A billiard ball moving at 5 m/s strikes a stationary ball of the same mass. After the collision, the original ball moves at a velocity of 4.35 m/s at an angle of 30 o below its original motion. Find the velocity and angle of the second ball after the collision.

A) 1.25 m/s at 31.2o
B) 1.44 m/s at 60.0o
C) 2.16 m/s at 30.0o
D) 2.47 m/s at 61.9o

Answers

Answer 1

90 degrees - 30 = 60 degrees

Velocity = (5m/s - 4.35m/s x cos(30)) / cos(60)

Velocity = 2.47 m/s

The answer is D) 2.47 m/s at 61.9 degrees

Answer 2

The velocity and angle of the second ball after the collision are (A.) 1.25 m/s at  31.2° below the horizontal. Option A

How to calculate velocity

Use the conservation of momentum and conservation of kinetic energy to solve this problem.

Let's denote the velocity and angle of the second ball after the collision as v₂ and θ₂, respectively.

Thus:

Conservation of momentum: [tex]m_1v_1 = m_1v_1'cos(30^o) + m_2v_2cos(\theta_2)[/tex]

Conservation of kinetic energy: [tex](1/2)m_1v_1^2 = (1/2)m_1v_1'^2 + (1/2)m_2v_2^2[/tex]

where m₁ and m₂ are the masses of the first and second balls, respectively.

Since the masses and initial velocity are the same, we can simplify the equations to:

m₁v₁ = m₁v₁'cos(30°) + m₂v₂cos(θ₂)

[tex]v_1^2 = v_1'^2 + v_2^2[/tex]

Substitute in the given values

[tex](1)(5) = (1)(4.35)cos(30^o) + (1)(v_2)cos(\theta_2)\\5^2 = 4.35^2 + v2^2\\v_2 = 1.25 m/s\\\theta2 = 31.2^o[/tex]

Therefore, the velocity and angle of the second ball after the collision are approximately 1.25 m/s at an angle of 31.2° below the horizontal.

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Related Questions

A technician builds an RLC series circuit which includes an AC source that operates at a fixed frequency and voltage. At the operating frequency, the resistance R is equal to the inductive reactance XL. The technician notices that when the plate separation of the parallel-plate capacitor is reduced to one-half its original value, the current in the circuit doubles. Determine the initial capacitive reactance in terms of the resistance R.

Answers

Answer:

Xc = (0.467 - 0.427j)R

Explanation:

Since the resistance in the circuit is R, the reactance of the inductor is XL and the reactance of the capacitor is XC, then the impedance of the circuit is

Z = √[R² + (XL - XC)²]

Since the inductive reactance XL equals the resistance R, we have that

Z = √[R² + (XL - XC)²]

Z = √[R² + (R - XC)²]

Thus, the current in the circuit is thus I = V/Z = V/√[R² + (R - XC)²]

Now, when the plate separation of the parallel plate capacitor is reduced to one-half its original value, the current doubles. Also, when the plate separation is reduced to half, the capacitance doubles since C ∝ 1/d where C is capacitance and d separation between the plates. Since the capacitance doubles, the new reactance XC' is twice the initial reactance XC. So, XC' = 2XC. Thus the new impedance is thus

Z' = √[R² + (R - XC')²]

Z' = √[R² + (R - 2XC)²]

The new current is I' = V/Z' = V/√[R² + (R - 2XC)²]

Since the current doubles, I' = 2I.

V/√[R² + (R - 2XC)²] = 2V/√[R² + (R - XC)²]

1/√[R² + (R - 2XC)²] = 2/√[R² + (R - XC)²]

√[R² + (R - XC)²] = 2√[R² + (R - 2XC)²]

squaring both sides, we have

[R² + (R - XC)²] = 4[R² + (R - 2XC)²]

expanding the brackets, we have

[R² + R² - 2RXC + XC²] = 4[R² + R² - 4RXC + 4XC²]

[2R² - 2RXC + XC²] = 4[2R² - 4RXC + 4XC²]

2R² - 2RXC + XC² = 8R² - 16RXC + 16XC²

collecting like terms, we have

16RXC - 2RXC + XC² - 16XC² = 8R² - 2R²

14RXC - 15XC² = 6R²

15XC² - 14RXC + 6R² = 0

Using the quadratic formula to find XC, we have

[tex]XC = \frac{-(-14R) +/- \sqrt{(-14R)^{2} - 4 X 15 X 6R^{2} } }{2 X 15}\\= \frac{-(-14R) +/- \sqrt{196R^{2} - 360R^{2} } }{30}\\ \\= \frac{14R +/- \sqrt{- 164R^{2} } }{30}\\ \\= \frac{14R +/- 12.81Ri }{30}\\\\= 0.467R +/- 0.427Ri[/tex]

Since it is capacitive, we take the negative part.

