What is the momentum of an object that has a mass of 0.03 grams and a velocity of 1200 m/s2

A bug flying horizontally at 0.65 m/s collides and sticks to the end of a uniform stick hanging vertically. After the impact, th

irina [24]

The angular momentum is defined as,

$L=I\omega$

Acording to this text we know for conservation of angular momentum that

$L_i=L_f$

Where $L_i$ is initial momentum

$L_f$ is the final momentum

How there is a difference between the stick mass and the bug mass, we define that

Mass of the bug= m

Mass of the stick=10m

At the point 0 we have that,

$L_i=mvl$

Where l is the lenght of the stick which is also the perpendicular distance of the bug's velocity

vector from the point of reference (O), and ve is the velocity

At the end with the collition we have

$L_f=(I_b+I_s)\omega$

Substituting

$L_f=(ml^2+\frac{10ml^2}{3})\omega$

$L_f=\frac{13}{10}ml^2w$

$m(0.65)l=\frac{13}{10}ml^2 \omega$

$\omega=\frac{1}{2l}$

Applying conservative energy equation we have

$\frac{1}{2}(I_b+I_s)\omega^2=mgh+10mgh'$

$\frac{1}{2}(ml^2+\frac{10ml^2}{3})(\frac{1}{2l})^2=mg(l-lcos\theta)+\frac{10}{2}mg(l-lcos\theta)$

Replacing the values and solving

$l=\frac{13}{0.54g}$

Substituting

l=\frac{13}{0.54(9.8)}

$l=2.45cm$

7 0

3 years ago

Explain why it is easier to climb a mountain on a zigzag path rather than one straight up the side.

GaryK [48]

Although a zig zag pattern going up a mountain means you walk further, the incline of the slope is a lot less so you don't have to work as hard.

8 0

3 years ago

The weight of an object is the force generated by Earth's gravity accelerating the object's _________. distance mass time destin

julsineya [31]

The weight of an object is the force of gravity between Earth's
mass and the object's mass.

The forces of gravity always come in equal, opposite pairs.
The Earth's weight on the object is the same as the object's
weight on the Earth, and when the object falls to Earth, Earth
falls to the object.

4 0

3 years ago

The central star of a planetary nebula emits ultraviolet light with wavelength 104nm. This light passes through a diffraction gr

Gala2k [10]

Answer: 31.33 degrees

Explanation:

The diffraction angles $\theta_{n}$ when we have a slit divided into $n$ parts are obtained by the following equation:

$dsin\theta_{n}=n\lambda$ (1)

Where:

$d$ is the width of the slit

$\lambda$ is the wavelength of the light

$n$ is an integer different from zero.

Now, the first-order diffraction angle is given when $n=1$ , hence equation (1) becomes:

$dsin\theta_{1}=\lambda$ (2)

Now we have to find the value of $\theta_{1}$ :

$sin\theta_{1}=\frac{\lambda}{d}$

$\theta_{1}=arcsin(\frac{\lambda}{d})$ (3)

We know:

$\lambda=104nm=104(10)^{-9}m$

In addition we are told the diffraction grating has 5000 slits per mm, this means:

$d=\frac{1mm}{5000}=\frac{1(10)^{-3}m}{5000}$

Substituting the known values in (3):

$\theta_{1}=arcsin(\frac{104(10)^{-9}m}{\frac{1(10)^{-3}m}{5000}})$

$\theta_{1}=arcsin(0.52)$

<u>Finally:</u>

$\theta_{1}=31.33\º$ >>>This is the first-order diffraction angle

4 0

3 years ago

When air resistance is ignored, _____ of the projectile affect(s) the range and maximum height of the projectile.

FromTheMoon [43]

When air resistance is ignored, initial velocity of the projectile affect the range and maximum height of the projectile.

Projectile is a missile designed to be fired from a rocket or gun.

A projectile is the object that is propelled by the application of an external force and then moves freely under the influence of gravity and air resistance.

The range is defined as the distance between the launch point and the point where the projectile hits the ground.

The height from the ground at the top most position of projectile is referred to as maximum height.

When air resistance is ignored, initial velocity of the projectile affect the range and maximum height of the projectile.

Learn more about maximum height click here brainly.com/question/6261898

#SPJ4

8 0

1 year ago