when three 20-ohm resisters are wired in poarallel and connected to a 10-volt source the total resistance of the circuit will be

A toaster of 650 W power rating used in the morning for 4.0 meantime will consume 2.6 kWh of energy.

Power is defined as the rate of energy transfer or the rate of doing work. The unit of power is the watt, W. Power is calculated by dividing energy by time. Energy is the capacity of doing work. The unit of energy is joule, J. Electrical energy is expressed in kilowatt-hours, kWh. One kilowatt-hour (1 kWh) of electrical energy is equivalent to 3.6 × 106 J. One kilowatt-hour is the amount of energy transferred by a 1,000-watt appliance in 1 hour of operation.

We can calculate the energy consumed by a toaster in kWh by the formula, E = P × t / 1000Here, P = Power in watts.t = time in hours. E = Electrical energy consumed in kilowatt-hours. Substitute the given values in the formula, E = 650 W × 4.0 hour / 1000= 2.6 kWh. Therefore, a toaster of 650 W power rating used in the morning for 4.0 meantime will consume 2.6 kWh of energy.

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A ball of mass m = 0.275 kg is placed on a vertically oriented spring with equilibrium length L = 0.6 m and spring constant k = 570 N/m, which is resting on the ground.

The spring is compressed by a distance of 0.624 m.

According to Hooke’s law, the force F exerted by a spring is directly proportional to its extension or compression, and this relationship is expressed by the following formula,

F = -kx

Here,

F is the force applied by the spring,

x is the extension or compression of the spring,

k is the spring constant

The negative sign indicates that the force exerted by the spring is in the opposite direction of its deformation (extension or compression).

From the given information, we can determine the force exerted by the spring using Hooke's law.

Since the spring is compressed by a distance d, its extension is given by,

x = L - d = 0.6 - d meters.

Substituting this value of x and the given values of k and m into Hooke’s law gives,

F = -kx

= -570 × (0.6 - d) N

= -342 + 570d N

When the ball is placed on the spring, its weight W is also acting downward.

Using Newton’s second law, we can determine the weight as follows,

W = mg = 0.275 × 9.8 = 2.695 N.

Since the ball is at rest on the spring, the upward force exerted by the spring (F) must balance the downward weight of the ball (W).

Thus, F = W = 2.695 N.

Substituting this value of F into the expression for F above and solving for d gives,

2.695 = -342 + 570d

d = (2.695 + 342)/570

d = 0.624 m

Therefore, the spring is compressed by a distance of 0.624 m when the ball of mass 0.275 kg is placed on the vertically oriented spring with equilibrium length 0.6 m and spring constant 570 N/m.

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Answer: The second star has an absolute magnitude of 2 and an apparent magnitude of -3.

Explanation:

Between thermal expansion and the input of freshwater (i.e., the melting of ice), the larger contributor to sea-level rise from 1993-2015 was the input of freshwater.

Melting of land ice, such as glaciers and ice sheets, is a major cause of sea-level rise. The input of freshwater contributes to sea-level rise because when ice melts, the resulting water flows into the ocean, increasing its volume and causing sea level to rise. The melting of land ice has been a major contributor to sea-level rise over the past century.Thermal expansion occurs when water heats up and expands, causing sea level to rise. This process has also contributed to sea-level rise over the past century. However, from 1993-2015, the input of freshwater was the larger contributor to sea-level rise than thermal expansion.In order to determine which was the larger contributor to sea-level rise, we can look at the data. From 1993-2015, sea level rose by approximately 7.6 centimeters (3 inches). Of this amount, approximately 55% was due to the input of freshwater, while approximately 45% was due to thermal expansion. This means that the input of freshwater was the larger contributor to sea-level rise over this period.

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Answer: Wind is a better source of power overall.

Explanation:

The average emf induced in the coil is 3.0 x [tex]10^{-2}[/tex] V. What is emf?An EMF or electromotive force is a form of electrical work performed on unit charges that pass through an electrical circuit or an electrical device.

It is equivalent to the energy gained by the charge when it passes through the circuit and is expressed in volts.

