A wooden sphere of mass 4.0 kg is completely immersed in water. A pushing force of 20. N is
applied.
21°
20 N
4.2 ms²
19⁰
At the moment shown in the diagram, the sphere is stationary and it experiences an
acceleration upwards and to the right as shown.
Calculate the size of the upwards force due to the water (upthrust) acting on the sphere.

A Wooden Sphere Of Mass 4.0 Kg Is Completely Immersed In Water. A Pushing Force Of 20. N Isapplied.2120

Answers

Answer 1

The size of the upwards force due to the water (upthrust) acting on the sphere is 12.64 N.

What is upthrust?

Buoyancy or upthrust, is an upward force exerted by a fluid on an object immersed in the fluid due to the weight of the object

Thus, upthrust is the upward force acting on an object immersed in a liquid.

Fu - Df = F(net_u)

where;

Fu is the upward forceDf is the downward force applied on the objectF(net_u) is the net upward force

Fu - F x sin(21) = ma x sin(19)

where;

m is the mass of the wooden spherea is the upward acceleration of the wooden sphere

Fu -  20 x sin(21) = (4 x 4.2) x sin(19)

Fu - 7.17 = 5.47

Fu = 12.64 N

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Related Questions

Ahmed wants to measure the weight of an object. What instrument should he use?

Answers

Answer:

a scale

Explanation:

you use it to weigh things

Answer:

its not a scale

Explanation:

When a stone and a feather is thrown down from a height under free fall. Which of the two will first reach the ground???​

Answers

Answer:

The stone

Explanation:

It will be the first to fall because of its mass which is heavier than that of the feather .

Determine the order of magnitude of the following small number by entering the appropriate exponent n below: 4,870×1021kg≈10nkg .(PLEASE LOOK AT THE SCREENSHOT FOR CORRECT NUMBERING)

Answers

To find the order of magnitude we move the decimal point until we have only one interger before it:

[tex]4870\times10^{21}=4.870\times10^{24}[/tex]

Now that we have in this form we conclude that the order of magnitude of the number is 24

Based on the circuit voltage and the wattage consumption,determine the approximate ampere rating of the followingappliances. Remember amps = watts divided by voltage.a = w÷ VRound to the nearest whole amp.1. AC Compressor on a 240 volt line and using 5,000 watts, amps =_____2. baseboard heater on a 120 volt line and using 1,200 watts, amps =_____3. vacuum cleaner on a 120 volt line and using 500 watts, amps =______4. blender on a 115 volt line and using 300 watts, amps5. toaster on a 120 volt line using 1,100 watts, amps =_____

Answers

Given:

1.

The voltage of AC compressor is V = 250 V

The power of the AC compressor is P = 5000 W

2.

The voltage of the baseboard heater is V = 120 V

The power of the baseboard heater is P = 1200 W

3.

The voltage of the vacuum cleaner is V = 120 V

The power of the vacuum cleaner is P = 500 W

4.

The voltage of the blender is V = 115 V

The power of the blender is P = 300 W

5.

The voltage of the toaster is 120 V

The power of the toaster is P = 1100 W

Required:

1. The approximate ampere rating of the AC compressor.

2. The approximate ampere rating of the baseboard heater.

3. The approximate ampere rating of the vacuum cleaner.

4. The approximate ampere rating of the blender.

5. The approximate ampere rating of the toaster.

Explanation:

1. The approximate ampere rating of the AC compressor can be calculated as

[tex]\begin{gathered} I\text{ = }\frac{P}{V} \\ =\frac{5000}{240} \\ =20.833\text{ A} \\ \approx21\text{ A} \end{gathered}[/tex]

2. The approximate ampere rating of the baseboard heater can be calculated as

[tex]\begin{gathered} I=\frac{1200}{120} \\ =\text{ 10 A} \end{gathered}[/tex]

3. The approximate ampere rating of the vacuum cleaner can be calculated as

[tex]\begin{gathered} I\text{ = }\frac{500}{120} \\ =4.2\text{ A} \\ \approx4\text{ A} \end{gathered}[/tex]

