Which image shows both potential and kinetic energy

A bell is rung. What best describes the density of air around the bell? The air density does not change. The air density increas

dezoksy [38]

I think the closest possible answer to this question is The air density increases and decreases repeatedly before returning to normal.Thank you for your question. Please don't hesitate to ask in Brainly your queries.

7 0

2 years ago

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Antifreeze is a "must have" in an Albertan winter because it helps keep your vehicle engine from freezing in cold temperatures.

Natasha_Volkova [10]

Products such as antifreeze are composed of organic compounds that are classified as <em>alcohols</em>. (a)

Maybe those other classes of chemicals also lower the freezing temperature of water, just like alcohol does. I don't know. But alcohol is what's used to make anti-freeze. I'm guessing alcohol must be cheaper, less toxic, and less corrosive inside the engines' cooling systems than any of that other stuff is.

4 0

2 years ago

You launch a cannonball at an angle of 35° and an initial velocity of 36 m/s (assume y = y₁=

velikii [3]

Answer:

Approximately $4.2\; {\rm s}$ (assuming that the projectile was launched at angle of $35^{\circ}$ above the horizon.)

Explanation:

Initial vertical component of velocity:

$\begin{aligned}v_{y} &= v\, \sin(35^{\circ}) \\ &= (36\; {\rm m\cdot s^{-1}})\, (\sin(35^{\circ})) \\ &\approx 20.6\; {\rm m\cdot s^{-1}}\end{aligned}$ .

The question assumed that there is no drag on this projectile. Additionally, the altitude of this projectile just before landing $y_{1}$ is the same as the altitude $y_{0}$ at which this projectile was launched: $y_{0} = y_{1}$ .

Hence, the initial vertical velocity of this projectile would be the exact opposite of the vertical velocity of this projectile right before landing. Since the initial vertical velocity is $20.6\; {\rm m\cdot s^{-1}}$ (upwards,) the vertical velocity right before landing would be $(-20.6\; {\rm m\cdot s^{-1}})$ (downwards.) The change in vertical velocity is:

$\begin{aligned}\Delta v_{y} &= (-20.6\; {\rm m\cdot s^{-1}}) - (20.6\; {\rm m\cdot s^{-1}}) \\ &= -41.2\; {\rm m\cdot s^{-1}}\end{aligned}$ .

Since there is no drag on this projectile, the vertical acceleration of this projectile would be $g$ . In other words, $a = g = -9.81\; {\rm m\cdot s^{-2}}$ .

Hence, the time it takes to achieve a (vertical) velocity change of $\Delta v_{y}$ would be:

$\begin{aligned} t &= \frac{\Delta v_{y}}{a_{y}} \\ &= \frac{-41.2\; {\rm m\cdot s^{-1}}}{-9.81\; {\rm m\cdot s^{-2}}} \\ &\approx 4.2\; {\rm s} \end{aligned}$ .

Hence, this projectile would be in the air for approximately $4.2\; {\rm s}$ .

8 0

1 year ago

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Which would melt first, germanium with a melting point of 1210 k or gold with a melting point of 1064oc?

iren2701 [21]

<span>Germanium To determine which melts first, convert their melting temperatures so they're both expressed on same scale. It doesn't matter what scale you use, Kelvin, Celsius, of Fahrenheit. Just as long as it's the same scale for everything. Since we already have one substance expressed in Kelvin and since it's easy to convert from Celsius to Kelvin, I'll use Kelvin. So convert the melting point from Celsius to Kelvin for Gold by adding 273.15 1064 + 273.15 = 1337.15 K So Germanium melts at 1210K and Gold melts at 1337.15K. Germanium has the lower melting point, so it melts first.</span>

8 0

3 years ago

Will give correct answer brainliest

lukranit [14]

Answer:

I THINK it’s A

Explanation:

Because all the other answers don’t make sense.

5 0

3 years ago