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3 years ago

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PLEASE HELP FAST The object distance for a convex lens is 15.0 cm, and the image distance is 5.0 cm. The height of the object is

9.0 cm. What is the height of the image?

1 answer:

IgorLugansk [536]3 years ago

4 0

Answer:

The image height is 3.0 cm

Explanation:

Given;

object distance, $d_o$ = 15.0 cm

image distance, $d_i$ = 5.0 cm

height of the object, $h_o$ = 9.0 cm

height of the image, $h_i$ = ?

Apply lens equation;

$\frac{h_i}{h_o} = -\frac{d_i}{d_o}\\\\ h_i = h_o(-\frac{d_i}{d_o})\\\\h_i = -9(\frac{5}{15} )\\\\h_i = -3 \ cm$

Therefore, the image height is 3.0 cm. The negative values for image height indicate that the image is an inverted image.

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qwelly [4]

Answer:

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Explanation:

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3 years ago

Two or more velocities add by ____?<br><br> Plz help

VladimirAG [237]

By vector addition.
In fact, velocity is a vector, with a magnitude intensity, a direction and a verse, so we can't simply do an algebraic sum of the two (or more velocities).
First we need to decompose each velocity on both x- and y-axis (if we are on a 2D-plane), then we should do the algebraic sum of all the components on the x- axis and of all the components on the y-axis, to find the resultants on x- and y-axis. And finally, the magnitude of the resultant will be given by
$R= \sqrt{(R_x)^2+(R_y)^2}$
where Rx and Rx are the resultants on x- and y-axis. The direction of the resultant will be given by
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3 years ago

Read 2 more answers

Describe the steps of aerobic cellular respiration and the amount of energy produced by each step.

Setler [38]

Aerobic cellular respiration have 3 parts, this is Glycolysis, Pyruvate Oxidation and Krebs cycle.

<h3>How is the aerobic breathing process?</h3>

Aerobic respiration consists of carrying out the process of degradation of organic molecules, reducing them to molecules with practically no releaseable energy. The products of the initial degradation of the organic molecule are combined with oxygen in the air and transformed into carbon dioxide and water.

In this case, Aerobic cellular respiration have 3 parts:

Glycolysis(yeilds 2ATP & 2NADH).
Pyruvate Oxidation(yeilds 2NADH).
Krebs cycle(yeilds 2GTP,2FADH2 & 6NADH).

So the total =4ATPs (2GTP equivalent to 2ATP) +10NADH (equivalent to 30ATPs) + 2FADH2(4 ATPs) =38 ATPs.

See more about Aerobic cellular respiration at brainly.com/question/22531444

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2 years ago

A 2.81 μF capacitor is charged to 1220 V and a 6.61 μF capacitor is charged to 560 V. These capacitors are then disconnected fro

fenix001 [56]

Answer:

756.88 Volts will be the potential difference across each capacitor.

Explanation:

$Q=C\times V$

Q = Charge on capacitor

C = Capacitance

V = Voltage across capacitor

Capacitance of first capacitor = $C_1=2.81 \mu F=2.81\times 10^{-6} F$

Charge of first capacitor = $Q_1$

Voltage across first capacitor = $V_1=1220 V$

$Q_1=C_1V_1$

$Q_1=2.81\times 10^{-6} F\times 1220 V=0.0034282 C$

Capacitance of first capacitor = $C_2=6.61\mu F=6.61\times 10^{-6} F$

Charge of second capacitor = $Q_2$

Voltage across first capacitor = $V_2=560 V$

$Q_2=C_2V_2$

$Q_1=6.61\times 10^{-6} F\times 560 V=0.0037016 C$

Both the capacitors are disconnected and positive plates are now connected to each other and the negative plates are connected to each other. These capacitors are connected in parallel combination.

Total charge = Q

$Q =Q_1+Q_2=0.0034282 C+ 0.0037016 C=0.0071298 C$

Total capacitance in parallel combination:

$C_p=C_1+C_2=2.81\times 10^{-6} F+6.61\times 10^{-6} F=9.42\times 10^{-6} F$

Potential across both capacitors = V

$V=\frac{Q}{C}=\frac{0.0071298 C}{9.42\times 10^{-6} F}=756.88 V$

756.88 Volts will be the potential difference across each capacitor.

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3 years ago

PLEASE HLEP IM BEING TIMED, I WILL GIVE BRANLIEST IF U R RITEEEEEEEEEEEEEEE

netineya [11]

D) igneous
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4 years ago