4. The approximate ampere rating of the blender can be calculated as

[tex]\begin{gathered} I\text{ =}\frac{300}{115} \\ =2.6\text{ A} \\ \approx3\text{ A} \end{gathered}[/tex]

5. The approximate ampere rating of the toaster can be calculated as

[tex]\begin{gathered} I\text{ =}\frac{1100}{120} \\ =9.2\text{ A} \\ \approx\text{ 9 A} \end{gathered}[/tex]

Final Answer:

1. The approximate ampere rating of the AC compressor is 21 A.

2. The approximate ampere rating of the baseboard heater is 10 A.

3. The approximate ampere rating of the vacuum cleaner is 4 A.

4. The approximate ampere rating of the blender is 3 A.

5. The approximate ampere rating of the toaster is 9 A.

What is held in orbit by the gravitational pull of earth

Answers

The international space station.

The moon.

All TV satellites.

All weather satellites.

All GPS satellites.

More than 4000 other artificial satellites.

Thousands of pieces of "space junk"

Answer:

The Moon.

Explanation:

The earths gravity holds the moon in place.

A basketball player jumps for a rebound and reaches a maximum height of 1.5 m. with what speed did he jump off the floor? How long was he in the air?

Answers

The final speed of the player can be given as,

[tex]v^2=u^2-2gh[/tex]

At the maximum height, the final speed of player is zero.

Plug in the known values,

[tex]\begin{gathered} (0m/s)^2=u^2-2(9.8m/s^2)(1.5\text{ m)} \\ u^2=29.4m^2s^{-2} \\ u=5.42\text{ m/s} \end{gathered}[/tex]

Thus, the speed with which it jump off the floor is 5.42 m/s.

The time for which the player was in player is,

[tex]t=\sqrt[]{\frac{2h}{g}}[/tex]

Plug in the known values,

[tex]undefined[/tex]

A 5.0 kg block moves in a straight line on a horizontal frictionless surface under the influence of a force that varies with position as shown in the figure. The scale of the figure's vertical axis is set by F = 20.0 N. How much work is done by the force as the block moves from the origin to x = 8.0 m?

Answers

The work done by the force to move the object from the origin to x = 8 m is 160 J.

The mass of the block, m = 5 kg

The block moves in a straight line on a frictionless surface.

A force of F = 20 N is acting on the object.

The work done by the force when the block moves from one place to another can be defined by the formula,

W = Fd where F is the force and d is the distance covered.

Now, we have F = 20 N and;

d = x = 8 m

Therefore, the work done by the object will be:

W = Fx

W = 20 × 8

W = 160 joules

The work done by the force to move the object to x = 8 m is 160 J.

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A girl walks 600 m north and then 800 m east. What is the displacement from her starting point?​

Answers

Displacement from her starting point is ​529m

What is Displacement?

"Displacement" describes a change in an object's location. It is a vector quantity since it has a magnitude and a direction. It looks like an arrow that leads from the beginning to the end.

Given :  A girl walks 600m north

           She then walks 800m east

To Find : Displacement from her starting point

Solution:                               North

                              West                        East

                                              South

North=600m

East =800m

We use Pythagoras Theorem

Displacement to be covered = [tex]\sqrt{800^{2} }[/tex]- [tex]\sqrt{600^{2} }[/tex]

                                                     529m

                                                   

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A total of 8.0 joules of work is done when a constant horizontal force of 2.0 newtons to the left is used to push a 3.0-kilogram box acrossa counter top. Determine the total horizontal distance the box moves.

Answers

ANSWER

4 meters

EXPLANATION

Given:

• Work = 8.0 J

,

• Force = 2.0 N

,

• Mass of the box = 3.0 kg

Unknown:

• Distance the box moves

Work is the product of the applied force in the direction of motion and the distance an object moves,

[tex]W=F\cdot d[/tex]

In this case, a box is pushed horizontally with a force of 2N producing a work of 8J. Solve the equation above for d,

[tex]d=\frac{W}{N}=\frac{8.0J}{2.0N}=4m[/tex]

Hence, the horizontal distance the box moves is 4 meters.