So, Xc = (0.467 - 0.427j)R

80 grams of iron at 100°C is dropped into 200 of water at 20°C contained in an iron vessel of mass 50 gram find the resulting temperature.​

Answers

Answer:

the resulting temperature is 23.37 ⁰C

Explanation:

Given;

mass of the iron, m₁ = 80 g = 0.08 kg

mass of the water, m₂ = 200 g = 0.2 kg

mass of the iron vessel, m₃ = 50 g = 0.05 kg

initial temperature of the iron, t₁ = 100 ⁰C

initial temperature of the water, t₂ = 20 ⁰C

specific heat capacity of iron, c₁ = 462 J/kg⁰C

specific heat capacity of water, c₂ = 4,200 J/kg⁰C

let the temperature of the resulting mixture = T

Apply the principle of conservation of energy;

heat lost by the hot iron = heat gained by the water

[tex]m_1c_1 \Delta t_1 = m_2c_2\Delta t_2\\\\m_1c_1 (100 - T) = m_2c_2 (T- 20)\\\\0.08 \times 462 (100-T) = 0.2 \times 4,200 (T-20)\\\\36.96 (100-T) = 840 (T-20) \\\\100 - T = 22.72 (T-20)\\\\100-T = 22.72 T - 454.4 \\\\554.4 = 23.72T\\\\T = \frac{554.4}{23.72} \\\\T = 23.37 \ ^0C[/tex]

Therefore, the resulting temperature is 23.37 ⁰C

what is the prefix notation of 0.0000738?​

Answers

Answer:

7.38 × 10-5

Explanation:

All numbers in scientific notation or standard form are written in the form m × 10n, where m is a number between 1 and 10 ( 1 ≤ |m| < 10 ) and the exponent n is a positive or negative integer.

To convert 0.0000738 into scientific notation, follow these steps:

Move the decimal 5 times to the right in the number so that the resulting number, m = 7.38, is greater than or equal to 1 but less than 10

Since we moved the decimal to the right the exponent n is negative

n = -5

Write in the scientific notation form, m × 10n

= 7.38 × 10-5

Therefore, the decimal number 0.0000738 written in scientific notation is 7.38 × 10-5 and it has 3 significant figures.

Answer = 7.38 × 10-5

The above answer is correct

An electric field E⃗ =5.00×105ı^N/C causes the point charge in the figure to hang at an angle. What is θ?

Answers

We have that the angle is

[tex]\theta=32.53[/tex]

From the Question we are told that

E⃗ =5.00×105ı^N/C

Generally the equation for Tension   is mathematically given

[tex]W=Tcos\theta[/tex]

Where

[tex]tan\theta=\frac{2.5*10^{-9}(5*10{5})}{2*10^{-3}(9.8)}[/tex]

[tex]\theta=32.53[/tex]

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A solid object is made of two materials, one material having density of 2 000 kg/m3 and the other having density of 6 000 kg/m3. If the object contains equal masses of the materials, what is its average density

Answers

Answer:

[tex]\rho_{avg}=4000kg/m^3[/tex]

Explanation:

From the question we are told that:

Density of Material 1 [tex]\rho_1=2000kg/m^3[/tex]

Density of Material 2 [tex]\rho_2=6000kg/m^3[/tex]

Generally the equation for Average density is mathematically given by

[tex]\rho_{avg}=frac{\rho _1+rho _2}{2}[/tex]

[tex]\rho_{avg}=\frac{2000+6000}{2}[/tex]

[tex]\rho_{avg}=4000kg/m^3[/tex]

In a large chemical factory, a feed pipe carries a liquid at a speed of 5.5 m/s. A pump pushes the liquid along at a gauge pressure of 140,000 Pa. The liquid travels upward 6.0 m and enters a tank at a gauge pressure of 2,000 Pa. The diameter of the pipe remains constant. At what speed does the liquid enter the tank

Answers

Answer:

v₂ = 15.24 m / s

Explanation:

This is an exercise in fluid mechanics

Let's write Bernoulli's equation, where the subscript 1 is for the factory pipe and the subscript 2 is for the tank.

