A blue train of mass 50 kg moves at 4 m/s toward a green train of 30 kg initially at rest. What is the initial momentum of the blue and green train combined?
A. 20 kgm/s
B. 50 kgm/s
C. 0 kgm/s
D. 200 kgm/s

Answers

Answer 1

The correct option is D.   The initial momentum of the blue and green train combined during the collision is 200 kgm/s.

Initial momentum of the blue and green train

Apply the principle of conservation of linear momentum as follows;

Pi = m1v1 + m2v2

where;

m1 is mass of blue trainm2 is mass of green trainv1 is velocity of blue trainv2 is velocity green trainPi is the initial momentum of the two trains

Pi = (50 x 4) + 30(0)

Pi = 200 kgm/s

Thus, the initial momentum of the blue and green train combined is 200 kgm/s.

Learn more about momentum here: https://brainly.com/question/7538238

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Related Questions

How do dog whistles work?

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The sound it emits comes from what is known as the ultrasonic range, a pitch that is so high humans can't hear it. Dogs can hear these sounds, however, as can cats and other animals. Because of this, the dog whistle is a favored training tool, though it may not be for every dog parent.

Two students on ice skates stand one behind the other. Student 2 pushes student 1 in the back; both students move away from each other. What law of motion is this. (Newton's laws)

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Answer:

forcing in act

Explanation:

Which of the following is not an example of energy transfer?
A. Riding a bike down a hill.
B. Sliding a table across a floor.
C. Holding a sign in the air.

Answers

Answer:

c.holding a sign in the air

Explanation:

because b is kinetic energy and a is also kinetic energy

C holding a sign in the air

What is the correct coefficient for 2H2 + O2 →2H2O

Answers

Explanation:

2forH2,1for02,and2forH20

Acceleration is the rate ot change of the velacity a -dejdt so it is the slope of the Velocity vs. Time graph Because it is dficult to drag the person in a consistent and reproducible way use the Expression Evakaator under the Special Features menu for this question lick Reset A and type in the hr on z t * t * t " t in the Expression Evaluator Click the Play button and let the simulation run roughly 5 sin ulation seconds before ressing the Pause but use the zoom buttons to a 쪄 the p s they the screen You should see 8 p at s ar l what you got in the previous question, but much smoother Look at the Postion vs Time. Velocity vs Time and Acceleration vs. Time piets h
a) the velocity is zero but the acceleration is negative
When the person is 8 to to the tight of the origin
b) the velocity is zero but the acceleration is positive
c) both the velocity and the acceleration are zero
d) both the velocity and the acceleraton are nonzero

Answers

Answer:

a) the body  is changing direction,

b)the body must go to the left and the acceleration to the right

c) the movement has not started.

d) all points of the motion

Explanation:

In this exercise you are asked to find in which position you have the following characteristics of the movement

a) The velocity is zero and the acceleration is negative

This is when the body reaches the end of the travel and turns around, in this case the speed is zero and the acceleration has the opposite direction to the movement.

In this case the body moves to the right and the acceleration is to the left, therefore the speed decreases

b) The velocity is zero, but the acceleration is positive

This occurs at the points where the speed is changing direction, specifically for this case the body must go to the left and the acceleration to the right

c) Both are zero

This only occurs where the body is stopped and the movement has not started.

d) both the velocity and the relation are nonzero.

This is at all points of the motion since the velocity is constantly changing as long as there is an acceleration

Let’s look at a radio-controlled model car. Suppose that at time t1=2.0st1=2.0s the car has components of velocity vx=1.0m/svx=1.0m/s and vy=3.0m/svy=3.0m/s and that at time t2=2.5st2=2.5s the components are vx=4.0m/svx=4.0m/s and vy=3.0m/svy=3.0m/s . Find (a) the components of average acceleration and (b) the magnitude and direction of the average acceleration during this interval.

