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4 years ago

11

Suppose two loaded train cars are moving toward one another, the first having a mass of 1.50 \times 10^5\, \mathrm{kg}1.50×10 5

kg and a velocity of (0.11 \, \mathrm{m/s})\, \hat{i}(0.11m/s) i ^ , and the second having a mass of 1.10 \times 10^5\, \mathrm{kg}1.10×10 5 kg and a velocity of (-0.21 \, \mathrm{m/s})\, \hat{i}(−0.21m/s) i ^ . What is their final velocity?

1 answer:

lara31 [8.8K]4 years ago

6 0

Answer:

0.2358

Explanation:

We need here apply the law momentum conservation, for objects who collide.

The equation is given by,

$m_1v_1+m_2v_2= (m_1+m_2)v_f$

Our values are giving by,

$m_1=1.5*10^5Kg\\v_2=0.11m/s\\m_2=1.1*10^5Kg\\v_2=-0.21m/s$

We replace in our equation and solve for V_f, then

$(1.5*10^5)(0.11)+(1.1*10^5)(-0.21)=(1.5*10^5+1.1*10^5)v_f$

$-6600=260000v_f$

$v_f=-\frac{6600}{260000}$

$v_f=-0.2538m/s$

Therefore their final velocity is -0.2358m/s (Negative symbol only indicate change in the direction)

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Consider an object with s=12cm that produces an image with s′=15cm. Note that whenever you are working with a physical object, t

Leni [432]

A. 6.67 cm

The focal length of the lens can be found by using the lens equation:

$\frac{1}{f}=\frac{1}{s}+\frac{1}{s'}$

where we have

f = focal length

s = 12 cm is the distance of the object from the lens

s' = 15 cm is the distance of the image from the lens

Solving the equation for f, we find

$\frac{1}{f}=\frac{1}{12 cm}+\frac{1}{15 cm}=0.15 cm^{-1}\\f=\frac{1}{0.15 cm^{-1}}=6.67 cm$

B. Converging

According to sign convention for lenses, we have:

- Converging (convex) lenses have focal length with positive sign

- Diverging (concave) lenses have focal length with negative sign

In this case, the focal length of the lens is positive, so the lens is a converging lens.

C. -1.25

The magnification of the lens is given by

$M=-\frac{s'}{s}$

where

s' = 15 cm is the distance of the image from the lens

s = 12 cm is the distance of the object from the lens

Substituting into the equation, we find

$M=-\frac{15 cm}{12 cm}=-1.25$

D. Real and inverted

The magnification equation can be also rewritten as

$M=\frac{y'}{y}$

where

y' is the size of the image

y is the size of the object

Re-arranging it, we have

$y'=My$

Since in this case M is negative, it means that y' has opposite sign compared to y: this means that the image is inverted.

Also, the sign of s' tells us if the image is real of virtual. In fact:

- s' is positive: image is real

- s' is negative: image is virtual

In this case, s' is positive, so the image is real.

E. Virtual

In this case, the magnification is 5/9, so we have

$M=\frac{5}{9}=-\frac{s'}{s}$

which can be rewritten as

$s'=-M s = -\frac{5}{9}s$

which means that s' has opposite sign than s: therefore, the image is virtual.

F. 12.0 cm

From the magnification equation, we can write

$s'=-Ms$

and then we can substitute it into the lens equation:

$\frac{1}{f}=\frac{1}{s}+\frac{1}{s'}\\\frac{1}{f}=\frac{1}{s}+\frac{1}{-Ms}$

and we can solve for s:

$\frac{1}{f}=\frac{M-1}{Ms}\\f=\frac{Ms}{M-1}\\s=\frac{f(M-1)}{M}=\frac{(-15 cm)(\frac{5}{9}-1}{\frac{5}{9}}=12.0 cm$

G. -6.67 cm

Now the image distance can be directly found by using again the magnification equation:

$s'=-Ms=-\frac{5}{9}(12.0 cm)=-6.67 cm$

And the sign of s' (negative) also tells us that the image is virtual.

H. -24.0 cm

In this case, the image is twice as tall as the object, so the magnification is

M = 2

and the distance of the image from the lens is

s' = -24 cm

The problem is asking us for the image distance: however, this is already given by the problem,

s' = -24 cm

so, this is the answer. And the fact that its sign is negative tells us that the image is virtual.

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To find the box's acceleration, we have to find first the net force acting on the box in the horizontal direction.

We have:

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So, the net force forward is

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Now we can find the acceleration by using Newton's second law of motion, which states that:

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where

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And solving for a, we find the acceleration:

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b)

The gravitational force on an object is the force with which the object is pulled towards the ground by the Earth.

It is given by

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In this problem we have

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So, the gravitational force on the box is

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c)

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In order to find the magnitude of this force, we apply Newton's second law of motion along the vertical direction.

We have two forces in this direction:

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So the net force is

$\sum F=N-W$

According to Newton's second law,

$\sum F=ma$

However, the box is at rest in the vertical direction, so the vertical acceleration is zero:

$a=0$

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