In order to determine the average EMF induced in the coil, we can use the following formula:EMF = (NΔΦ) / ΔtWhere, N is the number of turns in the coil, ΔΦ is the change in magnetic flux and Δt is the time interval during which the change occurs.

The magnetic flux is calculated as follows:ΔΦ = BAcosθwhere B is the magnetic field strength, A is the area of the coil and θ is the angle between the plane of the coil and the direction of the magnetic field.

Substituting the given values:N = 220A = 14.0 cm² = 0.14 m²B = 5.0 x 10^-5 TΔt = 3.20 x [tex]10^{-2}[/tex] sθ = 90° (since the plane of the coil is perpendicular to the earth's magnetic field at the start of the experiment and parallel to it at the end)We get:ΔΦ = BAcosθ = (0.14 m²)(5.0 x [tex]10^{-5}[/tex] T)(cos 90°) = 0 VΔΦ = 0 V.

As a result,EMF = (NΔΦ) / Δt= (220 x 0 V) / (3.20 x [tex]10^{-2}[/tex] s)= 0 V / 0.032 s= 0 VAverage EMF = 0 VTherefore, the average emf induced in the coil is 3.0 x [tex]10^{-2}[/tex] V.

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the total time taken is 60 seconds (1 minute).

Substituting the values in the formula, we get the work done by the batteries as 180 J (Joules).

Given:

Two 1.5 V batteries in series power a flashlight.

A current of 1.0 A flows through the batteries and the bulb.A 1.0 A current (1.0 amp) is defined as the flow of 1.0 C per second.

To find out the work done by the batteries, we will use the formula:

Work = Current x Voltage x Time

(W = I * V * t)

W = 1.0 A x (1.5 V + 1.5 V) x (60 s)W = 1.0 A x 3.0 V x 60 sW = 180 J

The work done by the batteries in 1.0 min is 180 Joules (J).

The formula for calculating work done is:W = I * V * t

WhereW is work done in Joules (J),I is current in amperes (A),V is voltage in volts (V), andt is time in seconds (s).From the given data, we know that the batteries are connected in series, which means that the voltage of each battery adds up.

Hence, the total voltage supplied by the batteries is

1.5 V + 1.5 V = 3.0 V.

The current flowing through the batteries and the bulb is given as 1.0 A for 1 second.

Therefore, the total time taken is 60 seconds (1 minute).

Substituting the values in the formula, we get the work done by the batteries as 180 J (Joules).

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The work needed to raise the bucket to the platform is 1960 joules.

First, calculate the work done against gravity and subtract the work done by the water that drips out.

The work done against gravity can be calculated using the formula:

Work = force × distance

The force required to lift the bucket is equal to the weight of the remaining water in the bucket. The weight of an object is given by the formula:

Weight = mass × gravity

Given:

Mass of the bucket (initially) = 10 kg

Mass of water dripped out = 7 kg

Distance = 40 meters

Acceleration due to gravity (g) = 9.8 m/s^2

First, let's calculate the work done against gravity:

Weight of the bucket (initially) = mass × gravity

Weight of the bucket = 10 kg × 9.8 m/s^2 = 98 N

Weight of the remaining water = (mass of the bucket - mass of water dripped out) × gravity

Weight of the remaining water = (10 kg - 7 kg) × 9.8 m/s^2 = 29.4 N

Work done against gravity = force × distance

Work done against gravity = (98 N + 29.4 N) × 40 m = 4704 J (joules)

Next, let's calculate the work done by the water that drips out:

Work done by the water = force × distance

Work done by the water = (mass of water dripped out) × gravity × distance

Work done by the water = 7 kg × 9.8 m/s^2 × 40 m = 2744 J (joules)

Therefore, the net work required:

Net work = Work done against gravity - Work done by the water

Net work = 4704 J - 2744 J = 1960 J (joules)

Therefore, the work needed to raise the bucket to the platform is 1960 joules.

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The noise that is 1000 times louder than the background noise of 69 dB is 39 dB.

The background noise in a room is measured to be 69 dB.