Answers

Answer:

[tex]a_x=6\ \text{m/s}^2[/tex] and [tex]a_y=0\ \text{m/s}^2[/tex]

Magnitude of accleration is [tex]6\ \text{m/s}^2[/tex] and the direction is [tex]0^{\circ}[/tex]

Explanation:

[tex]t_1=2\ \text{s}[/tex]

[tex]v_x=1\ \text{m/s}[/tex]

[tex]v_y=3\ \text{m/s}[/tex]

[tex]t_2=2.5\ \text{s}[/tex]

[tex]v_x=4\ \text{m/s}[/tex]

[tex]v_y=3\ \text{m/s}[/tex]

Average acceleration in the different axes

[tex]a_x=\dfrac{\Delta v_x}{\Delta t}\\\Rightarrow a_x=\dfrac{4-1}{2.5-2}\\\Rightarrow a_x=6\ \text{m/s}^2[/tex]

[tex]a_y=\dfrac{\Delta v_y}{\Delta t}\\\Rightarrow a_y=\dfrac{3-3}{2.5-2}\\\Rightarrow a_y=0\ \text{m/s}^2[/tex]

The components of the acceleration is [tex]a_x=6\ \text{m/s}^2[/tex] and [tex]a_y=0\ \text{m/s}^2[/tex]

The magnitude of acceleration

[tex]a=\sqrt{a_x^2+a_y^2}\\\Rightarrow a=\sqrt{6^2+0^2}\\\Rightarrow a=6\ \text{m/s}^2[/tex]

Direction

[tex]\theta=\tan^{-1}\dfrac{a_y}{a_x}\\\Rightarrow \theta=\tan^{-1}\dfrac{0}{6}\\\Rightarrow \theta=0^{\circ}[/tex]

The magnitude of accleration is [tex]6\ \text{m/s}^2[/tex] and the direction is [tex]0^{\circ}[/tex].

Formula One racers speed up much more quickly than normal passenger vehicles, and they also can stop in a much shorter distance. A Formula One racer traveling at 90m/s can stop in a distance of 110m. What is the magnitude of the car's acceleration as it slows during braking?

Answers

Answer:

The magnitude of the car's acceleration as it slows during braking is 36.81 m/s²

Explanation:

From the question, the given values are as follows:

Initial velocity, u = 90 m/s

final velocity, v = 0 m/s

distance, s = 110 m

acceleration, a = ?

Using the equation of motion, v² = u² + 2as

(90)² + 2 * 110 * a = 0

8100 + 220a = 0

220a = -8100

a = -8100/220

a = -36.81 m/s²

The value for acceleration is negative showing that car is decelerating to a stop. The magnitude of the car's acceleration as it slows during braking is therefore 36.81 m/s²

Please help I don’t get this give me answers please

Answers

Answer:

c

Explanation:

something that orbiys other things in space​

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Answer: well we all orbit the sun all the planets do so the

SuN

Explanation: two words common sense

a body of mass 20kg initially at rest is subjected to a force of 40N for 1sec calculate the change in kinetic energy showing the solution​

Answers

Answer:

Change in KE is 40 J

Explanation:

Recall that the impulse exerted on an object equal the change of momentum of the object (ΔP), which in time is defined as the product of the force exerted on it times the time the force was acting:

Change in momentum is:   ΔP = F * Δt

In our case,

ΔP = 40 N * 1 sec = 40 N s

Since the object was initially at rest, its initial momentum was zero, and the final momentum should then be 40 N s.

So, the initial KE was 0, and the final (KEf) can be calculated using:

KEf = 1 /(2 m) Pf^2 = 1 / (40) 40^2 = 40 J

So, the change in kinetic energy is:

KEf - KEi = 40 J - 0 j = 40 J

A motorcycle is following a car that is traveling at constant speed on a straight highway. Initially, the car and the motorcycle are both traveling at the same speed of 19.0 m/s , and the distance between them is 52.0 m . After t1 = 3.00 s , the motorcycle starts to accelerate at a rate of 4.00 m/s^2. The motorcycle catches up with the car at some time t2.

Required:
a. How long does it take from the moment when the motorcycle starts to accelerate until it catches up with the car?
b. How far does the motorcycle travel from the moment it starts to accelerate (at time t1) until it catches up with the car (at time t2)?

Answers

Answer:

a) 5.09 seconds

b) 107.07 meters

Explanation:

a) As we know

[tex]t_2- t_1 = \sqrt{\frac{2 X}{a} }[/tex]

Substituting the given values we get

[tex]t_2 - t_1 = \sqrt{\frac{2 * 52}{4} } \\t_2 - t_1 = 5.09[/tex]

It takes 5 .09 s for the motorcycle to accelerate until it catches up with the car

b)

[tex]X_{t`2} = v_i \sqrt{\frac{2X}{a} } + 0.5 a\sqrt{\frac{2X}{a} }\\X_{t`2} = (v_i + 0.5 a) \sqrt{\frac{2X}{a} }\\X_{t`2} = ( 19 + 2) \sqrt{\frac{2* 52}{4} }\\X_{t`2} = 21 * 5.09\\X_{t`2} = 107.07[/tex]

alex often draws his dream house​

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Answer:

hopefully alex quackity hahhaa

Explanation:

i hope this was free points and not an actual thing

Answer:

cool, cool for alex .....

