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kirza4 [7]
3 years ago
7

_____ are both ways of preserving current energy resources.

Physics
1 answer:
MakcuM [25]3 years ago
3 0

Answer:

4

Explanation:

I think if not. I am sorry

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An object moving in the xy-plane is subjected to the force f⃗ =(2xyı^ 3yȷ^)n, where x and y are in m
Svetllana [295]

The magnitude of the work done by force experience by the object is (2a²b + 3b²)J.

<h3>Work done by the force experienced  by the object</h3>

The magnitude of the work done by force experience by the object is calculated as follows;

W = f.d

where;

  • F is the applied force (2xyi + 3yj), where x and y are in meters
  • d is the displacement of the object = (a, b)

The work done by the force is determined from the dot product of the force and the displacement of the object.

F = (2xyi + 3yj).(a + b)

W = (2abi + 3bj).(ai + bj)

W = (2a²b + 3b²)J

Thus, the magnitude of the work done by force experience by the object is (2a²b + 3b²)J.

The complete question is below:

The particle moves from the origin to the point with coordinates (a, b) by moving first along the x-axis to (a, 0), then parallel to the y-axis.

How much work does the force do?

Learn more about work done here: brainly.com/question/8119756

5 0
2 years ago
An insulated rigid tank initially contains 1.4-kg saturated liquid water and water vapor at 200°C. At this state, 25 percent of
Elan Coil [88]

Solution:

Mass of liquid water and water vapor in the insulated tank initially = 1.4 kg

Temperature = 200 °C

And 25% of the volume by liquid water is steam.

State 1

$m=\frac{V}{v}$

$m=m_f+m_g$

$1.4=\frac{0.25V}{v_f}+\frac{0.75V}{v_g}$

$1.4=\frac{0.25V}{1.1565 \times 10^{-3}}+\frac{0.75V}{0.1274}$       (taking the value of $v_g$ and $v_g$ at 200°C  )

$V=6.304 \times 10^{-3}$

Now quality of vapor

$x=\frac{m_g}{m}$

  $=3.377 \times 10^{-3}$

Internal energy at state 1 can be found out by

$u_1=u_f+xu_{fg}$

    $=850.65+3.377\times10^{-3}\times 1744.65$

    = 856.54 kJ/kg

After heating with the resistor for 20 minutes, at state 2, the tank contains saturated water vapor $v_2=v_g \text { and }\ x=1$

Tank is rigid, so volume of tank is constant.

$v_g=v_2=\frac{V}{m}$

$v_g=\frac{6.304\times 10^{-3}}{1.4}$

$v_g=4.502 \times 10^{-3} \ m^3 /kg$

Now interpolate the value to get temperature at state 2 with specific volume value to get final temperature

$T_2=360+(374.14-360)\left(\frac{0.004502-0.006945}{0.003155-0.006945}\right)$

   = 369.11° C

Internal energy at state 2

$u_2=2154.9 \ kJ/kg$

Now power rating of the resistor

$P=\frac{m(u_2-u_1)}{t}$

$P=\frac{1.4(2154.9-856.54)}{20 \times 60}$

  = 1.51 kW

6 0
3 years ago
The electrons that are gained or lost in an ionic bond are called...?
Gala2k [10]

Answer:anion

Explanation:

4 0
3 years ago
There are two parallel conductive plates separated by a distance d and zero potential. Calculate the potential and electric fiel
taurus [48]

Answer:

The total electric potential at mid way due to 'q' is \frac{q}{4\pi\epsilon_{o}d}

The net Electric field at midway due to 'q' is 0.

Solution:

According to the question, the separation between two parallel plates, plate A and plate B (say)  = d

The electric potential at a distance d due to 'Q' is:

V = \frac{1}{4\pi\epsilon_{o}}.\frac{Q}{d}

Now, for the Electric potential for the two plates A and B at midway between the plates due to 'q':

For plate A,

V_{A} = \frac{1}{4\pi\epsilon_{o}}.\frac{q}{\frac{d}{2}}

Similar is the case with plate B:

V_{B} = \frac{1}{4\pi\epsilon_{o}}.\frac{q}{\frac{d}{2}}

Since the electric potential is a scalar quantity, the net or total potential is given as the sum of the potential for the two plates:

V_{total} = V_{A} + V_{B} = \frac{1}{4\pi\epsilon_{o}}.q(\frac{1}{\frac{d}{2}} + \frac{1}{\frac{d}{2}}

V_{total} = \frac{q}{4\pi\epsilon_{o}d}

Now,

The Electric field due to charge Q at a distance is given by:

\vec{E} = \frac{1}{4\pi\epsilon_{o}}.\frac{Q}{d^{2}}

Now, if the charge q is mid way between the field, then distance is \frac{d}{2}.

Electric Field at plate A, \vec{E_{A}} at midway due to charge q:

\vec{E_{A}} = \frac{1}{4\pi\epsilon_{o}}.\frac{q}{(\frac{d}{2})^{2}}

Similarly, for plate B:

\vec{E_{B}} = \frac{1}{4\pi\epsilon_{o}}.\frac{q}{(\frac{d}{2})^{2}}

Both the fields for plate A and B are due to charge 'q' and as such will be equal in magnitude with direction of fields opposite to each other and hence cancels out making net Electric field zero.

3 0
3 years ago
The half of the moon facing the sun is always lit, but the different phases happen
Marizza181 [45]

Answer:

the answer is the spinning of the moon lets us see different amounts of light

Explanation:

you wanna know why uh yes ok lets cut to the magic so when the moon.

3 0
3 years ago
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