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
agasfer [191]
3 years ago
15

Two cars collide at an icy intersection and stick together afterward. The first car has a mass of 1050 kg and was approaching at

6.00 m/s due south. The second car has a mass of 750 kg and was approaching at 25.0 m/s due west.
Physics
1 answer:
Soloha48 [4]3 years ago
7 0

Answer:

v = 11.0 m/s at 198.6° (18.6° south of west)

ΔKE = -145 kJ

Explanation:

I assume you want to find the final velocity and the change in kinetic energy.

Take east to be +x and north to be +y.

Momentum is conserved in the x direction:

(1050 kg) (0 m/s) + (750 kg) (-25.0 m/s) = (1050 kg + 750 kg) vₓ

vₓ = -10.4 m/s

Momentum is conserved in the y direction:

(1050 kg) (-6.00 m/s) + (750 kg) (0 m/s) = (1050 kg + 750 kg) vᵧ

vᵧ = -3.50 m/s

The magnitude of the final velocity is:

v² = (-10.4 m/s)² + (-3.50 m/s)²

v = 11.0 m/s

The direction of the final velocity is:

θ = atan(-3.50 m/s / -10.4 m/s)

θ = 198.6°

The initial kinetic energy is:

KE₀ = ½ (1050 kg) (6.00 m/s)² + ½ (750 kg) (25.0 m/s)²

KE₀ = 253,275 J

The final kinetic energy is:

KE = ½ (1800 kg) (11.0 m/s)²

KE = 108,682 J

The change in kinetic energy is:

ΔKE = 108,682 J − 253,275 J

ΔKE ≈ -145,000 J

You might be interested in
a girl that has a mass of 55kg is standing on a floor, how much supporting force does the floor have?
KengaRu [80]
A girl standing on a floor would have two opposite forces acting on it. These forces are the weight and the normal force. Since no other forces are acting and that the girl is at rest, then the weight must equate to the normal force. Therefore, the supporting force would be:

F = mg = 55kg (9.81 m/s^2) = 539.55 N
4 0
3 years ago
Why aren’t the Appalachian Mountains still as tall as the Himalayas?
stealth61 [152]

Answer:

mountains are limited in their theoretical height by several processes. First is isostasy: the bigger a mountain gets, the more it weighs down its tectonic plate, so it sinks lower. ... Bottom line: mountains can get taller than Mount Everest in earth gravity, like the Appalachians probably did—but not much taller.

3 0
2 years ago
Read 2 more answers
All galaxies follow the law of gravity. True or False
Juliette [100K]

The answer is true. All the galaxies in the universe follow the law of gravity.

<span>Based from the book, It's about Time: the Illusion of Einstein’s Time Dilation Explained, </span>

Einstein had explained that all the heavenly bodies in the universe follow the same scientific laws that are similar to our solar system. The stars and planets are held by the principles of inertia and gravity

8 0
3 years ago
What is the difference between weight and mass? mass and weight are both intrinsic properties of an object. if you multiply the
MatroZZZ [7]

if you multiply the mass of an object by the acceleration due to gravity, you will obtain the object's weight. mass is an intrinsic property of matter


looks like a good answer ...

4 0
3 years ago
Two point charges, a +45nC charge X and a +12nC charge Y are separated by a distance of 0.5m.
Gnoma [55]

A) Calculate the resultant electric field strength at the midpoint between the charges.

Qx is the charge at X and Qy is the charge at Y.

E at midpoint = k×Qx/0.25² - k×Qy/0.25²

k = 9×10⁹Nm²C⁻², Qx = 45nC, Qy = 12nC

E = 4752N/C

Well done.

B) Calculate the distance from X at which the electric field strength is zero.

Let D be some point between X and Y for which the net E field is 0.

Let d be the distance from X to D.

Set up the following equation:

E at D = k×Qx/d² - k×Qy/(0.5-d)² = 0

Do some algebra to solve for d:

k×Qx/d² = k×Qy/(0.5-d)²

Qx/d² = Qy/(0.5-d)²

Qx(0.5-d)² = Qyd²

(0.5-d)√Qx = d√Qy

0.5√Qx-d√Qx = d√Qy

d(√Qx+√Qy) = 0.5√Qx

d = (0.5√Qx)/(√Qx+√Qy)

Plug in Qx = 45nC, Qy = 12nC

d ≈ 330mm

C) Calculate the magnitude of the electric field strength at the point P on the diagram below.

First determine the angles of the triangle. The sides of the triangle are 0.3m, 0.4m, and 0.5m, so this is a right triangle where the angle between the 0.3m and 0.4m sides is 90°

∠Y = tan⁻¹(0.4/0.3) = 53.13°

∠X = 90-∠Y = 36.87°

Determine the horizontal component of E at P:

Ex = E from Qx × cos(∠X) - E from Qy × cos(∠Y)

Ex = k×Qx/0.4²×cos(36.87°) - k×Qy/0.3²×cos(53.13°)

Ex = 1305N/C

Determine the vertical component of E at P:

Ey = E from Qx × sin(∠X) - E from Qy × sin(∠Y)

Ey = k×Qx/0.4²×sin(36.87°) - k×Qy/0.3²×sin(53.13°)

Ey = 2479N/C

Use the Pythagorean theorem to determine the magnitude of E at P:

E = √(Ex²+Ey²)

E ≈ 2802N/C

4 0
3 years ago
Other questions:
  • What is the current in a 160V circuit if the resistance is 20Ω
    7·1 answer
  • A clever inventor has created a device that can launch water balloons with an initial speed of 85.0 m/s. Her goal is to pass a b
    8·1 answer
  • A typical laboratory centrifuge rotates at 3700 rpm . Test tubes have to be placed into a centrifuge very carefully because of t
    9·1 answer
  • Numeria
    9·2 answers
  • a race car accelerates uniformly from 18.5 m/s to 46.1m/s in 2.47 seconds detrrmine the acceleration of the car and distance tra
    12·1 answer
  • Explain how the particles of solids move
    12·1 answer
  • If a 8.1g ring is heated using 10.0 ), its temperature rises 21.7°C. Calculate the specific
    6·1 answer
  • Help me with both questions please?
    8·1 answer
  • A charged comb contains 1000 electrons. Calculate the charge on the comb.
    9·1 answer
  • Could someone please please help meee
    13·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!