27. Scientists have observed an increase in global temperatures over the past 100 years. Which phenomena do scientists believe contributes to the increase in temperatures? A. an increase in undersea volcanic activity B. a decrease in the distance between Earth and the Sun C. an increase in certain gases released during the use of fossil fuels D. a decrease in the amount of water on Earth due to overconsumption

Answers

The answer is letter C) An increase in certain gases released during the use of fossil fuels. Although the others do cause an increase in temperature, their scale cannot be compared to the one caused by fossil fuels.

An arrangement of two pulleys, as shown in the figure, is used to lift a 64.3-kg crate a distance of 4.36 m above the starting point. Assume the pulleys and rope are ideal and that all rope sections are essentially vertical.What is the change in the potential energy of the crate when it is lifted a distance of 4.36 m? (kJ)How much work must be done to lift the crate a distance of 4.36 m? (kJ)What length of rope must be pulled to lift the crate 4.36 m? (m)

Answers

Given data:

* The mass of the crate is m = 64.3 kg.

* The height of the crate is h = 4.36 m.

Solution:

(a). The potential energy of the crate at the initial state is zero (as the height of the crate is zero at the initial state), thus, the change in the potential energy of the crate is,

[tex]\begin{gathered} dU=\text{mgh}-0 \\ dU=mgh \end{gathered}[/tex]

where g is the acceleration due to gravity,

Substituting the known values,

[tex]\begin{gathered} dU=64.3\times9.8\times4.36 \\ dU=2747.4\text{ J} \\ dU=2.75\times10^3\text{ J} \\ dU=2.75\text{ kJ} \end{gathered}[/tex]

Thus, the change in the potential energy is 2.75 kJ.

(b). The work done to lift the crate is equal to the change in the potential energy of the crate.

Thus, the work done on the crate is 2.75 kJ.

(c). As the single is pulling the two ropes to increase the height of the crate, thus, the length of the rope pulled in terms of the height of the crate is,

[tex]l=2h[/tex]

Substituting the known values,

[tex]\begin{gathered} l=2\times4.36 \\ l=8.72\text{ m} \end{gathered}[/tex]

Thus, the length of the rope pulled to lift the crate is 8.72 meters.

wat is the mass of the car that has kinetic energy of 2400J and is moving with a speed of 20 m\s

Answers

Given,

The kinetic energy of the car, E=2400 J

The speed of the car, v=20 m/s

Kinetic energy is the energy that is possessed by an object due to its motion.

It is given by,

[tex]E=\frac{1}{2}mv^2[/tex]

Where m is the mass of the car.

On substituting the known values in the above equation,

[tex]\begin{gathered} 2400=\frac{1}{2}\times m\times20^2 \\ m=\frac{2\times2400}{20^2} \\ =\frac{4800}{400} \\ =12\text{ kg} \end{gathered}[/tex]

Thus the mass of the car is 12 kg

The Diagram shows the forces involved as a student slides a water bottle across the desk in front of them to their friend. Based on the image, in which direction is there friction? (ignore the selected answer it’s random)

Answers

Answer:

Left.

Step-by-step explanation:

The force of friction opposes the motion of an object, causing moving objects to lose energy and slow down. Therefore, the friction goes to the left.

Two toy cars collide in an inelastic collision. The first car has a mass of 10 kg
and a velocity of 4 m/s to the right. The second car has a mass of 8 kg and a
velocity of 6 m/s to the left.
The velocity of the cars after the collision is ____________ m/s.

Answers

The velocity of the cars after the inelastic collision is 0.44 m/s.

What is inelastic collision ?