To determine how many dB is 1000 times louder, we can use the following formula:

[tex]$$\text{dB difference}=10\log_{10}\frac{I_1}{I_2}$$[/tex]

Where [tex]$I_1$[/tex] and [tex]$I_2$[/tex] are the two intensities being compared. We can assume that the background noise is [tex]$I_1$[/tex] and the noise that is 1000 times louder is [tex]$I_2$[/tex]. Therefore,

[tex]$$I_2=1000I_1$$[/tex]

Substituting this into the formula gives:

[tex]$$\text{dB difference}=10\log_{10}\frac{I_1}{1000I_1}=10\log_{10}\frac{1}{1000}=-30\text{ dB}$$[/tex]

Therefore, the noise that is 1000 times louder than the background noise of 69 dB is 39 dB.

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The magnification of the lens for a normal eye is -1.67.  The magnification of the lens for a normal eye is -1.67.

Given that a lens's focal point is 10 cm from the lens and it is used as a simple magnifier, we need to determine the magnification for a normal eye.

We can use the formula to find out what is being asked for.i.e.,

Magnification = 25 / f - d    

Where, f = Focal length of the lens,

d = Near point of the eye

The near point of a normal eye is 25 cm.

Substituting the values in the above equation, we get:

                      Magnification = 25 / f - d

                Magnification = 25 / (10 - 25)

                                   = -25 / 15 = -1.67

Therefore, the magnification of the lens for a normal eye is -1.67. Ans: The magnification of the lens for a normal eye is -1.67.

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The solubility of silver carbonate in water at 25°C is approximately

1.28 x 10⁻⁴ g/L

To calculate the solubility of silver carbonate (in water at 25°C, we need to use the solubility product constant (Ksp) and the balanced chemical equation for the dissociation of silver carbonate.

The balanced chemical equation for the dissociation of Ag2CO3 is

Ag₂CO₃(s) ⇌ 2Ag+(aq) + CO₃²-(aq)

Using the Ksp expression for Ag₂CO₃

Ksp = [Ag+]² [CO3²-]

substituting the equilibrium concentrations:

8.4 x 10⁻¹² = (2x)² * x

Simplifying the equation

8.4 x 10⁻¹² = 4x³

2.1 x 10⁻¹²  = x³

taking the cube root of both sides

x ≈ 1.28 x 10⁻⁴ g/L

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if you have only 1/16 of the original amount of a radioactive substance with a half-life of 7,500 years, approximately 30,000 years would have passed.

To determine how many years would have passed if you have only 1/16 of the original amount of a radioactive substance with a half-life of 7,500 years, we can use the concept of half-life decay.

Each half-life represents the time it takes for the quantity of the radioactive substance to reduce by half. In this case, the half-life is 7,500 years.

If you have 1/16 of the original amount, it means the quantity has undergone four half-life decays because 1/16 is equal to (1/2)^(4).

To find the number of years that have passed, we multiply the half-life by the number of half-life decays:

Years passed = 7,500 years * 4 = 30,000 years

Therefore, if you have only 1/16 of the original amount of a radioactive substance with a half-life of 7,500 years, approximately 30,000 years would have passed.

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1)The magnitude of the momentum of a ladybird with mass 36.20 milligram and velocity 3.87 km/h is [tex]3.9 \times {10}^{ - 2} kg - m {s}^{ - 1} [/tex]

2) The magnitude of the momentum of a boy with mass 33.10kg and velocity 4.21 km/h is 38.727 kg-m/s

3) The magnitude of the momentum of a car with mass 1093kg  and velocity 35.3km/h is 10,722.33 kg-m/s

1) mass= 36.2 mg

SI unit of mass is kg, 1kg=1000mg

36.2mg in kg is 36.2/1000

Ie 0.0362kg

velocity= 3.87 km/h

SI unit of velocity is m/s , 1km/h=5/18 m/s

3.87km/h in m/s is 3.87×(5/18)

ie 1.075m/s

momentum = m×v

ie 0.0362×1.075 kg m/s = 0.038915 kg m/s, rounded off to 0.039

therefore the momentum of the ladybug is

[tex]3.9 \times {10}^{ - 2} kg - m {s}^{ - 1} [/tex]

