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Ghella [55]
3 years ago
13

The sports car has a weight of 4900 lblb and center of gravity at GG. If it starts from rest it causes the rear wheels to slip a

s it accelerates. The coefficients of static and kinetic friction at the road are μsμs = 0.5 and μkμkmu_k = 0.35, respectively. Neglect the mass of the wheelDetermine how long it takes for it to reach a speed of 10 ft/s.What are the normal reactions at each of the rear wheels on the road?What are the normal reactions at each of the front wheels on the road?

Engineering
1 answer:
Zinaida [17]3 years ago
3 0

Answer:

2.27secs

Explanation:

The car will take 2.27secs to reach a speed of 10 ft/s.

Kindly go through the attached files for a step by step solution that will show you how the answer was gotten and the normal reaction of both the front and rear wheels.

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Write multiple if statements:
lora16 [44]
Zrizorzlzfxxxgoxxxxpgxtoxxxhxuxyf
3 0
3 years ago
What are the two reasons for a clear cut
Inessa [10]

Answer:

to clear land for agriculture and settlement and to use or sell timber for lumber, paper products, or fuel.

3 0
3 years ago
Let A→=(150iˆ+270jˆ) mm , B→=(300iˆ−450jˆ) mm , and C→=(−100iˆ−250jˆ) mm . Find scalars r and s, if possible, such that R→=rA→+s
ioda

Answer: r = 0.8081; s = -0.07071

Explanation:

A = (150i + 270j) mm

B = (300i - 450j) mm

C = (-100i - 250j) mm

R = rA + sB + C = 0i + 0j

R = r(150i + 270j) + s(300i - 450j) + (-100i - 250j) = 0i + 0j

R = (150r + 300s - 100)i + (270r - 450s - 250)j = 0i + 0j

Equating the i and j components;

150r + 300s - 100 = 0

270r - 450s - 250 = 0

150r + 300s = 100

270r - 450s = 250

solving simultaneously,

r = 0.8081 and s = -0.07071

QED!

5 0
3 years ago
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: ");

  scanf("%d%d", &x, &y);

  B.x = x;

  B.y = y;

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

  printf("Euclidian Distance: %f\n", euclidean(A, B));

 

}

Sample output

8 0
3 years ago
In a much smaller model of the Gizmo apparatus, a 5 kg mass drops 86 mm (0.086 m) and raises the temperature of 1 gram of water
Orlov [11]

Answer:

The amount of energy transferred to the water is 4.214 J

Explanation:

The given parameters are;

The mass of the object that drops = 5 kg

The height from which it drops = 86 mm (0.086 m)

The potential energy P.E. is given by the following formula

P.E = m·g·h

Where;

m = The mass of the object = 5 kg

g = The acceleration de to gravity = 9.8 m/s²

h = The height from which the object is dropped = 0.086 m

Therefore;

P.E. = 5 kg × 9.8 m/s² × 0.086 m = 4.214 J

Given that the potential energy is converted into heat energy, that raises the 1 g of water by 1°C, we have;

The amount of energy transferred to the water = The potential energy, P.E. = 4.214 J.

6 0
2 years ago
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