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2 years ago

What’s the equation that links with total input, efficiency, energy and useful output energy transfer

1 answer:

8 0

Efficiency. The ratio of energy which was transferred to a useful form compared to the total energy initially supplied is called the efficiency of the device. Efficiencies can be written as decimals like 0.33 or percentages 33%. To convert a efficiency expressed as a decimal to a percentage you need to multiply by 100.

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Calculate the distance d from the center of the sun at which a particle experiences equal attractions from the earth and the sun

fredd [130]

Answer:

149.34 Giga meter is the distance d from the center of the sun at which a particle experiences equal attractions from the earth and the sun.

Explanation:

Mass of earth = m = $5.976\times 10^{24} kg$

Mass of Sun = M = 333,000 m

Distance between Earth and Sun = r = 149.6 gm = 1.496\times 10^{11} m[/tex]

1 giga meter = $10^{9} meter$

Let the mass of the particle be m' which x distance from Sun.

Distance of the particle from Earth = (r-x)

Force between Sun and particle:

$F=G\frac{M\times m'}{x^2}=G\frac{333,000 m\times m'}{x^2}$

Force between Sun and particle:

$F'=G\frac{mm'}{(r-x)^2}$

Force on particle is equal:

F = F'

$G\frac{333,000 m\times m'}{x^2}=G\frac{mm'}{(r-x)^2}$

$\frac{x}{r-x}=\sqrt{333,000}$ = ±577.06

Case 1:

$\frac{x}{r-x}=577.06$

x = $1.49\times 10^{11} m=149.34 Gm$

Acceptable as the particle will lie in between the straight line joining Earth and Sun.

Case 2:

$\frac{x}{r-x}=-577.06$

x = $1.49\times 10^{11} m=149.86 Gm$

Not acceptable as the particle will lie beyond on line extending straight from the Earth and Sun.

3 0

2 years ago

An inverted image of an object is viewed on a screen from the side facing a converging lens. An opaque card is then introduced c

Marta_Voda [28]

Answer:

d) The image would be dimmer

Explanation:

When inverted image of an object is viewed on a screen from the side facing a converging lens, the

lens focuses the diverging, and blurred light rays from the distance object through refraction of the rays two times.Then the rays is converge by the double bending cl at a focal point behind the lens inorder for a sharper image to be be observed.

But when an opaque card is then introduced to cover only the upper half of the lens then the image becomes dimmer because of the reduction in the light intensity on the screen by 50%

5 0

2 years ago

A skydiver is using wind to land on a target that is 50 m away horizontally. The skydiver starts from a height of 70 m and is fa

elena55 [62]

Answer:

Answer:

15.67 seconds

Explanation:

Using first equation of Motion

Final Velocity= Initial Velocity + (Acceleration * Time)

v= u + at

v=3

u=50

a= - 4 (negative acceleration or deceleration)

3= 50 +( -4 * t)

-47/-4 = t

Time = 15.67 seconds

6 0

2 years ago

A magnetic field of 0.080 T is in the y-direction. The velocity of wire segment S has a magnitude of 78 m/s and components of 18

Annette [7]

Answer:

Part a)

Induced EMF when length vector is along Z direction is 0.72 V

Part b)

Induced EMF when length vector is along Y direction is ZERO

Explanation:

As we know that the motional EMF induced in the wire is given as

$E = (v\times B). L$

As we know that

$v = 18\hat i + 24\hat j + 72\hat k$

$B = 0.080 \hat j$

$L = 0.50 \hat k$

now we have

$\vec v \times \vec B = 1.44\hat k - 5.76 \hat i$

so we have

$E = 1.44 (0.50) = 0.72 V$

If the length vector is along Y direction then we have

$L = 0.50 \hat j$

so again we have

$\vec v \times \vec B = 1.44\hat k - 5.76 \hat i$

so we have

EMF = 0

6 0

2 years ago

At time t, gives the position of a 3.0 kg particle relative to the origin of an xy coordinate system ( ModifyingAbove r With rig

Elena-2011 [213]

Complete Question

The complete Question is shown on the first uploaded image

Answer:

The torque acting on the particle is $\tau = 48t \r k$

The magnitude of the angular momentum increases relative to the origin

Explanation:

From the equation we are told that

The position of the particle is $\= r = 4.0 t^2 \r i - (2.0 t - 6.0 t^2 ) \r j$

The mass of the particle is $m = 3.0 kg$

The time is t

The torque acting on the particle is mathematically represented as

$\tau = \frac{ d \r l }{dt}$

where $\r l$ is change in angular momentum which is mathematically represented as

$\r l = m (\r r \ \ X \ \ \r v)$

Where X mean cross- product

$\r v$ is the velocity which is mathematically represented as

$\r v = \frac{d \r r }{dt}$

Substituting for $\r r$

$\r v = \frac{d }{dt} [ 4 t^2 \r i - (2t + 6t^2 ) \r j]$

$\r v = 8t \r i - (2 + 12 t) \r j$

Now the cross product of $\r r \ and \ \r v$ is mathematically evaluated as

$\r r \ \ X \ \ \r v = \left[\begin{array}{ccc}{\r i}&{\r j}&{\r k}\\{4t^2}&{-2t -6t^2}&0\\{8t}&{-2 -12t}&0\end{array}\right]$

$= 0 \r i + 0 \r j + (- 8t^2 -48t^3 + 16t^2 + 48t^3 ) \r k$

$\r r \ \ X \ \ \r v = 8t^2 \r k$

So the angular momentum becomes

$\r l = m (8t^2 \r k)$

Substituting for m

$\r l = 3 * (8t^2 \r k)$

$\r l =24t^2 \r k$

Substituting into equation for torque

$\tau = \frac{d}{dt} [24t^2 \r k]$

$\tau = 48t \r k$

The magnitude of the angular momentum can be evaluated mathematically as

$|\r l| = \sqrt{(24 t^2) ^2}$

$|\r l| = 24 t^2$

From the is equation we see that the magnitude of the angular momentum is varies directly with square of the time so it would relative to the origin because at the origin t= 0s and we move out from origin t increases hence angular momentum increases also

4 0

2 years ago