What is the average speed of an Olympic sprinter that runs 100 m in 9.88 s?

Answers

Answer:

speed = 10.1215 m/s

Explanation:

speed = distance / time

speed = 100 / 9.88 = 10.1215 m/s

Help plz with both I’ll mark brainliest

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The first one is to save your ears from the decibels
Save wars from decaibles and the second is C

What does Neil Degrasse Tyson mean when he says "Wolves domesticated humans" 15000 years ago?

Answers

Answer:

Explanation:

fufuu6u

An uncharged parallel-plate capacitor is connected through an open switch to a battery of voltage VV. The switch is closed and the capacitor is allowed to charge. As the capacitor is charged, energy is transferred to it from the battery. When the capacitor is fully charged, the energy stored in the capacitor is U1U1 . The energy stored in the capacitor when the stored charge is q02q02 is

Answers

Answer:

(1/2)U₁

Explanation:

An uncharged parallel-plate capacitor is connected through an open switch to a battery of voltage VV. The switch is closed and the capacitor is allowed to charge. As the capacitor is charged, energy is transferred to it from the battery. When the capacitor is fully charged, the energy stored in the capacitor is U1U1 . The energy stored in the capacitor when the stored charge is q₁/2 is

Solution:

A capacitor is an electrical device used to store electrical energy in an electric field. The energy stored in a capacitor is given by:

U = (1/2)QV; where U is the energy stored, Q is the charge and V is the voltage applied.

The energy stored in a fully charged capacitor with a charge q₁ and battery of voltage V is given as:

U₁ = (1/2)q₁V

If the stored charge is q₁/2, the energy stored (U₂) becomes:

U₂ = (1/2)(q₁ / 2)V

U₂ = (1/2)*  (1/2)q₁V

U₂ = (1/2)U₁

what is borh's postulates for the hydrogen atom​

Answers

Answer:

An atom has a number of stable orbits in which an electron can reside without the emission of radiant energy. ... Each orbit corresponds, to a certain energy level.

Explanation:

Hope it is helpful....

Plzzz answer this correctly

Answers

Answer:

D, the acceleration  of A is twice that of b.

Explanation: in four seconds b got to ten, in two seconds a got to 20. Going  10m/s faster in half the time is going twice the acceleration

How does the Law of Conservation of Energy (or energy transformation) relate to the home?

Answers

Answer:

"The law of conservation of energy states that energy can neither be created nor destroyed - only converted from one form of energy to another. This means that a system always has the same amount of energy, unless it's added from the outside. ... The only way to use energy is to transform energy from one form to another."

Explanation:

Brainliest?

3. Two cyclists that are 500 m apart start biking toward each other. They bike at speeds of 6 and

4 m/s.

How long does it take for them to reach each other?

a.

b. How far does the slower biker travel?

Answers

Answer:

A) 50 seconds

B) 200 m

Explanation:

They are 500 metres apart.

And one of the bike loves at 6 m/s while the other loves at 4 m/s.

A) Let distance of the 6 m/s bike before they meet be x.

Thus, time = x/6

Since time = distance/speed

For the second bike at 4 m/s, his distance covered before they meet will be 500 - x

Thus, time = (500 - x)/4

Now they will meet each other at the same time. Thus;

x/6 = (500 - x)/4

Cross multiply to get;

4x = 3000 - 6x

6x + 4x = 3000

10x = 3000

x = 3000/10

x = 300 m

Thus, time will be;

t = 300/6

t = 50 seconds

B) Distance covered by the slower bike is (500 - x)

Since from a above, x = 300

Thus; distance = 500 - 300 = 200 m

HURRY!!!
I need helppppp!!!

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Answer:

ok i think the answer would be C. or B. hope im right

true or false please help me now.
Calibration graphs can be used to determine unknown concentrations in electrochemical ​

Answers

Answer:

false

Explanation:

What kind of reasoning is most often used to form hypotheses?
inductive
deductive
detective
invective

Answers

Deductive reasoning

All magnetic fields result from the movement of
A. charged particles
B. electrons only
C. protons only
D. neutrons only

Answers

Don’t know sorry I’m just trying not a good person

This is the build up of substance such as pesticides in an organism and occurs when an organism absorb a substance at a rate faster than that at which the substance is lost

Answers

Answer:

which the substance is lost by catabolism and excretion.