          P₁ + ½ ρ v₁² + ρ g y₁ = P₂ + ½ ρ v₂² + ρ g y₂

They indicate the pressure in the factory P₁ = 140000 Pa, the velocity

v₁ = 5.5 m / s and the initial height is zero y₁ = 0

the tank is at a pressure of P2 = 2000 Pa and a height of y₂ = 6.0 m

          P₁ -P₂ + ρ g (y₁ -y₂) + ½ ρ v₁² = ½ ρ v₂²

           

let's calculate

         140,000 - 2000 + ρ 9.8 (0- 6) + ½ ρ 5.5² = ½ ρ v₂²

         138000 - ρ 58.8 + ρ 15.125 = ½ ρ v2²

         v₂² = 2 (138000 /ρ - 58.8 + 15.125)

         v₂ = [tex]\sqrt{\frac{276000}{\rho } - 43.675 }[/tex]

In the exercise they do not indicate what type of liquid is being used, suppose it is water with

           ρ = 1000 kg / m³

           v₂ = [tex]\sqrt{\frac{276000}{1000} - 43.675}[/tex]

           v₂ = 15.24 m / s

A 12.0 g sample of gas occupies 19.2 L at STP. what is the of moles and molecular weight of this gas?​

Answers

At STP, 1 mole of an ideal gas occupies a volume of about 22.4 L. So if n is the number of moles of this gas, then

n / (19.2 L) = (1 mole) / (22.4 L)   ==>   n = (19.2 L•mole) / (22.4 L) ≈ 0.857 mol

If the sample has a mass of 12.0 g, then its molecular weight is

(12.0 g) / n14.0 g/mol

monochromatic light of wavelength 500 nm is incident normally on a diffraction grating. if the third order maximum is 32. how many total number of maximuima can be seen

Answers

Answer:

The total number of maxima that can be seen is 11

Explanation:

Given the data in the question

wavelength λ = 500 nm = 5 × 10⁻⁷ m

if the third order maximum is 32

i.e m = 3 and θ = 32°

Now, we know that condition for diffraction maximum is as follows;

d × sinθ = m ×  λ

so we substitute in our given values

d × sin( 32° ) = 3 ×  5 × 10⁻⁷ m

d × sin( 32° ) = 1.5  × 10⁻⁶ m

d = [ 1.5  × 10⁻⁶ m ] / sin( 32° )

d = 2.83 × 10⁻⁶ m

Now, maxima n when θ = 90° will be;

sin( 90° ) = nλ / d

1 =  nλ / d

d =  nλ

n = d / λ

we substitute

n = [ 2.83 × 10⁻⁶ m ]  / [ 5 × 10⁻⁷ m ]

n = 5.66

so 5 is the max value

hence, total maxima value is;

⇒ 2n + 1 = 2( 5 ) + 1 = 10 + 1 = 11

Therefore, total number of maxima that can be seen is 11

On the sonometer shown below, a horizontal cord of length 5 m has a mass of 1.45 g. When the cord was plucked the wave produced had a frequency of 120 Hz and wavelength of 6 cm. (a) What was the tension in the cord? (b) How large a mass M must be hung from its end to give it this tension? ​

Answers

Answer:

(a) T = 0.015 N

(b) M = 1.53 x 10⁻³ kg = 1.53 g

Explanation:

(a) T = 0.015 N

First, we will find the speed of waves:

[tex]v =f\lambda[/tex]

where,

v = speed of wave = ?

f = frequency = 120 Hz

λ = wavelength = 6 cm = 0.06 m

Therefore,

v = (120 Hz)(0.06 m)

v = 7.2 m/s

Now, we will find the linear mass density of the coil:

[tex]\mu = \frac{m}{l}[/tex]

where,

μ = linear mass density = ?

m = mass = 1.45 g = 1.45 x 10⁻³ kg

l = length = 5 m

Thereforre,

[tex]\mu = \frac{1.45\ x\ 10^{-3}\ kg}{5\ m}\\\\\mu = 2.9\ x\ 10^{-4}\ kg/m[/tex]

Now, for the tension we use the formula:

[tex]v = \sqrt{\frac{T}{\mu}}\\\\7.2\ m/s = \sqrt{\frac{T}{2.9\ x\ 10^{-4}\ kg/m}}\\\\(51.84\ m^2/s^2)(2.9\ x\ 10^{-4}\ kg/m) = T[/tex]

T = 0.015 N

(b)

The mass to be hung is:

[tex]T = Mg\\\\M = \frac{T}{g}\\\\M = \frac{0.015\ N}{9.8\ m/s^2}\\\\[/tex]

M = 1.53 x 10⁻³ kg = 1.53 g

An electric field of 234,000 N/C points due west at a certain spot. What are the magnitude and direction of the force that acts on a charge of -7.25 µC at this spot?

