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2 years ago

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Assume that our computer stores decimal numbers using 16 bits - 10 bits for a sign/magnitude mantissa and 6 bits for a sign/magn

itude base-2 exponent. (The way we showed in class.) Show the internal representation of the following decimal quantities.
a. +7.5
b. -20.25
c. -1/64

1 answer:

madreJ [45]2 years ago

5 0

Explanation:

a) 7.5= 111.1×2°= 0.1111×2^3

which can also be written as

(1/2+1/4+1/8+1/16)×8

sign of mantissa:=0

Mantissa(9 bits): 111100000

sign of exponent: 0

Exponent(5 bits): 0011

the final for this is:011110000000011

b) -20.25= -10100.01×2^0= -0.1010001×2^5

sign of mantissa: 1

Mantissa(9 bits): 101000100

sign of exponent: 0

Exponent(5 bits): 00101

the final for this is:1101000100000101

c)-1/64= -.000001×2^0= -0.1×2^{-5}

sign of mantissa: 1

Mantissa(9 bits): 100000000

sign of exponent: 0

Exponent(5 bits): 00101

the final for this is:1100000000100101

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The box leaves position x=0 with speed v0. The box is slowed by a constant frictional force until it comes to rest at position x

Diano4ka-milaya [45]

Answer:

$Ff=m\times \dfrac{V_o^2}{2X_1}$

Explanation:

Given that

At X=0 V=Vo

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We know that

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$a=\dfrac{V_o^2}{2X_1}$

So the friction force on the box

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2 years ago

If 1 inch = 2.54 centimeters, how many inches are there in 2,540 centimeters?

NARA [144]

1,000 inches would be the answer

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2 years ago

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The deflection plates in an oscilloscope are 10 cm by 2 cm with a gap distance of 1 mm. A 100 volt potential difference is sudde

boyakko [2]

Explanation:

Given data:

Area A = 10 cm×2 cm = 20×10⁻⁴ m²

Distance d between the plates = 1 mm = 1×10⁻³m

Voltage of the battery is emf = 100 V

Resistance = 1025 ohm

Solution:

In RC circuit, the voltage between the plates is related to time t. Initially the voltage is equal to that of battery V₀ = emf = 100V. But After time t the resistance and capacitor changes it and the final voltage is V that is given by

$V = V_{0}(1-e^{\frac{-t}{RC} } )\\\frac{V}{V_{0} } = 1-e(^{\frac{-t}{RC} }) \\e^{\frac{-t}{RC} } = 1- \frac{V}{V_{0} }$

Taking natural log on both sides,

$e^{\frac{-t}{RC} } = 1- \frac{V}{V_{0} } \\\frac{-t}{RC} = ln(1-\frac{V}{V_{0} } )\\t = -RCln(1 - \frac{V}{V_{0} })$

$t = -RC ln (1-\frac{V}{V_{0} })$ (1)

Now we can calculate the capacitance by using the area of the plates.

C = ε₀A/d

= $\frac{(8.85*10^{-12))} (20*10^{-4}) }{1*10^{-3} }$

= 18×10⁻¹²F

Now we can get the time when the voltage drop from 100 to 55 V by putting the values of C, V₀, V and R in the equation (1)

$t = -RC ln (1-\frac{V}{V_{0} })$

= -(1025Ω)(18×10⁻¹² F) ln( 1 - 55/100)

= 15×10⁻⁹s

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2 years ago

An airplane propeller rotates with an angular speed of 260 rad/s though what angul does the propeller rotate in 5.0s? Give your

ch4aika [34]

Answer:

1300 rad ( $74522^{\circ}$ )

Explanation:

The angular speed of an object is the rate of change of angular position.

It is given by:

$\omega=\frac{\theta}{t}$

where

$\omega$ is the angular speed

$\theta$ is the angular displacement of the object

t is the time elapsed

For the propeller rotating in this problem, we have

$\omega=260 rad/s$ is the angular speed

t = 5.0 s is the time elapsed

Therefore, the angular displacement is:

$\theta=\omega t=(260)(5.0)=1300 rad$

And since $2\pi rad = 360^{\circ}$ , the angle in degrees is

$\theta=\frac{1300}{2\pi}(360)=74522^{\circ}$

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2 years ago

Two lasers, one red (with wavelength 633.0 nm) and the other green (with wavelength 532.0 nm), are mounted behind a 0.150-mm sli

Orlov [11]

Answer:

a.3.20m

b.0.45cm

Explanation:

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#Substitute our variable values in the minima equation to obtain $\theta$ :

$\theta=sin^-^1 (\frac{3\times 6.33\times 10^-^7}{0.00015})\\\\\theta=0.01266rad$

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$tan(\theta)=y/L$ #where $y=4.05cm$

$L=y/tan(\theta)=3.20$

Hence the screen is 3.20m from the split.

b. To find the closest minima for green(the fourth min will give you the smallest distance)

#Like with a above, the minima equation will be defined as:

$sin \theta=\frac{m\lambda}{\alpha}$ , where $m=4$ given that it's the minima with the smallest distance.

$sin \theta=\frac{4\lambda}{\alpha}\\\theta=sin^-^1 (\frac{4\times 6.33\times 10^-^7}{0.00015})\\\\\theta=0.01688rad$

#we then use $tan(\theta)=y/L$ to calculate $L$ =4.5cm

Then from the equation subtract $y_3$ from $y$ :

$4.50cm-4.05cm=0.45cm$

Hence, the distance $\bigtriangleup y$ is 0.45cm

8 0

2 years ago