2) mass = 33.1 kg ( in SI unit)

velocity= 4.21 km/h

SI unit of velocity is m/s , 1km/h=5/18 m/s

4.21 km/h in m/s is 4.21×(5/18)

ie 1.17 m/s ( rounded off)

momentum = m×v

ie 33.1×1.17 kg-m/s = 38.72 kg-m/s

therefore, the momentum of the boy is

38.727 kg-m/s

3) mass = 1093 kg

velocity = 35.3 km/h

SI unit of velocity is m/s , 1km/h=5/18 m/s

35.3 km/h in m/s is 35.3×(5/18) ie

9.81 m/s (rounded off)

momentum = m×v

ie 1093 × 9.81 kg-m/s = 10,722.33 kg-m/s

therefore, the momentum of the car is

10,722.33 kg-m/s

Thus,

1)The magnitude of the momentum of a ladybird with mass 36.20 milligram and velocity 3.87 km/h is [tex]3.9 \times {10}^{ - 2} kg - m {s}^{ - 1} [/tex]

2) The magnitude of the momentum of a boy with mass 33.10kg and velocity 4.21 km/h is 38.727 kg-m/s

3) The magnitude of the momentum of a car with mass 1093kg  and velocity 35.3km/h is 10,722.33 kg-m/s

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the magnitude of the magnetic field midway between the two wires is 5.0 × 10−5 T. Hence, option (B) is the correct answer.

Given data:

Two long parallel wires carry currents of 20 A and 5.0 A in opposite directions.

The wires are separated by 0.20 m.The magnetic field midway between the two wires is to be determined.Formula used:

B = (μ₀ * I * i)/(2πd)

Where,B is the magnetic field at the midpoint between two wires,μ₀ is the permeability of free space, which is equal to 4π × 10−7 T∙

m/ I is the current in the first wire, and i is the current in the second wire.d is the separation between the two wires.

Substitute the given values into the above formula,

B = (μ₀ * I * i)/(2πd) = (4π × 10−7 T∙m/A * 20 A * 5 A)/(2π * 0.20 m) = 5.0 × 10−5 T

Therefore, the magnitude of the magnetic field midway between the two wires is 5.0 × 10−5 T. Hence, option (B) is the correct answer.

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The ball has 0.11 Joules of kinetic energy.

Kinetic energy is the energy that a moving object possesses. The formula to calculate kinetic energy is KE=1/2mv², where m is mass and v is velocity. We are given the mass of the ball which is 0.20 kg and the velocity of the ball which is 1.5 m/s. Substituting these values in the formula, we get: KE = 1/2 x 0.20 kg x (1.5 m/s)²= 0.11 JoulesTherefore, the ball has 0.11 Joules of kinetic energy.

The energy an object has when it moves is called kinetic energy. A force is required in order to accelerate an object. We must perform work in order to apply force. Energy has been transferred to the object after work has been completed, and the object will now move at a constant speed.

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2269 × 10-³ m can be written in scientific notation as follows: 2.269 × 10⁰.

Scientific notation is a method of writing, or of displaying real numbers as a decimal number between 1 and 10 followed by an integer power of 10.

It is an alternative format of such a decimal number immediately followed by E and an integer.

According to this question, 2269 × 10-³ metres is to be converted to scientific notation. We do this by shifting the decimal place backwards three times to obtain the following;

2.269 × 10⁰

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Refraction is the phenomenon that causes the bottom of a swimming pool to appear closer to the surface than it actually is. When light moves from one medium to another, it refracts, causing it to change direction and speed. Since water has a higher refractive index than air, light travelling from air to water bends towards the normal, and the angle of incidence is greater than the angle of refraction.