Answers

Answer:

F = 1.69 N

Explanation:

Given that,

Electric field, E = 234,000 N/C

Charge, Q = -7.25 µC

We need to find the electric force acting on the charge. It can be given as follows :

[tex]F=qE\\\\F=7.25\times 10^{-6}\times 234000\\\\F=1.69\ N[/tex]

As the charge is negative, the force will act in the opposite direction of electric field. Hence, the electric force is 1.69 N.

A cardiac pacemaker can be affected by a static magnetic field as small as 1.7 mT. How close can a pacemaker wearer come to a long, straight wire carrying 25 A

Answers

Answer:

The distance is 2.94 mm.

Explanation:

Magnetic field, B = 1.7 mT

Current, I = 25 A

Let the distance is d.

The magnetic field is given by

[tex]B = \frac{\mu o}{4\pi}\times \frac{2I}{r}\\\\1.7\times 10^{-3} = 10^{-7}\times \frac{2\times 25}{r}\\\\r = 2.94\times 10^{-3} m \\\\ r = 2.94 mm[/tex]

Consider a neutron star with a mass equal to the sun, a radius of 19 km, and a rotation period of 1.0 s. What is the speed of a point on the equator of the star

Answers

Answer:

120 km/s

Explanation:

Given data :

Radius of the star is r = 19 km

Rotational time period of the star is T = 1 s

Therefore, we know that the velocity of the star is given by :

[tex]$V=\frac{2\pi r}{T}$[/tex]

[tex]$V=\frac{2 \times 3.14 \times 19\times 10^3}{1}$[/tex]

V = 119380.52 m/s

Therefore, the velocity of the point on the equator of the star is = 120 km/s

At a distance of 14,000 km from the center of Planet Z-99, the acceleration due to gravity is 32 m/s2. What is the acceleration due to gravity at a point 28,000 km from the center of this planet

Answers

A body of mass m feels a gravitational force due to the planet of

F = GmM/R ² = ma

where

• G = 6.67 × 10⁻¹¹ N•m²/kg² is the universal gravitational constant

• M is the mass of the planet

• R is the distance between the body and the planet's center

• a is the acceleration due to gravity

Solving for a gives

a = GM/R ²

Notice that 28,000 km is twice 14,000 km. The equation says that the acceleration varies inversely with the square of the distance. So if R is changed to 2R, we have a new acceleration of

GM/(2R)² = 1/4 × GM/R ² = a/4

so the acceleration of the body at 28,000 km from the planet's center would be (32 m/s²)/4 = 8 m/s².

dujevduxjehhsusheheh
m=100g
F-?​

Answers

Answer:

Force = mass × acceleration

[tex]F =(100 \times 1000) \times 10 \\ = 1 \times {10}^{6} \: newtons[/tex]

A negative slope on the velocity vs. time graph indicates a negative
acceleration.
A True
B. False

Answers

A. true

means decelerating

found in brainly itself

shirleywashington

Ambitious

2.3K answers

14M people helped

Explanation :

We know that the slope of velocity -time graph gives the acceleration. Acceleration of an object is defined as the rate of change of velocity i.e.

Suppose a driver suddenly applies brakes. In this case the initial velocity of his or her vehicle is more and the final velocity is less.

So, the acceleration is negative in this case i.e. the object is decelerating.

A negative slope on the velocity versus time graph indicates that an object is not accelerating. This statement is false as the object is decelerating.

A penny of mass 3.10 g rests on a small 20.0 g block supported by a spinning disk with radius of 12.0 cm. The coefficients of friction between block and disk are 0.850 (static) and 0.575 (kinetic) while those for the penny and block are 0.395 (kinetic) and 0.495 (static). What is the maximum rate of rotation in revolutions per minute that the disk can have, without the block or penny sliding on the disk

Answers

Answer:

do this Q yourself because i havent read the chapter

The maximum rate of rotation in revolutions per minute that the disk can have, without the block or penny sliding on the disk is 63 rpm.

How to solve

This is calculated using the coefficient of static friction between the penny and block, which is 0.495.

