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4 years ago

6

Elias serves a volleyball at a velocity of 16 m/s. The mass of the volleyball is 0.27 kg. What is the height of the volleyball a

bove the gym floor when its total mechanical energy is 41.70 J? Round to the nearest tenth\

2 answers:

evablogger [386]4 years ago

7 0

Ingenuity says that the answer is 2.7 m

givi [52]4 years ago

4 0

Hello! The anwser is 2.7! I hope this helps! ^^

You might be interested in

Consider a motor that exerts a constant torque of 25.0 nâ‹…m to a horizontal platform whose moment of inertia is 50.0 kgâ‹…m2. A

Crazy boy [7]

Answer:

W = 1884J

Explanation:

This question is incomplete. The original question was:

<em>Consider a motor that exerts a constant torque of 25.0 N.m to a horizontal platform whose moment of inertia is 50.0kg.m^2 . Assume that the platform is initially at rest and the torque is applied for 12.0rotations . Neglect friction. </em>

<em> How much work W does the motor do on the platform during this process? Enter your answer in joules to four significant figures.</em>

The amount of work done by the motor is given by:

$W=\Delta K$

$W= 1/2*I*\omega f^2-1/2*I*\omega o^2$

Where I = 50kg.m^2 and ωo = rad/s. We need to calculate ωf.

By using kinematics:

$\omega f^2=\omega o^2+2*\alpha*\theta$

But we don't have the acceleration yet. So, we have to calculate it by making a sum of torque:

$\tau=I*\alpha$

$\alpha=\tau/I$ => $\alpha = 0.5rad/s^2$

Now we can calculate the final velocity:

$\omega f = 8.68rad/s$

Finally, we calculate the total work:

$W= 1/2*I*\omega f^2 = 1883.56J$

Since the question asked to "<em>Enter your answer in joules to four significant figures.</em>":

W = 1884J

3 0

3 years ago

You are standing in a building whose height (40m) you throw a ball downward at a angle of -30 at a speed of (10m/s) acceleration

jeyben [28]

Answer: 3.41 s

Explanation:

Assuming the question is to find the time $t$ the ball is in air, we can use the following equation:

$y=y_{o}+V_{o}sin \theta t-\frac{1}{2}gt^{2}$

Where:

$y=0m$ is the final height of the ball

$y_{o}=40 m$ is the initial height of the ball

$V_{o}=10 m/s$ is the initial velocity of the ball

$t$ is the time the ball is in air

$g=9.8 m/s^{2}$ is the acceleration due to gravity

$\theta=30\°$

Then:

$0 m=40 m+(10 m/s)(sin(30\°))t-\frac{1}{2}9.8 m/s^{2}t^{2}$

$0 m=40 m+5m/s t-4.9 m/s^{2}t^{2}$

Multiplying both sides of the equation by -1 and rearranging:

$4.9 m/s^{2}t^{2}-5m/s t-40 m=0$

At this point we have a quadratic equation of the form $at^{2}+bt+c=0$ , which can be solved with the following formula:

$t=\frac{-b \pm \sqrt{b^{2}-4ac}}{2a}$

Where:

$a=4.9$

$b=-5$

$c=-40$

Substituting the known values:

$t=\frac{-(-5) \pm \sqrt{(-5)^{2}-4(4.9)(-40)}}{2(4.9)}$

Solving the equation and choosing the positive result we have:

$t=3.41 s$ This is the time the ball is in air

5 0

3 years ago

Water initially at 200 kPa and 300°C is contained in a piston–cylinder device fitted with stops. The water is allowed to cool at

netineya [11]

Answer:

Δu=1300kJ/kg

Explanation:

Energy at the initial state

$p_{1}=200kPa\\t_{1}=300^{o}\\u_{1}=2808.8kJ/kg(tableA-5)$

Is saturated vapor at initial pressure we have

$p_{2}=200kPa\\x_{2}=1(stat.vapor)\\v_{2}=0.8858m^3/kg(tableA-5)$

Process 2-3 is a constant volume process

$p_{3}=100kPa\\v_{3}=v_{2}=0.8858m^{3}/kg\\u_{3}=1508.6kJ/kg(tableA-5)$

The overall in internal energy

Δu=u₁-u₃

We replace the values in equation

Δu=u₁-u₃

$=2808.8kJ/kg-1508.6kJ/kg\\=1300kJ/kg$

Δu=1300kJ/kg

3 0

3 years ago

1. The statement about Newton's reflecting telescope contains 1 error. Find the error, write it, and

Jlenok [28]

Answer:

i. The error is the rough convex mirror.

ii. This should be replaced with a smooth convex morror.

Explanation:

Reflection is dependent on the surface involved and has two types; diffuse and specular. When the surface is rough, diffused reflection is observed. The surface causes a distortion of the incident light (the rays would be reflected at different angles to one another) after reflection. This makes some rays to interfere with one another. While specular reflection is observed with a smooth surface.

In the statement, the rough convex mirror would produce a distorted reflection which would produce diffused reflection. The effect is that few or no rays (depending on the degree of how rough the surfce is) would be reflected to the other smooth, flat diagonal mirror.

8 0

3 years ago

"In a pure substance all the particles must be identical; therefore pure substances are composed only of elements." Do you agree

kenny6666 [7]

I do not agree with the statement.
The "substance" can be a compound. It's "pure"
as long as there's nothing else in it but its name.

'Pure' water is 100% H₂O with nothing else in it.
'Pure' table salt is 100% NaCl with nothing else in it.
'Pure' carbon dioxide is 100% CO₂ with nothing else in it.

These example substances are all compounds, not elements.

6 0

3 years ago