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3 years ago

13

skater spins over a point at a speed of 3.0 rotations per second then the momentum of inertia is 0.60 kg.M2, what is its angular

momentum?

1 answer:

laiz [17]3 years ago

8 0

Answer:

$L=11.3\ kg-m^2/s$

Explanation:

Given that,

Angular speed of a skater, $\omega=3\ rot/s=18.84\ rad/s$

The moment of inertia of the skater, I = 0.6 kg-m²

We need to find the angular momentum of the skater. The formula for the angular momentum of the skater is given by :

$L=I\omega$

Substitute all the values,

$L=0.6\times 18.84\\\\L=11.3\ kg-m^2/s$

So, its angular momentum is equal to $11.3\ kg-m^2/s$ .

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The CERN particle accelerator is circular with a circumference of 7.0 km.

Contact [7]

Answer:

$a_c=2.0196\times 10^{13}\ m/s^2$

$F=3.37273\times 10^{-14}\ N$

Explanation:

m = Mass of proton = $1.67\times 10^{-27}\ kg$

v = Speed of proton = 0.5c = $0.5\times 3\times 10^8=1.5\times 10^8\ m/s$

Circumference of the colider is 7 km

$P=2\pi r\\\Rightarrow r=\frac{P}{2\pi}\\\Rightarrow r=\frac{7000}{2\pi}\ m$

$a_c=\frac{v^2}{r}\\\Rightarrow a_c=\frac{\left(1.5\times 10^8\right)^2}{\frac{7000}{2\pi}}\\\Rightarrow a_c=2.0196\times 10^{13}\ m/s^2$

Centripetal acceleration is $2.0196\times 10^{13}\ m/s^2$

$F_c=ma_c\\\Rightarrow F_c=1.67\times 10^{-27}\times 2.0196\times 10^{13}\\\Rightarrow F=3.37273\times 10^{-14}\ N$

Force on protons is $3.37273\times 10^{-14}\ N$

8 0

3 years ago

The wavelength of violet light is about 425 nm (1 nanometer = 1 × 10−9 m). what are the frequency and period of the light waves?

BigorU [14]

1. Frequency: $7.06\cdot 10^{14} Hz$

The frequency of a light wave is given by:

$f=\frac{c}{\lambda}$

where

$c=3\cdot 10^{-8} m/s$ is the speed of light

$\lambda$ is the wavelength of the wave

In this problem, we have light with wavelength

$\lambda=425 nm=425\cdot 10^{-9} m$

Substituting into the equation, we find the frequency:

$f=\frac{c}{\lambda}=\frac{3\cdot 10^{-8} m/s}{425\cdot 10^{-9} m}=7.06\cdot 10^{14} Hz$

2. Period: $1.42 \cdot 10^{-15}s$

The period of a wave is equal to the reciprocal of the frequency:

$T=\frac{1}{f}$

The frequency of this light wave is $7.06\cdot 10^{14} Hz$ (found in the previous exercise), so the period is:

$T=\frac{1}{f}=\frac{1}{7.06\cdot 10^{14} Hz}=1.42\cdot 10^{-15} s$

4 0

3 years ago

Read 2 more answers

Ocean waves pass through two small openings, 20.0 m apart, in a breakwater. You're in a boat 70.0 m from the breakwater and init

Klio2033 [76]

Answer:

λ = 5.65m

Explanation:

The Path Difference Condition is given as:

δ= $(m+\frac{1}{2})\frac{lamda}{n}$ ;

where lamda is represent by the symbol (λ) and is the wavelength we are meant to calculate.

m = no of openings which is 2

∴δ= $\frac{3*lamda}{2}$

n is the index of refraction of the medium in which the wave is traveling

To find δ we have;

δ= $\sqrt{70^2+(33+\frac{20}{2})^2 }-\sqrt{70^2+(33-\frac{20}{2})^2 }$

δ= $\sqrt{4900+(\frac{66+20}{2})^2}-\sqrt{4900+(\frac{66-20}{2})^2}$

δ= $\sqrt{4900+(\frac{86}{2})^2 }-\sqrt{4900+(\frac{46}{2})^2 }$

δ= $\sqrt{4900+43^2}-\sqrt{4900+23^2}$

δ= $\sqrt{4900+1849}-\sqrt{4900+529}$

δ= $\sqrt{6749}-\sqrt{5429}$

δ= 82.15 -73.68

δ= 8.47

Again remember; to calculate the wavelength of the ocean waves; we have:

δ= $\frac{3*lamda}{2}$

δ= 8.47

8.47 = $\frac{3*lamda}{2}$

λ = $\frac{8.47*2}{3}$

λ = 5.65m

3 0

3 years ago

3. Take sugar, oil, corn syrup, a glass and water. Pour the water in the glass and then add each of the above the substances one

hoa [83]

Here are the observations

<u>S</u><u>u</u><u>g</u><u>a</u><u>r</u><u>:</u><u>-</u>

Sugar is soluble in water
so It will dissolve in water .

<u>C</u><u>o</u><u>r</u><u>n</u><u> </u><u>s</u><u>y</u><u>r</u><u>u</u><u>p</u><u>:</u><u>-</u>

Corn syrup is also basically a sugar.
It will dissolve in water too .
If we shake the mixture in glass then corn syrup will be dissolved.

<u>O</u><u>i</u><u>l</u><u>:</u><u>-</u>

Oil is not soluble in water
Hence it won't dissolve in water.
It will float over water and make two layers

7 0

2 years ago

Thermal Conductors don't have to be hot to transfer heat, explain a situation when a ice cube would still transfer heat to anoth

barxatty [35]

An ice cube would transfer heat to another object whose temperature
is lower than zero°C (32°F).

A block of "dry ice" is sitting there at a temperature of -78°C (-109°F).
An ice cube helps to melt dry ice nice and fast.

If you could find a block of solid nitrogen, its temperature would be
63K (-210°C, -346°F). An ice cube would transfer heat to that baby
so fast that it would instantly boil.

6 0

3 years ago