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4 years ago

I really need to know what is on the physics SOL please someone tell me

1 answer:

5 0

The term sol is used by planetary astronomers to refer to the duration of a solar day on Mars.[7] A mean Martian solar day, or "sol", is 24 hours, 39 minutes, and 35.244 seconds.[6]

“Sol” is often used as a direct replacement for “Day” when concerning Mars. Mission duration for Mars missions is measured in Sols, so saying “Today is Sol xyz” would be normal, but I’m not sure if anyone would say “what a wonderful Sol tomorrow is going to be”.

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How could you determine if a chemical is an element or a compound?

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Answer:

An element is a pure substance and is found on the periodic table. If you have more than one element chemically bonded, it is a compound. ... H2O is a compound

Explanation:

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3 years ago

What item produces a magnetic field in the electromagnetic

ikadub [295]

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Hope this helps,

kwrob

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4 years ago

The manipulated variable is plotted on the horizontal axis

tatuchka [14]

True.

Manipulated variable or also called the controlled variables are variables in which you regulate. Manipulate or as said control. By this you want to certain the outcome of a certain experiment. Making it close to which you want it or desired it to be, possibly. Characteristics could be: 
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4 years ago

When density increases, what happens to the number of molecules in a volume of air?

solniwko [45]

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4 years ago

a ball is thrown with a speed of 17.7 m/s at an angle of 49.8° above the horizontal. how much time does the ball need to reach a

julsineya [31]

Answer:

The ball needs approximately $0.41\; \rm s$ to reach that height for the first time.

Explanation:

The initial speed of the ball $17.7\; \rm m \cdot s^{-1}$ . However, what would be the initial vertical speed of this ball?

The angle of elevation is $\theta = 49.8^\circ$ . Consider the initial speed of this ball as the length of the hypotenuse of a right triangle. If the angle of elevation is one of the two acute angles of this triangle, the initial vertical speed of this ball would be the leg opposite to that angle.

$\displaystyle \sin \theta = \frac{\text{opposite}}{\text{hypotenuse}}$ .

$\displaystyle \sin \theta = \frac{v(\text{vertical, initial})}{v(\text{initial})}$ .

Therefore:

$\begin{aligned}&v(\text{vertical, initial})\\ &= v(\text{initial}) \cdot \sin\theta \\ &= 17.7\; \rm m \cdot s^{-1} \times \sin \left(49.8^\circ\right) \approx 13.5\; \rm m \cdot s^{-1}\end{aligned}$ .

Let $t$ denote the time (in seconds) required for the ball to reach a height of $4.7\; \rm m$ .

Let $g$ denote the acceleration because of gravity (typically $g \approx 9.81\; \rm m \cdot s^{-2}$ near the surface of the earth.) The height of the ball $t$ seconds after it was thrown would be: $\displaystyle \frac{1}{2}\, g \cdot t^{2} + v(\text{vertical, initial}) \cdot t$ .

Assume that $g \approx 9.81\; \rm m \cdot s^{-2}$ . Set the value of this expression for height to $4.7\; \rm m$ and solve for $t$ :

$\displaystyle \frac{1}{2} \times 9.81 \, t^{2} + 13.5 \, t = 4.7$ .

Either $t \approx 0.41$ or $t \approx 2.3$ will satisfy this equation. Both of these two values are reasonable. The first value for $t$ ( $0.41\; \rm s$ ) is the time required for the ball to reach a height of $4.7\; \rm m$ for the first time. The second value ( $2.3\; \rm s$ ) is the time required for the ball to come under that height on its way back to the ground. The question seems to be asking only for the first (the smaller one) of these two times.

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3 years ago