What does Kinetic Energy depend on

What is a positive impact of artificial fertilizers

dybincka [34]

<span>The top benefit of pesticides is their effectiveness against pests that would otherwise decimate crops large and small. </span>

5 0

2 years ago

Is the star moving toward earth, away from earth, or is there not enough information provided to determine its motion?

ivanzaharov [21]

If a star is moving towards Earth, shift towards the blue end of the spectrum, this is called blue shift. If the star is moving away from Earth the light from that star will be red and is called red shift .

The faster a star moves towards the earth, the more its light is shifted to higher frequencies. In contrast, if a star is moving away from the earth, its light is shifted to lower frequencies on the color spectrum

if a star is moving towards Earth, it appears to emit light that is shorter in wavelength compared to a source of light that isn't moving. Because shorter wavelengths correspond to a shift towards the blue end of the spectrum, this is called blue shift.

If the star is moving away from Earth, its light will lose energy to reach Earth, therefore the light from that star will be red and is called red shift

learn more about blue shift :

brainly.com/question/5368237?referrer=searchResults

#SPJ4

8 0

1 year ago

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

2 years ago

A tennis player tosses a tennis ball straight up and then catches it after 1.25 s at the same height as the point of release.

Alenkasestr [34]

Answer:

A. 9.8 m/s²

B. Zero

C. 6.125 m/s

D. 1.91 m

Explanation:

From the question given above, the following data were obtained:

Time (T) spent in the air = 1.25 s

A. Determination of the acceleration of the ball.

From the description given in question above, the motion of the tennis ball is motion under gravity. Hence, the ball will experience an acceleration due to gravity of 9.8 m/s²

B. Determination of the velocity at maximum height.

Maximum height is the greatest point reached by the tennis ball above the ground. At maximum height, the velocity of the tennis ball is zero since it has no further force to propel it upward.

C. Determination of the initial velocity of the ball.

We'll begin by calculating the time taken to reach the maximum height. This can be obtained as follow:

Time (T) spent in the air = 1.25 s

Time (t) to reach the maximum height =?

T = 2t

1.25 = 2t

Divide both side by 2

t = 1.25 / 2

t = 0.625 s

Finally, we shall determine the initial velocity of the ball. This can be obtained as follow:

Time (t) to reach the maximum height = 0.625 s

Acceleration due to gravity (g) = 9.8 m/s²

Final velocity (v) = 0 (at maximum height)

Initial velocity (u) =?

v = u – gt (since the ball is going against gravity)

0 = u – (9.8 × 0.625)

0 = u – 6.125

Collect like terms

0 + 6.125 = u

u = 6.125 m/s

Thus, the initial velocity of the ball is 6.125 m/s

D. Determination of the maximum height.

Acceleration due to gravity (g) = 9.8 m/s²

Final velocity (v) = 0 (at maximum height)

Initial velocity (u) = 6.125 m/s

Maximum height (h) =?

v² = u² – 2gh (since the ball is going against gravity)

0² = 6.125² – (2 × 9.8 × h)

0 = 37.52 – 19.6h

Collect like terms

0 – 37.52 = – 19.6h

– 37.52 = – 19.6h

Divide both side by – 19.6

h = – 37.52 / – 19.6

h = 1.91 m

Thus, the maximum height reached by the ball is 1.91 m

3 0

2 years ago

Block with m = 300 grams oscillates at the end of a linear spring with k = 6.5 N/m. (assume this is a top-down view and the bloc

masha68 [24]

Explanation:

Given that,

Mass of the block, m = 300 g = 0.3 kg

Linear spring constant, k = 6.5 N/m

(a) Let T is the period of the block's motion. It is given by :

$T=\dfrac{2\pi}{\omega}$

where

$\omega=\sqrt{\dfrac{k}{m}}$ = angular frequency

Also, $\omega=\sqrt{\dfrac{k}{m}}$

$T=\dfrac{2\pi}{\sqrt{\dfrac{k}{m}}}$

$T=\dfrac{2\pi}{\sqrt{\dfrac{6.5}{0.3}}}$

T = 1.34 seconds

(b) The maximum acceleration of the block is, $a_{max}=2\ m/s^2$

The maximum acceleration is given by :

$a_{max}=\omega^2A$

A is the amplitude of the motion,

$A=\dfrac{a_{max}}{\omega^2}$

$A=\dfrac{a_{max}}{(\sqrt{\dfrac{k}{m}})^2}$

$A=\dfrac{ma_{max}}{k}$

$A=\dfrac{0.3\times 2}{6.5}$

A = 0.09 meters

Hence, this is the required solution.

3 0

2 years ago