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solmaris [256]
3 years ago
12

WILL MARK BRAINLIST. HELP ASAP

Physics
1 answer:
Pachacha [2.7K]3 years ago
3 0

Answer:

Less than the distance

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One form of energy can be _____ another type of energy.
kotegsom [21]

We can answer this one very quickly.  From the <em>Law of Conservation of Energy</em>, we know that "Energy can't be created or destroyed.".

So that only leaves us one way to complete the sentence in this question:

"One form of energy can be <em>transformed into</em> another type of energy. "  <em>(B)</em>

4 0
3 years ago
Name TWO WEAKNESSES of the model pictured below
Nikolay [14]

Answer:

Here are a few:

1) The orbital radius of these planets is ridiculously small an in no way representative of their actual radii.

2) The planets will only line up like that once every 5200 years, making this very unrepresentative of their usual relations - although this does make their order in distance from the sun.

3) The nebulae, comet, lens flare,  and other junk in the background is incorrect.

4) If this is meant as a representation of the planets, then Pluto should not be there as it is now considered a planetoid.

5) The planets are incorrectly scaled both to each other and to the sun.

7 0
3 years ago
If the caffeine concentration in a particular brand of soda is 2.97 mg/oz, 2.97 mg/oz, drinking how many cans of soda would be l
Ray Of Light [21]

Explanation:

The given data is as follows.

     Concentration of caffeine = 2.97 mg/oz

     Number of oz in a can = 12 oz

Therefore, the concentration of caffeine in one can is calculated as follows.

                 = (12 \times 2.97) mg

                 = 35.64 mg

                 = 35.64 \times 10^{-3} g

Since, it is given that lethal dose is 10.0 g. Hence, number of cans are calculated as follows.

     No. of cans = \frac{\text{Lethal dose}}{\text{concentration in one can}}

                         = \frac{10 g}{35.64 \times 10^{-3} g}

                         = 280.58

                         = 281 (approx)

Thus, we can conclude that 281 cans of soda would be lethal.

5 0
3 years ago
If 35 mg of Oxygen is mixed with 1000 g of water at 25 degrees Celsius, a solution forms.
Serjik [45]

Answer:

The solute is oxygen

The solvent is water

Explanation:

A solvent is any chemical substance that dissolves other chemical substances, while a solute refers to any chemical substance that dissolves in other chemical substances. The best way to know when a chemical substance dissolves in another chemical substance is when a solid or gas dissolves in water. The solid or the gas can now be referred to as the solute and it will be shown to be in the the aqueous state, while the solvent is usually shown to be in the liquid state in any chemical equation.

Let us use the particular example of the dissolution of oxygen gas in water as shown below;

O2(g) + H2O(l)⇄O2(aq) + H2O(l)

The aqueous oxygen is the solute while the liquid water is the solvent.

Also, the substance having a smaller mass must be the solute and the substance having the larger mass must be the solvent.

3 0
3 years ago
Two planets P1 and P2 orbit around a star S in circular orbits with speeds v1 = 40.2 km/s, and v2 = 56.0 km/s respectively. If t
Readme [11.4K]

Answer: 3.66(10)^{33}kg

Explanation:

We are told both planets describe a circular orbit around the star S. So, let's approach this problem begining with the angular velocity \omega of the planet P1 with a period T=750years=2.36(10)^{10}s:

\omega=\frac{2\pi}{T}=\frac{V_{1}}{R} (1)

Where:

V_{1}=40.2km/s=40200m/s is the velocity of planet P1

R is the radius of the orbit of planet P1

Finding R:

R=\frac{V_{1}}{2\pi}T (2)

R=\frac{40200m/s}{2\pi}2.36(10)^{10}s (3)

R=1.5132(10)^{14}m (4)

On the other hand, we know the gravitational force F between the star S with mass M and the planet P1 with mass m is:

F=G\frac{Mm}{R^{2}} (5)

Where G is the Gravitational Constant and its value is 6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}

In addition, the centripetal force F_{c} exerted on the planet is:

F_{c}=\frac{m{V_{1}}^{2}}{R^{2}} (6)

Assuming this system is in equilibrium:

F=F_{c} (7)

Substituting (5) and (6) in (7):

G\frac{Mm}{R^{2}}=\frac{m{V_{1}}^{2}}{R^{2}} (8)

Finding M:

M=\frac{V^{2}R}{G} (9)

M=\frac{(40200m/s)^{2}(1.5132(10)^{14}m)}{6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}} (10)

Finally:

M=3.66(10)^{33}kg (11) This is the mass of the star S

4 0
3 years ago
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