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ss7ja [257]
3 years ago
13

An object is 10 cm from thé mirror, its height is 1 cm and thé focal length is 5 cm. What is thé distance from thé mirror? S1= _

___
__cm and what is the image height? H1= _____cm.

An object is 10 cm from the mirror it’s height is 3 cm and thé focal length is 2 cm. What is the distance from the image t the mirror?
S1= ______cm.


An object is 10 cm from the mirror, its height is 3 cm and thé focal length is 2.0 cm. What is the image height? (Indicate the object orientation by including the + or - sign with answer. H1= ____cm.
Physics
1 answer:
Viefleur [7K]3 years ago
5 0
Note: I assume the mirror is concave, so that its focal length is positive (it is not specified in the text)

1a) We can use the mirror equation to find the distance of the image from the mirror:
\frac{1}{f}= \frac{1}{p}+ \frac{1}{q}
where 
f=5 cm is the focal length
p=10 cm is the distance of the object from the mirror
q is the distance of the image from the mirror.

Rearranging the equation, we find
\frac{1}{q}= \frac{1}{f}- \frac{1}{p}= \frac{1}{5}- \frac{1}{10}= \frac{1}{10 cm}
so, the distance of the image from the mirror is q=10 cm.

1b) The image height is given by the magnification equation:
\frac{h_i}{h_o}=- \frac{p}{q}
where h_i is the heigth of the image and h_o=1 cm is the height of the object. By rearranging the equation and using p and q, we find
h_i=-h_o  \frac{p}{q}=-(1 cm) \frac{10 cm}{10 cm}=-1 cm
and the negative sign means the image is inverted.

2) As before, we can find the distance of the image from the mirror by using the mirror equation:
\frac{1}{f}= \frac{1}{p}+ \frac{1}{q}
Rearranging it, we find
\frac{1}{q}= \frac{1}{f}- \frac{1}{p}= \frac{1}{2}- \frac{1}{10}= \frac{4}{10 cm}
so, the distance of the image from the mirror is
q= \frac{10}{4}cm= 2.5 cm

3) As before, we find the distance of the image from the mirror by using the mirror equation:
\frac{1}{f}= \frac{1}{p}+ \frac{1}{q}
Rearranging it, we find
\frac{1}{q}= \frac{1}{f}- \frac{1}{p}= \frac{1}{2}- \frac{1}{10}= \frac{4}{10 cm}
so, the distance of the image from the mirror is
q= \frac{10}{4}cm= 2.5 cm

And now we can use the magnification equation to find the image height:
\frac{h_i}{h_o}=- \frac{p}{q}
Rearranging it, we find
h_i=-h_o \frac{p}{q}=-(3cm) \frac{10 cm}{2.5 cm}=-12 cm
and the negative sign means the image is inverted.
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Explanation:

According to newton's second law,

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From the question,

F+F₁+F₂ = ma................ Equation 1

Where F = The force generated from the engine, F₁ = Force exerted by the wind, F₂ = Force exerted due to the water, m = mass of the boat, a = acceleration of the boat.

Given: F = 4080 N , F₁ = -680 N(east), F₂ = -1160 N(east). m = 7660 kg

substitute into equation 1

4080-680-1160 = 7660(a)

2240 = 7660a

Therefore,

a = 2440/7660

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Answer:

solution:

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&

q_x = -k*(dT/dx)

<u>Case (1)  </u>

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Case (2)

dT/dx= (-10+30)/0.35==> 80 K/m

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Case (2)

dT/dx= (-10+30)/0.35==> 80 K/m

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Case (3)

q_x  =-50*(160)*10^3==>-8 kW

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Case (4)

q_x  =-50*(-80)*10^3==>4 kW

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Case (5)

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note:

all graph are attached

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