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3 years ago

8

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

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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A copper (Young's modulus 1.1 x 1011 N/m2) cylinder and a brass (Young's modulus 9.0 x 1010 N/m2) cylinder are stacked end to en

Nina [5.8K]

We have that for the Question, it can be said that the amount by which the length of the stack decreases is

$dl'=3.621*10^{-4}m$

From the question we are told

A copper (<em>Young's modulus </em>1.1 x 1011 N/m2) cylinder and a brass (Young's modulus 9.0 x 1010 N/m2) cylinder are stacked end to end, as in the drawing. Each <em>cylinder </em>has a radius of 0.24 cm.

A compressive force of F = 7900 N is applied to the right end of the brass cylinder. Find the amount by which the length of the stack <em>decreases</em>.

Generally the equation for <em>copper </em>cylinder is mathematically given as

$dl=\frac{Flo}{yA}$

$dl=\frac{7900*3*10^-^2}{1.1*10^{11}*\pi(0.24*10^{-2})^2}$

$dl=1.19064778*10^-^4$

Generally the equation for brass<em> </em>cylinder is mathematically given as

$dl=\frac{7900*5*10^-^2}{9*10^{10}*\pi(0.24*10^{-2})^2}$

$dl=2.43*10^{-4}$

Therefore Total change in length

$dl'=1.191*10^-^4+(2.43*10^{-4})$

$dl'=3.621*10^{-4}m$

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3 years ago

If the world pulls down on an apple, what is the reaction force? Explain.

-Dominant- [34]

The Apple pulling up on the world

8 0

3 years ago

Read 2 more answers

An a.c. supply is connected to a wire stretched between the poles of a magnet. Which way does the wire move?

snow_lady [41]

The wire vibrates back and forth between the poles of the magnet.
The frequency of the vibration is the frequency of the AC supply.

7 0

2 years ago

Variations in the resistivity of blood can give valuable clues about changes in various properties of the blood. Suppose a medic

Elena-2011 [213]

Answer:

Answer:

1.1 x 10^9 ohm metre

Explanation:

diameter = 1.5 mm

length, l = 5 cm

Potential difference, V = 9 V

current, i = 230 micro Ampere = 230 x 10^-6 A

radius, r = diameter / 2 = 1.5 / 2 = 0.75 x 10^-3 m

Let the resistivity is ρ.

Area of crossection

A = πr² = 3.14 x 0.75 x 0.75 x 10^-6 = 1.766 x 10^-6 m^2

Use Ohm's law to find the value of resistance

V = i x R

9 = 230 x 10^-6 x R

R = 39130.4 ohm

Use the formula for the resistance

$R=\rho \frac{l}{A}$

$\rho =\frac{RA}{l}$

$\rho =\frac{39130.4\times 0.05}{1.766\times 10^{-6}}$

ρ = 1.1 x 10^9 ohm metre

Explanation:

7 0

3 years ago

21. A toy car starts from rest and begins to

Bond [772]

Answer:

v = 66 m/s

Explanation:

Given that,

The initial velocity of a car, u = 0

Acceleration of the car, a = 11 m/s²

We need to find the final velocity of the toy after 6 seconds.

Let v is the final velocity. It can be calculated using first equation of motion. It is given by :

v = u +at

v = 0 + 11 m/s² × 6 s

v = 66 m/s

So, the final velocity of the car is 66 m/s.

8 0

3 years ago