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S_A_V [24]
2 years ago
7

A ray diagram without the produced image is shown.

Physics
2 answers:
tia_tia [17]2 years ago
5 0

Answer: Real and smaller than the object.

Explanation:

It is a convex lens which converges light. The object lies beyond 2F. The light rays would converge and focus at point between F and 2F. As can be seen in the diagram given in the question. The image would be formed where the two rays are intersecting between F and 2F. The image formed would be <u>real, inverted and smaller </u>than the object. Hence, the correct option is real and smaller than the object.

Mashutka [201]2 years ago
3 0
The answer is real and smaller than the object. The image point of the top of the object is the point where the two refracted rays intersect. Tracing the entire image having the same distance from the mirror as the image of the top of the object and with the bottom on the principal axis. Hence, a real inverted image will be formed for an object outside the focal point. 
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A parallel-plate capacitor is constructed of two horizontal 12.0-cm-diameter circular plates. A 1.0 g plastic bead, with a charg
marissa [1.9K]

Answer:

Please find the answer in the explanation

Explanation:

Given that A 1.0 g plastic bead, with a charge of -6.0 nC, is suspended between the two plates by the force of the electric field between them.

Since it is suspended, it must have been repelled by the bottom negative plate and trying to be attracted to the top plate.

We can therefore conclude that the upper plate, is positively charged

B.) The charge on the positive plate of parallel-plate capacitor is constructed of two horizontal 12.0-cm-diameter circular plates must be less than 6.0 nC

3 0
2 years ago
A two-resistor voltage divider employing a 2-k? and a 3-k? resistor is connected to a 5-V ground-referenced power supply to prov
vesna_86 [32]

Answer:

circuit sketched in first attached image.

Second attached image is for calculating the equivalent output resistance

Explanation:

For calculating the output voltage with regarding the first image.

Vout = Vin \frac{R_{2}}{R_{2}+R_{1}}

Vout = 5 \frac{2000}{5000}[/[tex][tex]Vout = 5 \frac{2000}{5000}\\Vout = 5 \frac{2}{5} = 2 V

For the calculus of the equivalent output resistance we apply thevenin, the voltage source is short and current sources are open circuit, resulting in the second image.

so.

R_{out} = R_{2} || R_{1}\\R_{out} = 2000||3000 = \frac{2000*3000}{2000+3000} = 1200

Taking into account the %5 tolerance, with the minimal bound for Voltage and resistance.  

if the -5% is applied to both resistors the Voltage is still 5V because the quotient  has 5% / 5% so it cancels. to be more logic it applies the 5% just to one resistor, the resistor in this case we choose 2k but the essential is to show that the resistors usually don't have the same value. applying to the 2k resistor we have:

Vout = 5 \frac{1900}{4900}\\Vout = 5 \frac{19}{49} = 1.93 V

Vout = 5 \frac{2100}{5100}\\Vout = 5 \frac{21}{51} = 2.05 V

R_{out} = R_{2} || R_{1}\\R_{out} = 1900||2850= \frac{1900*2850}{1900+2850} = 1140

R_{out} = R_{2} || R_{1}\\R_{out} = 2100||3150 = \frac{2100*3150 }{2100+3150 } = 1260

so.

V_{out} = {1.93,2.05}V\\R_{1} = {1900,2100}\\R_{2} = {2850,3150}\\R_{out} = {1140,1260}

4 0
2 years ago
Mr. Chang's class participated in a science experiment where they placed three substances outside in the sunlight to determine w
Montano1993 [528]

Answer:

B or D but im pretty sure it is D

Explanation:

When molecules are left in the sun, it heats up. When molecules heat up, the begin to vibrate rapidly. The sun is not constant as it could get blocked by clouds, so it would, at times, slow down the movement of the molecules. The answer is most likely D.

7 0
2 years ago
Jake has noticed his best friend Alison cheating at chess several times this week. If she gets caught, she will be kicked off th
tensa zangetsu [6.8K]
He should confront her about it and if after that point she continues report it to the chess team
8 0
2 years ago
Read 2 more answers
You are in a hot-air balloon that, relative to the ground, has a veloc- ity of 6.0 m/s in a direction due east. You see a hawk m
Ann [662]

Answer:

6.32 m/s 18.43° northeast

Explanation:

We express the velocity of hawk as:

v_{Hawk}=v_{balloon}+v_{HawkRelativetoBalloon}=6 x+2 y

We consider positive x towards east and positive y due north. So the magnitude is simply the square root of the square components:

|v_{hawk}|=\sqrt[]{6^2+2^2}=\sqrt{40}≈6.32 m/s

And the angle with respect to the east should be with:

arctan(\frac{2}{6} )=18.43 \°

8 0
2 years ago
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