Answer:
A low frequency sound (say, 50 hz) might sound like a low rumble, while a high frequency sound (say 12,000 hz), might sound more like a "sizzle". ... The higher the frequency, the shorter the distance between each successive compression (or rarification) in the incoming sound wave. This distance is called the "wavelength".
Explanation:
The number of cycles per second (Hertz) (abbreviated as, hz) of anything that oscillates is called the "frequency". The electricity of an AC wall outlet is said to have a frequency of 60 Hertz as it cycles negative then positive 60 times each second.
Sound is an oscillating wave, but it has a broad range of frequencies. A low frequency sound (say, 50 hz) might sound like a low rumble, while a high frequency sound (say 12,000 hz), might sound more like a "sizzle". A person with normal hearing can hear all the way up to about 20,000 hz.
Sound is actually more like "compression" waves, rather than waves on the ocean. When something makes a sound, the air is compressed or rarified in waves that travel out from that source in all directions. When those compressed or rarified areas of air hit your eardrum, it vibrates in sympathy with those compression waves in the air and allows you to hear.
The higher the frequency, the shorter the distance between each successive compression (or rarification) in the incoming sound wave. This distance is called the "wavelength". Sound travels at about 750 miles/hour, so the compression waves between 100 hz and 20,000 hz have wavelengths that range between several feet (for the 100 hz sound) to a fraction of an inch (for the 20,000 hz sound).
Speech also has a range of frequencies, but it mostly limited to the range between a hundred (or so) hz and 8,000 (or so) hz. The frequencies that make up vowel sounds are typically lower frequencies, while the consonant sounds (at least the parts that help us hear which consonant sound was spoken) tend to be higher frequency sounds.
People with even moderately good hearing up to about 3,000 hz can understand speech fairly well. Wired telephones typically do not transmit sound above 3,500 hz.
When people have a hearing loss, they usually have less ability to hear some frequencies than to hear other frequencies. This creates a distortion that can make it difficult to understand speech. Usually, people with hearing loss have poorer hearing in high frequencies than in low frequencies, but some people may lose lower frequencies, or even middle frequencies, while having less loss in other frequencies.
Hearing aids now have the ability to be tuned to amplify different frequencies by differing amounts, so that it fills in where you need the amplification without giving you too much sound in frequencies where you hear better.
The low frequency wave has longer wavelength and less energy compare to high frequency waves.
What is frequency of wave?The frequency of a repeated event is its number of instances per unit of time. It differs from angular frequency and is sometimes referred to as temporal frequency for clarification. The unit of frequency is hertz (Hz).
A wave's wavelength and frequency are inversely proportional. Accordingly, waves with high frequencies have a short wavelength, whereas waves with low frequencies have a longer wavelength.
Very, very short wavelengths characterize light waves. While blue and purple light have even shorter waves with wavelengths around 400 or 500 nanometers (nm), red light waves have wavelengths about 700 nm. With wavelengths ranging from millimeters to kilometers, radio waves, another kind of electromagnetic radiation, have far longer waves than light.
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What are five of the most important things you learned in physics? ASAP
Answer:
The definitions of mass, energy, time and space as used in Physics are circular
The Conservation of Mass and Energy
The definition of zero is not sufficiently defined in Physics
Newton’s Laws of Motion are related by Calculus
Knowing the terms of Physics is most important.
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Sound travels slower in colder air than it does in warmer air. Make a claim about the effect of air temperature on the speed of sound.
Answer: The temperature of the air affects the speed of sound in the due to the fact that colder contains air molecules with low kinetic energy. For warm air the kinetic energy of the molecules is high causing them to move in rapid motion. Because of these vibrations and collisions sound waves moves in a faster rate.
Explanation:
How is temperature and viscosity related?
Answer:
With an increase in temperature, there is typically an increase in the molecular interchange as molecules move faster in higher temperatures. The gas viscosity will increase with temperature. ... With high temperatures, viscosity increases in gases and decreases in liquids, the drag force will do the same.
3. What is the net force acting on the box in the diagram below? (1 mark)
Answer: Fnet = 40 N on the right
Explanation: The net force is the summation of forces acting on an object. Add the forces in the same direction and subtract to the opposing force moving in the opposite side.
