1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
maw [93]
3 years ago
5

You are a traffic accident investigator. You have arrived at the scene of an accident. Two cars of equal mass (1,000 kg each) we

re involved in a rear-end accident at a stop sign. Here is what you know: Car 1 approached the intersection from the top of a 25-meter hill. Car 2 was on a flat stretch of road directly in front of Car 1. At the bottom of the hill, Car 1 was going 20 m/s; Car 2 was going 30 m/s before it stopped at the stop sign. There were no skid marks left by Car 2. The collision occurred at the stop sign, where Car 2 had stopped. After the collision, both cars were moving together in the same direction at 10 m/s before slowly rolling to a stop. You must now push Car 2, using 800 N of force, 10 meters off to the side of the road so no one else gets hurt.
1. What was the potential energy of Car 1 at the top of the hill? (3 points)




2. What was the kinetic energy of each car before braking? (4 points)




3. A. How much energy did Car 1 lose from the top to the bottom of the hill? (3 points)




B. Where do you suppose that energy went? (2 points)




4. How much work was done to bring Car 2 to a stop? (4 points)




5. If Car 2 came to a stop in 15 seconds, how much power (in watts) did it take to stop Car 2 at the stop sign by applying force to the brakes? (3 points)




6. What is the final combined momentum of the cars right after the accident? (6 points)




7. Assuming that there are no other nonconservative forces involved, what impulse was given to each car during the collision? (6 points)
A. Car 1 (3 points)




B. Car 2 (3 points)




8. How much energy did Car 1 lose in the collision? (3 points)




9. How much energy did Car 2 gain in the collision? (3 points)




10. Which of the two laws covered in this unit - Law of Conservation of Momentum or Law of Conservation of Energy - is obeyed in this problem? Explain your reasoning. (5 points)




11. How much work did it take to move Car 2 off to the side of the road? (3 points)




12. If it took you 40 seconds to move Car 2 off the road, how much power did you expend? (3 points)




13. Which simple machine does a car have that helped you move Car 2? Explain. (2 points)
Physics
1 answer:
kumpel [21]3 years ago
4 0
6. Is 50 m/s that’s the answer
You might be interested in
5.54 Two kilograms of air within a piston–cylinder assembly WP execute a Carnot power cycle with maximum and minimum temper atur
Step2247 [10]

Answer:

A) 60%

B) p2 = 1237.2 kPa

   v2 = 0.348 m^3

C) w1-2 = w3-4 = 1615.5 kJ

   Q2-3 = 60 kJ

Explanation:

A) calculate thermal efficiency

  Л = 1 - \frac{Tl}{Th}  

where Tl = 300 k

            Th = 750 k

hence thermal efficiency ( Л ) = [1 - ( 300 / 750 )] * 100 = 60%

B) calculate the pressure and volume at the beginning of the isothermal expansion

calculate pressure ( P2 )  :

= P3v3 = mRT3  ----- (1)

v3 = 0.4m , mR = 2* 0.287, T3 = 750

hence P3 = 1076.25

next equation to determine P2

Qex = p3v3 ln( p2/p3 )

60 = 1076.25 * 0.4 ln(p2/p3)

hence ; P2 = 1237.2 kpa

calculate volume ( V2 )

p2v2 = p3v3

v2 = p3v3 / p2

   = (1076.25 * 0.4 ) /  1237.2  

  = 0.348 m^3

C) calculate the work and heat transfer for each four processes

work :

W1-2 = mCv( T2 - T1 )

        = 2*0.718 ( 750 - 300 ) = 1615.5 kJ

W3-4 = 1615.5 kJ

heat transfer

Q2-3 = W2-3 = 60KJ

Q3-4 = 0

D ) sketch of the cycle on p-V coordinates

attached below  

6 0
3 years ago
A car is traveling at 15 m/sm/s . Part A How fast would the car need to go to double its kinetic energy
GREYUIT [131]

Answer:

21.21 m/s

Explanation:

Let KE₁ represent the initial kinetic energy.

Let v₁ represent the initial velocity.

Let KE₂ represent the final kinetic energy.

Let v₂ represent the final velocity.

Next, the data obtained from the question:

Initial velocity (v₁) = 15 m/s

Initial kinetic Energy (KE₁) = E

Final final energy (KE₂) = double the initial kinetic energy = 2E

Final velocity (v₂) =?