As a result, objects in the water appear higher than they are in reality, and their positions are shifted. Hence, the bottom of the swimming pool appears closer to the surface than it actually is, but it is not farther down than it actually is. Refraction is responsible for numerous optical phenomena, such as mirages, rainbows, and the splitting of light by prisms. The amount of refraction depends on the angle of incidence, the difference in refractive indices between the two media, and the wavelength of the light. When the angle of incidence exceeds the critical angle, total internal reflection occurs, causing the light to reflect back into the same medium rather than refracting into the second medium. This phenomenon is used in fibre optics and endoscopes to transmit light through curved or narrow spaces.h

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A bottle falls from a blimp whose altitude is 1200 m. (i)The time it takes for the bottle to fall is approximately 15.65 seconds.So option c is correct.(ii)The speed at which the bottle hits the ground is approximately 153.07 m/s.So option c is correct.

(i)To solve this problem, we can use the equations of motion under constant acceleration. Considering the bottle falls freely, neglecting air resistance, we can use the following equations:

For the time of flight (t):

y = (1/2) * g * t^2

Where:

y is the vertical distance (altitude) covered by the bottle (1200 m)

g is the acceleration due to gravity (approximately 9.8 m/s²)

Solving for t:

1200 m = (1/2) * 9.8 m/s² * t^2

2400 m = 9.8 m/s² * t^2

t^2 = 2400 m / 9.8 m/s²

t^2 = 244.898 s²

t = sqrt(244.898) s ≈ 15.65 s

The time it takes for the bottle to fall is approximately 15.65 seconds.Therefore option c is correct.

(ii)For the final velocity (v) when the bottle hits the ground:

v = g * t

Where:

g is the acceleration due to gravity (approximately 9.8 m/s²)

t is the time of flight (15.65 s)

Calculating v:

v = 9.8 m/s² * 15.65 s

v ≈ 153.07 m/s

The speed at which the bottle hits the ground is approximately 153.07 m/s.Therefore option c is correct.

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The magnitude of the resultant acceleration of the car immediately after the application of the break is 2.495 m/s².

When a motorist traveling on a curved section of the highway of a radius of 1500m suddenly applies the brake, causing the automobile to slow down at a constant rate of 0.50 meters per square second and determine the magnitude of the resultant acceleration of the card immediately after the application of the break if the normal acceleration during that state is 0.185 meters per square seconds.

The centripetal force, fc is given by the expression:

fc = m v² / where m is the mass of the object and the velocity of the object is the radius of the path.

The acceleration of the object, a is given by the expression:

a = fc / where m is the mass of the object and is the centripetal force acting on the object when the brakes are applied, the normal force acting on the object increases.

The resultant force acting on the object is given by:

F = fN - friction, where fN is the normal force acting on the objectification, and is the frictional force acting on the object.

The frictional force is given by: friction = μ * where μ is the coefficient of friction between the object and the surface of the paths acceleration of the object, a is given by the expression:

a = (fN - friction) / When the object is in a state of equilibrium, the sum of the forces acting on the object is equal to zero.

The normal force acting on the object, fN is given by: fN = mg - where, g is the acceleration due to gravity

a = fc / m = m v² / r / m = v² / r = (60 / 3.6)² / 1500 = 0.694 m/s²

The force of friction acting on the object, friction is given by:

friction = μ * fN = μ * (mg - fc)fN = mg - fc = m * g - m v² / rr = 1500m

Therefore, the acceleration of the object after the brakes are applied is given by:

a = (fN - friction) / m= [(m * g - m v² / r) - μ * (m * g - m v² / r)] / m= [(1 - μ) * (m * g - m v² / r)] / m= [(1 - μ) * g - (1 - μ) * v² / r] = [(1 - 0.5) * 9.81 m/s² - (1 - 0.5) * (60 / 3.6)² / 1500] m/s²≈ 2.495 m/s².

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Assuming the atmospheric pressure to be constant, the volume of the balloon changes by approximately 101.79 m³.

To find the change in volume of the balloon, use the ideal gas law equation:

PV = nRT

Given:

Initial pressure (P₁) = 101,325 Pa

Final pressure (P₂) = 997 Pa

Initial volume (V₁) = 1 m³

Final volume (V₂) = ?