When some of the kinetic energy of a colliding object or system is wasted, the collision is said to be inelastic. In a perfectly inelastic collision, the colliding particles stay together and the most kinetic energy is lost. Such situations involve the use of wasted kinetic energy to bind the two bodies together. The conservation of momentum and energy is typically used to tackle collision-related problems.

m 1 = 10kg

m 2 = 8kg

u 1 = 4m/s

u 2 = 6m/s

Total mass of the combined system,

M=m 1+m 2 = 18kg

​Let the velocity of the combined system after the collision be v

Applying conservation of momentum before and after the collision :

m 1 x  u 1 − m 2 x u 2 =Mv

10 x 4 - 8 x 6 = 18v          

v = 8/18 = 0.44 m/s

The velocity of the cars after the inelastic collision is 0.44 m/s.

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A mass on a spring vibrates in simple harmonic motion at a frequency of 4.0 Hz and an amplitude of 4.0cm. If a timer is started when it's displacement is a maximum (hence x=4cm when t=0), what is the speed of the mass when t=3s ?

Answers

Since the frequency is 4 Hz, it completes one cycle in 1/4 seconds (its period, T, is 1/4 sec).

f = 1/T

Let's draw a graph of the displacement of the mass along time.

This pattern repeats, with each crest (high point) at every 1/4 second. This means that at t = 3, which is a multiple of 1/4, the mass is at its high point.

Notice that when the spring reaches its high or low point, it changes direction.

At this instant when it changes direction, its speed momentarily becomes 0.

That means that each multiple of 1/8 seconds, the speed is momentarily 0.

The speed of the mass is 0 at t = 3.

When a 10 V battery is connected to a resistor, 5 A of current flows through the resistor. What is the resistor's value?

Answers

Given data

*The value of battery voltage is V = 10 V

*The current flows through the resistor is I = 5 A

The formula for the resistor is given by the Ohm's law as

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

Substitute the values in the above expression as

[tex]\begin{gathered} R=\frac{10}{5} \\ =2\text{ ohm} \end{gathered}[/tex]

A car initially traveling at 3.2 m/s accelerated uniformly to a speed of 14.9 m/s over a distance of 60 meters. How much time does it take for the car to reach 9 m/s? Report your answer to 1 decimal place. Do not type in the unit or the computer will mark your answer incorrect.

Answers

Given

Initial velocity, u=3.2 m/s

Final speed , v=14.9 m/s

Distance travelled, s=60 m

To find

Time taken to reach 9 m/s

Explanation

By equation of kinematics,

[tex]\begin{gathered} v^2=u^2+2as \\ \Rightarrow14.9^2=3.2^2+2a\times60 \\ \Rightarrow222.01=10.24+12a \\ \Rightarrow a=17.65\text{ m/s}^2 \end{gathered}[/tex]

Let the time taken to reach 9 m/s be t

Thus

taking v=9 m/s here we have

[tex]\begin{gathered} v=u+at \\ \Rightarrow9=3.2+17.65t \\ \Rightarrow t=0.3\text{ s} \end{gathered}[/tex]

Conclusion

The required time is 0.3 s

7. Which of the following measurement tools would you need to
determine the temperature of boiling water?

Answers

Hi I’m not sure what ur asking do you mind clarifying ? Maybe try inserting a photo

Answer:Laboratory thermometer

Explanation:

The image shows street lights powered by solar panels. Which sequence shows the energy transformations taking place in these lights?

Picture of three solar panels street light on a sunny day with blue background

A.
gravitational potential energy → vibrational energy → chemical potential energy
B.
radiant energy → chemical potential energy → motion energy
C.
radiant energy → electric energy → radiant energy
D.
sound energy → chemical potential energy → radiant energy
E.
gravitational potential energy → motion energy → radiant energy
Reset Next

Answers

The sequence that shows the energy transformations taking place in these lights are radiant energyelectric energyradiant energy.

What is law of conservation of conservation of energy?

The principle or law of conservation of energy states that energy can neither be created nor destroyed but can be converted from one form to another.

Based on this law, the energy of a substance can be converted from one form to another

For example, energy can be converted as follows;

potential energy to kinetic energychemical energy to electric energyelectrical energy to sound energyetc,

The sequence of energy that takes place on street lights powered by solar panels is given as follows;

Radiant energy (light energy from sun) to electrical energy (converted by photo voltaic cell of the panels) to radiant energy (light given by the street lights).

