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Makovka662 [10]

4 years ago

5

A 2.0-kg block travels around a 0.40-m radius circle with an angular speed of 16 rad/s. The circle is parallel to the xy plane a

nd is centered on the z axis, 0.60 m from the origin. What is the magnitude of the component in the xy plane of the angular momentum around the origin?

1 answer:

fgiga [73]4 years ago

3 0

To solve this problem it is necessary to apply the concept related to the Angular Moment.

By definition the angular momentum is given as

$L=mvr$

Where

L= Angular momentum

m= mass

v= linear velocity

Recall that line speed can also be expressed as a function of angular velocity as

$v = \omega R \rightarrow$ where $\omega$ is the angular velocity and R the Radius

For the case described in the statement we have that there are two distances in different components, therefore the net radius would be given by

$r^2 = R^2+d^2$

Where

R= Radius of the block

d= displacement in z-axis

Replacing with our values we have,

$r^2=0.4^2+0.6^2$

$r=0.7211m$

Replacing in the angular momentum equation, depending on the angular velocity we have to

$L=m(\omega R)r$

$L=(2)(16*0.4)(0.7211)$

$L = 9.23008Kg.m/s$

Therefore the magnitude of the component in the xy plane of the angular momentum around the origin is $9.23008Kg.m/s$

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Given light with frequencies of: 5.5 x 1014 Hz, 7 x 1014 Hz and 8 x 1014 Hz, calculate the wavelengths in free space. What color

dolphi86 [110]

Answer:

Explanation:

Given light with frequencies of 5.5 x 10¹⁴ Hz, 7 x 10¹⁴ Hz and 8 x 10¹⁴ Hz

we now, $\lambda = \dfrac{c}{\nu}$

now

1) $\lambda_1 =\dfrac{3\times 10^8}{5.5\times 10^{14}}$

= 545.45 nm

2) $\lambda_2 = \dfrac{3\times 10^8}{7\times 10^{14}}$

= 428.57 nm

3) $\lambda_3 = \dfrac{3\times 10^8}{8\times 10^{14}}$

= 375 nm

the color related to wavelength are

545.45 nm - green

428.57 nm - blue

375 nm - violet

8 0

3 years ago

A uniform soda can of mass 0.140 kg is 12.0 cm tall and filled with 0.354 kg of soda (Fig. 9-41). Then small holes are drilled i

Ugo [173]

Answer:

Explanation:

a ) The centre of mass of a can filled completely with any liquid of any mass lies at the center of the can or at height h /2 where h is height of the can . So centre of mass of can of mass .140 kg of height 12 cm will lie at height 6 cm . Similarly center of mass of its content will also lie at height 6 cm that is center point of the content .

Hence overall center of mass of can along with its content will lie at height 6 cm initially .

b )

After the can loses all the soda , can becomes empty . Hence its center of mass will lie at its center again .

c ) Its center of mass remains unchanged as the soda drains out completely .

d ) When soda starts draining out of the can , its lower part becomes heavier than the upper part . Hence its center of mass moves downwards . Its height is lowered . As height of soda is lowered in the can , center of mass also is lowered . It reaches its lowest point when soda in the can reaches half the height or when the can is half filled .

After that center of mass again starts rising upwards until the can is completely empty .

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3 years ago

Why do water and table salt have very different properties?

Alexandra [31]

Answer:

D)

Explanation:

6 0

3 years ago

Which of the following cart configurations could produce the X versus T graph shown

Xelga [282]

Answer:

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Explanation:

8 0

3 years ago

What an object is placed 8 mm from a concave spherical mirror a clear image can be projected on the screen 16 mm in front of me

alexgriva [62]

Concept: The magnification of spherical mirror can be defined by two ways.

(i) In terms of the height of the object and image.

The magnification of the spherical mirror is defined as the ratio of the height of the image' $h_{i}$ ' to the height of the object ' $h_{o}$ '. It is denoted by letter 'm'.

Mathematically, it can be written as

$m= \frac{h_{i}}{h_{o}} ------------(1)$

(ii) In terms of the object's and image's distances.

The magnification of the spherical mirror is defined as the negative ratio of the image distance' $d_{i}$ ' to the object distance ' $d_{o}$ '.

Mathematically, it can be written as

$m= - \frac{d_{i}}{d_{o}} ------------(2)$

Now, from equation (1) and (2) we have,

$m = \frac{h_{i}}{h_{o}} = - \frac{d_{i}}{d_{o}} -----------(3)$

Given: Spherical Concave Mirror,

We will consider positive sign for object's and image's distance because both are in front of the mirror.

Object distance $(d_{o}) = + 8 mm$ .

Image distance $(d_{i}) = + 16 mm$

Object's height $(h_{o}) = + 4 mm$

Image's height $(h_{i}) =?$

Now, apply equation (3)

$\frac{h_{i}}{h_{o}} = - \frac{d_{i}}{d_{o}}$

$Or, \frac{h_{i}}{4 mm} = - \frac{+16 mm}{+8 mm}$

Or, hi = - 8 mm

Here; negative sign means, the image will be inverted.

The image's height will be 8 mm.

4 0

3 years ago