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wariber [46]
4 years ago
11

The period-luminosity relation is critical in finding distances with

Physics
1 answer:
muminat4 years ago
5 0

D hopefully this helps

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4application of energy from water
RSB [31]
<span><span>Fuel Extraction and Production – Water is a critical resource for the drilling and mining of natural gas, coal, oil, and uranium. In many cases, fuel extraction also produces wastewater, as with natural gas and oil wells and coal slurry ponds.
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Fuel Refining and Processing – Oil, uranium, and natural gas all require refining before they can be used as fuels – a process that uses substantial amounts of water.
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Fuel Transportation – Water is used to transport coal through slurries — pipelines of finely ground coal mixed with water — and to test energy pipelines for leaks.[1]</span><span>Emissions Control – Many thermoelectric power plants emit sulfur, mercury, particulates, carbon dioxide, and other pollutants, and require pollution control technologies. These technologies also require significant amounts of water to operate.</span></span>
5 0
3 years ago
A centrifuge rotor rotating at 10,000 rpm is shut off and is eventually brought to rest by a frictional force of 1.20m n. if the
Pani-rosa [81]
<span>Answer: The moments of inertia are listed on p. 223, and a uniform cylinder through its center is: I = 1/2mr2 so I = 1/2(4.80 kg)(.0710 m)2 = 0.0120984 kgm2 Since there is a frictional torque of 1.20 Nm, we can use the angular equivalent of F = ma to find the angular deceleration: t = Ia -1.20 Nm = (0.0120984 kgm2)a a = -99.19 rad/s/s Now we have a kinematics question to solve: wo = (10,000 Revolutions/Minute)(2p radians/revolution)(1 minute/60 sec) = 1047.2 rad/s w = 0 a = -99.19 rad/s/s Let's find the time first: w = wo + at : wo = 1047.2 rad/s; w = 0 rad/s; a = -99.19 rad/s/s t = 10.558 s = 10.6 s And the displacement (Angular) Now the formula I want to use is only in the formula packet in its linear form, but it works just as well in angular form s = (u+v)t/2 Which is q = (wo+w)t/2 : wo = 1047.2 rad/s; w = 0 rad/s; t = 10.558 s q = (125.7 rad/s+418.9 rad/s)(3.5 s)/2 = 952.9 radians But the problem wanted revolutions, so let's change the units: q = (5528.075087 radians)(revolution/2p radians) = 880. revolutions</span>
6 0
3 years ago
A wire that is 0.86 meters long is moved perpendicularly through a constant magnetic field of strength 0.035 newtons/amp·meter a
avanturin [10]
<span>The answer to your question is choice: D</span>
8 0
3 years ago
Read 2 more answers
A student releases a marble from the top of a ramp. The marble increases
Fed [463]

Answer:

Vf = 69.56 cm/s

Explanation:

In order to find the final speed of the ramp, we will use the equations of motion. First we use second equation of motion to find out the acceleration of marble:

s = Vi t + (1/2)at²

where,

s = distance traveled = 160 cm

Vi = Initial Speed = 0 cm/s (since, marble starts from rest)

t = time interval = 4.6 s

a = acceleration = ?

Therefore,

160 cm = (0 cm/s)(4.6 s) + (1/2)(a)(4.6 s)²

a = (320 cm)/(4.6 s)²

a = 15.12 cm/s²

Now, we use first equation of motion:

Vf = Vi + at

Vf = 0 cm/s + (15.12 cm/s²)(4.6 s)

<u>Vf = 69.56 cm/s</u>

7 0
3 years ago
A circuit consists of a series combination of 6.50 −kΩ and 4.50 −kΩ resistors connected across a 50.0-V battery having negligibl
Vlada [557]

Answer:

Part A: 16.1 V

Part B: 20.5 V

Part C: 21.5%

Explanation:

The voltmeter is in parallel with the 4.5-kΩ resistor and the combination is in series with the 6.5-kΩ resistor. The equivalent resistance of the parallel combination is given as

\dfrac{1}{R_E}=\dfrac{1}{4.50}+\dfrac{1}{10.0}

R_E=\dfrac{4.50\times10.0}{4.50+10.0} = 3.10

Part A

The voltmeter reading is the potential difference across the parallel combination. This is found by using the voltage-divider rule.

V_1 = \dfrac{3.10}{3.10+6.50}\times50.0 = \dfrac{3.10}{9.60}\times50.0 = 16.1 \text{ V}

Part B

Without the voltmeter, the potential difference across the 4.5-kΩ resistor is found using the same rule as above:

V_2 = \dfrac{4.50}{4.50+6.50}\times50.0 = \dfrac{4.50}{11.0}\times50.0 = 20.5 \text{ V}

Part C

The error in % is given by

\dfrac{20.5-16.1}{20.5}\times100\% = \dfrac{4.4}{20.5}\times100\% = 21.5\%

4 0
3 years ago
Read 2 more answers
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