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Cerrena [4.2K]
3 years ago
6

A solid cylinder with a radius of 10 cm and a mass of 3.0 kg is rotating about its center with anangular speed of 3.5 rad/s. Wha

t is its kinetic energy?
Physics
1 answer:
Lynna [10]3 years ago
3 0

Answer:

kE=0.0735 J

Explanation:

Given that

Radius ,R=10 cm = 0.1 m

Mass ,m= 3 kg

Angular speed ,ω = 3.5 rad/s

We know that moment of inertia for solid sphere given as

I=\dfrac{2}{5}mR^2

Kinetic energy

KE=\dfrac{1}{2}I\omega^2

Now by putting the values

KE=\dfrac{1}{2}\times \dfrac{2}{5}mR^2\times \omega^2

KE=\dfrac{1}{2}\times \dfrac{2}{5}\times 3\times 0.1^2\times 3.5^2\ J

kE=0.0735 J

Therefore the kinetic energy will be 0.0735 J

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PLS HELP
qwelly [4]

Answer:

B

Explanation:

6 0
3 years ago
A freshly prepared sample of radioactive isotope has an activity of 10 mCi. After 4 hours, its activity is 8 mCi. Find: (a) the
Maurinko [17]

Answer:

(a). The decay constant is 1.55\times10^{-5}\ s^{-1}

The half life is 11.3 hr.

(b). The value of N₀ is 2.38\times10^{11}\ nuclei

(c). The sample's activity is 1.87 mCi.

Explanation:

Given that,

Activity R_{0}=10\ mCi

Time t_{1}=4\ hours

Activity R= 8 mCi

(a). We need to calculate the decay constant

Using formula of activity

R=R_{0}e^{-\lambda t}

\lambda=\dfrac{1}{t}ln(\dfrac{R_{0}}{R})

Put the value into the formula

\lambda=\dfrac{1}{4\times3600}ln(\dfrac{10}{8})

\lambda=0.0000154\ s^{-1}

\lambda=1.55\times10^{-5}\ s^{-1}

We need to calculate the half life

Using formula of half life

T_{\dfrac{1}{2}}=\dfrac{ln(2)}{\lambda}

Put the value into the formula

T_{\dfrac{1}{2}}=\dfrac{ln(2)}{1.55\times10^{-5}}

T_{\dfrac{1}{2}}=44.719\times10^{3}\ s

T_{\dfrac{1}{2}}=11.3\ hr

(b). We need to calculate the value of N₀

Using formula of N_{0}

N_{0}=\dfrac{3.70\times10^{6}}{\lambda}

Put the value into the formula

N_{0}=\dfrac{3.70\times10^{6}}{1.55\times10^{-5}}

N_{0}=2.38\times10^{11}\ nuclei

(c). We need to calculate the sample's activity

Using formula of activity

R=R_{0}e^{-\lambda\times t}

Put the value intyo the formula

R=10e^{-(1.55\times10^{-5}\times30\times3600)}

R=1.87\ mCi

Hence, (a). The decay constant is 1.55\times10^{-5}\ s^{-1}

The half life is 11.3 hr.

(b). The value of N₀ is 2.38\times10^{11}\ nuclei

(c). The sample's activity is 1.87 mCi.

4 0
3 years ago
A perfectly flexible cable has length L, and initially it is at rest with a length Xo of it hanging over the table edge. Neglect
zaharov [31]

Answer:

X=X_o+\dfrac{1}{2}gt^2

Explanation:

Given that

Length = L

At initial over hanging length = Xo

Lets take the length =X after time t

The velocity of length will become V

Now by energy conservation

\dfrac{1}{2}mV^2=mg(X-X_o)

So

V=\sqrt{2g(X-X_o)}

We know that

\dfrac{dX}{dt}=V

\dfrac{dX}{dt}=\sqrt{2g(X-X_o)}

\sqrt{2g}\ dt=(X-X_o)^{-\frac{1}{2}}dX

At t= 0 ,X=Xo

So we can say that

X=X_o+\dfrac{1}{2}gt^2

So the length of cable after time t

X=X_o+\dfrac{1}{2}gt^2

6 0
3 years ago
How is the setting in chapter one significant to a possible theme in chapter one of mice and men?
ra1l [238]
In Steinbecks of nice and men, the setting at the start of chapter one is “ a few miles south of Soledad” Lenny and George have been walking for what seems like four miles. George is aggravated having been dropped if by a truck driver who said they didn’t have far to go
4 0
4 years ago
Which statement about the dark nebulae is true?(A) They can be penetrated only with shorter waves, such as UV and x-ray.(B) They
TiliK225 [7]

Answer: (B) They can be penetrated only with longer wavelengths such as radio and infrared.

Explanation:

A dark nebula is a type of interstellar cloud with such a high density that it "hides" the light of the background stars.

It should be noted that these nebulae are composed mostly of hydrogen (the star's primary fuel), and the absence of light (in the visible range of the electromagnetic spectrum) is caused by the interstellar solid dust that constitutes the cloud, which is why they are also called absorption nebulae.

However, <u>if powerful telescopes that detect at long wavelengths, such as infrared and X-rays, it is possible to observe through these nebulae and even observe the existence of young stars.</u>

8 0
3 years ago
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