The maximum angular velocity of the disk is when the force of static friction is just sufficient to prevent the penny from sliding.

This force is equal to the mass of the penny multiplied by the acceleration due to gravity, multiplied by the coefficient of static friction.

The angular velocity of the disk is then calculated from this force and the radius of the disk.

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What are the major sources of energy utilized during a 100 meter race, a 1000 meter race, and a marathon

Answers

Answer:

The energy from food and then from plants and then from sun.

As sun is the ultimate source of energy.

Explanation:

Distance = 100 m, 1000m, marathon

As the distance is covered by the person, so the muscular energy is used and thus the energy comes form out food.

As we know that the energy can neither be created nor be destroyed it can transform from one form to another.

So, the energy form the food which we consume is converted into the kinetic energy as we run.

Consider a wave along the length of a stretched slinky toy, where the distance between coils increases and decreases. What type of wave is this

Answers

"Longitudinal wave" is the wave where the difference between the coils increases as well as decreases.

Generating waves whenever the form of communication being displaced in a similar direction as well as in the reverse way of the wave's designated points, could be determined as Longitudinal waves.A wave running the length of something like a Slinky stuffed animal, which expands as well as reduces the spacing across spindles, produces a fine image or graphic.

Thus the above answer is correct.

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A bicycle wheel has a diameter of 63.4 cm and a mass of 1.86 kg. Assume that the wheel is a hoop with all of the mass concentrated on the outside radius. The bicycle is placed on a stationary stand and a resistive force of 123 N is applied tangent to the rim of the tire. What force is required if you shift to a 5.60-cm-diameter sprocket?

Answers

Answer:

Njfjrhrjrkrirkehrbrhrrhrhehrhrhejejebrbrhrbrbbbrhje

What is the result of (305.120 + 267.443) x 0.50? How many answers can be written based on the principle of significant digits?

Answers

Answer:

The answer is 286.2815.

A spaceship travels 360km in one hour. Express its speed in m/s

Answers

Answer:

Spaceship speed is 36000 km/h

So, in 1 hour spaceship travel 36000 km

Or we can say that in 60×60 second spaceship travel 36000 km

Therefore in 1 sec spaceship travel

=

= 10 km/s

Answer:

Explanation:

360 km/hr(1000 m/km) / 3600 s/hr) = 100 m/s

A positive statement is:________. a. reflects oneâs opinions. b. can be shown to be correct or incorrect. c. a value judgment. d. based upon an optimistic judgment.

Answers

Answer:

b

Explanation:

A dog statue is standing in the bed of a pickup truck. The bed is coated with ice, causing the force of friction between the statue and the truck to be zero. The truck is initially at rest, and then accelerates to the right, moving along a flat road. As seen from a stationary observer (watching the truck move to the right), the dog statue Group of answer choices moves to the right, but not as quickly as the truck is moving to the right, causing it to slide towards the back of the truck. does not move left or right, but the back of the truck moves towards the statue. moves to the left, as the truck moves to the right, causing the statue to slide towards the back of the truck. moves to the right at the same rate as the truck, so it doesn't slide.

Answers

Answer:

Moves towards left

Explanation:

When the truck is moving towards right then there is pseudo force acting on the fog statue which is acting left wards.

A seen from the stationary observer, the dog statue moves towards left.

Suppose 2.10 C of positive charge is distributed evenly throughout a sphere of 1.30-cm radius. 1) What is the charge per unit volume for this situation

Answers

Answer:

[tex]\rho=2.28\times 10^5\ C/m^3[/tex]

Explanation:

Given that,

Charge, Q = 2.1 C

The radius of sphere, r = 1.3 cm = 0.013 m

We need to find the charge per unit volume for this situation. It can be calculated a follows:

[tex]\rho=\dfrac{Q}{\dfrac{4}{3}\pi r^3}\\\\\rho=\dfrac{2.1}{\dfrac{4}{3}\pi \times (0.013)^3}\\\\\rho=2.28\times 10^5\ C/m^3[/tex]

So, the charge per unit volume is [tex]2.28\times 10^5\ C/m^3[/tex].

the product 17.10 ✕

Answers

Explanation:

pls write the full question

if a voltage v(t) = 150sinwt in volts at an operating frequency of 100hz. determine the I. dc value ii.rms value iii.period of the voltage signal​

Answers

Answer:

(a) The dc value is 95.5 volts

(b) The rms value is 106.1 volts

(c) The period is 0.01s

Explanation:

Given

[tex]V(t) = 150\sin(wt)[/tex]

