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Lostsunrise [7]
3 years ago
5

A 5.0-Ω resistor and a 9.0-Ω resistor are connected in parallel. A 4.0-Ω resistor is then connected in series with this parallel

combination. An ideal 6.0-V battery is then connected across the series-parallel combination of the three resistors.(a) What is the current through the 4.0-Ω resistor?
Physics
1 answer:
lbvjy [14]3 years ago
8 0
The 4ohm resistor is not parallel to any devices, so the current through it is the same as the total current through the circuit.

Apply I = V/R
I = total current, V = battery voltage, R = total resistance

Calculate the total resistance R of the circuit first.

R will be the sum of the 4ohm resistor and the resistance of the 5&9ohm parallel resistors. To find the resistance of the pair, you apply the “sum of inverses” rule:

R = 1/(1/5 + 1/9) = 3.21ohm

Now add the 4ohm resistor:
R = 3.21 + 4 = 7.21ohm

Now let’s calculate I = V/R:
Given values:
V = 6.0V
R = 7.21ohm
Plug in and solve for I:
I = 6.0/7.21

I = 0.83A
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A continuous spectrum contains all the wavelengths 

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pickupchik [31]

Answer:

finding Cepheid variable and measuring their periods.

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5 0
3 years ago
Calculate the critical angle for light going from Glycerine to air.
Georgia [21]
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\theta_c = \arcsin ( \frac{n_2}{n_1} )
where n1 and n2 are the refractive indices of the two mediums. If we susbtitute the refractive index of glycerine and air in the formula, we find the critical angle for this case:
\theta_c = \arcsin ( \frac{1.00}{1.473} )=42.8^{\circ}
6 0
3 years ago
An astronaut finds herself in a predicament in which she has become untethered from her shuttle. She figures that she could get
Blizzard [7]

In order to solve the problem, it is necessary to apply the concepts related to the conservation of momentum, especially when there is an impact or the throwing of an object.

The equation that defines the linear moment is given by

mV_i = (m-m_O)V_f - m_OV_O

where,

m=Total mass

m_O = Mass of Object

V_i = Velocity before throwing

V_f = Final Velocity

V_O = Velocity of Object

Our values are:

m_1=5.3kgm_2=7.9kg\\m_3=10.5kg\\m_A=75kg\\m_{Total}=m=98.7Kg

Solving to find the final speed, after throwing the object we have

V_f=\frac{mV_0+m_TV_O}{m-m_O}

We have three objects. For each object a launch is made so the final mass (denominator) will begin to be subtracted successively. In addition, during each new launch the initial speed will be given for each object thrown again.

That way during each section the equations should be modified depending on the previous one, let's start:

A) 5.3Kg\rightarrow 15m/s

V_{f1}=\frac{mV_0+m_TV_O}{m-m_O}

V_{f1}=\frac{(98.7)*0+5.3*15}{98.7-5.3}

V_{f1}=0.8511m/s

B) 7.9Kg\rightarrow 11.2m/s

V_{f2}=\frac{mV_{f1}+m_TV_O}{m-m_O}

V_{f2}=\frac{(98.7)(0.8511)+(7.9)(11.2)}{98.7-5.3-7.9}

V_{f2} = 2.0173m/s

C) 10.5Kg\rightarrow 7m/s

V_{f3}=\frac{mV_{f2}+m_TV_O}{m-m_O}

V_{f3}=\frac{(98.7)(2.0173)+(10.5)(7)}{98.7-5.3-7.9-10.5}

V_{f3} = 3.63478m/s

Therefore the final velocity of astronaut is 3.63m/s

7 0
3 years ago
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VARVARA [1.3K]
Good morning.

We have that:

\mathsf{V = a\cdot t} , since we have rest in the inicial time.

The acceleration can be found with Newton's Law:

\mathsf{F = m\cdot a\iff a = \dfrac{F}{m}}

Now we put the acceleratin in the velocity equation:

\mathsf{V = \dfrac{F}{m} \cdot t}

We want the force, so, let's isolate F:

\mathsf{V\cdot m = F\cdot t}\\ \\ \\ \boxed{\mathsf{F = \dfrac{V\cdot m}{t}}}

3 0
3 years ago
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