Fnet= F1 + F2 - F3
= 120 N + 60 N - 140 N
= 180 N - 140 N
= 40 N on the right
5) Consider pushing a 50.0 kg box through a 5.00m displacement on both a flat surface and up a
ramp inclined to the horizontal by 15.0°. In both cases, you apply a force of 100. N parallel to the
surface (parallel to the floor or parallel to the ramp). Calculate the work done by:
a) the gravitational force as the box is pushed across the flat ground
b) the gravitational force as the box is pushed up the ramp
c) the force you apply as the box is pushed across the flat ground
d) the force you apply as the box is pushed up the ramp
Explanation:
Work equals the force times the parallel distance.
a) Force gravity is in the downward direction. The box is moving on flat ground, so there's no displacement in that direction.
W = Fd
W = (50 kg) (-9.8 m/s²) (0 m)
W = 0 J
b) This time, there is a displacement in the vertical direction.
W = Fd
W = (50 kg) (-9.8 m/s²) (5.00 m sin 15.0°)
W = -634 J
c) The force is parallel to the displacement.
W = Fd
W = (100. N) (5.00 m)
W = 500. J
d) Again, the force is parallel to the displacement.
W = Fd
W = (100. N) (5.00 m)
W = 500. J
A force of 30N acts through a distance of 4m in the direction of the force, what is the work done
Answer:
120J
Explanation:
Work done = Force (N) X Distance (m)
Answer:
Work = 120Nm or 120J
Explanation:
Work is said to be done when a force moves a body over a distance, in the direction of force applied.
Work done = force x distance (in the direction of the force)
Therefore work = force x displacement
As displacement is distance in a specific or specified direction.
From the parameters given
Force = 30N
Displacement = 4m.
Therefore work = force x displacement
= 30N x 4m
Work = 120Nm or 120J
Note :J is joules which is the S. I unit of work
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What effect does Earth’s spherical shape have on the amount of sunlight that reaches the equator?
A.
The latitudes near the equator receive direct sunlight, which causes a high-pressure system.
B.
The latitudes near the equator receive indirect sunlight, which causes a low-pressure system.
C.
The latitudes near the equator receive direct sunlight, which causes a low-pressure system.
D.
The latitudes near the equator receive indirect sunlight, which causes a high-pressure system.
A
Explanation:
The earth is spherical. So it's middle part is bulgjng outside. So more sunlight will be incident on the latitude near the equator. This will heat the air and it will rise up. This will cause high pressure difference and polar disturbances.
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The Earth's spherical shape ensures that the equator receives the most direct sunlight, resulting in higher temperatures and a low-pressure system due to the rising warm air. This leads to the high precipitation in the equatorial region and plays a significant role in global atmospheric and oceanic circulation patterns.
The spherical shape of the Earth influences the way sunlight is distributed across different latitudes. At the equator, the sun's rays are most direct, resulting in a higher concentration of solar energy and higher temperatures. This intense sunlight is focused on a smaller area, leading to the warm conditions that are characteristic of equatorial regions. Due to the heat at the equator, the air warms up, becomes less dense and creates a low-pressure system.
As a consequence of the heat and low-pressure conditions, there is significant convection, which causes air to rise. This is part of the reason why the equatorial region, also known as the Intertropical Convergence Zone, is often characterized by high precipitation levels. The rising warm air can lead to cloud formation and frequent rainfall. In contrast, at higher latitudes, the sun's rays strike the surface at a more oblique angle, spreading the energy over a wider area and resulting in cooler temperatures and different atmospheric conditions.
Therefore, the correct answer to the question is: C. The latitudes near the equator receive direct sunlight, which causes a low-pressure system.
2. A man walks 38 m west from home, rest, then walks 20 m east. What is the man's displacement?
Answer:18m west
Explanation:
Since they are moving in opposite direction we subtract to get the displacement
Displacement =38-20
Displacement =18m in west direction
How does slope on this graph indicate the amount of density?
Density comparison
Question 3 options:
A higher density is indicated by a slope that is not steep (yellow line)
The line in the middle (red line) has the highest density
The steeper slope (green line) indicates a higher density
Final answer:
The slope of a line on the graph represents the amount of density. A steeper slope indicates a higher density, while a flatter slope indicates a lower density.
Explanation:
The slope of a line on the graph represents the amount of density in this scenario. A steeper slope indicates a higher density, while a flatter slope indicates a lower density.
