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Katen [24]
3 years ago
14

Proper design of automobile braking systems must account for heat buildup under heavy braking. Part A Calculate the thermal ener

gy dissipated from brakes in a 1600 kg car that descends a 15 ∘ hill. The car begins braking when its speed is 95 km/h and slows to a speed of 40 km/h in a distance of 0.34 km measured along the road.
Physics
1 answer:
AVprozaik [17]3 years ago
8 0

Answer:

1838216 J

Explanation:

95 km/h = 26.39 m/s

40 km/h = 11.11 m/s

Initial kinetic energy

= .5 x 1600 x(26.39)²

= 557145.67 J

Final kinetic energy

= .5 x 1600 x ( 11.11)²

= 98745.68 J

Loss of kinetic energy

= 458400 J

Loss of potential energy

= mg x loss of height

= 1600 x 9.8 x 340 sin 15

= 1379816 J

Sum of Loss of potential energy and Loss of kinetic energy

=  1379816 + 458400

= 1838216 J

This is the work done by the friction . So this is heat generated.

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.3.3: Populating a vector with a for loop. Write a for loop to populate vector userGuesses with NUM_GUESSES integers. Read integ
Reptile [31]

Answer:

#include <iostream>

#include <vector>

using namespace std;

int main() {

  const int NUM_GUESSES = 3;

  vector<int> userGuesses(NUM_GUESSES);

  int i = 0;

int uGuess = 0;

for(i = 0; i <= userGuesses.size() - 1; i++){

 cin >> uGuess;

 userGuesses.at(i) = uGuess;

}

cout << endl;

  return 0;

}

Explanation:

First inbuilt library were imported. Then inside the main( ) function, 3 was assigned to NUM_GUESSES meaning the user is to guess 3 numbers. Next, a vector was defined with a size of NUM_GUESSES.

Then a for-loop is use to receive user guess via cin and each guess is assigned to the vector.

8 0
3 years ago
A 25.0-kg child is standing at the edge of a horizontal merry-go-round with a radius of 2.40 m and a moment of inertia of 356 kg
motikmotik

Answer:

\omega_{f}=1.634\ rad/s  

Explanation:

given,  

diameter of merry - go - round = 2.40 m  

moment of inertia = I = 356 kg∙m²

speed of the merry- go-round = 1.80 rad/s

mass of child = 25 kg  

initial angular momentum of the system  

L_i = I\omega_i  

L_i =356\times 1.80  

L_i =640.8\ kg.m^2/s  

final angular momentum of the system  

L_f = (I_{disk}+mR^2)\omega_{f}  

L_f = (356 + 25\times 1.2^2)\omega_{f}  

L_f= (392)\omega_{f}  

from conservation of angular momentum  

L_i = L_f  

640.8= (392)\omega_{f}  

\omega_{f}=1.634\ rad/s  

8 0
2 years ago
Read 2 more answers
According to this equation, F=ma, how much force is needed to accelerate an 82-kg runner at 7.5m/s2?
Murljashka [212]
You multiply the mass by the acceleration 82*7.5=615; that's what I would do
5 0
2 years ago
Read 2 more answers
A decrease in the magnitude of velocity is called
vagabundo [1.1K]
Any change in the speed or direction of motion is called "acceleration". You'll hear "deceleration" used for slowing down but that's not technically correct.
6 0
3 years ago
A pilot is flying a small airplane at 40 m/s in a circular path with a radius of 200 m. The pilot has a mass of 80.5 kg
g100num [7]
Try This Method(substitute your values)
Centripetal force = Mass X Velocity^2/ Radius 
<span>F= mv^2/r </span>
<span>so 80.5 X 40^2/200 </span>
<span>=644N</span>
4 0
2 years ago
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