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3 years ago

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A meter stick is held vertically with one end on the floor and is then allowed to fall. Find the speed of the other end when it

hits the floor, assuming that the end on the floor does not slip. (Hint: Use conservation of energy.)

1 answer:

Ilia_Sergeevich [38]3 years ago

5 0

Below is the attachment of the solutions.

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the answer is gravitational force hope this helps u stay safe

Explanation:

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Which of the following processes can produce either rain or snow

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Bergeron–Findeisen Process.

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A camera lens focuses on an object 75.0 cm from the lens. The image forms 3.50 cm behind the lens. What is the magnification of

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Explanation:

Refraction by a Convex Lens

Refraction bends the light downward upon entering the glass because the bottom part of the ray hits the slow medium first.

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Image formation depends upon bending light rays with lenses.

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Magnification:Transverse &Angular

The linear magnification or transverse magnification is the ratio of the image size to the object size. If the image and object are in the same medium it is just the image distance divided by the object distance.

Using the Gaussian form of the lens equation, a negative sign is used on the linear magnification equation as a reminder that all real images are inverted. If the image is virtual, the image distance will be negative, and the magnification will therefore be positive for the erect image.

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In this equation V is the vergence, n is the index of refraction, and u is used for the angle. Note that in this expression for the magnification, the minus sign doesn't appear. In this type of setting, the Cartesian sign convention is typically used and the object distance has a negative value. That takes care of the minus sign that is put in above in the Gaussian form.

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3 years ago

A pendulum of length L is suspended from the ceiling of an elevator. When the elevator is at rest the period of the pendulum is

Stolb23 [73]

<h2>Option A is the correct answer.</h2>

Explanation:

When an elevator moves upward with consonant acceleration a, the overall acceleration on the body is given by

a' = a + g

So acceleration of pendulum is a + g.

We have equation for period of simple pendulum

$T=2\pi \sqrt{\frac{l}{a'}}$

In normal case a' = g here a' is more.

From the equation we can see that period of simple pendulum is inversely proportional to square root of acceleration.

Since acceleration increases period decreases.

Option A is the correct answer.

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3 years ago

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A pin-supported, vertically-oriented 1-m long thin rod is struck by a pellet at m down from the pin at the top. The mass of the

Natasha_Volkova [10]

Answer:

the angular velocity of the rod immediately after being struck by the pellet, provided that the pellet gets lodged in the rod is = 0.5036 k` rad/s

Explanation:

Using the conservation of momentum of approach.

From the question; the pellet is hitting at a distance of 0.4 m down from the point of rotation of the rod.

So, the angular momentum of the system just before the collision occurs with respect to the axis of the rotation is expressed by the formula:

$L_i ^ { ^ \to } = mp ( r_y } ^ { ^ \to } * v_{pi} ^ { ^ \to } )$ ----- equation (1)

The position vector can now be :

$x ^ { ^ \to }$ = $- 0.4 \ j \ m$

Also, given that :

$v_{p,i} ^ { ^ \to } = (280 \ i - 350 \ j) \ m/s$

Replacing the value into above equation (1); we have:

$L_i ^ { ^ \to } =0.012 ((- \ 0.4 \ j) *(280 \ i - 350 \ j ))$

$L_i ^ { ^ \to } =0.012 * 112 \ k$ (by using cross product )

$L_i ^ { ^ \to } = 1.344 k` \ \ kg m^2 s^{-1}$

However; the moment of inertia of the rod about the axis of rotation is :

$I_{rod} = \frac{1}{3}m_rl^2 \\ \\ I_{rod} = \frac{1}{3}*8*1^2 \\ \\ I_{rod} = \frac{8}{3} \ \ kg \ m^2$

Also, the moment of inertia of the pellet about the axis of rotation is:

$I_{pellet} = m_pr_y^2 \\ \\ I_{pellet} = 0.012 *0.4^2 \\ \\ I_{pellet} = 1.92*10^{-3} kg . m^2$

So, the moment of inertia of the rod +pellet system is:

$I = I_{rod}+I_{pellet}$

$I =( \frac{8}{3}+1.92 *10^{-3} )kg. m^2$

$I = 2.6686 \ kg. m^2$

The final angular momentum is :

$L_f ^ {^ \to} = I \omega { ^ {^ \to} } = 2.6686 \ \omega ^ {^ \to}$

The angular velocity of the rod $\omega$ is determined by equating the angular momentum just before the collision with the final angular momentum (i.e after the collision).

So;

$L_f ^ {^ \to} = L_i ^ { ^ \to}$

$2.6686 \omega ^ {^ \to} = 1.344 \ k ^ {^ \to}$

$\omega ^ {^ \to} = \frac{1.344 \ k`}{2.6686}$

= 0.5036 k` rad/s

Hence; the angular velocity of the rod immediately after being struck by the pellet, provided that the pellet gets lodged in the rod is = 0.5036 k` rad/s

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3 years ago