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2 years ago

15

A 2 kg stone is tied to a 0.5 m string and swung around a circle at a constant angular velocity of 12 rad/s. the angular momentu

m of the stone about the center of the circle is:

1 answer:

Illusion [34]2 years ago

3 0

Starting from the angular velocity, we can calculate the tangential velocity of the stone:
$v=\omega r= (12 rad/s)(0.5 m)= 6 m/s$

Then we can calculate the angular momentum of the stone about the center of the circle, given by
$L=mvr$
where
m is the stone mass
v its tangential velocity
r is the radius of the circle, that corresponds to the length of the string.

Substituting the data of the problem, we find
$L=(2 kg)(6 m/s)(0.5 m)=6 kg m^2 s^{-1}$

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A uniform meterstick of mass 0.20 kg is pivoted at the 40 cm mark. where should one hang a mass of 0.50 kg to balance the stick?

Tcecarenko [31]

The weight of the meterstick is:
$W=mg=0.20 kg \cdot 9.81 m/s^2 = 1.97 N$
and this weight is applied at the center of mass of the meterstick, so at x=0.50 m, therefore at a distance
$d_1 = 0.50 m - 0.40 m=0.10 m$
from the pivot.
The torque generated by the weight of the meterstick around the pivot is:
$M_w = W d_1 = (1.97 N)(0.10 m)=0.20 Nm$

To keep the system in equilibrium, the mass of 0.50 kg must generate an equal torque with opposite direction of rotation, so it must be located at a distance d2 somewhere between x=0 and x=0.40 m. The magnitude of the torque should be the same, 0.20 Nm, and so we have:
$(mg) d_2 = 0.20 Nm$
from which we find the value of d2:
$d_2 = \frac{0.20 Nm}{mg}= \frac{0.20 Nm}{(0.5 kg)(9.81 m/s^2)}=0.04 m$

So, the mass should be put at x=-0.04 m from the pivot, therefore at the x=36 cm mark.

4 0

2 years ago

A resultant vector is 8.00 units long and makes an angle of 43.0 degrees measured �� – �� with respect to the positi

Komok [63]

Answer:

223 degree

Explanation:

We are given that

Magnitude of resultant vector= 8 units

Resultant vector makes an angle with positive -x in counter clockwise direction

$\theta=43^{\circ}$

We have to find the magnitude and angle of the equilibrium vector.

We know that equilibrium vector is equal in magnitude and in opposite direction to the given vector.

Therefore, magnitude of equilibrium vector=8 units

x-component of a vector= $v_x=vcos\theta$

Where v=Magnitude of vector

Using the formula

x-component of resultant vector= $v_x=8cos43=5.85$

y-component of resultant vector= $v_y=vsin\theta=8sin43=5.46$

x-component of equilibrium vector= $v_x=-5.85$

y-component of equilibrium vector= $-v_y=-5.46$

Because equilibrium vector lies in III quadrant

$\theta=tan^{-1}(\frac{v_x}{v_y})=tan^{-1}(\frac{-5.46}{-5.85})=43^{\circ}$

The angle $\theta'$ lies in III quadrant

In III quadrant ,angle = $\theta'+180^{\circ}$

Angle of equilibrium vector measured from positive x in counter clock wise direction=180+43=223 degree

4 0

1 year ago

Determine the magnitude and sense (direction) of the current in the 500-latex: \omega ω resistor when i = 30 ma.

VARVARA [1.3K]

Complete Question:

Check the circuit in the file attached to this solution

Answer:

Total current = 0.056 A(From left to right)

Explanation:

Let the current in loop 1 be I₁ and the current in loop 2 be I₂

Applying KVL to loop 1

30 - (I₁ - I₂)500 + I₂R + 15 = 0

45 - 500I₁ - 500I₂ + RI₂ = 0

I₁ = 30mA = 0.03 A

45 - 500(0.03) - 500I₂ + RI₂ = 0

30 -500I₂ + RI₂ = 0...............(1)

Applying kvl to loop 2

-RI₂ - 15 + 10 - 400I₁ = 0

-RI₂ = 5 + 400*0.03

RI₂ = -17 ................(2)

Put equation (2) into (1)

30 -500I₂ -17 = 0

-500I₂ = 13

I₂ = -13/500

I₂ = -0.026 A

The total current in the 500 ohms resistor = I₁ - I₂ = 0.03+0.026

Total current = 0.056 A

The current will flow from left to right

5 0

2 years ago

An astronaut holds a rock 100m above the surface of Planet X . The rock is then thrown upward with a speed of 15m/s , as shown i

Butoxors [25]

Answer: $5 m/s^{2}$

Explanation:

The described situation is is related to vertical motion (and free fall). So, we can use the following equation that models what happens with this rock:

$y=y_{o}+V_{o}sin\theta t-\frac{1}{2}gt^{2}$ (1)

Where:

$y=0m$ is the rock's final height

$y_{o}=100 m$ is the rock's initial height

$V_{o}=15 m/s$ is the rock's initial velocity

$\theta=90\°$ is the angle at which the rock was thrown (directly upwards)

$t=10 s$ is the time

$g$ is the acceleration due gravity in Planet X

Then, isolating $g$ and taking into account $sin(90\°)=1$ :

$g=(-\frac{2}{t^{2}})(y-y_{o}-V_{o}t)$ (2)

$g=(-\frac{2}{(10 s)^{2}})(0 m-100 m-(15 m/s)(10 s))$ (3)

Finally:

$g=5 m/s^{2}$ (4) This is the acceleration due gravity in Planet X

7 0

1 year ago

Romeo lanza suavemente guijarros a la ventana de julieta y quiere que los guijarros golpeen la ventana solo con con un component

Yuki888 [10]

Answer:

$5.219\,\frac{m}{s}$

Explanation:

Las condiciones del problema requieren el cálculo de la rapidez inicial de los guijarros. Se sabe que el componente vertical de la rapidez final es cero. Por tanto, el tiempo se determina a continuación: (The conditions of this problems require the calculation of the initial speed of the peebles. It is known that vertical component of the final speed is zero. Therefore, the time is determined herein:).

$(0\,\frac{m}{s})^{2} = v_{o,y}^{2} - 2\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (4.5\,m)$

$v_{o,y} = 9.395\,\frac{m}{s}$

$0\,\frac{m}{s} = 9.395\,\frac{m}{s} - \left(9.807\,\frac{m}{s^{2}} \right)\cdot \Delta t$

$\Delta t = 0.958\,s$

Además, se determina el componente horizontal de la rapidez inicial (Likewise, the horizontal component of the initial speed is determined):

$v_{o,x} = \frac{5\,m}{0.958\,s}$

$v_{o,x} = 5.219\,\frac{m}{s}$

El guijarro tiene una rapidez de $5.219\,\frac{m}{s}$ cuando golpea la ventana (The peeble has a speed of $5.219\,\frac{m}{s}$ when it hits the window).

6 0

2 years ago