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leva [86]
4 years ago
9

Use the ray diagrams below the answer the following question. Where should an object (O) be placed to produce an image that is u

pright, virtual, and smaller?

Physics
1 answer:
Lelu [443]4 years ago
3 0

Answer:

the position of an object with a virtual image must be in between the points 2F' and F'

Explanation:

Due to the geometric optics laws, you know that, in order to generate a virtual image, you need that the projections of divergent rays cross in some point at the same region in which the object O is present (left or right of the lens).

By observing the fourth ray diagram you can notice that a virtual image has been generated. Then, the position of an object with a virtual image must be in between the points 2F' and F', as you can see in the fourth ray diagram.

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A plane is flying a circular path at a speed of 55.0 m/ s, with a radius of 18.3 m. The centripetal force needed to maintain thi
Nadusha1986 [10]

The plane has a centripetal acceleration <em>a</em> of

<em>a</em> = <em>v</em> ²/<em>r</em>

where <em>v</em> is the plane's tangential speed and <em>r</em> is the radius of the circle. By Newton's second law,

<em>F</em> = <em>mv</em> ²/<em>r</em>

Solve for the mass <em>m</em> :

<em>m</em> = <em>Fr</em>/<em>v</em> ² = (3000 N) (18.3 m) / (55.0 m/s)² ≈ 18.1 kg

7 0
3 years ago
When an automobile moves with constant speed down a highway, most of the power developed by the engine is used to compensate for
uysha [10]

Answer:

F=5449 N

Explanation:

Work done is a product of force and displacement ie

Work done, W, = Force*Displacement

Power, P, is Work done/Time

P=\frac {W}{t}=\frac {FS}{t} where P is power, W is work done, F is force, S is displacement and t is time

In this case, F is the frictional force. Converting the power from hp to W, we multiply by 746 hence P=746*168=125328  W

Since displacement/time is velocity, then

P=FV where V is velocity in m/s

Making F the subject

F=\frac {P}{V}

F=\frac {125328}{23}=5449.043478  N

F=5449 N

7 0
3 years ago
The common electrical wall receptacle voltage in North America is often referred to as 120 volts AC. One hundred twenty volts is
Vika [28.1K]

Answer:

120 volts is the root mean square (rms) average of the voltage as it varies with time.

Explanation:

A. The average voltage over many weeks of time (false)

Reason: Average AC voltage over one cycle is cycle (from one peak to other) is zero and so over many weeks of time  it is zero.

B. The peak voltage from an AC wall receptacle (false)

Reason: The peak voltage of an AC source in North America is zero.

C. The arithmetic mean of the voltage as it varies with time (false)

Reason: Arithmetic mean AC voltage over one cycle is cycle (from one peak to other) is zero and so over many weeks of time  it is zero.

D.  One-half the peak voltage (false)

Peak voltage =170 Volts

One-half the peak voltage = 85 volts

E. The root mean square (rms) average of the voltage as it varies with time (True)

Reason:

The peak voltage and root mean square voltage are related by:

V_{rms}=\frac{V_{p}}{\sqrt{2} }\\\\V_{rms}=\frac{170}\sqrt{2}V_{rms}\\\\V_{rms}=120 Volts

Average value of voltage over one cycle is zero, so instead of calculating average voltage for AC peak voltage is first squared and the mean is calculated.

5 0
3 years ago
Properties of helium
postnew [5]

Answer:

Helium has many unique properties: low boiling point, low density, low solubility, high thermal conductivity and inertness, so it is use for any application which can explioit these properties. Helium was the first gas used for filling balloons and dirigibles

7 0
3 years ago
Two large, flat metal plates are separated by a distance that is very small compared to their height and width. The conductors a
mote1985 [20]

Answer:

E=0

Explanation:

Electric field due to each thin sheet of charge=\sigma/2\varepsilon

let us say the right plate has positive charge density \varepsilonand left sheet has a negative charge density -\varepsilon .

In the region between the plates,the electric field due to each plate is in same direction,

E=\sigma/2\varepsilon-(-\sigma/2\varepsilon)

E=\sigma/\varepsilon

in the region outside the plates, the field due to the plates is in opposite directions

E=-\sigma/2\varepsilon-(-\sigma/2\varepsilon)

E=-\sigma/2\varepsilon+\sigma/2\varepsilon

E=0

4 0
3 years ago
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