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lisov135 [29]
3 years ago
5

Find the equivalent resistance, the current supplied by the battery and the current through each resistor when the specified res

istors are connected across a 2-V battery. Part (a) uses two resistors with resistance values that can be set with the animation sliders, and you can use the animation to verify your calculation. In part (b), three resistors are specified.
(a) Two resistors are connected in series across a 2-V battery. Let R1 = 10 ? and R2 = 9 ?.
Req = ?
I = A
?V1 = V
?V2 = V

(b) Add a third resistor to the circuit in series. Let R1 = 10 ?, R2 = 9 ?, and R3 = 3 ?.
Req = ?
I = A
?V1 = V
?V2 = V
?V3 = V
Physics
1 answer:
egoroff_w [7]3 years ago
4 0

Answer with Explanation:

We are given that

Potential difference =2V

a.R_1=10 ohm,R_2=9 ohm

We know that in series

R_{eq}=R_1+R_2

R_{eq}=10+9=19 ohm

I=\frac{V}{R}

I=\frac{2}{19}=0.105 A

V_1=IR_1

V_1=\frac{2}{19}\times 10=1.05 V

V_2=\frac{2}{19}\times 9=0.95 V

b.R_1=10 ohm , R_2=9 ohm ,R_3=3 ohm

R_{eq}=10+9+3=22\Omega

R_{eq}=22\Omega

I=\frac{2}{22}=\frac{1}{11}=0.09 A

V_1=10(0.09)=0.9 V

V_2=9(0.09)=0.81 V

V_3=3(0.09)=0.27 V

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Answer: a highly unpleasant or unhealthy smell or vapor.

Explanation: not really an explanation sorry

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A net force of 25N causes an object to accelerate at 4m/s^2. what is the mass of the object?
Doss [256]

Answer:

6.25 kg

Explanation:

Fnet=ma

m=Fnet/a

m=25/4

m=6.25kg

3 0
3 years ago
Cual es la fuerza electrica sobre el electrón (-1.6 x 10¹⁹c) de un atomo de hidrógeno ejercida por el protón (1.6 x 10¹⁹c)? Supó
kkurt [141]

Answer:

La  fuerza eléctrica es -8.2*10⁻⁸ N

Explanation:

El enunciado correcto es: <em>¿Cuál es la fuerza eléctrica sobre el electrón (-1.6 x 10⁻¹⁹c) de un átomo de hidrógeno ejercida por el protón (1.6 x 10⁻¹⁹c)? Supóngase que la distancia entre el electrón y el protón es de 5.3 x 10⁻¹¹ m</em>

Entre dos o más cargas aparece una fuerza denominada fuerza eléctrica. Su valor depende del valor de las cargas y de la distancia que las separa, mientras que su signo depende del signo de cada carga. Las cargas del mismo signo se repelen entre sí, mientras que las de distinto signo se atraen.

La fuerza eléctrica con la que se atraen o repelen dos cargas puntuales en reposo es directamente proporcional al producto de las mismas e inversamente proporcional al cuadrado de la distancia que las separa:

F=K*\frac{q1*q2}{d^{2} }

donde:

  • F es la fuerza eléctrica de atracción o repulsión. En el Sistema Internacional (S.I.) se mide en Newtons (N).
  • q1 y q2 son lo valores de las dos cargas puntuales. En el S.I. se miden en Culombios (C).
  • d es el valor de la distancia que las separa. En el S.I. se mide en metros (m).
  • K es una constante de proporcionalidad llamada constante de la ley de Coulomb. Depende del medio en el que se encuentren las cargas. Para el vacío K tiene un valor aproximadamente de 9*10⁹ \frac{N*m^{2} }{C^{2} }.