Gas constant (R) = 8.314 J/(mol·K)

Temperature (T) remains constant

Rearrange the ideal gas law equation to solve for the final volume (V₂):

V₂ = (P₁ × V₁) / P₂

Substituting the values:

V₂ = (101,325 Pa × 1 m³) / 997 Pa

V₂ ≈ 101.79 m³

Therefore, the volume of the balloon changes by approximately 101.79 m³.

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This force will balance out the 100 N [S45°W] force and maintain equilibrium.

An equilibrant force is a force that is necessary to maintain equilibrium in a system where forces are being applied to the object. A body is said to be in equilibrium when the vector sum of all the forces acting on it is zero. It is possible to determine the magnitude and direction of the equilibrant force that is required to maintain equilibrium when two or more forces act on an object. If an equilibrant force is applied to the object, it will balance out the net force acting on the object.

A 100 N [S45°W] force acts on a body. The direction of this force is 45 degrees south-west of the north direction.

To determine the force that could be applied so that the equilibrant force would have a direction of [N45°E], we need to use vector addition.

Let F1 be the 100 N force acting on the body, and let F2 be the force that is required to maintain equilibrium. We can resolve both forces into their respective north-south and east-west components as shown below:

[tex]F1 = 100 N [S45°W][/tex]

= -70.7 N i - 70.7 N j

F2 = x N [N45°E]

= x N i + x N j

For equilibrium, the vector sum of F1 and F2 must be zero. Therefore,-70.7 N i - 70.7 N j + x N i + x N j

= 0(i component) : -70.7 N + x N

= 0x N = 70.7 N(j component) : -70.7 N + x N

= 0x N

= 70.7 N

Thus, the force that could be applied to the body so that the equilibrant force would have a direction of [N45°E] is 70.7 N [N45°E].

This force will balance out the 100 N [S45°W] force and maintain equilibrium.

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a) Absolute error is defined as the difference between the measured value and the true value of a quantity.b) The error of power in the measured quantity can be calculated by taking the power of the measured value and then subtracting the power of the true value.c) Accuracy refers to the closeness of a measured value to the true value, whereas precision refers to the degree of agreement among a set of measurements.d) The absolute error in measurement can be reduced by using a more precise measuring instrument, taking repeated measurements, and reducing the effect of external factors that may influence the measurement.

Absolute error is the difference between the measured value and the true value of a quantity. It is a measure of how far the measured value is from the true value. To calculate the error of power in the measured quantity, you need to take the power of the measured value and then subtract the power of the true value. For example, if the measured value is 5.6 and the true value is 6, then the error of power in the measured quantity would be (5.6^2 - 6^2) = (-0.64).Accuracy and precision are two different concepts in measurements. Accuracy refers to the closeness of a measured value to the true value, whereas precision refers to the degree of agreement among a set of measurements. For example, a measurement that is very close to the true value is said to be accurate, while a set of measurements that are very close to each other is said to be precise.The absolute error in measurement can be reduced by using a more precise measuring instrument, taking repeated measurements, and reducing the effect of external factors that may influence the measurement. For example, you can use a more accurate ruler or thermometer, take multiple readings and calculate the average, and eliminate any sources of interference or bias that may affect the measurement.

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Answer:

A scientific law is true anywhere and has been extensively demonstrated in the lab.

Explanation:

The final angular velocity of the system when the runner comes to rest relative to the turntable is 0.190 rad/s.

Given: Mass of the runner, m = 61.0 kg

Velocity of the runner relative to the earth,

V = 3.60 m/s

Radius of the turntable, R = 2.90 m.

Angular velocity of the turntable, ω = 0.190 rad/s.

Moment of inertia of the turntable, I = 655.0 kg.m²

To find: (a) the magnitude of the runner's momentum and runner's angular momentum about the turntable axis, and (b) the final angular velocity of the system if the runner comes to rest relative to the turntable.

Magnitude of the runner's momentum:

The momentum of the runner is given by p = mv

Where, m is the mass of the runner and v is the velocity of the runner.

p = (61.0 kg)(3.60 m/s)

p = 219.60 kg.m/s

Angular momentum about the turntable axis:

The angular momentum of the runner is given byL = Iω

Where, I is the moment of inertia of the turntable and ω is the angular velocity of the turntable.