Thus, the velocity (v₂) with which the car we travel in order to double it's kinetic energy can be obtained as follow:

KE = ½mv²

NOTE: Mass (m) = constant (since we are considering the same car)

KE₁/v₁² = KE₂/v₂²

E /15² = 2E/v₂²

E/225 = 2E/v₂²

Cross multiply

E × v₂² = 225 × 2E

E × v₂² = 450E

Divide both side by E

v₂² = 450E /E

v₂² = 450

Take the square root of both side.

v₂ = √450

v₂ = 21.21 m/s

Therefore, the car will travel at 21.21 m/s in order to double it's kinetic energy.

8 0
3 years ago
Que significa gobernar​
kow [346]

Answer: conducir la política, acciones y asuntos de (un estado, organización o personas).

7 0
3 years ago
Creates an image that appears upside down behind the focal point
zheka24 [161]
An image that appears upside down behind the focal point is an image that is reflected on a concave mirror. Mirrors reflect different kinds of images based on the placement of an object that is reflected towards it. There are two kinds of mirrors, concave and a convex mirrors, the latter makes objects seem smaller and farther than where it is exactly.
8 0
3 years ago
Initially, a 2.00-kg mass is whirling at the end of a string (in a circular path of radius 0.750 m) on a horizontal frictionless
drek231 [11]

Answer:

v_f = 15 \frac{m}{s}

Explanation:

We can solve this problem using conservation of angular momentum.

The angular momentum \vec{L} is

\vec{L}  = \vec{r} \times \vec{p}

where \vec{r} is the position and \vec{p} the linear momentum.

We also know that the torque is

\vec{\tau} = \frac{d\vec{L}}{dt}  = \frac{d}{dt} ( \vec{r} \times \vec{p} )

\vec{\tau} =  \frac{d}{dt}  \vec{r} \times \vec{p} +   \vec{r} \times \frac{d}{dt} \vec{p}

\vec{\tau} =  \vec{v} \times \vec{p} +   \vec{r} \times \vec{F}

but, as the linear momentum is \vec{p} = m \vec{v} this means that is parallel to the velocity, and the first term must equal zero

\vec{v} \times \vec{p}=0

so

\vec{\tau} =   \vec{r} \times \vec{F}

But, as the only horizontal force is the tension of the string, the force must be parallel to the vector position measured from the vertical rod, so

\vec{\tau}_{rod} =   0

this means, for the angular momentum measure from the rod:

\frac{d\vec{L}_{rod}}{dt} =   0

that means :

\vec{L}_{rod} = constant

So, the magnitude of initial angular momentum is :

| \vec{L}_{rod_i} | = |\vec{r}_i||\vec{p}_i| cos(\theta)

but the angle is 90°, so:

| \vec{L}_{rod_i} | = |\vec{r}_i||\vec{p}_i|

| \vec{L}_{rod_i} | = r_i * m * v_i

We know that the distance to the rod is 0.750 m, the mass 2.00 kg and the speed 5 m/s, so:

| \vec{L}_{rod_i} | = 0.750 \ m \ 2.00 \ kg \ 5 \ \frac{m}{s}

| \vec{L}_{rod_i} | = 7.5 \frac{kg m^2}{s}

For our final angular momentum we have:

| \vec{L}_{rod_f} | = r_f * m * v_f

and the radius is 0.250 m and the mass is 2.00 kg

| \vec{L}_{rod_f} | = 0.250 m * 2.00 kg * v_f

but, as the angular momentum is constant, this must be equal to the initial angular momentum

7.5 \frac{kg m^2}{s} = 0.250 m * 2.00 kg * v_f

v_f = \frac{7.5 \frac{kg m^2}{s}}{ 0.250 m * 2.00 kg}

v_f = 15 \frac{m}{s}

8 0
3 years ago
Other questions:
  • Col. Joe Kittinger of the United States Air Force crossed the Atlantic Ocean in nearly 86 hours. The distance he traveled was 57
    8·1 answer
  • A woman performs 2000 J of work in order to push a cart full of groceries 50 meters. How much force did she apply to the cart?
    13·1 answer
  • You have two balls of equal size and smoothness, and you can ignore air resistance. One is heavy, the other is much lighter. You
    9·1 answer
  • Which of the following statements about ionization energy is true?. A. In a period, elements farther to the right generally have
    7·1 answer
  • In what region would solar cookers be useful
    5·1 answer
  • What was the main cause of the shift to a postindustrial society?
    9·1 answer
  • Vector ⃑ has a magnitude of 9.0 m and points 40° east of north. Vector ሬ⃑ has a magnitude of 15.0 m and points 30° north of west
    8·1 answer
  • If it requires 8.0 J of work to stretch a particular spring by 2.0 cm from its equilibrium length, how much more work will be re
    14·1 answer
  • Which best describes the surface of a concave mirror?
    12·2 answers
  • Please help me I will give a brainless
    6·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!