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1 year ago

An object of the same mass has three different weights at different times.

Physics

1 answer:

Sindrei [870]1 year ago

5 0

Answer:

See below

Explanation:

Possible.....mass is the amount of matter an object has ....weight is the result of gravity on an object.... different gravity ( Earth , Moon or Mars etc)

results in different weights .

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An object is placed in front of a convex lens of a length 10cm. What is the nature of the image formed if the object distance is

Lady_Fox [76]

${\mathfrak{\underline{\purple{\:\:\: Given:-\:\:\:}}}} \\ \\$

$\:\:\:\:\bullet\:\:\:\sf{Focal\:length=10\:cm}$

$\:\:\:\:\bullet\:\:\:\sf{Object \ distance = -15\:cm}$

$\\$

${\mathfrak{\underline{\purple{\:\:\:To \:Find:-\:\:\:}}}} \\ \\$

$\:\:\:\:\bullet\:\:\:\sf{Nature \: of \:the\:image}$

$\\$

${\mathfrak{\underline{\purple{\:\:\: Solution:-\:\:\:}}}} \\ \\$

<h3>☯ <u>By using formula of Lens</u> </h3>

$\\$

$\dashrightarrow\:\: {\boxed{\sf{\dfrac{1}{u} + \dfrac{1}{v} = \dfrac{1}{f}}}}$

$\\$

$\dashrightarrow\:\: \sf{\dfrac{1}{v}-\dfrac{1}{-15}=\dfrac{1}{10}}$

$\\$

$\dashrightarrow\:\: \sf{\dfrac{1}{v}+\dfrac{1}{15}=\dfrac{1}{10}}$

$\\$

$\dashrightarrow\:\: \sf{\dfrac{1}{v} = \dfrac{1}{10} - \dfrac{1}{15}}$

$\\$

$\dashrightarrow\:\: \sf{\dfrac{1}{v} = \dfrac{1}{30}}$

$\\$

$\dashrightarrow\:\: \sf{ v = 30 \ cm}$

$\\$

<h3>☯ <u>Now, Finding the magnification </u></h3>

$\\$

$\dashrightarrow\:\: \sf{ m = \dfrac{-30}{-15}}$

$\\$

$\dashrightarrow\:\: \sf{m = -2}$

$\\$

<h3>☯ <u>Hence</u>, $\\$ </h3>

$\:\:\:\:\star\:\:\:\sf{Image \ distance = 30 \ cm}$

$\:\:\:\:\star\:\:\:\sf{Nature = Real \ \& \ inverted}$

3 0

2 years ago

A 25 n object requires a 5.0 n to start moving over a horizontal surface. what is the coefficient of static friction?

olganol [36]

Static friction is the friction that exists between two or more solids that are not moving with a relative speed. To calculate the static friction coefficient we use the formula Fs=us × n where Fs is the static friction , us is the coefficient of static friction and the n is the normal force.
thus the coefficient of static friction will be 5 N÷ 25 N = 0.2
Hence 0.2 is the coefficient of static friction

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3 years ago

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Using 6400 km as the radius of Earth, calculate how high above Earth’s surface you would have to be in order to weigh 1/16th of

FrozenT [24]

Gravity obeys the inverse square law. At 6400 km above the center of the Earth (Earth's surface) you weigh x. Twice that reduces your weight to 1/4th. Four times that height reduces your weight to 1/16th. 4 times 6400 km is 25,600 km. But that is above the center of the earth, and the question requests the height above the surface, so we deduct 6400 km to arrive at our final answer: 19,200 km.

Incidentally, it doesn't exactly work the opposite way. At the center of the Earth the mass would be equally distributed around you, and you would therefore be weightless.

6 0

3 years ago

Astronomers suspect that a galaxy’s type can be affected both by the conditions in the protogalactic cloud from which it forms (

Katen [24]

Answer:

The items here are describing either a condition in a later interacton or a protogalactic cloud. The results matching with spiral and elliptical galaxy are:

For spiral galaxy are options 6,3,2 and 5.

and for elliptical galaxy are options 4 and 1.

Explanation:

Here it is given that astrnomers suspect that types of galaxy can be affected both by the conditions which occurs due to protogalactic cloud and then from it forms the initial conditions and then by the later interactions with the other galaxies.

so, both types of galaxies are matched with their respective items given:

A. Spiral galaxy:

2. A galaxy collision results tostripping of gas.

3. The protogalactic cloud rotates in a very slow motion.

5. The density of protogalactic cloud is very high.

6. when the protogalactic cloud shrinks cloud forms very rapidly.

B. Elliptical galaxy:

1. The protogalactic cloud has high angular momentum.

4. Most of the protogalactic gases settles down into a disk.

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3 years ago

Seema knows the mass of a basketball. What other information is needed to find the ball’s potential energy? the volume and heigh

Wewaii [24]

It is because the potential energy is similar to MgH.

When it comes to MgH, it means mass, gravity and height respectively.

By using the value of acceleration, seema will find the potential energy of a ball.

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3 years ago

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