The fragment of an asteroid or any interplanetary material is known as a

What two factors are a part of thermohaline circulation

Margarita [4]

Answer:

These deep-ocean currents are driven by differences in the water's density, which is controlled by temperature (thermo) and salinity (haline). This process is known as thermohaline circulation.

Explanation:

7 0

3 years ago

What are the factors of evaporation and give explanation

Ira Lisetskai [31]

Temperature and rate of evaporation are proportional to each other. Surface area: As the surface area increases, the rate of evaporation increases. The surface area and rate of evaporation are proportional to each other. Humidity: The rate of evaporation decreases with an increase in humidity.

7 0

3 years ago

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How does the temperature affect the phase of water

Sever21 [200]

Answer:

Temperature affects phase change by slowing down the movement in between the atoms, thus causing a change in kinetic energy, which in turn causes the atoms to undergo forms of combining or a type of disepersion.

Explanation:

Kinetic energy while being the reason phase changes are constant, Kinetic Energy can be caused by other means. Pressure and temperature can affect many other states kinetic energy, which in turn can affect each state of matter. Making a group of atoms or compounds compacts will force the atoms to move closer together thus with a lower net kinetic energy energy. Reducing temperature also works along the same lines. Colder temperatures can slow down atomic movements which in turn will naturally make each atom move close to each other.

With all of the information provided, it is only feasible that pressure and temperature are directly corresponding with the matter and atomic phase change

8 0

2 years ago

A simple harmonic oscillator consists of a block (m = 0.50 kg) attached to a spring (k = 128 N/m). The block is pulled a certain

Zinaida [17]

Answer:

0.5 m

14.00595

8 m/s, 0.0625 s

5.71314 m/s

Explanation:

k = Spring constant = 128 N/m

A = Amplitude

E = Energy in spring = 16 J

Energy in spring is given by

$E=\dfrac{1}{2}kA^2\\\Rightarrow A=\sqrt{\dfrac{2E}{k}}\\\Rightarrow A=\sqrt{\dfrac{2\times 16}{128}}\\\Rightarrow A=0.5\ m$

The amplitude is 0.5 m

Time period is given by

$T=2\pi\sqrt{\dfrac{m}{k}}\\\Rightarrow T=2\pi\sqrt{\dfrac{0.5}{128}}\\\Rightarrow T=0.39269\ s$

Number of oscillations is given by

$N=\dfrac{5.5}{0.39269}\\\Rightarrow N=14.00595$

The number of oscillations is 14.00595

For maximum speed

$\dfrac{1}{2}mv^2=16\\\Rightarrow v=\sqrt{\dfrac{16\times 2}{0.5}}\\\Rightarrow v=8\ m/s$

The maximum speed is 8 m/s

For a distance of 0.5 m which is the amplitude

$Time=\dfrac{Distance}{Speed}\\\Rightarrow Time=\dfrac{0.5}{8}\\\Rightarrow Time=0.0625\ s$

The time taken would be 0.0625 s

The maximum kinetic energy is equal to the mechanical energy

$\dfrac{1}{2}mv^2+\dfrac{1}{2}kx^2=16$

At x = 0.35 m

$v=\sqrt{\dfrac{16-\dfrac{1}{2}kx^2}{\dfrac{1}{2}m}}\\\Rightarrow v=\sqrt{\dfrac{16-\dfrac{1}{2}128\times 0.35^2}{\dfrac{1}{2}0.5}}\\\Rightarrow v=5.71314\ m/s$

The speed of the block is 5.71314 m/s

4 0

3 years ago

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A soccer ball is kicked and left

Vedmedyk [2.9K]

Answer:

Explanation:

Considering that this is parabolic motion, we know that the time the ball is in the air begins the instant it leaves the ground, reaches up to its max height, and then begins falling until it reaches the ground. Duh, right? Some important things happen during this trip. There are a few things we need to know in order to even begin the problem. Parabolic motion has x and y coordinates because it is 2-dimmensional; the acceleration in the x dimension is not the same as the acceleration in the y dimension; the velocity of an object at its max height is always 0; the time it takes to reach its max height (where the max height is half the distance the object travels) is half the time it takes to make the whole trip. Yikes. That's a lot to know and much to remember! Don't you just LOVE physics!?

For a. the hang time is the time the ball was in the air. Some of that stuff we talked about above is pertinent to solving this problem. We know that the velocity of the ball is 0 at its max height, and we also know that if we find the time it takes to reach its max height, we can double that number to find how long it was in the air for the whole trip. Use the one-dimensional equation

$v=v_0+at$ to find out how long it took to reach the max height. Even though we don't yet know the max height, we DO know that the velocity at that point is 0. BUT before we do that, since we are working in the y-dimension only, it would behoove us (benefit us) to find the velocity particular to this dimension. We are going to answer c. first, then backtrack.

c. wants the initial vertical velocity. That is found in the magnitude of the "blanket" or generic velocity times the sin of the angle, namely:

$V_y=25sin(45)$ so

$V_y=$ 18 m/s Now we can use that as the initial upwards velocity in part a:

$v=v_0+at$ and filling in:

0 = 18 + (-9.8)t and

-18 = -9.8t so

t = 1.8 seconds. But remember, this is only half the time it was in the air. The whole trip, then, takes 2(1.8) which is

t = 3.6 seconds

That's a and c. Now for b:

b. asks for the x component of the velocity:

$V_x=Vcos\theta$ which works out to be the same as the vertical velocity, since the sin and cos of 45 degrees is the same:

$V_x=25cos45$ and

$V_x=$ 18 m/s

Onto d:

d. wants the max height. Remember, it took 1.8 seconds to get to the max height, so using yet another one-dimensional equation:

Δx = v₀t + $\frac{1}{2}at^2$ where Δx is the displacement, v₀ is the initial upwards velocity, a is the pull of gravity, and t is the time it takes to reach that max height (Δx, our unknown). Filling in:

Δx = $18(1.8)+\frac{1}{2}(-9.8)(1.8)^2$ and if you do the rounding correctly, you'll end up with this:

Δx = 32 - 16 so

the max height, Δx, is 16 meters.

e. wants the range. That translates to the distance the ball traveled. This is found in a glorified version of d = rt, where d is displacement, r is velocity, and t is...well, time (that doesn't change):

Δx = vt so

Δx = 18(3.6) remember that the ball was in the air for a total of 3.6 seconds, so

Δx = 65 meters.

Phew!!!!! That's a lot! I suggest you learn your physics or this will make you insane by the end of the course!

6 0

3 years ago