For example, in the given graph, the steeper slope (green line) indicates a higher density, while the slope of the yellow line is not steep, indicating a higher density as well.
Therefore, the statement 'The steeper slope (green line) indicates a higher density' is correct.
superposition of a crest and trough produces
Answer:
pure constructive interference
Explanation:
The crests of two waves and the troughs are aligned. The superposition produces pure constructive interference. Because the disturbances add pure constructive interference, it produces a wave that has twice the amplitude of the individual waves but has the same wavelength.
How much pressure is applied to the ground
by a 93 kg man who is standing on square
stilts that measure 0.04 m on each edge?
Answer in units of Pa.
003 (part 2 of 2) 10.0 points
What is this pressure in pounds per square
inch?
Answer in units of lb/in.
Answer:
Pressure applied by the man= 285103.125 [tex]Pa[/tex] or 41.35 [tex]lb/in^{2}[/tex]
Explanation:
Pressure is defined as the perpendicular force applied per unit area.
i.e. [tex]Pressure=\frac{Force}{Area}[/tex]
Now, [tex]Force= mg[/tex]
where, [tex]m[/tex] = mass of the body(man) = 93 kg
[tex]g[/tex] = acceleration due to gravity of Earth = 9.81 [tex]m/{s^{2}}[/tex]
[tex]Area[/tex] covered is equal to the area of both stilts(a man generally stands on two feet)
therefore [tex]Area=2(0.04)^{2}[/tex] [tex]m^{2}[/tex]
and putting in the values, we get,
[tex]Pressure=\frac{93\times9.81}{2\times(0.04)^{2}}Nm^{-2}=285103.125Nm^{-2}[/tex]
Now we need to convert to our required units:
[tex]1Nm^{-2}=1Pa\\1Pa=0.000145038lb/in^{2}[/tex]
(We can get the above result by individually converting kg to lb and meters to inches respectively)
Using the above relations we get,
[tex]Pressure=285103.125Pa=0.000145038\times285103.125lb/in^{2}=41.35lb/in^{2}[/tex]
Based on the formula for pressure, the pressure applied on the ground is 570206.25 N/m² or 82.7 pound per square inch
What is pressure?Pressure is defined as the perpendicular force applied per unit area.
Pressure = Force/AreaAlso,
Force = mgwhere,
mass of the body = 93 kg
g = 9.81 m/s²
Force = 93 * 9,81
Force = 912.33 N
Area of square stilt = (0.04m)²
Area = 1.6 * 10⁻³ m²
Then;
Pressure = 912.33 N/1.6 * 10⁻³ m²
Pressure = 570206.25 N/m²
Converting to pounds per square inch:
1 pound per square inch = 6894.76 N/m²
570206.25 N/m² = 570206.25 N/m² * 1 pound per square inch/6894.76 N/m²
Pressure = 82.7 pound per square inch
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Which statement describes Newton’s conception of the Solar System?
1.) the Sun at the center, with the Earth and other planets orbiting the Sun
2.) the Earth orbiting the Sun, with other planets orbiting the Earth
3.) the Earth at the center, with the Sun and other planets orbiting Earth
4.) the Sun orbiting the Earth, with other planets orbiting the Sun
PLEASE HELP I NEED THIS ANSWER QUICK!
Answer:
1). the sun at the center, with the earth and other planets orbiting the sun.
Explanation:
Newton believes that it's gravity between the sun and the other planets even at the large distances that cause them to orbit around the sun. This force build up by the masses of the sun and the planets which keeps the planets in their orbit.
Answer:the sun at the center, with the earth and other planets orbiting the sun.
Explanation:
A 20-newton weight is attached to a spring.
causing it to stretch, as shown in the diagram.
What is the spring constant of this spring?
Answer:
40 N/m
Explanation:
The diagram attached is used to answer the question
We know from Hooke's law that extension is directly proportional to the applied force hence
F=kx where x is extension, F is applied force and k is the spring constant. Making k the subject of the formula then
[tex]k=\frac {F}{k}[/tex]
From the attached diagram extension is given by subtracting unstretched spring from stretched spring hence extension, x=1-0.5=0.5m
Substituting 20 N for F and 0.5 m for x then
[tex]k=\frac {20}{0.5}=40 N/m[/tex]
A battery is marked. As having a voltage of 12V and is connected to a circuit that has a resistance of 5 ohms.if the terminal voltage is 8.75V. What is the internal resistance
Answer:
1.857 Ohms
Explanation:
The marked voltage of the battery is it's E.M.F. (Electromotive force). It is the maximum voltage the battery can supply because, ideally internal resistance of a cell or battery should be zero. Let us denote the E.M.F. as E.
