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3 years ago

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Two resistors connected in series (R1, R2) are connected to two resistors that are connected in parallel (R3,R4). The series-par

allel combination is connected to a battery. Each resistor has a resistance of 10.00 Ohms. The wires connecting the resistors and battery have negligible resistance. A current of 2.00 A runs through resistor R1. What is the voltage supplied by the voltage source?

1 answer:

Debora [2.8K]3 years ago

6 0

Explanation:

R1=R2=R3=R4=10 ohms

V=IR

V=2.10

V=20v

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Which statement best explains the difference between greenhouse gases and other atmospheric gases?(1 point)

Dmitriy789 [7]

Greenhouse gases trap thermal energy and reflect the sun’s harmful radiation back to Earth is the answer

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2 years ago

You have a 3.00-liter container filled with N₂ at 25°C and 4.45 atm pressure connected to a 2.00-liter container filled with Ar

LuckyWell [14K]

Answer : The final pressure in the two containers is, 2.62 atm

Explanation :

Boyle's Law : It is defined as the pressure of the gas is inversely proportional to the volume of the gas at constant temperature and number of moles.

$P\propto \frac{1}{V}$

Thus, the expression for final pressure in the two containers will be:

$PV=P_1V_1+P_2V_2$

$P=\frac{P_1V_1+P_2V_2}{V}$

where,

$P_1$ = pressure of N₂ gas = 4.45 atm

$P_2$ = pressure of Ar gas = 2.75 atm

$V_1$ = volume of N₂ gas = 3.00 L

$V_2$ = volume of Ar gas = 2.00 L

P = final pressure of gas = ?

V = final volume of gas = (4.45 + 2.75) L = 7.2 L

Now put all the given values in the above equation, we get:

$P=\frac{(4.45atm)\times (3.00L)+(2.75atm)\times (2.00L)}{7.2L}$

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8 0

2 years ago

True [87]

Answer:

$\Delta T=3.615^{\circ}C$ is the drop in the water temperature.

Explanation:

Given:

mass of ice, $m_i=14.7\ g=0.0147\ kg$

mass of water, $m_w=324\ g=0.324\ kg$

Assuming the initial temperature of the ice to be 0° C.

<u>Apply the conservation of energy:</u>

Heat absorbed by the ice for melting is equal to the heat lost from water to melt ice.

<u>Now from the heat equation:</u>

$Q_i=Q_w$

$m_i.L=m_w.c_w.\Delta T$ ......................(1)

where:

$L=$ latent heat of fusion of ice $=333.55\ J.g^{-1}$

$c_w=$ specific heat of water $=4.186\ J.g^{-1}.^{\circ}C^{-1}$

$\Delta T=$ change in temperature

Putting values in eq. (1):

$14.7 \times 333.55=324\times 4.186\times \Delta T$

$\Delta T=3.615^{\circ}C$ is the drop in the water temperature.

8 0

2 years ago

A series circuit has a capacitor of 0.25 × 10⁻⁶ F, a resistor of 5 × 10³ Ω, and an inductor of 1H. The initial charge on the cap

viktelen [127]

Answer:

$q = (3 + e^{-4000 t} - 4 e^{-1000 t})\times 10^{-6}$

at t = 0.001 we have

$q = 1.55 \times 10^{-6} C$

at t = 0.01

$q = 2.99 \times 10^{-6} C$

at t = infinity

$q = 3 \times 10^{-6} C$

Explanation:

As we know that they are in series so the voltage across all three will be sum of all individual voltages

so it is given as

$V_r + V_L + V_c = V_{net}$

now we will have

$iR + L\frac{di}{dt} + \frac{q}{C} = 12 V$

now we have

$1\frac{d^2q}{dt^2} + (5 \times 10^3) \frac{dq}{dt} + \frac{q}{0.25 \times 10^{-6}} = 12$

So we will have

$q = 3\times 10^{-6} + c_1 e^{-4000 t} + c_2 e^{-1000 t}$

at t = 0 we have

q = 0

$0 = 3\times 10^{-6} + c_1 + c_2$

also we know that

at t = 0 i = 0

$0 = -4000 c_1 - 1000c_2$

$c_2 = -4c_1$

$c_1 = 1 \times 10^{-6}$

$c_2 = -4 \times 10^{-6}$

so we have

$q = (3 + e^{-4000 t} - 4 e^{-1000 t})\times 10^{-6}$

at t = 0.001 we have

$q = 1.55 \times 10^{-6} C$

at t = 0.01

$q = 2.99 \times 10^{-6} C$

at t = infinity

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5 0

2 years ago

A driver in a 2000 kg Porsche wishes to pass to pass a slow-moving school bus on a four-lane road. What is the average power in

Kryger [21]

The average power is $3.0\cdot 10^6 W$

Explanation:

First of all, we calculate the work done to accelerate the car; according to the work-energy theorem, the work done is equal to the change in kinetic energy of the car:

$W=K_f -K_i= \frac{1}{2}mv^2-\frac{1}{2}mu^2$

where :

$K_f = \frac{1}{2}mv^2$ is the final kinetic energy of the car, with

m = 2000 kg is the mass of the car

v = 60 m/s is the final speed of the car

$K_i = \frac{1}{2}mu^2$ is the initial kinetic energy of the car, with

u = 30 m/s is initial speed of the car

Soolving:

$W=\frac{1}{2}(2000)(60)^2 - \frac{1}{2}(2000)(30)^2=2.7\cdot 10^6 J$

Now we can find the power required for the acceleration, which is given by

$P=\frac{W}{t}$

where

t = 9 s is the time elapsed

Solving:

$P=\frac{2.7\cdot 10^6}{9}=3.0\cdot 10^6 W$

Learn more about power:

brainly.com/question/7956557

#LearnwithBrainly

8 0

3 years ago