Explanation:

Calculate the radiative and collisional energy losses (in keV/micron) for a 1.9 MeV electron in lead and determine the rad./coll. ratio. (b) Plexiglas is often used to shield high-energy beta emitters rather than lead, even though lead is a better shield against the bremsstrahlung photons. Both shields will stop the high-energy beta, so why is Plexiglas used instead of lead?

Answers

Answer:

Explanation:

During an energy transfer, the collision loss for an electron can be determined by using the formula:

[tex]Q = \dfrac{4mME }{(m+M)^2}[/tex]

However; from the total stopping power & power loss of the electron;

[tex]\dfrac{radiational \ energy \ loss}{colisional \ energy \ loss } = \dfrac{ZE}{800}[/tex]

where;

Z = atomic no. for lead = 82

E = 1.9 MeV

radiational energy loss = collisional energy loss  [tex]=\dfrac{82 \times 1.9}{800}[/tex]

= 0.19475

b)

Normally, the traditional lead shielding in its pure shape contains high brittleness. However, the functionality of this carbon group chemical element is useful for protection because it has an excessive density.

Initially, the conventional lead protection however reduces the mild clarity at the same moment as plexiglass is useful for light transmittance and readability.

Moreover, the traditional lead with its high density and thickness reduces observation features, in the meantime, the plexiglass is a whole lot higher than the stated.

Finally, plexiglass contains a high dimensional balance with an excessive dielectric constant.

Plzz help me with this
I’ll give brainliest

Answers

Answer:

B. Objects with more mass have more gravitational force acting upon them.

Answer:

Should be A but it can be B as well.

Matter is made of small particles to small to be seen. Which of these best describe evidence of this statement? 1. Tara’s crayons melted when she left them under the sun. 2. Kerris glass of water overflowed when she added ice. 3. Sams basketball expands as he pumps air into it. 4. Stephanie dropped a vase and it broke into pieces.

Answers

Answer:

Explanation:

I think the answer is statement no 3.

Hope it helps.

Answer:

1 Tara's crayons melted when she left them under the sun

Galileo
o did not believe friction existed
o believed that friction stopped objects in motion
o believed that friction kept objects in motion
О
assumed that in a frictionless environment objects would never move

Answers

Answer:

object would move but it could be difficult to slow down or stop.

Thermodynamic Processes

Two moles of a monatomic ideal gas at (5 MPa, 5 L) is expanded isothermally until the volume is doubled (step 1). Then it is cooled isochorically until the pressure is 1 MPa (step 2). The temperature drops in this process. The gas is now compressed isothermally until its volume is back to 5 L, but its pressure is now 2 MPa (step 3). Finally, the gas is heated isochorically to return to the initial state (step 4). (a) Draw the four processes in the pV plane. (b) Find the total work done by the gas.

Answers

Answer:

A. Part a is the attachment

B. total work = 10.4kj

Explanation:

[tex]workdone=nRT1ln\frac{Vb}{Va}[/tex]

T1 = constant temperature

nRT1 = PaVa = PbVb

We write equation as

[tex]workdone =(PaVa)ln\frac{Vb}{Va}[/tex]

5ma = Pa, 5L = Va, Vb = 10L(temperature is doubled)

[tex]w1 = workdone =(5mpa*5L)ln\frac{10L}{5L}[/tex]

W1 = 25 ln2

W1 = 25 x 0.693

= 17.327kj

The isochoric expansion has no change in volume. So,

W2 = 0

Isothermal compression

[tex]w3=nRT3ln\frac{Vd}{Vc}[/tex]

T3 = constant temperature

nRT3 = PcVc = PdVd

[tex]workdone=(PcVc)ln\frac{Vd}{Vc}[/tex]

Pc = 1mpa Vc = 10L Vd = 5L

[tex]w3=(1)(10)ln\frac{5L}{10L}[/tex]

= 10x-0.693

= -6.93kj

Isochoric compression has no change in volume. Workdone w4 = 0

Total workdone = w1 + w2 + w3 + w4

= 17.33 + 0 + (-6.93) + 0

= 10.4kj

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