answer.
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
Olin [163]
1 year ago
8

A 225 kg red bumper car is moving at 3.0 m/s. It hits a stationary 180 kg blue bumper car. The red and blue bumper cars combine

and hit a stationary green bumper car of mass 150 kg. The red, blue, and green bumper cars all combine.
What is the final velocity of the combined bumper cars?
A .93m/s
B 1.2 m/s
C 1.7m/s
D 2.7m/s
Please Help!!!
Physics
2 answers:
Illusion [34]1 year ago
6 0

Answer: B. 1.2 m/s

Explanation:

First we would find the velocity with which red car and blue car move when they combine after first collision.

Using law of conservation of momentum,

Initial momentum = final momentum

Mass of red Car, Mr = 225 kg

Mass of blue car, Mb = 180 kg

Mass of green car, Mg = 150 kg

Initial velocity of red car, Ur = 3.0 m/s

Initial velocity of blue car, Ub = 0

Initial velocity of green car, Ug = 0

collision 1:

Mr Ur + Mb Ub = (Mr+Mb) V

where, V is the final velocity with which red and blue car moves.

225 × 3 + 0 = (225 + 180 ) V

⇒ V = 675/ 405 = 1.67 kg.m/s

collision 2:

These cars  collide with the green car. Let the final velocity with which the three cars move together after collision: V'

(Mr+Mb) V + Mg Ug = (Mr+Mb+Mg) V'

675 + 0 = 555 V'

⇒V' = 1.2 m/s

Thus, the correct option is B.

Alex Ar [27]1 year ago
4 0

Given


m1(mass of red bumper): 225 Kg


m2 (mass of blue bumper): 180 Kg


m3(mass of green bumper):150 Kg


v1 (velocity of red bumper): 3.0 m/s


v2 (final velocity of the combined bumpers): ?




The law of conservation of momentum states that when two bodies collide with each other, the momentum of the two bodies before the collision is equal to the momentum after the collision. This can be mathemetaically represented as below:


Pa= Pb


Where Pa is the momentum before collision and Pb is the momentum after collision.


Now applying this law for the above problem we get


Momentum before collision= momentum after collision.


Momentum before collision = (m1+m2) x v1 =(225+180)x 3 = 1215 Kgm/s


Momentum after collision = (m1+m2+m3) x v2 =(225+180+150)x v2

=555v2

Now we know that Momentum before collision= momentum after collision.


Hence we get


1215 = 555 v2


v2 = 2.188 m/s


Hence the velocity of the combined bumper cars is 2.188 m/s

You might be interested in
A baseball of mass m = 0.49 kg is dropped from a height h1 = 2.25 m. It bounces from the concrete below and returns to a final h
Brilliant_brown [7]

Answer:

Explanation:

Impulse = change in momentum

mv - mu , v and u are final and initial velocity during impact at surface

For downward motion of baseball

v² = u² + 2gh₁

= 2 x 9.8 x 2.25

v = 6.64 m / s

It becomes initial velocity during impact .

For body going upwards

v² = u² - 2gh₂

u² = 2 x 9.8 x 1.38

u = 5.2 m / s

This becomes final velocity after impact

change in momentum

m ( final velocity - initial velocity )

.49 ( 5.2 - 6.64 )

= .7056 N.s.

Impulse by floor in upward direction

= .7056 N.s

6 0
1 year ago
A pet-store supply truck moves at 25.0 m/s north along a highway. inside, a dog moves at 1.75 m/s at an angle of 35.0° east of
koban [17]

<u>Answer:</u>

 Velocity of the dog relative to the road = 26.04 m/s 3.15⁰ north of east.

<u>Explanation:</u>

  Let the east point towards positive X-axis and north point towards positive Y-axis.

  Speed of truck = 25 m/s north = 25 j m/s

  Speed of dog = 1.75 m/s at an angle of 35.0° east of north = (1.75 cos 35 i + 1.75 sin 35 j)m/s

                          = (1.43 i + 1.00 j) m/s

    Velocity of the dog relative to the road = 25 j + 1.43 i + 1.00 j = 1.43 i + 26.00 j

    Magnitude of velocity = 26.04 m/s

    Angle from positive horizontal axis = 86.85⁰

 So Velocity of the dog relative to the road = 26.04 m/s 86.85⁰ east of north = 26.04 m/s 3.15⁰ north of east.

