleration. Gravitational acceleration (G) is equal to 9.8 m/s2, the Earth's radius is 6.37 106 m, and the gravitational constant (G) is 6.673 1011 Nm2/kg2. The Earth has a mass of 5.96 1024 kg after rearranging the equation and entering all the numbers.

To learn more about earth please visit-
brainly.com/question/14042561
#SPJ4

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A planet's distance from the sun is 2.0x10^11 m. what is the orbital period?
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Answer:

The square of the orbital period of a planet is directly proportional to the cube of the semimajor axis of its orbit.

Explanation:

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4 0
2 years ago
An object initially at rest experiences an acceleration of 1.90 ­m/s² for 6.60 s then travels at that constant velocity for anot
borishaifa [10]

Answer:

The magnitude of the object's average velocity is 9.74 m/s (9.80 m/s without any intermediate rounding).

Explanation:

Hi there!

The average velocity is calculated as the displacement of the object divided by the time it takes the object to do that displacement.

The displacement is calculated as the distance between the final position of the object and the initial position. <u>In this problem</u>, the displacement is equal to the traveled distance because the object travels only in one direction:

a.v = Δx/t

Where:

a.v = average velocity.

Δx = displacement = final position - initial position

t = time

So, let's find the distance traveled while the object was accelerating. For that, we will use this equation:

x = x0 + v0 · t + 1/2 · a · t²

Where:

x = position of the object at time t.

x0 = initial position.

v0 = initial velocity.

t = time.

a = acceleration.

In this case, since the object is initially at rest, v0 = 0. If we place the origin of the frame of reference at the point where the object starts moving, then x0 = 0. So, the equation of the position of the object after a time t will be:

x = 1/2 · a · t²

x = 1/2 · 1.90 m/s² · (6.60 s)²

x = 41.4 m

The object traveled 41.4 m during the first 6.60 s.

Now, let's find the rest of the traveled distance.

When the velocity is constant, a = 0. Then, the equation of position will be:

x = x0 + v · t

Let's place now the origin of the frame of reference at the point where the object starts traveling at constant velocity so that x0 = 0:

x = v · t

The velocity reached by the object during the acceleration phase is calculated as follows:

v = v0 + a · t   (v0 = 0 because the object started from rest)

v = 1.90 m/s² · 6.60 s

v = 12.5 m/s

Then, the distance traveled by the object at a constant velocity will be:

x = 12.5 m/s · 8.50 s

x = 106 m

The total traveled distance in 15.1 s is (106 m + 41.4 m) 147 m.

Then the displacement will be:

Δx = final position - initial position

Δx = 147 m - 0 = 147 m

and the average velocity will be:

a.v = Δx/t

a.v = 147 m / 15.1 s

a.v = 9.74 m/s

The magnitude of the object's average velocity is 9.74 m/s (9.80 m/s without any intermediate rounding).

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<h2>MARK BRAINLIEST</h2>

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