Thus, the sequence that shows the energy transformations taking place in these lights are radiant energyelectric energyradiant energy.

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Answer:

C

Explanation:

I did the test

Which part of the rock cycle can only occur because of thermal convection?

1. Magma changing to igneous rock
2. Igneous rock changing to metamorphic rock
3. Metamorphic rock changing to sedimentary rock
4. Sediment changing to sedimentary rock

Answers

Answer:

magma changing to igneous rock

Explanation:

In rock formation, there are several processes involved which could include sedimentation, melting, crystallization, etc but there is a part of the rock cycle that can only occur because of thermal convection which is magma changing to igneous rock.

This occurs when the heat from inside the earth and cooler temperature moves the rock in its liquid state which is known as convection and can also be defined as a cycle of heat transfer from where the hot material rises and the cool material condenses.

A small object of mass 0.500 kg is attached by a 0.440 m-long cord to a pin set into the surface of a frictionless table top. The object moves in a circle on the horizontal surface with a speed of 5.34 m/s.What is the magnitude of the radial acceleration of the object? What is the tension in the cord?

Answers

Given data:

* The mass of the object attached is m = 0.5 kg.

* The radius of the circle is r = 0.44 m.

* The speed of the object moving in circular motion is v = 5.34 m/s.

Solution:

(a). The radial acceleration of the object is also known as the centripetal acceleration of the object.

The value of centripetal acceleration in terms of the velocity of the object is,

[tex]a_c=\frac{v^2}{r}[/tex]

Substituting the known values,

[tex]\begin{gathered} a_c=\frac{5.34^2}{0.44} \\ a_c=64.8ms^{-2} \end{gathered}[/tex]

Thus, the radial acceleration of the object is 64.8 meters per second squared.

(b). The tension in the chord is equivalent to the centripetal force acting on the object which helps it to move in the circular motion.

Thus, the tension acting on the chord is,

[tex]F=ma_c[/tex]

Substituting the known values,

[tex]\begin{gathered} F=0.5\times64.8 \\ F=32.4\text{ N} \end{gathered}[/tex]

Thus, the tension acting in the chord is 32.4 N.

How long does it take to  stop a 1000 kg object moving at 20 m/s with a force 5000N? 2500 N, 1000 N, 500 N, 400 N, 200 N, 100 N

Answers

ANSWER

[tex]\begin{gathered} 5000N\Rightarrow4s \\ 2500N=8s \end{gathered}[/tex]

EXPLANATION

To find the time taken to stop the object, we first have to find the acceleration of the object.

Since the force is working to stop the object (slow down the object), it means that the object is decelerating (slowing down).

To find the acceleration (deceleration), apply Newton's second law of motion:

[tex]F=ma[/tex]

where F = force; m = mass; a = acceleration

Therefore, for a force of 5000N, we have that:

[tex]\begin{gathered} 5000=1000\cdot a \\ \Rightarrow\frac{5000}{1000}=a \\ \Rightarrow a=5m\/s^2 \end{gathered}[/tex]

Now, we can apply Newton's equation of motion to find the time taken to stop the object:

[tex]v=u-at[/tex]

where v = final velocity = 0 m/s; u = initial velocity = 20 m/s; t = time taken

Note: the negative sign indicates deceleration

Hence, the time taken for a force of 5000N to stop the object is:

[tex]\begin{gathered} 0=20-5\cdot t \\ \Rightarrow5t=20 \\ \Rightarrow t=\frac{20}{5} \\ t=4s \end{gathered}[/tex]

For a force of 2500N, the deceleration is:

[tex]\begin{gathered} 2500=1000\cdot a \\ a=\frac{2500}{1000} \\ a=2.5m\/s^2 \end{gathered}[/tex]

Hence, the time taken for a force of 2500N to stop the object is:

[tex]\begin{gathered} 0=20-2.5\cdot t \\ \Rightarrow2.5t=20 \\ t=\frac{20}{2.5} \\ t=8s \end{gathered}[/tex]

