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Mazyrski [523]
3 years ago
13

How does mechanical energy relate to the work an object can do

Physics
1 answer:
kifflom [539]3 years ago
6 0
I am 11 year old and u just needed to login in so I can't answer ur questions
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What 2 things do you need to make sound? help ASAP
inessss [21]

Answer:

- A vibrating object

- a medium to travel

HOPE IT HELPS :)

PLEASE MARK IT THE BRAINLIEST!

8 0
4 years ago
Steam at 400C has a specific volume of 0.02m3/kg. Determine the pressure of the steam based on a) the ideal gas equation b) the
nikklg [1K]

Answer:

by ideal gas pressure = 15529.475 kPa

by compressibility chart pressure = 12576 kPa

by steam tables Pressure = 12517 kPa

Explanation:

given data

temperature T = 400°C = 673 K

volume v = 0.02 m³/kg

to find out

pressure by ideal gas, compressibility chart and steam tables

solution

we know here by table

gas constant R is 0.4615 kJ/ kg-K

and critical temp Tc = 647.1 K

and critical pressure Pc = 22064 kPa

so by ideal gas pressure is

pressure = R×T / v

pressure = 0.4615 × 673  / 0.02

pressure = 15529.475 kPa

and

by compressibility chart

temperature reduce is = T/ Tc

temperature reduce Tr = 673 / 647.1

Tr = 1.040 K

so pseudo reduce volume is here

reduce volume Vr = v / ( RTc/Pc)

reduce volume Vr =\frac{0.02}{\frac{461.5(647.1)}{22064*10^{3} } }

0.02 / ( 461.5(647.1) / 22064×10³)

reduce volume = 1.48

and we know by compressibility chart

reduce pressure Pr is 0.57

so

pressure = Pr × Pc

pressure = 0.57 × 22064 ×  10³

pressure = 12576 kPa

and

from steam table

pressure is 12.5 MPa at 673 K and 0.020030 m³/kg

pressure is 15 MPa at 673 K and 0.015671 m³/kg

so

pressure P is

\frac{0.02 - 0.020030}{0.015671 - 0.020030} = \frac{ P - 12.5}{15 - 12.5}

so

Pressure = 12517 kPa

4 0
3 years ago
A loop of wire is placed in a uniform magnetic field. Let ø be the angle between the direction of b and the normal. At which val
zaharov [31]

Answer:

B. 0 degrees

Explanation:

The magnetic flux through a certain area enclosed by a loop of wire immersed in a region with magnetic field is given by:

\Phi = B A cos \theta

where

B is the strength of the magnetic field

A is the area enclosed by the coil

\theta is the angle between the direction of the magnetic field and the direction of the normal to the coil

From the equation above, we see that the magnetic flux is maximum when

cos \theta=1

So when the field is parallel to the normal to the coil (so, when it is perpendicular to the coil), therefore when:

\theta=0^{\circ}

So, the correct option is

B. 0 degrees

5 0
4 years ago
A rocket is fired with an initial VELOCITY OF 100m/s at an angle of 55° above the horizontal, It explodes On the mountain Side 1
GuDViN [60]

Answer

688.32m and 277.44m

Explanation :

⠀

\large{\maltese{\textsf{\underline{To find :-}}}}

The X and Y coordinates of the rocket relative of firing

⠀

⠀

\large{\maltese{\textsf{\underline{Given :-}}}} \\ \\ \sf velocity (v_i) = 100m{s}^{-1} \\ \sf angle ({\theta}_{1}) = 55.0{\degree} \\ \sf time (t) = 12s

⠀

⠀

\Large{\maltese{\textsf{\underline{\underline{Step by Step Solution:-}}}}}

⠀

⠀

<u>The</u><u> </u><u>horizontal</u><u> </u><u>range</u><u> </u><u>of</u><u> </u><u>projectile</u><u> </u><u>at</u><u> </u><u>x</u><u>.</u><u> </u>

⠀

\sf \large{x = v_{xi}} \times t \\ \\ \sf \large{x = v_i \times \cos {\theta}_{i} \times t}

⠀

⠀

\large\textsf{\underline{Now substituting the required values}}

⠀

⠀

\sf x = 300 \times \cos 55{\degree} \times 12 \\ \\ \sf x = 100 \times 0.5756 \times 12 \\ \\ {\underline{\boxed{\bold{ x = 688.32m}}}}

⠀

⠀

The vertical position of projectile at y.

⠀

⠀

\sf \large  y = v_{yi} \times t -  (\frac{1}{2}  \times g  \times {t}^{2}) \\  \\   \sf  \large y = v_i \times  \cos \theta  \times t -  \frac{1}{2} g {t}^{2}

⠀

⠀

\textsf{ \large {\underline{Now substituting the required values}}  }

⠀

⠀

\sf y = 100 \times  \cos55{ \degree} \times 12 -  \frac{1}{2}   \times 9.80 \times  {12}^{2} \\  \\  \sf  y = 100 \times 0.8192 \times 12 - 0.5 \times 9.8 \times 144 \\  \\  \sf y = 983.04 - 705.6 \\  \\  \underline{ \boxed{ \bold{y = 277.44m}}}

⠀

⠀

⠀

<h3><u>Henceforth</u><u>,</u><u> </u><u>the</u><u> </u><u>distance</u><u> </u><u>at</u><u> </u><u>horizon</u><u> </u><u>is</u><u> </u><u>6</u><u>8</u><u>8</u><u>.</u><u>3</u><u>2</u><u>m</u><u> </u><u>and</u><u> </u><u>at</u><u> </u><u>vertical</u><u> </u><u>is</u><u> </u><u>2</u><u>7</u><u>7</u><u>.</u><u>4</u><u>4</u><u>m</u><u>.</u></h3>

8 0
2 years ago
A scientist conducted research and plotted his results in the histogram shown below. Which of the
ludmilkaskok [199]

Answer:

Median

Explanation:

  • we commonly use mean as the measure of the center of the data.
  • But Median is better measure than mean.

<u>Extra</u><u> </u><u>information</u><u>:</u><u>-</u>

<u>Median</u><u> </u><u>formula</u><u>:</u><u>-</u>

\boxed{\begin{minipage}{6cm}$\bigstar$\:\:\sf Median = l + $\sf\dfrac{\frac{n}{2}-C.f.}{f}\times h\\\\Here: \\1)\:n = \sum f =\\2)\:l=Lower\:limit\:of\:median\:class=\\3)\:C.f.=Cumulative\:frequency\:of\:class\\preceeding\:the\:median\:class=\\4)\:f= frequency\:of\:median\:class=\\5)\:h= Class\:interval = \end{minipage}}

6 0
3 years ago
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