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Murljashka [212]
3 years ago
12

A small candle is 34 cm from a concave mirror having a radius of curvature of 22 cm. Where will the image of the candle be locat

ed? Follow the sign conventions.
Physics
1 answer:
xxTIMURxx [149]3 years ago
5 0

Answer:

16.26 cm in front of the mirror

Explanation:

Using,

1/f = 1/u+1/v....................... Equation 1

Where f = focal focal length of the concave mirror, u = object distance, v = image distance.

make v the subject of the equation

v = fu/(u-f)................... Equation 2

Note: The focal length of a concave mirror is positive

Using the real- is- positive convention

Given: f = 22/2 = 11 cm, u = 34 cm.

Substitute into equation 2

v = (34×11)/(34-11)

v = 374/23

v = 16.26 cm.

The image will be formed 16.26 cm in front of the mirror.

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just for anyone looking for the answer i just took the test and the answer is tornadoes

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3 years ago
Estimate the Joule-Thomson coefficient of refrigerant-134a at 0.70 MPa and 50°C. Assume the second state will be selected for a
leva [86]

Answer:

\mu = 0.018\,\frac{^{\textdegree}C}{kPa}

Explanation:

The Joule-Thomson coefficient is the ratio of the change of temperature to the change of pressure under isoenthalpic conditions:

\mu = \left(\frac{\Delta T}{\Delta P}\right)_{h}

Initial and final properties are:

T_{1} = 50^{\textdegree}C, P_{1}=700\,kPa, h_{1}=288.54\,\frac{kJ}{kg}. Superheated Vapor.

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\mu = 0.018\,\frac{^{\textdegree}C}{kPa}

5 0
4 years ago
Three people are pushing a 500 kg of box in the same direction. applied forces are 30 n, 20 n, and 10 n respectively. if the acc
GREYUIT [131]
The total force applied by the three people is:
F=30 N+20 N+10 N=60 N
This force is pushing toward the direction of the motion, while the  frictional force F_f points in the opposite direction.
We can write Newton's second law applied to the block: the resultant of the two forces must be equal to the product between the block's mass and its acceleration
F-F_f = ma
We know the mass of the block, m=500 kg, and the acceleration, a=0.02 m/s^2, so we can find the friction:
F_f = F-ma=60 N - (500 kg)(0.02 m/s^2)=50 N
8 0
4 years ago
A flock of ducks is trying to migrate south for the winter, but they keep being blown off course by a wind blowing from the west
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The ducks' flight path as observed by someone standing on the ground is the sum of the wind velocity and the ducks' velocity relative to the wind:

ducks (relative to wind) + wind (relative to Earth) = ducks (relative to Earth)

or equivalently,

\vec v_{D/W}+\vec v_{W/E}=\vec v_{D/E}

(see the attached graphic)

We have

  • ducks (relative to wind) = 7.0 m/s in some direction <em>θ</em> relative to the positive horizontal direction, or

\vec v_{D/W}=\left(7.0\dfrac{\rm m}{\rm s}\right)(\cos\theta\,\vec\imath+\sin\theta\,\vec\jmath)

  • wind (relative to Earth) = 5.0 m/s due East, or

\vec v_{W/E}=\left(5.0\dfrac{\rm m}{\rm s}\right)(\cos0^\circ\,\vec\imath+\sin0^\circ\,\vec\jmath)

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\vec v_{D/E}=v(\cos270^\circ\,\vec\imath+\sin270^\circ\,\vec\jmath)

Then by setting components equal, we have

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We only care about the direction for this question, which we get from the first equation:

\left(7.0\dfrac{\rm m}{\rm s}\right)\cos\theta=-5.0\dfrac{\rm m}{\rm s}

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or approximately 136º or 224º.

Only one of these directions must be correct. Choosing between them is a matter of picking the one that satisfies <em>both</em> equations. We want

\left(7.0\dfrac{\rm m}{\rm s}\right)\sin\theta=-v

which means <em>θ</em> must be between 180º and 360º (since angles in this range have negative sine).

So the ducks must fly (relative to the air) in a direction 224º relative to the positive horizontal direction, or about 44º South of West.

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Answer:

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Explanation:

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