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2 years ago

A kickball is kicked straight up at a speed of 22.4m/s. how high does it go

Physics

1 answer:

Kaylis [27]2 years ago

3 0

Answer:

the kickball goes straight up 25.6 meters

Explanation:

We are in the case of motion under constant (uniform) acceleration. This is an object with initial velocity pointing strictly upwards, under th acceleration of gravity which slows down its motion. The projectile keeps losing velocity as it moves upward until its velocity is zero when it stops moving upwards, and starts falling also under accelerated motion and gaining downward velocity as time goes by.

We use he formula for displacement under accelerated motion (due to gravity "g" opposing the initial velocity):

$y-y_0=v_0\,t-\frac{1}{2} g\,t^2$

and also the formula for the velocity in terms of time:

$v(t)=v_0-g\,t$

Using that the initial velocity is 22.4 m/s, g = 9.8 m/s^2 , and the fact that we want the time "t" at which the velocity v(t) = 0 (zero), we solve for "t" in the second equation, and use it to substitute for "t" in the first one:

$0=v_0-g\,t\\t=\frac{v_0}{g} \\\\\\y-y_0=v_0\,t-\frac{1}{2} g\,t^2\\y-y_0=v_0\,(\frac{v_0}{g})-\frac{1}{2} \,g\,(\frac{v_0}{g})^2\\y-y_0=\frac{v_0^2}{g} -\frac{1}{2}\,\frac{v_0^2}{g}\\y-y_0=\frac{1}{2}\,\frac{v_0^2}{g}\\y-y_0=\frac{1}{2}\,\frac{22.4^2}{9.8}\,\,m\\\\y-y_0=25.6\,\,m$

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Answer:

c is the answer

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2 years ago

Which statement is not a good practice when working inside a computer case?

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Answer:

a. be sure to hold expansion cards by the edge connectors

Explanation:

Removal of loose jewelry is a good safety practice. Also not touching a microchip with a magnetized screwdriver is also a good practice.

But holding expansion cards by the edge connectors is not a good practice, so it is the odd one in the question. Therefore answer option a provides the correct and best answer to the question

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2 years ago

010) Identify the true statement. Group of answer choices The height of waves is determined by wind strength and fetch. Wave bas

AlekseyPX

Answer:

The height of the wave is determined by the wind strength and fetch.

Explanation:

The height of the wave is determined by the wind strength and fetch.

The more the strength and the more the fetch size the more will be the height of the wave.

Remember as the wave approaches the coast its wavelength decreases and the wave height increases, whereas when the wave goes away from the coast its wavelength increases and height decreases.

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2 years ago

When you urinate, you increase pressure in your bladder to produce the flow. For an elephant, gravity does the work. An elephant

weqwewe [10]

Answer:

a) v = 1.19 m / s , b) P₁ = 0.922 10⁵ Pa

Explanation:

1) Let's use the fluid continuity equation

Q = A v

The area of a circle is

A = π r2 = π d²/4

v = Q / A = Q 4 / pi d²

v = 0.006 4/π 0.08²

v = 1.19 m / s

2) write Bernoulli's equation, where point 1 is the bladder and point 2 is the urine exit point

P₁ + ½ rho v₁² + rho g y₁ = P₂ + ½ rho v₂² + rho g y₂

The exercise tell us

P₂ = 1.0013 105 Pa

v₁ = 0

y₁ = 1 m

y₂=0

Rho (water) = 1000 kg / m³

P₁ + rho y₁ = P₂ + ½ rho v₂²

P₁ = P₂ + ½ rho v₂² - rho g y₁

P₁ = 1.013 10⁵ + ½ 1000 (1.19)² - 1000 9.8 1

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P₁ = 0.922 10⁵ Pa

8 0

2 years ago

A 0.200-kg mass attached to the end of a spring causes it to stretch 5.0 cm. If another 0.200-kg mass is added to the spring, th

ziro4ka [17]

Answer:

A: 4 times as much

B: 200 N/m

C: 5000 N

D: 84,8 J

Explanation:

In the first question, we have to caculate the constant of the spring with this equation:

$m*g=k*x$

Getting the k:

$k=\frac{m*g}{x} =\frac{0,2[kg]*9,81[\frac{m}{s^{2} } ]}{0,05[m]} =39,24[\frac{N}{m}]$

Then we can calculate how much the spring stretch whith the another mass of 0,2kg:

$x=\frac{m*g}{k} =\frac{0,4[kg]*9,81[\frac{m}{s^{2} } ]}{39,24[\frac{N}{m}]} =0,1[m]\\$

The energy of a spring:

$E=\frac{1}{2}*k*x^{2}$

For the first case:

$E=\frac{1}{2} *39,24[\frac{N}{m}]*(0,05[m])^{2} =0,049 [J]$

For the second case:

$E=\frac{1}{2} *39,24[\frac{N}{m}]*(0,1[m])^{2} =0,0196 [J]$

If you take the relation E2/E1 = 4.

We have the next facts:

x=0,005 m

E = 0,0025 J

Using the energy equation for a spring:

$E=\frac{1}{2}*k*x^{2}$ ⇒ $k=\frac{E*2}{x^{2} } =\frac{0,0025[J]*2}{(0,005[m])^{2} } =200[\frac{N}{m} ]$

The potential energy of the diver will be equal to the kinetic energy in the moment befover hitting the watter.

$E=W*h=500[N]*10[m]=5000[J]$

Watch out the units in this case, the 500 N reffer to the weighs of the diver almost relative to the earth, thats equal to m*g.

The work is equal to the force acting in the direction of the motion. so we have to do the diference beetwen angles to obtain the effective angle where the force is acting: 47-15=32 degree.

The force acting in the direction of the ramp will be the projection of the force in the ramp, equal to F*cos(32). The work will be:

W=F*d=F*cos(32)*d=10N*cos(32)*10m=84,8J

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2 years ago