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dalvyx [7]
3 years ago
5

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
Physics
1 answer:
madreJ [45]3 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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If we use the equation:
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Which element is a metalloid?
Ann [662]

Answer:

The metalloids are located on the right side of the periodic table in a "step-like" arrangement.

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3 0
1 year ago
A bicyclist is finishing his repair of a flat tire when a friend rides by with a constant speed of 3.6 m/s . Two seconds later t
dybincka [34]

Answer:

A) t = 7.0 s    

B) x = 25 m  

C) v = 10 m/s

Explanation:

The equations for the position and velocity of an object traveling in a straight line is given by the following expressions:

x = x0 + v0 · t + 1/2 · a · t²

v = v0 + a · t

Where:

x = position at time t

x0 = initial position

v0 = initial velocity

t = time

a = acceleration

v = velocity at time t

A)When both friends meet, their position is the same:

x bicyclist = x friend

x0 + v0 · t + 1/2 · a · t² = x0 + v · t

If we place the center of the frame of reference at the point when the bicyclist starts following his friend, the initial position of the bicyclist will be 0, and the initial position of the friend will be his position after 2 s:

Position of the friend after 2 s:

x = v · t

x = 3.6 m/s · 2 s = 7.2 m

Then:

1/2 · a · t² = x0 + v · t       v0 of the bicyclist is 0 because he starts from rest.

1/2 · 2.0 m/s² · t² = 7.2 m + 3.6 m/s · t

1  m/s² · t² - 3.6 m/s · t - 7.2 m = 0

Solving the quadratic equation:

t = 5.0 s

It takes the bicyclist (5.0 s + 2.0 s) 7.0 s to catch his friend after he passes him.

B) Using the equation for the position, we can calculate the traveled distance. We can use the equation for the position of the friend, who traveled over 7.0 s.

x = v · t

x = 3.6 m/s · 7.0 s = 25 m

(we would have obtained the same result if we would have used the equation for the position of the bicyclist)

C) Using the equation of velocity:

v = a · t

v = 2.0 m/s² · 5.0 s = 10 m/s

8 0
3 years ago
How much heat is needed to bring 12.0 g of water from 28.3 °C to 43.87 °C, if the specific heat capacity of water is 4.184 J/(g•
Vika [28.1K]

Answer:

Heat capacity, Q = 781.74 Joules

Explanation:

Given the following data;

Mass = 12g

Initial temperature = 28.3°C

Final temperature = 43.87°C

Specific heat capacity of water = 4.184J/g°C

To find the quantity of heat needed?

Heat capacity is given by the formula;

Q = mcdt

Where;

Q represents the heat capacity or quantity of heat.

m represents the mass of an object.

c represents the specific heat capacity of water.

dt represents the change in temperature.

dt = T2 - T1

dt = 43.87 - 28.3

dt = 15.57°C

Substituting into the equation, we have;

Q = 12*4.184*15.57

Q = 781.74 Joules

7 0
3 years ago
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