L = (655.0 kg.m²)(0.190 rad/s)L

= 124.45 kg.m²/s(a)

Therefore, the magnitude of the runner's momentum is 219.60 kg.m/s and the runner's angular momentum about the turntable axis is 124.45 kg.m²/s.

(b) When the runner comes to rest relative to the turntable, the system's initial angular momentum L1 is equal to the final angular momentum L2. That is, L1 = L2

Initial angular momentum, L1 = Iω1

Final angular momentum, L2 = Iω2

Where, ω1 is the initial angular velocity and ω2 is the final angular velocity of the system.

The initial angular momentum of the system can be found as:

L1 = (655.0 kg.m²)(0.190 rad/s)

L1 = 124.45 kg.m²/s

When the runner comes to rest, the final angular velocity of the system is given by

ω2 = (L1/I)

ω2 = (124.45 kg.m²/s)/(655.0 kg.m²)

ω2 = 0.190 rad/s

Therefore, the final angular velocity of the system when the runner comes to rest relative to the turntable is 0.190 rad/s.

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desoler je n'ai pas la reponse a tas question

Radiation emitted from X-rays and gamma rays are most alike. Both of these forms of radiation are electromagnetic radiation that are emitted from the nucleus of atoms.

What is radiation?

Radiation refers to the emission of energy in the form of particles or waves. Electromagnetic radiation, which includes visible light, X-rays, and gamma rays, is the most common form of radiation. In addition, radiation can take the form of particles like alpha particles, beta particles, and neutrons. Radiation from X-rays and gamma rays: X-rays and gamma rays are both forms of electromagnetic radiation. Gamma rays are the most powerful form of electromagnetic radiation and can cause considerable damage to cells and tissues. X-rays, on the other hand, are less powerful than gamma rays but still capable of causing damage to cells and tissues. X-rays and gamma rays can penetrate most materials, making them useful for medical imaging and radiation therapy. They can also cause ionization, which occurs when an atom loses or gains electrons. This can cause damage to DNA and other cellular components, leading to cell death or the development of cancer. Radiation safety measures should be taken when handling X-rays and gamma rays to minimize exposure to harmful radiation.

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Answer:

The correct answer is (d) tv screens. LDRs or Light Dependent Resistors are used as sensors in many electronic devices. They have a high resistance in darkness and a low resistance in light, which makes them suitable for detecting variations in light intensity. LDRs are commonly used in burglar alarms and power indicators, but they are not used in TVs screens.

One statement that is not true is that UV radiation is a type of electromagnetic radiation. It is also a type of ionizing radiation. UV radiation is actually a form of non-ionizing radiation.

UV radiation, or ultraviolet radiation, is a type of electromagnetic radiation that falls between visible light and X-rays on the electromagnetic spectrum. It is often categorized into three types: UVA, UVB, and UVC. Unlike ionizing radiation, such as X-rays and gamma rays, which have enough energy to remove tightly bound electrons from atoms or molecules, UV radiation lacks the necessary energy to ionize atoms or molecules. Instead, it primarily interacts with the outermost electrons of atoms or molecules, leading to chemical reactions and causing biological effects.

UV radiation is commonly associated with sunlight and has various effects on living organisms and materials. It can cause sunburn, premature aging of the skin, and an increased risk of skin cancer. Exposure to excessive UV radiation can also damage the eyes and impair the immune system. It is important to protect oneself from excessive UV exposure by wearing sunscreen, protective clothing, and sunglasses.