4 0
1 year ago
Three arrows are shot horizontally. They have left the bow and are traveling parallel to the ground. Air resistance is negligibl
timurjin [86]

Answer:

F₁ = F₂ = F₃ = 0 N

Explanation:

given,

Arrow 1 mass = 80 g   speed = 10 m/s

Arrow 2 mass = 80 g   speed = 9 m/s

Arrow 3 mass = 90 g   speed = 9 m/s

Horizontal Force:- F₁ , F₂ and F₃

There is no air resistance.

If Air resistance is zero then the horizontal acceleration of the arrow also equal to zero.

We know,

According to newton's second law

        F = m a

If Acceleration is equal to zero

Then Force is also equal to zero.

Hence, F₁ = F₂ = F₃ = 0 N

4 0
1 year ago
Mari places a marble at the top of a ramp and lets it go. It rolls down. At the bottom of the ramp, the marble bumps into a bloc
tankabanditka [31]

Mari placeing more blocks around the perimeter

5 0
1 year ago
Read 2 more answers
1. California sea lions can swim as fast as 40.0 km/h. Suppose a sea lion begins to chase a fish at this speed when the fish is
Pie

#1

In order to chase the fish the distance traveled by sea lion in time t must be equal to the distance of sea lion from the fish and distance traveled by fish in the same time.

So here we can say let say sea lion chase the fish in time "t"

then here we have

d_1 = d_2 + L

here

d1 = distance covered by sea lion in time t

d2 = distance covered by fish in the same time t

L = distance between fish and sea lion initially = 60 m

d_1 = v_1 * t

d_1 = (40*\frac{5}{18})*t = \frac{100}{9}*t

d_2 = (16*\frac{5}{18})*t = \frac{40}{9}*t

\frac{100}{9}*t = \frac{40}{9}*t + 60

\frac{100}{9}*t - \frac{40}{9}*t = 60

\frac{60}{9}*t = 60

t = 9 s

So it will take 9 s to chase the fish by sea lion

# 2

velocity of truck on road = 25 m/s along North

velocity of dog inside the truck = 1.75 m/s at 35 degree East of North

v_{dt} = 1.75 cos35\hat j + 1.75sin35 \hat i

v_{dt} = 1.43 \hat j + 1 \hat i

we can write the relative velocity as

v_d - v_t = 1.43 \hat j + 1 \hat i

v_d = v_t + (1.43 \hat j + 1 \hat i)

now plug in the velocity of truck in this

v_d = 25 \hat j + (1.43 \hat j + 1 \hat i)

v_d = 26.43 \hat j + 1 \hat i

so it is given as

v_d = \sqrt{26.43^2 + 1^2} = 26.44 m/s

direction will be given as

\theta = tan^{-1}\frac{v_x}{v_y}

\theta = tan^{-1}\frac{1}{26.43} = 2.2 degree

so with respect to ground dog velocity is 26.44 m/s towards 2.2 degree East of North

8 0
2 years ago
Read 2 more answers
Other questions:
  • A 4-lb ball b is traveling around in a circle of radius r1 = 3 ft with a speed (vb)1 = 6 ft&gt;s. if the attached cord is pulled
    12·1 answer
  • Consider the waveform expression. y (x, t) = ym sin (0.333x + 5.36 + 585t) The transverse displacement (y) of a wave is given as
    7·1 answer
  • Geological evidence based on several radiometric techniques has provided a scientifically well-accepted age for the Earth. Repre
    11·1 answer
  • When light energy hits the retina, the retinal changes from a _____ to a _____ configuration.
    5·1 answer
  • "A block of metal weighs 40 N in air and 30 N in water. What is the buoyant force on the block due to the water? The density of
    14·2 answers
  • A microprocessor scans the status of an output I/O device every 20 ms. This is accomplished by means of a timer alerting the pro
    8·1 answer
  • The deuterium nucleus starts out with a kinetic energy of 1.24 × 10-13 joules, and the proton starts out with a kinetic energy o
    12·1 answer
  • If an irregularly shaped object (such as a wrench) is dropped from rest in a classroom and feels no air resistance, it will:
    6·1 answer
  • A p-type Si sample is used in the Haynes-Shockley experiment. The length of the sample is 2 cm, and two probes are separated by
    5·1 answer
  • A 6V radio with a current of 2A is turned on for 5 minutes. Calculate the energy transferred in joules
    12·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!