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3 years ago

What is the range of a 32-bit unsigned integer?

Engineering

1 answer:

Katyanochek1 [597]3 years ago

4 0

Answer:

0 to 4294967295

Explanation:

Unsigned integers have only positive numbers and zero. Their range goes from zero to (2^n)-1.

In the case of a 32-bit unsigned integer this would be.

(2^32) - 1 = 4294967296 - 1 = 4294967295

So the range goes from 0 to 4294967295 or form zero to about 4.3 billion.

The minus one term is because the zero takes one of the values.

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Two identical billiard balls can move freely on a horizontal table. Ball a has a velocity V0 and hits balls B, which is at rest,

Lyrx [107]

Answer:

Velocity of ball B after impact is $0.6364v_0$ and ball A is $0.711v_0$

Explanation:

$v_0$ = Initial velocity of ball A

$v_A=v_0\cos45^{\circ}$

$v_B$ = Initial velocity of ball B = 0

$(v_A)_n'$ = Final velocity of ball A

$v_B'$ = Final velocity of ball B

$e$ = Coefficient of restitution = 0.8

From the conservation of momentum along the normal we have

$mv_A+mv_B=m(v_A)_n'+mv_B'\\\Rightarrow v_0\cos45^{\circ}+0=(v_A)_n'+v_B'\\\Rightarrow (v_A)_n'+v_B'=\dfrac{1}{\sqrt{2}}v_0$

Coefficient of restitution is given by

$e=\dfrac{v_B'-(v_A)_n'}{v_A-v_B}\\\Rightarrow 0.8=\dfrac{v_B'-(v_A)_n'}{v_0\cos45^{\circ}}\\\Rightarrow v_B'-(v_A)_n'=\dfrac{0.8}{\sqrt{2}}v_0$

$(v_A)_n'+v_B'=\dfrac{1}{\sqrt{2}}v_0$

$v_B'-(v_A)_n'=\dfrac{0.8}{\sqrt{2}}v_0$

Adding the above two equations we get

$2v_B'=\dfrac{1.8}{\sqrt{2}}v_0\\\Rightarrow v_B'=\dfrac{0.9}{\sqrt{2}}v_0$

$\boldsymbol{\therefore v_B'=0.6364v_0}$

$(v_A)_n'=\dfrac{1}{\sqrt{2}}v_0-0.6364v_0\\\Rightarrow (v_A)_n'=0.07071v_0$

From the conservation of momentum along the plane of contact we have

$(v_A)_t'=(v_A)_t=v_0\sin45^{\circ}\\\Rightarrow (v_A)_t'=\dfrac{v_0}{\sqrt{2}}$

$v_A'=\sqrt{(v_A)_t'^2+(v_A)_n'^2}\\\Rightarrow v_A'=\sqrt{(\dfrac{v_0}{\sqrt{2}})^2+(0.07071v_0)^2}\\\Rightarrow \boldsymbol{v_A'=0.711v_0}$

Velocity of ball B after impact is $0.6364v_0$ and ball A is $0.711v_0$ .

5 0

3 years ago

A force of 16,000 will cause a 1 1 bar of magnesium to stretch from 10 to 10.036 . Calculate the modulus of elasticity in . (Ent

Bad White [126]

The modulus of elasticity is 44.4GPa

E<u>xplanation:</u>

Given

original length L1=10cm

New length L2=10.036 cm

Force F=16000N

$dimensions\ of\ the\ bar\ is\ 1 cm\times1cm\\Area=1cm\times1cm=1cm^2=0.0001m^2$

Stress of the bar σ=Force/Area

$=16000/0.0001=16\times 10^7N/m^2$

Change in length ΔL=L2-L1=10.036-10=0.036 cm

Strain is obtained by dividing the change in length by original length

Strain ε=ΔL/L1

=0.036/10=0.0036

Modulus of elasticity=stress/strain

=σ/ε

$=16\times10^7/0.0036\\=4.44\times10^\ 10$ Pa

$=44.4GPa$

5 0

3 years ago

Why is sssniperwolf so fine

RSB [31]

Bc she’s just a baddie

5 0

3 years ago

Read 2 more answers

What is the name of the type of rocker arm stud that does not require a valve adjustment?

-Dominant- [34]

Answer:

Jesel Premium Stud rockers

Explanation:

This specific stud rocker does not need a valve adjustment. It does not need stud girdles or pushed guide plates. It can be adjusted on a steel stanchion post.

(have a good day and I hope that this explanation fits what you needed to know!)

4 0

3 years ago

_______________ is an effective way to manage waste in a shop.

Murrr4er [49]

Answer:

To reduce waste.

Explanation:

3 0

3 years ago