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Naddik [55]
3 years ago
10

A railroad car with a mass of 11 000 kg collides and couples with a second car of mass 19 000 kg that is initially at rest. The

first car is moving with a speed of 4.5 m/s prior to the collision. a. What is the initial momentum of the first car? b. If external forces can be ignored, what is the final velocity of the two railroad cars after they couple?
Physics
1 answer:
Alex787 [66]3 years ago
7 0

Answer:

a) 49500 Kg.m/s

b)  1.65 m/s

Explanation:

Given:

Mass of the car, m₁ = 11000 kg

Mass of the second car, m₂ = 19000 kg

Initial Speed of the first car, u₁ = 4.5 m/s

Initial velocity of the second car , u₂ = 0

Now,

Momentum = Mass × Velocity

Initial momentum = m₁u₁

Thus

Initial momentum P₁ = 11000 × 4.5 = 49500 kg-m/sec

b)By using the concept of momentum conservation

Initial momentum = Final momentum

m₁u₁ + m₂u₂ = ( m₁ + m₂ )v

Where, v is the velocity after collision

thus,

49500 + 19000 × 0 = ( 11000 + 19000 ) × v

or

V = \frac{\textup{49500}}{\textup{30000}}

or

v = 1.65 m/s

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Wavelength = (speed) / (frequency) = (460 m/s) / (230/sec) = <em>2 meters</em>


3 0
3 years ago
A football punker attempts to kick the football so that it lands on the ground 67.0 m from where it is kicked and stays in the a
Flauer [41]

To solve this problem we will apply the linear motion kinematic equations. We will find the two components of velocity and finally by geometric and vector relations we will find both the angle and the magnitude of the vector. In the case of horizontal speed we have to

v_x = \frac{x}{t}

v_x = \frac{67}{4.5}

v_x = 14.89m/s

The vertical component of velocity is

-h = v_y t -\frac{1}{2} gt^2

Here,

h = Height

g = Gravitational acceleration

t = Time

v_y = Vertical component of velocity

-1.23 = v_y(4.5)-\frac{1}{2} (9.8)(4.5)^2

-1.23= 4.5v_y - 99.225

v_y = 21.77m/s

The direction of the velocity will be given by the tangent of the components, then

tan\theta = \frac{v_y}{v_x}

\theta = tan^{-1} (\frac{21.77}{14.89})

\theta = 55.59\°

The magnitude is given vectorially as,

|V| = \sqrt{v_x^2+v_y^2}

|V| = \sqrt{14.89^2 +21.77^2}

|V| = 26.37m/s

Therefore the angle is 55.59° and the velocity is 26.37m/s

6 0
3 years ago
Savion listed the steps involved when nuclear power plants generate electricity.
OverLord2011 [107]

Answer:

Reorder the steps so that step 4 appears before step 3

Explanation:

In a nuclear power plant, we have;

1) Nuclear reaction between the radio active species and the particles takes place to generate energy in the nucleus of atoms

2) The nuclear energy in the atom is converted into radiant energy, which is the energy found in light, and thermal (heat) energy

3) The produced radiant and thermal energy is released as heat and light

4) With the produced heat, steam is generated

5) The generated steam turns the steam turbines and produced mechanical energy

6) The produced mechanical energy is then converted into electrical energy in the electrical generator of the power plant

To correct Savion's error, Step 4) the light and heat should be released before step 3) the released heat can be used to generate steam, we therefore reorder the steps so that step 4 appears before step 3.

4 0
3 years ago
Read 2 more answers
Please, pretty please help me!
Andrews [41]

Answer:

umm

Explanation:

1223565 578633 =334675

8 0
3 years ago
Read 2 more answers
10.
myrzilka [38]

Answer:

<em>The new period of oscillation is D) 3.0 T</em>

Explanation:

<u>Simple Pendulum</u>

A simple pendulum is a mechanical arrangement that describes periodic motion. The simple pendulum is made of a small bob of mass 'm' suspended by a thin inextensible string.

The period of a simple pendulum is given by

T=2\pi \sqrt{\frac{L}{g}}

Where L is its length and g is the local acceleration of gravity.

If the length of the pendulum was increased to 9 times (L'=9L), the new period of oscillation will be:

T'=2\pi \sqrt{\frac{L'}{g}}

T'=2\pi \sqrt{\frac{9L}{g}}

Taking out the square root of 9 (3):

T'=3*2\pi \sqrt{\frac{L}{g}}

Substituting the original T:

T'=3*T

The new period of oscillation is D) 3.0 T

4 0
3 years ago
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