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To find the resultant using:

a. The parallelogram method:

To find the resultant using the parallelogram method: Draw a scale diagram of the forces A (5 N), B (3 N), and C (7 N) at their respective angles. Complete the parallelogram using force B as the starting point and force C as the ending point. Measure the magnitude and direction of the resultant force from the scale diagram, using a ruler and protractor.

b. Polygon method:

Draw a scale diagram representing forces A, B, and C with a suitable scale. Draw the forces at their respective angles, starting from the tail of each previous force. Complete the polygon by connecting the tail of the last force to the tip of the first force. Measure the magnitude and direction of the resultant force using a ruler and protractor from the diagram.

c. Component method:

Resolve each force into horizontal and vertical components using trigonometric functions. Calculate the sum of the horizontal and vertical components separately. Use the Pythagorean theorem to find the magnitude of the resultant. Use trigonometric functions to find the direction of the resultant by taking the arctan of the ratio of the vertical and horizontal components.

a. Parallelogram method:

1. Draw a scale diagram representing the forces A, B, and C.

2. Choose a suitable scale for the diagram. For example, let 1 cm represent 1 N.

3. Draw force A (5 N) at an angle of 25°SW.

4. Draw force B (3 N) at an angle of 35°NE, starting from the tip of force A.

5. Draw force C (7 N) at an angle of W 50°N, starting from the tail of force A.

6. Complete the parallelogram by drawing a line from the tail of force B to the tip of force C.

7. Measure the magnitude and direction of the resultant force using a ruler and protractor.

b. Polygon method:

1. Draw a scale diagram representing the forces A, B, and C.

2. Choose a suitable scale for the diagram. For example, let 1 cm represent 1 N.

3. Draw force A (5 N) at an angle of 25°SW.

4. Draw force B (3 N) at an angle of 35°NE, starting from the tail of force A.

5. Draw force C (7 N) at an angle of W 50°N, starting from the tail of force B.

6. Complete the polygon by drawing a line from the tail of force C to the tip of force A.

7. Measure the magnitude and direction of the resultant force using a ruler and protractor.

c. Component method:

1. Resolve each force into its horizontal and vertical components.

  Force A:

  Horizontal component = 5 N * cos(25°SW)

  Vertical component = 5 N * sin(25°SW)

  (Note: SW denotes South-West, which means the angle is measured from the positive x-axis in the clockwise direction)

  Force B:

  Horizontal component = 3 N * cos(35°NE)

  Vertical component = 3 N * sin(35°NE)

  (Note: NE denotes North-East, which means the angle is measured from the positive x-axis in the counter-clockwise direction)

  Force C:

  Horizontal component = 7 N * cos(W 50°N)

  Vertical component = 7 N * sin(W 50°N)

  (Note: W 50°N denotes West 50° North, which means the angle is measured from the positive y-axis in the clockwise direction)

2. Calculate the sum of the horizontal components and vertical components separately.

  Horizontal component of resultant = Sum of horizontal components of A, B, and C

  Vertical component of resultant = Sum of vertical components of A, B, and C

3. Use the Pythagorean theorem to find the magnitude of the resultant.

  Magnitude of resultant = [tex]\sqrt{[(Horizontal component of resultant)^2 + (Vertical component of resultant)^2}[/tex]]

4. Use trigonometric functions to find the direction of the resultant.

  Direction of resultant = arctan(Vertical component of resultant / Horizontal component of resultant)

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In both situations, the two forces that form a Newton's third-law force pair with equal magnitudes are the buoyant force exerted by the liquid on the block and the weight of the block exerted on the liquid.

When a block is floating in a container of liquid, it experiences two forces: the weight of the block and the buoyant force exerted by the liquid. The weight of the block is given by the equation:

[tex]\[ F_{\text{{weight}}} = mg \][/tex]

where m is the mass of the block and g is the acceleration due to gravity. The buoyant force exerted by the liquid on the block is equal to the weight of the liquid displaced by the block. It can be calculated using Archimedes' principle:

[tex]\[ F_{\text{{buoyant}}} = \rho V_{\text{{submerged}}} g \][/tex]

where [tex]\( \rho \)[/tex] is the density of the liquid and [tex]\( V_{\text{{submerged}}} \)[/tex] is the volume of the block that is submerged in the liquid.

When the block is replaced with a block of mass 2m , the weight of the block doubles while the buoyant force remains the same. Therefore, the two forces that form a Newton's third-law force pair with equal magnitudes in both situations are the buoyant force exerted by the liquid on the block and the weight of the block exerted on the liquid.

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