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Sergeu [11.5K]
3 years ago
5

A 1200-kg car moving at 20 km/h is accelerated

Engineering
1 answer:
Grace [21]3 years ago
4 0

Answer:

Force on the car will be 4800 N and time required to cover this distance 13.75  sec

Explanation:

We have given mass of the car = 1200 kg

Initial velocity u = 20 km/h

Final velocity v = 75 km/h

Acceleration a=4m/sec^2

From the first equation of motion we know that

v = u+at, here v is final velocity, u is initial velocity, a is acceleration and t is time

So 75=20+4\times t

t = 13.75 sec

From second law of motion we know that F=ma

So force F=1200\times 4=4800N

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In this exercise, you will write a Point structure that represents a space in two-dimensional space. This Point should have both
Afina-wow [57]

Answer:

Check the explanation

Explanation:

Points to consider:

We need to take the input from the user

We need to find the manhatan distance and euclidian using the formula

(x1, y1) and (x2, y2) are the two points

Manhattan:

|x_1 - x_2| + |y_1 - y_2|

Euclidian Distance:

\sqrt{(x1 - yl)^2 + (x2 - y2)^2)}

Code

#include<stdio.h>

#include<math.h>

struct Point{

  int x, y;

};

int manhattan(Point A, Point B){

  return abs(A.x - B.x) + abs(A.y- B.y);

}

float euclidean(Point A, Point B){

  return sqrt(pow(A.x - B.x, 2) + pow(A.y - B.y, 2));

}

int main(){

  struct Point A, B;

  printf("Enter x and Y for first point: ");

  int x, y;

  scanf("%d%d", &x, &y);

  A.x = x;

  A.y = y;

  printf("Enter x and Y for second point: ");

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  B.x = x;

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  printf("Manhattan Distance: %d\n", manhattan(A, B));

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}

Sample output

8 0
3 years ago
Why do we need technical sketching?
Scrat [10]

Technical Drawings give a better understanding of what is needed and required in the project.

Explanation:

8 0
2 years ago
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An ideal gas initially at 300 K and 1 bar undergoes a three-step mechanically reversible cycle in a closed system. In step 12, p
Veseljchak [2.6K]

Answer:

Ts =Ta E)- 300(

569.5 K

5

Q12-W12 = -4014.26

Mol

AU2s = Q23= 5601.55

Mol

AUs¡ = Ws¡ = -5601.55

Explanation:

A clear details for the question is also attached.

(b) The P,V and T for state 1,2 and 3

P =1 bar Ti = 300 K and Vi from ideal gas Vi=

10

24.9x10 m

=

P-5 bar

Due to step 12 is isothermal: T1 = T2= 300 K and

VVi24.9 x 10x-4.9 x 10-3 *

The values at 3 calclated by Uing step 3l Adiabatic process

B-P ()

Since step 23 is Isochoric: Va =Vs= 4.99 m* and 7=

14

Ps-1x(4.99 x 103

P-1x(29x 10)

9.49 barr

And Ts =Ta E)- 300(

569.5 K

5

(c) For step 12: Isothermal, Since AT = 0 then AH12 = AU12 = 0 and

Work done for Isotermal process define as

8.314 x 300 In =4014.26

Wi2= RTi ln

mol

And fromn first law of thermodynamic

AU12= W12 +Q12

Q12-W12 = -4014.26

Mol

F'or step 23 Isochoric: AV = 0 Since volume change is zero W23= 0 and

Alls = Cp(L3-12)=5 x 8.311 (569.5 - 300) = 7812.18-

AU23= C (13-72) =5 x 8.314 (569.3 - 300) = 5601.53

Inol

Now from first law of thermodynamic the Q23

AU2s = Q23= 5601.55

Mol

For step 3-1 Adiabatic: Since in this process no heat transfer occur Q31= 0

and

AH

C,(T -Ts)=x 8.314 (300- 569.5)= -7842.18

mol

AU=C, (T¡-T)= x 8.314 (300

-5601.55

569.5)

mol

Now from first law of thermodynamie the Ws1

J

mol

AUs¡ = Ws¡ = -5601.55

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Hell no,cause i’m not
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