Consider a fluid of density 3.43 g⋅cm−3 flowing through a pipe of varying cross-section. The diameter of the pipe in one section is 9.1 cm, while the diameter in a second section is 12.6 cm. When the diameter of the pipe is 9.1 cm, the flow speed of the fluid is 339 cm⋅s−1 and the pressure is 2.93 × 105 Pa.A)Calculate the flow speed (in m⋅s−1) of the fluid when the diameter of the pipe is 12.6 cm. B)Calculate the pressure (in × 105 Pa) when the pipe has a diameter of 12.6 cm

Answers

Given that the pipe has varying cross-sections.

The diameter of one section is d1 = 9.1 cm and the diameter of second section is d2 = 12.6 cm.

Also, the fluid has the density,

[tex]\rho=3.43gcm^{-3}[/tex]

The area of the cross-section for the first section is

[tex]\begin{gathered} A_1=\frac{\pi(d1)^2}{4} \\ =\frac{\pi(9.1)^2}{4}cm^2 \end{gathered}[/tex]

The area of the cross-section for the second section is

[tex]\begin{gathered} A_2=\frac{\pi(d2)^2}{4} \\ =\frac{\pi(12.6)^2}{4}cm^2 \end{gathered}[/tex]

The flow speed for the first section is v1 = 339 cm s^-1

The flow speed for the second section will be v2.

(a) The flow speed for the second section can be calculated as

[tex]\begin{gathered} A_1v1=A_2_{}v2 \\ v2=\frac{A_1v1}{A_2} \\ =\frac{\pi(9.1)^2\times339\times4}{4\times\pi\times(12.6)^2} \\ =\text{ 176.82 cm/s} \\ =1.7682\text{ m/s} \end{gathered}[/tex]

(b) The pressure for first section is p1 = 2.93 x 10^5 Pa

The pressure for the second section will be p2.

The pressure for the second section can be calculated by the formula,

[tex]\begin{gathered} p2=p1+\frac{1}{2}\rho\mleft\lbrace(v1)^2-(v2\mright)^2\} \\ =2.93\times10^5+\frac{1}{2}\times3.43\mleft\lbrace(339)^2-(176.82)^2\mright\rbrace \\ =4.36\text{ }\times10^5\text{ Pa} \end{gathered}[/tex]

DQuestion 111 ptsAn object has a mass of 12 kg. Assume the acceleration due to gravity is 10 m/s². If it is lifted to aheight of 20 m, what is its gravitational potential energy?2400 JoulesO 2400 NewtonsO 200 JoulesO 120 Newtons

Answers

Given:

• Mass, m = 12 kg

,

• Acceleration due to gravity, g = 10 m/s²

,

• Height, h = 20 m

Let's find the gravitational potential energy.

To find the gravitational potential energy, apply the formula:

[tex]GPE=m*g*h[/tex]

Where GPE is the gravitational potential energy.

Thus, we have:

[tex]\begin{gathered} GPE=12*10*20 \\ \\ GPE=2400\text{ J} \end{gathered}[/tex]

Therefore, the gravitational potential energy is 2400 Joules.

ANSWERl

Electric charges exist within atoms. Is this true or false?

Answers

This statement is true because within atoms we have protons and electrons, which have positive and negative charge, respectively.

Therefore, the answer is true.

Answer: True

Explanation:

Aced the test

Power equals work multiplied by time.Question 9 options:TrueFalse

Answers

The power in terms of work done and time is defined as,

[tex]P=\frac{W}{t}[/tex]

Thus, power is is workdone per unit time.

Hence, given statement is false.

A cylinder of gas at room temperature has a pressure . To p_{1} what temperature in degrees Celsius would the temperature have to be increased for the pressure to be 1.5p_{1} ,

Answers

In order to calculate the temperature, we need to know that temperature and pressure are directly proportional, that is, if the pressure increases, the temperature (in Kelvin) also increases in the same proportion.

