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Vanyuwa [196]
3 years ago
15

A railroad car of mass M moving at a speed v1 collides and couples with two coupled railroad cars, each of the same mass M and m

oving in the same direction at a speed v2.
(a) What is the speed vf of the three coupled cars after the collision in terms of v1 and v2? vf =

(b) How much kinetic energy is lost in the collision? Answer in terms of M, v1, and v2. ?KE =
Physics
1 answer:
Tatiana [17]3 years ago
7 0

Answer:

Explanation:

Given

mass of each railroad is M

first railroad is moving with velocity v_1 and other two's with v_2

conserving Momentum

Mv_1+2Mv_2=3M v_f

where v_f is the final velocity of combined system

v_f=\frac{v_1+2v_2}{3}

(b)Initial Kinetic Energy

=\frac{1}{2}Mv_1^2+\frac{1}{2}2Mv_2^2

=\frac{1}{2}Mv_1^2+Mv_2^2

Final Kinetic Energy

=\frac{1}{2}3Mv_f^2=\frac{1}{6}M(v_1^2+4v_2^{2}+4v_{1}v_2)

Loss in Kinetic Energy

=\frac{1}{2}Mv_1^2+Mv_2^2-\frac{1}{6}M(v_1^2+4v_2^{2}+4v_{1}v_2)

=M(v_1-v_2^2)^2

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A baseball hit just above the ground leaves the bat 27 m/s at 45° above the horizontal. A) How far away does the ball strike the
Sedbober [7]

Answer:

A) The ball hits the ground 74.45 m far from the hitting position.

B) Maximum height of the ball = 18.57 m

Explanation:

There are two types of motion in this horizontal and vertical motion.

We have velocity = 27 m/s at 45° above the horizontal

Horizontal velocity = 27cos45 = 19.09 m/s

Vertical velocity = 27sin45 = 19.09 m/s

Time to reach maximum height,

           v = u + at

           0 = 19.09 - 9.81 t

            t = 1.95 s

So total time of flight = 2 x 1.95 = 3.90 s

A) So the ball travels at 19.09 m/s for 3.90 seconds.

     Horizontal distance traveled = 19.09 x 3.90 = 74.45 m

     So the ball hits the ground 74.45 m far from the hitting position.

B) We have vertical displacement

              S = ut + 0.5 at²

              H = 19.09 x 1.95 - 0.5 x 9.81 x 1.95² = 18.57 m

    Maximum height of the ball = 18.57 m

6 0
3 years ago
According to diagonal rule, the orbital with lowest energy in the given following is ________.
Kaylis [27]
The orbital with the lowest energy is 3s.
5 0
3 years ago
Calculate the air pressure in the pressurized tank, if h1 = 0. 18 m, h2 = 0. 2m and h3 = 0. 25m. The density of the mercury, wat
fomenos

The air pressure in the pressurized tank will be 24014.88 N/m²,196.2 N/m²,2084.625  N/m².

<h3 /><h3>What is pressure?</h3>

The force applied perpendicular to the surface of an item per unit area across which that force is spread is known as pressure.

It is denoted by P. The pressure relative to the ambient pressure is known as gauge pressure.

Pressure is found as the product of the density,acceleraton due to gravity and the height.

P₁=ρ₁gh₁

P₁=13,600 kg/m³×9.81 (m/s²)×0.18 m

P₁=24014.88 N/m²

P₂=ρ₂gh₂

P₂= 1000 kg/m³×9.81 (m/s²)×00.2 m

P₂=196.2 N/m²

P₃=ρ₃gh₃

P₃=850 kg/m³×9.81 (m/s²)×0.25

P₃=2084.625  N/m²

Hence,the air pressure in the pressurized tank will be 24014.88 N/m²,196.2 N/m²,2084.625  N/m².

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6 0
2 years ago
Suppose that you are holding a pencil balanced on its point. If you release the pencil and it begins to fall, what will be the a
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Answer:

The angular acceleration of the pencil<em> α  = 17 rad·s⁻²</em>

Explanation:

Using Newton's second angular law or torque to find angular acceleration, we get the following expressions:

    τ = I α                              (1)

    W r = I α                          (2)

The weight is that the pencil has is,

   sin 10 = r / (L/2)

   r = L/2(sin(10))

 

The shape of the pencil can be approximated to be a cylinder that rotates on one end and therefore its moment of inertia will be:

    I = 1/3 M L²

Thus,

   mg(L / 2)sin(10) = (1/3 m L²)(α) 

   α(f) = 3/2(g) / Lsin(10)

   α  = 3/2(9.8) / 0.150sin(10)

  <em> α  = 17 rad·s⁻²</em>

Therefore, the angular acceleration of the pencil<em> </em>is<em> 17 rad·s⁻²</em>

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When you changed from low to high power, how did the change affect the working distance of the lens?
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The working distance gets shorter as the magnification gets bigger. In order to focus, the high-power objective lens must be significantly nearer to the specimen than the low-power lens. Magnification is negatively correlated with working distance.

Magnification change The magnification of a specimen is increased by switching from low power to high power. The magnification of an image is determined by multiplying the magnification of the objective lens by the magnification of the ocular lens, or eyepiece.

The geometry of the optical system connects the magnifying power, or how much the thing being observed seems expanded, and the field of view, or the size of the object that can be seen.

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