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3 years ago

5. The load across a battery consists of two resistors, with values of 15 Ohm and 45 Ohm connected in series.

Physics

1 answer:

Hunter-Best [27]3 years ago

8 0

a) $60 \Omega$

The total resistance of resistors connected in series is equal to the sum of the individual resistances:

$R=R_1 + R_2 + .... + R_n$

In this problem, we have two resistors connected in series:

$R_1 = 15 \Omega$

$R_2 = 45 \Omega$

So the total resistance of the circuit is

$R=R_1 + R_2 = 15 \Omega + 45 \Omega = 60 \Omega$

b) 6.0 V

The relationship between voltage, current and resistance of a circuit is given by Ohm's law

$V=RI$

where

V is the voltage

R is the total resistance

I is the current

In this circuit, we have

$R=60 \Omega$ is the total resistance

I = 0.1 A is the current

So, the voltage of the battery is

$V=(60 \Omega)(0.1 A)=6.0 V$

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In rural areas, water is often extracted from underground by pumps. Consider an underground water source whose free surface is 6

stealth61 [152]

Answer:

W = 9533.09 Watt

Explanation:

given,

diameter of pipe inlet, d₁ = 10 cm

r₁ = 5 cm

diameter of pipe outlet, d₂ = 15 cm

r₂= 7.5 cm

head upto water level is to rise = 60 + 5

= 65 m

flow rate = 0.015 m³/s

we know

A₁ v₁ = A₂ v₂ = Q

π r₁² v₁ = π r₂² v₂ = 0.015

$v_1= \dfrac{r_2^2}{r_1^2} v_2$

$v_1= \dfrac{7.5^2}{5^2} v_2$

$v_1= 2.25 v_2$

$v_2 = \dfrac{0.015}{\pi r_2^2}$

$v_2 = \dfrac{0.015}{\pi 0.075^2}$

v₂ = 0.848 m/s

v₁ = 1.908 m/s

Applying Bernoulli's equation

$P_p = \dfrac{1}{2}\rho (v_2^2-v_1^2)+ \rho g h$

$P_p= \dfrac{1}{2}\times 1000\times (0.848^2-1.908^2)+ 1000\times 9.8\times 65$

$P_p= 635539.32 Pa$

P_p is the pump pressure

Power of the pump

W = P_p x Q

W = 635539.32 x 0.015

W = 9533.09 Watt

6 0

4 years ago

Physics question 28 plz help me

Alexus [3.1K]

Answer:

a. I = 30 A

b. E = 1080000 J = 1080 KJ

c. ΔT = 12.86°C

d. Cost = $ 4.32

Explanation:

The current in the coil is given by Ohm's Law:

$V = IR\\I = \frac{V}{R}$

where,

I = current = ?

V = Voltage = 120 V

R = Resistance = 4 Ω

Therefore,

$I = \frac{120\ V}{4 \Omega}\\$

I = 30 A

The energy can be calculated as:

$E = VIt\\E = (120\ V)(30\ A)(5\ min)(\frac{60\ s}{1\ min})\\$

E = 1080000 J = 1080 KJ

For the increase in the temperature of water:

$E = mC\Delta T\\$

where,

m = mass of water = 20 kg

C = specific heat of water = 4.2 KJ/kg.°C

Therefore,

$1080\ KJ = (20\ kg)(4.2\ KJ/kg.^oC)\Delta T$

ΔT = 12.86°C

First, we will calculate the total energy consumed:

$E=(Power)(Time)\\E=VI(Time)\\E = (120\ V)(30\ A)(0.5\ h/d)(30\ d)\\E = 54000\ Wh\\E = 54 KWh$

Now, for the cost:

$Cost = (Unit\ Cost)(Energy)\\Cost = (\$ 0.08\KWh)(54\ KWh)$

Cost = $ 4.32

7 0

3 years ago

A bubble, located 0.200 m beneath the surface in a glass of beer, rises to the top. The air pressure at the top is 1.01 ???? 10

SIZIF [17.4K]

Answer:

the ratio of the bubble’s volume at the top to its volume at the bottom is 1.019

Explanation:

given information

h = 0.2 m

$P_{0}$ = 1.01 x $10^{5}$ Pa

$P_{1} V_{1} = P_{2} V_{2}$

$\frac{V_{2} }{V_{1}} = \frac{P_{1} }{P_{2}}$

$P_{1}$ = $P_{0}$ + ρgh, ρ = 1000 kg/ $m^{3}$

$P_{1}$ = 1.01 x $10^{5}$ Pa + (1000 x 9.8 x 0.2) = 1,0296 x $10^{5}$ Pa

$P_{2}$ = $P_{0}$ = $10^{5}$ Pa

thus,

$\frac{V_{2} }{V_{1}} = 1,0296 x [tex]10^{5}$ / $10^{5}$ = 1.019

8 0

3 years ago

A lunar eclipse occurs only when the Moon is new. A lunar eclipse occurs only when the Moon is new. True False

Fittoniya [83]

False. False.

A lunar eclipse occurs only when the moon is full.

A solar eclipse occurs only when the moon is new.

7 0

4 years ago

An LC circuit is built with a 20 mH inductor and an 8.0 PF capacitor. The capacitor voltage has its maximum value of 25 V at t =

Margaret [11]

Answer:

a) the required time is 0.6283 μs

b) the inductor current is 0.5 mA

Explanation:

Given the data in the question;

The capacitor voltage has its maximum value of 25 V at t = 0

i.e V $_m$ = V₀ = 25 V

we determine the angular velocity;

ω = 1 / √( LC )

ω = 1 / √( ( 20 × 10⁻³ H ) × ( 8.0 × 10⁻¹² F) )

ω = 1 / √( 1.6 × 10⁻¹³ )

ω = 1 / 0.0000004

ω = 2.5 × 10⁶ s⁻¹

a) How much time does it take until the capacitor is fully discharged for the first time?

V $_m$ = V₀sin( ωt )

we substitute

25V = 25V × sin( 2.5 × 10⁶ s⁻¹ × t )

divide both sides by 25 V

sin( 2.5 × 10⁶ × t ) = 1

( 2.5 × 10⁶ × t ) = π/2

t = 1.570796 / (2.5 × 10⁶)

t = 0.6283 × 10⁻⁶ s

t = 0.6283 μs

Therefore, the required time is 0.6283 μs

b) What is the inductor current at that time?

$I$ (t) = V₀√(C/L) sin(ωt)

{ sin(ωt) = 1 )

$I$ (t) = V₀√(C/L)

we substitute

$I$ (t) = 25V × √( ( 8.0 × 10⁻¹² F ) / ( 20 × 10⁻³ H ) )

$I$ (t) = 25 × 0.00002

$I$ (t) = 0.0005 A

$I$ (t) = 0.5 mA

Therefore, the inductor current is 0.5 mA

8 0

3 years ago