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katrin2010 [14]
4 years ago
9

A star has a spectrum with lines indicating the presence of ionized helium. This star should be classified as a(n) _____ type st

ar.
O
A
G
M
Physics
2 answers:
Oduvanchick [21]4 years ago
4 0
It should be o,not 100% sure tho

scoundrel [369]4 years ago
4 0

Answer:

A star has a spectrum with lines indicating the presence of ionized helium. This star should be classified as a(n) <u>O</u> type star.

Explanation:

Helium is a noble has. Its formation and ionization requires high temperature which is possible only in core of stars. Stars classified based on temperatures. Type O stars have very high temperatures (more than 30,000 K). At these temperatures, Helium gets ionized. Thus, the spectrum of O-type stars indicates the presence of ionized helium more than the neutral.

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Intramolecular forces is a strong bond that helps to bond atoms together while intermolecular forces are weak bond that are present between molecules.

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If gas costs $3.65 per gallon at a local service station, how many cubic feet of gas can a customer buy with $40 ?
svet-max [94.6K]
Hi, thank you for posting your question here at Brainly. 

For consistency you must convert gallons to cubic foot. The conversions are:

7.481 gal = 1 ft3

Then, $3.65 per gallon becomes $27.3 per ft3. If you buy $40 worth of gas, you get $40/$27.3 = 1.46 ft3 of gas. 

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3 0
3 years ago
The combination of an applied force and a friction force produces a constant total torque of 35.5 N · m on a wheel rotating abou
Cerrena [4.2K]

Answer:

a) I = 19.799\,kg\cdot m^{2}, b) T = -3.405\,N\cdot m, c) n_{T} \approx 54.842\,rev

Explanation:

a) The net torque is:

T = I\cdot \alpha

Let assume a constant angular acceleration, which is:

\alpha = \frac{\omega-\omega_{o}}{t}

\alpha = \frac{10.4\,\frac{rad}{s} - 0\,\frac{rad}{s} }{5.80\,s}

\alpha = 1.793\,\frac{rad}{s^{2}}

The moment of inertia of the wheel is:

I = \frac{T}{\alpha}

I = \frac{35.5\,N\cdot m}{1.793\,\frac{rad}{s^{2}} }

I = 19.799\,kg\cdot m^{2}

b) The deceleration of the wheel is due to the friction force. The deceleration is:

\alpha = \frac{\omega-\omega_{o}}{t}

\alpha = \frac{0\,\frac{rad}{s} - 10.4\,\frac{rad}{s}}{60.4\,s}

\alpha = - 0.172\,\frac{rad}{s^{2}}

The magnitude of the torque due to friction:

T = (19.799\,kg\cdot m^{2})\cdot (-0.172\,\frac{rad}{s^{2}} )

T = -3.405\,N\cdot m

c) The total angular displacement is:

\theta_{T} = \theta_{1} + \theta_{2}

\theta_{T} = \frac{(10.4\,\frac{rad}{s} )^{2}-(0\,\frac{rad}{s} )^{2}}{2\cdot (1.793\,\frac{rad}{s^{2}} )} + \frac{(0\,\frac{rad}{s} )^{2}-(10.4\,\frac{rad}{s} )^{2}}{2\cdot (-0.172\,\frac{rad}{s^{2}} )}

\theta_{T} = 344.580\,rad

The total number of revolutions of the wheel is:

n_{T} = \frac{\theta_{T}}{2\pi}

n_{T} = \frac{344.580\,rad}{2\pi}

n_{T} \approx 54.842\,rev

5 0
3 years ago
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