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solniwko [45]
3 years ago
11

A disc with a mass of 1 kg moves horizontally to the right with a speed of 7 m/s on a table with negligible friction when it col

lides with a second disc with a mass of 9 kg. The second disc is moving horizontally to the right with a speed of 1 m/s at the moment of impact. The two discs stick together upon impact. What is the speed of the composite body immediately after the collision? (round your answer to the nearest tenth)

Physics
1 answer:
Elodia [21]3 years ago
7 0

Answer:

1.6 m/s

Explanation:

First you need to find the momentums of each disc by multiplying their velocities with mass.

disc 1: 7*1= 7 kg m/s

disc 2: 1*9= 9 kg m/s

Second, you need to find the total momentum of the system by adding the momentums of each sphere.

9+7= 16 kg m/s

Because momentum is conserved, this is equal to the momentum of the composite body.

Finally, to find the composite body's velocity, divide its total momentum by its mass. This is because mass*velocity=momentum

16/10=1.6

The velocity of the composite body is 1.6 m/s.

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A compound microscope is equipped with two objective lenses (10x and 45x) and has a 10x ocular lens. The highest magnification a
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Answer:

450X

Explanation:

When a specimen is been viewed, both

objective and ocular lenses works together so that the object is magnified.

From the question,objective lenses are;

1)10x

2)45x

ocular lens= 10x

Highest magnification

= 10X ocular × 45X objective

=450X

This implies that the image that was viewed will appear 450 times the actual size.

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3 years ago
What is the mass of an object moving with 80N of force and an acceleration of 8 m/s2?
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Answer:

10 Kg

Explanation:

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6 0
3 years ago
Find the moments of inertia Ix, Iy, I0 for a lamina that occupies the part of the disk x2 y2 ≤ 36 in the first quadrant if the d
Tasya [4]

Answer:

I(x)  = 1444×k ×{\pi}

I(y)  = 1444×k ×{\pi}

I(o) = 3888×k ×{\pi}  

Explanation:

Given data

function =  x^2 + y^2 ≤ 36

function =  x^2 + y^2 ≤ 6^2

to find out

the moments of inertia Ix, Iy, Io

solution

first we consider the polar coordinate (a,θ)

and polar is directly proportional to a²

so p = k × a²

so that

x = a cosθ

y = a sinθ

dA = adθda

so

I(x) = ∫y²pdA

take limit 0 to 6 for a and o to \pi /2 for θ

I(x) = \int_{0}^{6}\int_{0}^{\pi/2} y²p dA

I(x) = \int_{0}^{6}\int_{0}^{\pi/2} (a sinθ)²(k × a²) adθda

I(x) = k  \int_{0}^{6}a^(5)  da ×  \int_{0}^{\pi/2}  (sin²θ)dθ

I(x) = k  \int_{0}^{6}a^(5)  da ×  \int_{0}^{\pi/2}  (1-cos2θ)/2 dθ

I(x)  = k ({r}^{6}/6)^(5)_0 ×  {θ/2 - sin2θ/4}^{\pi /2}_0

I(x)  = k × ({6}^{6}/6) × (  {\pi /4} - sin\pi /4)

I(x)  = k ×  ({6}^{5}) ×   {\pi /4}

I(x)  = 1444×k ×{\pi}    .....................1

and we can say I(x) = I(y)   by the symmetry rule

and here I(o) will be  I(x) + I(y) i.e

I(o) = 2 × 1444×k ×{\pi}

I(o) = 3888×k ×{\pi}   ......................2

3 0
3 years ago
What is the difference between absolute strength and relative strength? absolute strength is the type of strength the average pe
Orlov [11]
Absolute strength measures strength adjusted for your body size, while relative strength measurses maximum strength exerted in a single effort. Hopefully that helps wasn't really sure what you were asking seemed like you had answered your own question.
8 0
3 years ago
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