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3 years ago

8: As a grade-school project, students construct a mobile representing some of the major food groups.

Physics

1 answer:

Airida [17]3 years ago

4 0

Answer:

m1=0.72kg m2=0.48kg m3=0.98

Explanation:

12*m1=18*m2, so 2*m1=3*m2

6*(m1+m2)=24*.03, so 2*m1+2*m2=2.4

so, 5*m2=2.4, m2=0.48 then, m1=0.72

(m1+m2+0.3+0.2)*18=31*m3

1.7*18=31*m3

m3=0.98

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Assemblage is an additive process where artists construct work by putting together objects and attaches them in some way. The correct option among all the options that are given in the question is the second option or option "b". This kind of artistry actually gives a three dimensional impression to the work that is done by the artist.

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3 years ago

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A person throws a pumpkin at a horizontal speed of 4.0 — off a cliff. The pumpkin travels 9.5 m horizontally

emmainna [20.7K]

Complete Question

A person throws a pumpkin at a horizontal speed of 4.0 m/s off a cliff. The pumpkin travels 9.5m horizontally before it hits the ground. We can ignore air resistance.What is the pumpkin's vertical displacement during the throw? What is the pumpkin's vertical velocity when it hits the ground?

Answer:

The pumpkin's vertical displacement is $H = 27 .67 \ m$

The pumpkin's vertical velocity when it hits the ground is $v_v__{f}} = 23.298 \ m/s$

Explanation:

From the question we are told that

The horizontal speed is $v_h = 4 m/s$

The horizontal distance traveled is $d = 9.5 \ m$

The horizontal distance traveled is mathematically represented as

$S = v_h * t$

Where t is the time taken

substituting values

$9.5 = 4 * t$

=> $t = \frac{9.5}{4}$

$t = 2.38 \ sec$

Now the vertical displacement is mathematically represented as

$H = v_v t + \frac{1}{2} a_v t^2$

now the vertical velocity before the throw is zero

$H = 0 + \frac{1}{2} (9.8) * (2.375)^2$

$H = 27 .67 \ m$

Now the final vertical velocity is mathematically represented as

$v_v__{f}} = v_v + at$

substituting values

$v_v__{f}} = 0 + (9.8)* (2.375)$

$v_v__{f}} = 23.298 \ m/s$

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3 years ago

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Answer:

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What are gamma rays and what are its uses??<br>

sukhopar [10]

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2 years ago

A camera with a 50.0-mm focal length lens is being used to photograph a person standing 3.00 m away. (a) How far from the lens m

kirill [66]

a) 50.8 mm

b) The whole image (1:1)

c) It seems reasonable

Explanation:

To project the image on the film, the distance of the film from the lens must be equal to the distance of the image from the lens. This can be found by using the lens equation:

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

where

f is the focal length of the lens

p is the distance of the object from the lens

q is the distance of the image from the lens

In this problem:

f = 50.0 mm = 0.050 m is the focal length (positive for a convex lens)

p = 3.00 m is the distance of the person from the lens

Therefore, we can find q:

$\frac{1}{q}=\frac{1}{f}-\frac{1}{p}=\frac{1}{0.050}-\frac{1}{3.00}=19.667m^{-1}\\q=\frac{1}{19.667}=0.051 m=50.8 mm$

Here we need to find the height of the image first.

This can be done by using the magnification equation:

$\frac{y'}{y}=-\frac{q}{p}$

where:

y' is the height of the image

y = 1.75 m is the height of the real person

q = 50.8 mm = 0.0508 m is the distance of the image from the lens

p = 3.00 m is the distance of the person from the lens

Solving for y', we find:

$y'=-\frac{qy}{p}=-\frac{(0.0508)(1.75)}{3.00}=-0.0296 m=-29.6mm$

(the negative sign means the image is inverted)

Therefore, the size of the image (29.6 mm) is smaller than the size of the film (36.0 mm), so the whole image can fit into the film.

This seems reasonable: in fact, with a 50.0 mm focal length, if we try to take the picture of a person at a distance of 3.00 m, we are able to capture the whole image of the person in the photo.

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3 years ago