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Murljashka [212]
3 years ago
15

8. What is the rate of change in an object’s position?

Physics
2 answers:
adelina 88 [10]3 years ago
4 0
A similar but separate notion is that of velocity, which the rate of change<span> of </span>position<span>. Example . If p(t) is the </span>position<span> of an </span>object<span> moving on a number line at time t (measured in minutes, say), then the average </span>rate of change<span> of p(t) is the average velocity of the </span>object<span>, measured in units per minute.</span>
Sloan [31]3 years ago
4 0
<span> the </span>rate<span> at which an object </span><span>changes position. but if you need a number or a formula i need to know if its on a hill. what panet. what the object is. etc. etc. etc.

</span>
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A 15-ft3 tank contains oxygen initially at 14.7 psia and 80°F. A paddle wheel within the tank is rotated until the pressure insi
Nikolay [14]

Explanation:

Equation for energy balance will be as follows.

         \Delta E_{system} = E_{in} - E_{out}

        \Delta U = W_{in} - Q_{out}

Hence,    W_{in} = Q_{out} + mC_{v} (T_{2} - T_{1})

Therefore, we will calculate the final temperature as follows.

            \frac{P_{1}V}{T_{1}} = \frac{P_{2}V}{T_{2}}

       T_{2} = \frac{20 psia}{14.7}(638 R)

                   = 868.03 R

Now, we will calculate the mass as follows.

             m = \frac{P_{1}V}{RT_{1}}

                 = \frac{14.7 psia \times 15 ft^{3}}{0.3353 psi ft^{3}/lbm R \times 638 R}

                 = 1.031 lbm

Hence,

        W_{in} = Q_{out} + mC_{v} (T_{2} - T_{1})

Putting the values into the above equation as follows.

            W_{in} = Q_{out} + mC_{v} (T_{2} - T_{1})

    W_{in} = 20 Btu + 1.031 lbm (\frac{0.160 Btu}{lbm R})(735 - 540)R

            W_{in} = 655.2 Btu

Thus, we can conclude that work done by paddle wheel is 655.2 Btu.

6 0
3 years ago
In which one or more of the following is the earth assumed to be the center of the universe?
sladkih [1.3K]
Can you put the answers, if there are any?
Once you do, Ill respond with a answer asap!
:)
4 0
3 years ago
Read 2 more answers
A string of length 100 cm is held fixed at both ends and vibrates in a standing wave pattern. The wavelengths of the constituent
azamat

The wavelengths of the constituent travelling waves CANNOT be 400 cm.

The given parameters:

  • <em>Length of the string, L = 100 cm</em>

<em />

The wavelengths of the constituent travelling waves is calculated as follows;

L = \frac{n \lambda}{2} \\\\n\lambda = 2L\\\\\lambda = \frac{2L}{n}

for first mode: n = 1

\lambda = \frac{2\times 100 \ cm}{1} \\\\\lambda = 200 \ cm

for second mode: n = 2

\lambda = \frac{2L}{2} = L = 100 \ cm

For the third mode: n = 3

\lambda = \frac{2L}{3} \\\\\lambda = \frac{2 \times 100}{3} = 67 \ cm

For fourth mode: n = 4

\lambda = \frac{2L}{4} \\\\\lambda = \frac{2 \times 100}{4} = 50  \ cm

Thus, we can conclude that, the wavelengths of the constituent travelling waves CANNOT be 400 cm.

The complete question is below:

A string of length 100 cm is held fixed at both ends and vibrates in a standing wave pattern. The wavelengths of the constituent travelling waves CANNOT be:

A. 400 cm

B. 200 cm

C. 100 cm

D. 67 cm

E. 50 cm

Learn more about wavelengths of travelling waves here: brainly.com/question/19249186

5 0
2 years ago
A ball is projected upward at time t=0.0s, from a point on a roof 90m above the ground. The ball rises, then falls and strikes t
chubhunter [2.5K]
As v becomes zero at the highest point, i prefer considering different travelling directions so it will become less complicated.
dont forget to add the total time up .

also to master the skills, write down the "uvsat" may help (thats the way i found it easier to handle problems)

4 0
3 years ago
Consider a 2.54-cm-diameter power line for which the potential difference from the ground, 19.6 m below, to the power line is 11
tiny-mole [99]

Answer:

The line charge density is 1.59\times10^{-4}\ C/m

Explanation:

Given that,

Diameter = 2.54 cm

Distance = 19.6 m

Potential difference = 115 kV

We need to calculate the line charge density

Using formula of potential difference

V=EA

V=\dfrac{\lambda}{2\pi\epsilon_{0}r}\times\pi r^2

\lambda=\dfrac{V\times2\epsilon_{0}}{r}

Where, r = radius

V = potential difference

Put the value into the formula

\lambda=\dfrac{115\times10^{3}\times2\times8.8\times10^{-12}}{1.27\times10^{-2}}

\lambda=1.59\times10^{-4}\ C/m

Hence, The line charge density is 1.59\times10^{-4}\ C/m

4 0
3 years ago
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