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sdas [7]
3 years ago
7

Train cars are coupled together by being bumped into one another. Suppose two loaded train cars are moving toward one another, t

he first having a mass of 130,000 kg and a velocity of 0.300 m/s, and the second having a mass of 115,000 kg and a velocity of -0.120 m/s. (The minus indicates direction of motion.) What is their final velocity
Physics
1 answer:
frozen [14]3 years ago
6 0

Answer:

Their final velocity is 0.103 m/s

Explanation:

Given;

mass of the first car, m₁ = 130,000 kg

mass of the second car, m₂ = 115,000 kg

initial velocity of the first car, u₁ = 0.300 m/s

initial velocity of the second car, u₂ = -0.120 m/s

Let their final velocity, = v (since they were coupled together, they will have common final velocity)

Apply the principle of conservation of linear momentum;

Total momentum before collision = Total momentum after collision

m₁u₁ + m₂u₂ = v(m₁ + m₂)

(130,000 x 0.3) + (115,000 x -0.12) = v(130,000 + (115,000)

39000 -13800 = 245,000v

25200 = 245,000v

v = 25200 / 245,000

v = 0.103 m/s

Therefore, their final velocity is 0.103 m/s

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A moving curling stone, A, collides head on with stationary stone, B. Stone B has a larger mass than stone A. If friction is neg
Kitty [74]

Answer:

The correct answer is option 'c': Smaller stone rebounds while as larger stone remains stationary.

Explanation:

Let the velocity and the mass of the smaller stone be 'm' and 'v' respectively

and the mass of big rock be 'M'

Initial momentum of the system equals

p_i=mv+0=mv

Now let after the collision the small stone move with a velocity v' and the big roch move with a velocity V'

Thus the final momentum of the system is

p_f=mv'+MV'

Equating initial and the final momenta we get

mv=mv'+MV'\\\\m(v-v')=MV'.....i

Now since the surface is frictionless thus the energy is also conserved thus

E_i=\frac{1}{2}mv^2

Similarly the final energy becomes

E_f=\frac{1}{2}mv'^2+\frac{1}{2}MV'^2\

Equating initial and final energies we get

\frac{1}{2}mv^2=\frac{1}{2}mv'^2+\frac{1}{2}MV'^2\\\\mv^2=mv'^2+MV'^2\\\\m(v^2-v'^2)=MV'^2\\\\m(v-v')(v+v')=MV'^2......(ii)

Solving i and ii we get

v+v'=V'

Using this in equation i we get

v'=\frac{v(m-M)}{(M-m)}=-v

Thus putting v = -v' in equation i  we get V' = 0

This implies Smaller stone rebounds while as larger stone remains stationary.

4 0
3 years ago
Pls help i begg youuuuu
Anna [14]
Pitch is the sensation of certain frequencies to the ear. High frequency = high pitch, low frequency = low pitch. 

f = c(speed of the wave) /  <span>λ (wavelength)

1. 343m/s / 0.77955m = 439.99 Hz   
     This corresponds to pitch A 

2. 343m/s / 0.52028m = 659.26 Hz
</span>     This corresponds to pitch E 
<span>
3. 343m/s / 0.65552m = 523.349 Hz
    </span>This corresponds to pitch C

4. using f = c /  λ
  λ = c / f<span>
     = 343m/s / 587.33 = 0.583999 m = 0.584 m

</span>
3 0
3 years ago
How much power is needed to lift the 200-N object to a height of 4 m in 4 s?
Vitek1552 [10]

Answer: 2000 watts

Explanation:

Given that,

power = ?

Weight of object = 200-N

height = 4 m

Time = 4 s

Power is the rate of work done per unit time i.e Power is simply obtained by dividing work by time. Its unit is watts.

i.e Power = work / time

(since work = force x distance, and weight is the force acting on the object due to gravity)

Then, Power = (weight x distance) / time

Power = (200N x 4m) / 4s

Power = 8000Nm / 4s

Power = 2000 watts

Thus, 2000 watts of power is needed to lift the object.

3 0
3 years ago
How much is 1.75 inches in cm?
saul85 [17]

Answer:

4.44 cm

Explanation:

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3 0
3 years ago
Read 2 more answers
In the final situation below, the 8.0 kg box has been launched with a speed of 10.0 m/s across a frictionless surface. Find the
Murljashka [212]

Answer:

the energy of the spring at the start is 400 J.

Explanation:

Given;

mass of the box, m = 8.0 kg

final speed of the box, v = 10 m/s

Apply the principle of conservation of energy to determine the energy of the spring at the start;

Final Kinetic energy of the box = initial elastic potential energy of the spring

K.E = Ux

¹/₂mv² = Ux

¹/₂ x 8 x 10² = Ux

400 J = Ux

Therefore, the energy of the spring at the start is 400 J.

8 0
3 years ago
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