In reality we get a lower voltage called the Terminal Voltage (Voltage across the terminals of the battery). Let us call that V. This is due to the voltage drop caused by the internal resistance (say 'r').
So when the circuit is connected a current I flows through a resistance
R(= 5 ohms, given).
Now, potential difference = Terminal Voltage = V = 8.75 Volts
Hence, [tex]I=\frac{V}{R} =\frac{8.75}{5}=1.75 Amperes[/tex]
Now, by conservation of energy we can say that,
E.M.F. = Terminal Voltage + Voltage drop due to the battery's internal resistance
or, [tex]E=V+Ir[/tex]
or, [tex]r=\frac{E-V}{I}=\frac{12-8.75}{1.75}=1.857[/tex] Ohms
what is 1/12+7/9 hj
Explanation:
[tex] \frac{1}{12} + \frac{7}{9} \\ \\ = \frac{1 \times 3}{12 \times 3} + \frac{7 \times 4}{9 \times 4} \\ \\ = \frac{3}{36} + \frac{28}{36} \\ \\ = \frac{3 + 28}{36} \\ \\ = \frac{31}{36} \\ \\ \huge \purple{ \boxed{\therefore \: \frac{1}{12} + \frac{7}{9} = \frac{31}{36} }} \\ [/tex]
Answer:
31/36
Explanation:
do the math you will see
Calculate the kinetic energy in joules of a 10 g bullet moving at 300 m/s.
____J
1. 900
2. 450
3. 15
4. 45
Answer:450joules
Explanation:
Let mass=m velocity =v
Mass given=10g=(10/1000)kg=0.01kg
Velocity given =300m/s
Kinetic energy =(m x v^2)/2
Kinetic energy =(0.01x300^2)/2
Kinetic energy =(0.01 x 90000)/2
Kinetic energy =900/2
Kinetic energy =450joules
4) Consider a separate rocket, also in deep space with a mass of 30.0 kg. If the rocket is
observed to be travelling at v= 5.00 m/s at t= 3.00 s and then travelling at û = -2.00 m/s at t= 14.0 s
with constant acceleration, calculate:
a) the acceleration, a.
b) the force F, acting on the rocket from the thrusters,
Answer:
(a) -0.636 m/s²
(b) -19.08 N
Explanation:
Given:
Mass of the rocket (m) = 30.0 kg
Initial velocity of rocket (v) = 5.00 m/s
Initial time of rocket (t₁) = 3.00 s
Final velocity of the rocket (u) = -2.00 m/s
Final time of rocket (t₂) = 14.0 s
(a)
Acceleration is given as the rate of change of velocity. Therefore,
[tex]a=\frac{u-v}{t_2-t_1}\\\\a=\frac{-2.00-5.00}{14.00-3.00}\ m/s^2\\\\a=\frac{-7.00}{11.00}\ m/s^2\\\\a=-0.636\ m/s^2[/tex]
Therefore, the acceleration of the rocket is 0.636 m/s².
(b)
From Newton's second law, we know that, force acting on a body is equal to the product of its mass and acceleration.
So, the force acting on the rocket is given as:
[tex]F=ma\\\\F=(30.0\ kg)(-0.636\ m/s^2)\\\\F=-19.08\ N[/tex]
Therefore, the force acting on the rocket is -19.08 N.
The negative sign implies the force acts in the direction opposite to motion.
67. Race Car A race car is slowed with a constant accel-
eration of 11 m/s2 opposite the direction of motion.
a. If the car is going 55 m/s, how many meters will it
travel before it stops?
b. How many meters will it take to stop a car going
twice as fast?
Answer:
a. 137.5 m
b. 550 m
Explanation:
Accelerated Motion
If an object is changing its velocity at a constant rate, it has a uniformly accelerated motion. When the object is moving in one fixed axis, then the sign of the acceleration is negative if the object is braking, and positive if the object is increasing its speed.