So, first let's convert the temperature from Celsius to Kelvin, by adding 273 units:

[tex]\begin{gathered} K=C+273 \\ K=20+273 \\ K=293 \end{gathered}[/tex]

Then, let's calculate the proportion:

[tex]\begin{gathered} \frac{P_1}{T_1}=\frac{P_2}{T_2} \\ \frac{p_1}{293}=\frac{1.5p_1}{T_2} \\ \frac{1}{293}=\frac{1.5}{T_2} \\ T_2=1.5\cdot293 \\ T_2=439.5\text{ K} \end{gathered}[/tex]

Now, converting back to Celsius, we have:

[tex]\begin{gathered} C=K-273 \\ C=439.5-273 \\ C=166.5\text{ \degree{}C} \end{gathered}[/tex]

So the temperature would be 166.5 °C.

having been given a newly discovered mineral, carry out a method in the physics lab on how to determine its specific gravity

Answers

ANSWER

Weigh a piece of the material, drop it in water and weigh the amount of water displaced. Specific gravity is the quotient between the mass of the mineral and the mass of water.

EXPLANATION

The specific gravity is the ratio of the mineral's density to the density of water at 23°C,

[tex]specific\text{ }gravity=\frac{density\text{ }of\text{ }mineral}{density\text{ }of\text{ }water}[/tex]

Density, denoted with the greek letter ρ, is the ratio between the mass and volume of the substance,

[tex]\rho=\frac{m}{V}[/tex]

If the mineral is a solid piece, we can weigh it - so we know the mass, and can find its volume by dropping it into water. Then, if we weigh the amount of water displaced, we have the mass of water as well - by Archimedes principle, the volume of water will be the volume of the object,

[tex]specific\text{ }gravity=\frac{\frac{m_{mineral}}{V}}{\frac{m_{water}}{V}}=\frac{m_{mineral}}{m_{water}}[/tex]

Use the data that you have for the air-filled tube. The tube is closed at one end.• Make a drawing of the standing wave in the tube when you have two, three, and four nodes or peaks. These will correspond to the first, second, and third overtones.• For each standing wave, label the nodes and the antinodes.• Label and record the distance in meters of the half wavelength (λ/2) for each standing wave.• Explain how the standing wave occurs.

Answers

Part 1)

When a standing wave oscilates in an air-filled tube closed at one end and open at the other, the closed end is a node and the open end is an antinode. Then, the diagram of the wave inside the tube looks like the following:

Part 2)

The nodes are the points where the two curves intersect (darker regions) and the antinodes are the clearer regions.

Part 3)

In each wave on the diagram, the distance between two consecutive nodes or two consecutive antinodes is the same as half the wavelength.

Part 4)

The standing waves arise from the combination of reflection and interference of waves inside the air column when the wavelengths match the length of the tube in such a way that the open end is an antinode and the closed end is a node.

The ammeter in the figure below reads 1.0A. Calculate the magnitude of the currents I1 and I2.

Answers

Given:

• Ammeter reading, I = 1.0 A

,

• R1 = 2.0 Ω

,

• R2 = 4.0 Ω

,

• R3 = 5.0 Ω

Let's find the magnitude of the currents in I1 and I2.

Apply Kirchhoff's Current Law.

We have:

[tex]I_1+I_2+I_3=0[/tex]

From Kirchoff's junction rule:

[tex]I_1-I_2-I_3=1A[/tex]

Thus, we have:

[tex]\begin{gathered} \frac{1}{R_{eq}}=\frac{1}{2}+\frac{1}{4}+\frac{1}{5} \\ \\ \frac{1}{R_{eq}}=\frac{10+5+4}{20}=\frac{19}{20} \\ \\ R_{eq}=\frac{20}{19}=1.1\text{ ohms} \end{gathered}[/tex]

Apply Ohm's law to find the voltage in the circuit:

[tex]\begin{gathered} V=IR \\ \\ V=1.0*1.1 \\ \\ V=1.1\text{ V} \end{gathered}[/tex]

To find I1, apply Kirchoff's loop law:

[tex][/tex]

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