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netineya [11]
3 years ago
12

Three resistors have resistance of R, 2R, and 3R, are connected to a 12 volt energy source as shown in the circuit diagram. The

current is 2 A. What is the potential difference across the resister with resistance R?
a. 2 volts
b. 4 volts
c. 6 volts
Physics
1 answer:
Kipish [7]3 years ago
7 0

Answer:

2volts

Explanation:

Let us assume the resistor are in series

Equivalent resistance = R + 2R + 3R

Equivalent resistance = 6R

Since different voltage flows through the resistors

V = IR

Given V = 12V

I = 2A

Assume R = 1ohms

V = IR

V = 2 * 1

V = 2volts

Hence the potential difference across the resister with resistance R is 2volts

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Jenny is studying a compound that has two oxygen atoms and one nitrogen atom. Which statement describes the subscripts she will
STALIN [3.7K]

Answer:

She will use one 2 as a subscript.

Explanation:

Nitrogen is a chemical element with the symbol "N" and has an atomic number of 7. Thus, it is found in group (5) of the periodic table and as such it has 5 electrons in its outermost shell. Therefore, nitrogen has two (5) valence electrons.

On the other hand, oxygen has an atomic number of 8 and with the symbol "O."

When nitrogen and oxygen react chemically, they produce a compound known as nitrogen dioxide NO_{2}

In this scenario, Jenny is studying a compound that has two oxygen atoms and one nitrogen atom. Therefore, the statement which describes the subscripts she will use to write the chemical formula is, she will use one 2 as a subscript.

N + O_{2} ------>  NO_{2}

Where: 2 represents the subscript of oxygen.

7 0
3 years ago
Suppose that a star has a spectrum that includes red, blue, and violet lines spaced in the pattern of the lines from hydrogen bu
ladessa [460]

Answer:

It can be concluded that the star is moving away from the observer.

Explanation:

Spectral lines will be shifted to the blue part of the spectrum if the source of the observed light is moving  toward the observer, or to the red part of the spectrum when is moving away from the observer (that is known as the Doppler effect).

The wavelength at rest for this case is 434 nm and 410 nm (\lambda_{0} = 434nm, \lambda_{0} = 410nm)

Redshift: \lambda_{measured}  >  \lambda_{0}

Blueshift: \lambda_{measured}  <  \lambda_{0}

Since, \lambda_{measured} (444nm) is greater than \lambda_{0} (434 nm) and \lambda_{measured} (420nm) is greater than \lambda_{0} (410 nm), it can be concluded that the star is moving away from the observer

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3 years ago
A car initially traveling at 17.1 mph comes to rest in 9.7s what was its acceleration in this time?
ra1l [238]

Answer:

a=-.78m/s^2

Explanation:

Δv=at

  • Δv is the difference in velocity before and after a given time.
  • a is the acceleration of the object during this time.
  • t is time

(v_f-v_i)=at is another way to write this equation.

  • The Δ symbol represents "the difference between the initial and final values of a magnitude or vector", so Δv=(v_f-v_i)

v_f-v_i=at\\\frac{at}{t}=\frac{v_f-v_i}{t}\\a=\frac{v_f-v_i}{t}

  • I rearranged this equation to solve for a, but this is a step that you don't need to take, it's just good to get in the habit of doing this.
  • Plug in the given values. Note that our final velocity is 0, because the car travels until at <em>rest</em>.

a=\frac{v_f-v_i}{t}\\a=\frac{(0)-[(17.1\frac{miles}{hour} )(\frac{hour}{3600s})(\frac{1609.34m}{mile})]}{9.7s}

  • Our initial velocity is in mph, something not in standard units, so if not changed, you will get an incorrect answer. What you need to do is cancel out the units your prior value had using division and multiplication, and at the same time multiply and divide the correct numbers and units into your equation. Or look up a converter.

a=\frac{(0)-[(17.1\frac{miles}{hour} )(\frac{hour}{3600s})(\frac{1609.34m}{mile})]}{9.7s}\\a=\frac{0m/s-7.6m/s}{9.7s} \\a=\frac{-7.6m/s}{9.7s}

  • if you converted correctly, your answer for v_f will be ≅ 7.6m/s.
  • Now divide. Notice that the units for acceleration are m/s^2 or <em>meters per second, per second</em>.

a=\frac{-7.6m/s}{9.7s}\\a=-.78m/s^2

  • Our final answer is <em>negative </em>because the car is <em>slowing down</em>. Do not square this answer as the square symbol only applies to the units, not the magnitude.
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how height does a 2.5 Kg object rises when 255J of work done against gravity A 637.5 B 20.5 C 15m D 10.2 m​
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Answer:

D 10.2

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