En este caso:

  • F=?
  • K= 9*10⁹ \frac{N*m^{2} }{C^{2} }
  • q1= -1.6*10⁻¹⁹ C
  • q2= 1.6*10⁻¹⁹ C
  • d= 5.3*10⁻¹¹ m

Reemplazando:

F=9*10^{9} \frac{N*m^{2} }{C^{2} }*\frac{(-1.6*10^{19} C)*(1.6*10^{19} C)}{(5.3*10^{-11} )^{2} }

Resolviendo:

F= -8.2*10⁻⁸ N

<u><em>La  fuerza eléctrica es -8.2*10⁻⁸ N</em></u>

6 0
3 years ago
The gas tank is made from A-36 steel and has an inner diameter of 1.50 m. If the tank is designed to withstand a pressure of 5 M
Tanzania [10]

Answer:

a. t = 23mm, b. t = 20mm

Explanation:

Obtain the value of yield strength in tension for A- 36 steel from table ‘Average Mechanical Properties of typical Engineering Materials’, which is σ(y) = 250 MPa.

a.

Assume that the thin wall analysis is valid, Calculate the hoop stress

σ(1) = pd/2t, where p is the pressure in the tank, d is the internal diameter of the tank and t is the thickness.

Substitute 5MPa for p and 1.5m for d

σ(1) = 5 x 10⁶x 1.5/2t

σ(1) = (3.75 x 10⁶)/t

Calculate the longitudinal stress

σ(2) = pd/4t

σ(2) = (5 x 10⁶ x 1.5)/4t

Apply Maximum Shear stress theory which states that failure occurs when the maximum shear stress from a combination of principal stresses <u>equals or exceeds</u> the value obtained for the shear stress at yielding in the uni-axial tensile test. Hence

τ(abs.max) ≤ τ (allowed)

τ (abs.max) ≤ σy /2FS, where FS is the factor of safety

Substitute σ(1)/2 for τ (abs.max) as both the principal stresses have same sign.

σ(1)/2 ≤ σy/2FS

3.75 x 10⁶/2t = 250 x 10⁶/2 x 1.5

T = 0.0225m = 22.5mm = 23mm to the nearest millimeter

Hence the required minimum thickness using the maximum shear stress theory is t = 23mm

b.

Apply maximum distortion energy theorem

σ²(allowed) =σ²(1) – σ(1) x σ(2) + σ²(2)

σ²(y)(allowed)/FS = (3.75 x 10⁶/t)² – (3.75 x 10⁶/t) x (1.875 x 10⁶/t) + (1.875 x 106/t)²

250 x 10⁶/1.5 = (3.2476 x 10⁶)/t

t = 3.2476/166.67

t = 0.0195 m = 19.50 mm = 20mm to the nearest millimeter

Hence, the required minimum thickness using the maximum distortion energy theory is t = 20 mm

7 0
3 years ago
A gas expands at a constant pressure of 3 atm from a volume of 0.02 cubic meters to 0.10 cubic meters. In the process it experie
Nonamiya [84]

a) 2.4\cdot 10^4 J

For a gas transformation occuring at a constant pressure, the work done by the gas is given by

W=p(V_f -V_i)

where

p is the gas pressure

V_f is the final volume of the gas

V_i is the initial volume

For the gas in the problem,

p=3 atm = 3\cdot 1.013\cdot 10^5 Pa = 3.039\cdot 10^5 Pa is the pressure

V_i = 0.02 m^2 is the initial volume

V_f = 0.10 m^3 is the final volume

Substituting,

W=(3.039\cdot 10^5 Pa)(0.10 m^3-0.02m^2)=24312 J = 2.4\cdot 10^4 J

b) 3.24\cdot 10^5 J

The heat absorbed by the gas can be found by using the 1st law of thermodynamics:

\Delta U = Q-W

where

\Delta U is the change in internal energy of the gas

Q is the heat absorbed

W is the work done

Here we have

\Delta U = 3.0\cdot 10^5 J

W=2.4\cdot 10^4 J

So we can solve the equation to find Q:

Q=\Delta U + W = 3.0\cdot 10^5 J +2.4\cdot 10^4 J = 3.24\cdot 10^5 J

And this process is an isobaric process (=at constant pressure).

8 0
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