[tex]f = 100Hz[/tex]

Solving (a): The dc value

[tex]V(t) = 150\sin(wt)[/tex] implies that

[tex]V_{max} = 150[/tex]

So, the dc value is:

[tex]V = \frac{2}{\pi} * V_{max}[/tex]

[tex]V = \frac{2}{\pi} * 150[/tex]

[tex]V = \frac{300}{\pi}[/tex]

[tex]V = 95.5V[/tex]

Solving (b): The RMS value

This is calculated as:

[tex]V_{rms} = \frac{1}{\sqrt 2} * V_{max}[/tex]

So, we have:

[tex]V_{rms} = \frac{1}{\sqrt 2} * 150[/tex]

[tex]V_{rms} = \frac{150}{\sqrt 2}[/tex]

[tex]V_{rms} = 106.1V\\[/tex]

Solving (c): The period

This is calculated as:

[tex]T = \frac{1}{f}[/tex]

So, we have:

[tex]T = \frac{1}{100Hz}[/tex]

[tex]T = 0.01s[/tex]

fraternity hazing is acceptable because it is an initational rite to the brotherhood

Answers

Answer:

that is the right answer

Explanation:

A navy diver hears the underwater sound wave from an exploding ship across the harbor. They immediately lift their head out of the water. The sound wave from the explosion propagating through the air reaches the diver 4.00 s later. The sound velocity is 1440 m/s in water How far away is the ship?

Answers

Answer:

s = 1800 m = 1.8 km

Explanation:

The distance, the speed, and the time of reach of the sound are related by the following formula:

[tex]s = vt[/tex]

where,

s = distance

v = speed

t = time

FOR WATER:

[tex]s = v_wt[/tex] ---------------------- eq (1)

where,

s = distance between ship and diver = ?

[tex]v_w[/tex] = speed of sound in water = 1440 m/s

t = time taken by sound in water

FOR AIR:

[tex]s = v_a(t+4\ s)[/tex] ---------------------- eq (2)

where,

s = distance between ship and diver = ?

[tex]v_a[/tex] = speed of sound in water = 344 m/s

t + 4 s = time taken by sound in water

Comparing eq (1) and eq (2),because distance remains constant:

[tex]v_wt=v_a(t+4\ s)\\\\(1440\ m/s)t = (344\ m/s)(t+4\ s)\\(1440\ m/s - 344\ m/s)t=1376\ m\\t = \frac{1376\ m}{1096\ m/s}[/tex]

t = 1.25 s

Now using this value in eq (1):

[tex]s = (1440\ m/s)(1.25\ s)[/tex]

s = 1800 m = 1.8 km

The electric power delivered to your home has 110 Volts.

All outlets in your kitchen are connected in parallel.

You always have the lights on for 660 Watts, the toaster has a resistance of 440 Ohms, the stove uses power at the rate of 2200 Watts, and the Blender has a resistance of 220 Ohms.

If all these appliances are running at he same time how much total current are you using?

Answers

Answer:

I = 26.80 A

Explanation:

From Ohm's law,

V = IR ............ 1

P = IV ............ 2

where V is the value of the voltage, I is the current, R is the resistance and P is the power.

i. The power of light switched on = 660 Watts.

ii. The power of the toaster = [tex]\frac{V^{2} }{R}[/tex]

               = [tex]\frac{(110)^{2} }{440}[/tex]

               = 27.5 Watts

iii. The power of the stove = 2200 Watts.

iv. The power of the blender = [tex]\frac{V^{2} }{R}[/tex]

                = [tex]\frac{(110)^{2} }{220}[/tex]

                = 55 Watts

Total power of the appliances = 660 + 27.5 + 2200 + 55

                                                   = 2942.5 Watts

So that,

P = IV

I = [tex]\frac{P}{V}[/tex]

 = [tex]\frac{2942.5}{110}[/tex]

 = 26.75

I = 26.75 A

The total current being used when all the appliances are running at the same time is 26.80 A.

Do all substances conduct heat ?Why/ Why not ?​

Answers

Answer:

no, all substances doesnot conduct heat

Answer:

No, all substances do not conduct heat easily because it depends on the nature of the substance. Some are good conductors of heat and some are bad. Therefore, it depends on their characteristics and their ability to conduct heat.

The bad conductors of heat are water, air, plastic, wood, etc.

Gold, Silver, Copper, Aluminium, Iron, etc. are good heat conductors as well as electrical conductors.

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