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Xelga [282]
2 years ago
10

The base ( foundation ) of building is made wider. Why ?​

Physics
1 answer:
attashe74 [19]2 years ago
7 0

Answer: It is increased to exert the pressure off the bywalls and on large area (brainliest pls)

Explanation: ...

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Except for mass, length, time, and electric charge, EVERY quantity is
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A graph labeled velocity versus time with horizontal axis time (seconds) and vertical axis velocity (meters per second). A blue
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B

Explanation:

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A 5.0 cm object is 6.0 cm from a convex lens, which has a focal length of 7.0 cm.
Reil [10]

Answer : The distance of the image from the lens is, -42 cm

The height of the image is, 35 cm

Solution :

First we have to calculate the image distance.

Formula used :

\frac{1}{f}=\frac{1}{p}+\frac{1}{q}

where,

f = focal length = 7 cm

p = object distance = 6 cm

q = image distance = ?

Now put all the given values in the above formula, we get the distance of the image from the lens.

\frac{1}{7cm}=\frac{1}{6cm}+\frac{1}{q}

q=-42cm

Therefore, the distance of the image from the lens is, 42 cm and the negative sign indicates that the image is virtual.

Now we have to calculate the height of the image.

Formula used :

\frac{h}{h'}=\frac{p}{q}

where,

h = height of object = 5 cm

h' = height of image = ?

Now put all the given values in this formula, we get the height of the image.

\frac{5cm}{h'}=\frac{6cm}{-42cm}=-35cm

Therefore, the height of the image is, 35 cm and the negative sign indicates that the image is inverted.

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3 years ago
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Which of the following is NOT one of the types of precipitation?
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Answer:

d . ................. .........

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Halley's comet moves about the Sun in an elliptical orbit, with its closest approach to the Sun being 0.59 A.U. and its greatest
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Answer:

Explanation:

Given

Halley's closest distance from sun is r_1=0.59\ A.U.

Greatest distance is r_2=35\ A.U.

Comet's speed at closest approach is v_1=47\ km/s

As there is no external torque so angular momentum of comet about the sun is conserved

L_1=L_2

mr_1^2\times \omega _1=mr_2^2\times \omega _2

where \omega =angular\ velocity

This can be written as \omega =\frac{v}{r}

Therefore  

mr_1^2\times \frac{v_1}{r_1}=mr_2^2\times \frac{v_2}{r_2}

r_1\times v_1=r_2\times v_2

0.59\times 47=35\times v_2

v_2=0.79\ km/s  

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