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Masteriza [31]
2 years ago
15

Drivers must be careful when driving close to cyclists and should keep at least ___ feet apart when passing cyclists on the road

Engineering
1 answer:
faust18 [17]2 years ago
4 0

Answer:

at least 8 feet

Explanation:

You might be interested in
Give me some examples of fragile structures.
Anvisha [2.4K]

Answer:

i don't know if this help tell me if i am wrong

Explanation:

Gravity is the force that pulls all elements of matter together. Matter refers to things you can physically touch. The more matter there is, the greater the amount of gravity or force. This means that the Earth or other planets have a great deal of pull and that everything on Earth is pulled back to Earth.

Some examples of the force of gravity include:

The force that holds the gases in the sun.

The force that causes a ball you throw in the air to come down again.

The force that causes a car to coast downhill even when you aren't stepping on the gas.

The force that causes a glass you drop to fall to the floor.

3 0
3 years ago
Fill in the empty function so that it returns the sum of all the divisors of a number, without including it. A divisor is a numb
tekilochka [14]

Answer:

// Program is written in C++

// Comments are used to explain some lines

// Only the required function is written. The main method is excluded.

#include<bits/stdc++.h>

#include<iostream>

using namespace std;

int divSum(int num)

{

// The next line declares the final result of summation of divisors. The variable declared is also

//initialised to 0

int result = 0;

// find all numbers which divide 'num'

for (int i=2; i<=(num/2); i++)

{

// if 'i' is divisor of 'num'

if (num%i==0)

{

if (i==(num/i))

result += i; //add divisor to result

else

result += (i + num/i); //add divisor to result

}

}

cout<<result+1;

}

6 0
3 years ago
Interpret the assembly program below: MOV R3,R0;
Reika [66]

Answer:

Explanation:

1.  With the operands R0, R1, the program would compute AND operation and ADD operation .

2. The operands could truly be signed 2's complement encoded (i.e Yes) .

3. The overflow truly occurs when two numbers that are unsigned were added and the result is larger than the capacity of the register, in that situation, overflow would occur and it could corrupt the data.

 When the result of an operation is smaller in magnitude than the smallest value represented by the data type, then arithmetic underflow will occur.

7 0
3 years ago
Assume a program requires the execution of 50 x 106 FP instructions, 110 x 106 INT instructions, 80 x 106 L/S instructions, and
Pavlova-9 [17]

Answer:

Part A:

1.3568*10^{-5}=\frac{5300* New\  CPI_1+11660*1+8480*4+1696*2}{2*10^9\ Hz} \\ New\ CPI_1=-4.12

CPI cannot be negative so it is not possible to for program to run two times faster.

Part B:

1.3568*10^{-5}=\frac{5300*1+11660*1+8480*New\ CPI_3+1696*2}{2*10^9\ Hz} \\ New\ CPI_3=0.8

CPI reduced by 1-\frac{0.8}{4} = 0.80=80%

Part C:

New Execution Time=\frac{5300*0.6+11660*0.6+8480*2.8+1696*1.4}{2*10^9\ Hz}=1.81472*10^{-5}\ s

Increase in speed=1-\frac{1.81472*10^{-5}}{2.7136*10^{-5}} =0.33125= 33.125\%

Explanation:

FP Instructions=50*106=5300

INT  Instructions=110*106=11660

L/S  Instructions=80*106=8480

Branch  Instructions=16*106=1696

Calculating Execution Time:

Execution Time=\frac{\sum^4_{i=1} Number\ of\ Instruction*\ CPI_{i}}{Clock\ Rate}

Execution Time=\frac{5300*1+11660*1+8480*4+1696*2}{2*10^9\ Hz}

Execution Time=2.7136*10^{-5}\ s

Part A:

For Program to run two times faster,Execution Time (Calculated above) is reduced to half.

New Execution Time=\frac{2.7136*10^{-5}}{2}=1.3568*10^{-5}\ s

1.3568*10^{-5}=\frac{5300* New\  CPI_1+11660*1+8480*4+1696*2}{2*10^9\ Hz} \\ New\ CPI_1=-4.12

CPI cannot be negative so it is not possible to for program to run two times faster.

Part B:

For Program to run two times faster,Execution Time (Calculated above) is reduced to half.

New Execution Time=\frac{2.7136*10^{-5}}{2}=1.3568*10^{-5}\ s

1.3568*10^{-5}=\frac{5300*1+11660*1+8480*New\ CPI_3+1696*2}{2*10^9\ Hz} \\ New\ CPI_3=0.8

CPI reduced by 1-\frac{0.8}{4} = 0.80=80%

Part C:

New\ CPI_1=0.6*Old\ CPI_1=0.6*1=0.6\\New\ CPI_2=0.6*Old\ CPI_2=0.6*1=0.6\\New\ CPI_3=0.7*Old\ CPI_3=0.7*4=2.8\\New\ CPI_4=0.7*Old\ CPI_4=0.7*2=1.4

New Execution Time=\frac{\sum^4_{i=1} Number\ of\ Instruction*\ CPI_{i}}{Clock\ Rate}

New Execution Time=\frac{5300*0.6+11660*0.6+8480*2.8+1696*1.4}{2*10^9\ Hz}=1.81472*10^{-5}\ s

Increase in speed=1-\frac{1.81472*10^{-5}}{2.7136*10^{-5}} =0.33125= 33.125\%

8 0
3 years ago
A cylinder of aluminum-magnesium alloy 0.5 m long is subjected to an elastic tensile stress of 10.2 MPa. The measured elastic el
elena55 [62]

Answer:

E= 15 GPa.

Explanation:

Given that

Length ,L = 0.5 m

Tensile stress ,σ = 10.2 MPa

Elongation ,ΔL = 0.34 mm

lets take young modulus  = E

We know that strain ε given as

\varepsilon =\dfrac{\Delta L}{L}

\varepsilon =\dfrac{0.34}{0.5\times 1000}

\varepsilon =0.00068

We know that

\sigma = \varepsilon  E\\\\E=\dfrac{10.2}{0.00068}\\E= 15000\ MPa\\E=15\ GPa

Therefore the young's modulus will be 15 GPa.

8 0
3 years ago
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