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2 years ago

What 3 points must an object possess to be considered a planet

1 answer:

4 0

<span>1. It must be an object which independently orbits the Sun (this means moons can't be considered planets, since they orbit planets)
2. It must have enough mass that its own gravity pulls it into a spheroidal shape.
3. </span><span>It must be large enough to "dominate" its orbit (i.e. its mass must be much larger than anything else which crosses its orbit).</span>

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The lowest note on a grand piano has a frequency of 27.5 Hz. The entire string is 2.00 m long and has a mass of 400 g. The vibra

Norma-Jean [14]

Answer:

1456 N

Explanation:

Given that

Frequency of the piano, f = 27.5 Hz

Entire length of the string, l = 2 m

Mass of the piano, m = 400 g

Length of the vibrating section of the string, L = 1.9 m

Tension needed, T = ?

The formula for the tension is represented as

T = 4mL²f²/ l, where

T = tension

m = mass

L = length of vibrating part

F = frequency

l = length of the whole part

If we substitute and apply the values we have Fri. The question, we would have

T = (4 * 0.4 * 1.9² * 27.5²) / 2

T = 4368.1 / 2

T = 1456 N

Thus, we could conclude that the tension needed to tune the string properly is 1456 N

4 0

2 years ago

Rock X is released from rest at the top of a cliff that is on Earth. A short time later, Rock Y is released from rest from the s

frosja888 [35]

Answer:

C) True. S increases with time, v₁ = gt and v₂ = g (t-t₀) we see that for the same t v₁> v₂

Explanation:

You have several statements and we must select which ones are correct. The best way to do this is to raise the problem.

Let's use the vertical launch equation. The positive sign because they indicate that the felt downward is taken as an opponent.

Stone 1

y₁ = v₀₁ t + ½ g t²

y₁ = 0 + ½ g t²

Rock2

It comes out a little later, let's say a second later, we can use the same stopwatch

t ’= (t-t₀)

y₂ = v₀₂ t ’+ ½ g t’²

y₂ = 0 + ½ g (t-t₀)²

y₂ = + ½ g (t-t₀)²

Let's calculate the distance between the two rocks, it should be clear that this equation is valid only for t> = to

S = y₁ -y₂

S = ½ g t²– ½ g (t-t₀)²

S = ½ g [t² - (t²- 2 t to + to²)]

S = ½ g (2 t t₀ - t₀²)

S = ½ g t₀ (2 t -t₀)

This is the separation of the two bodies as time passes, the amount outside the Parentheses is constant.

For t <to. The rock y has not left and the distance increases

For t> = to. the ratio (2t/to-1)> 1 therefore the distance increases as time

passes

Now we can analyze the different statements

A) false. The difference in height increases over time

B) False S increases

C) Certain s increases with time, v₁ = gt and V₂ = g (t-t₀) we see that for the same t v₁> v₂

3 0

3 years ago

Which of the following was NOT one of Ghana's chief trading exports?

valina [46]

The answer is b. ivory

3 0

2 years ago

You are presently taking a weather observation. The sky is full of wispy cirrus clouds estimated to be about 10 km overhead. If

marshall27 [118]

Answer:

x = 2000 Km

Explanation:

Given

y = 10 km

Slope: 1 : 200

x = ?

We can apply the formula

y / x = 1 / 200 ⇒ x = 200*y = 200*10 Km

⇒ x = 2000 Km

7 0

3 years ago

A car (mass 1200 kg, speed 100 km/h) and a truck (mass 2800 kg, speed 50 km/h) are moving in the same direction along a highway.

Sloan [31]

Answer:

Speed of the wreck after the collision is 65 km/h

Explanation:

When a car hits truck and sticks together, the collision would be totally inelastic. Since the both the vehicles locked together, they have the same final velocity.

Mass of car $m_{1}=1200 kg$