The initial speed vo, final speed vt, acceleration a, and distance traveled x are related by the formula
[tex]v_f^2=v_o^2+2ax[/tex]
a. The Race Car A has an initial speed of 55 m/s and it's said to stop. We must find at what distance it goes to vf=0. This means that the above formula becomes
[tex]0=v_o^2+2ax[/tex]
Solving for x
[tex]\displaystyle x=-\frac{v_o^2}{2a}[/tex]
The acceleration is [tex]-11\ m/s^2[/tex], negative because it's against the movement. Thus
[tex]\displaystyle x=-\frac{55^2}{2\times (-11)}[/tex]
[tex]x=137.5\ m[/tex]
b. If the car is going twice as fast (v0=110 m/s), then
[tex]\displaystyle x=-\frac{110^2}{2\times (-11)}[/tex]
[tex]x=550\ m[/tex]
Light in a vacuum travels at a constant speed of 3x10^8 m/s. If the moons average distance from the earth is 38776106 km how long would it take for a beam of light to travel from earth to the moon and back to earth again?
Answer: 258.3 s
Explanation:
The speed [tex]s[/tex] is given by the following equation:
[tex]s=\frac{D}{t}[/tex]
Where:
[tex]s=3(10)^{8} m/s[/tex] is the speed of light in vacuum
[tex]D=2(38776106 km \frac{1000 m}{1 km})=7.75(10)^{10} m[/tex] is the double of the distance between Earth and Moon, since the beam of light travels from Earth to the Moon and back to Earth again.
[tex]t[/tex] is the time it takes to the beam of light to travel the mentioned distance
Isolating [tex]t[/tex] and solving with the given information:
[tex]t=\frac{D}{s}[/tex]
[tex]t=\frac{7.75(10)^{10} m}{3(10)^{8} m/s}[/tex]
Finally:
[tex]t=258.3 s \approx 258 s[/tex]
A fully loaded Saturn V rocket has a mass of 2.92 x 106 kg. Its engines have a thrust of 3.34 x 107 N.
Complete Question:
A fully loaded Saturn V rocket has a mass of 2.92 x 106 kg. Its engines have a thrust of 3.34 x 107 N. (8 marks)
a) What is the downward force of gravity on the rocket at blast-off?
b) What is the unbalanced force on the rocket at blast-off?
c) What is the acceleration of the rocket as it leaves the launching pad?
d) As the rocket travels upwards, the engine thrust remains constant, but the mass of the rocket decreases. Why?
e) Does the acceleration of the rocket increase, decrease, or remain the same as the engines continue to fire?
Answer:
a) [tex]-2.8616 \times 10^{7} N[/tex] is the downward force of gravity on the rocket at blast-off.
b) [tex]4.784 \times 10^{6} N[/tex] is the unbalanced force on the rocket at blast-off
c) [tex]1.638 \mathrm{m} / \mathrm{s}^{2}[/tex] is the acceleration of the rocket as it leaves the launching pad
d) Because the propellant here is burned up, hence the mass of the rocket seems to be varied (total mass of all its parts). Thereby, the mass decreases when the rocket moves upward.
e) The acceleration of the rocket increases when engines continue to fire
Explanation:
Given:
Mass (m) = [tex]2.92 \times 10^{6} \mathrm{kg}[/tex]
a) In physics, weight can be defined as the applied force on a body by gravity. It is the product of mass (m) and gravity [tex]\left(g=9.8 \mathrm{m} / \mathrm{s}^{2}\right)[/tex]
[tex]\text { weight }(W)=m \times g=2.92 \times 10^{6} \times(-9.8)=28.616 \times 10^{6}=-2.8616 \times 10^{7} N[/tex]
The negative sign indicates the downward force of gravity.
b) To find the unbalanced force on the rocket at blast-off,
Accelerating force,
[tex]F_{a}=F+W=3.34 \times 10^{7}+\left(-2.8616 \times 10^{7}\right)=(3.34-2.8616) \times 10^{7}[/tex]
[tex]F_{a}=0.4784 \times 10^{7}=4.784 \times 10^{6} N[/tex]
c) Newton’s second law of motion states that the object’s acceleration depends on two variable:
Directly proportionate to the object’s force existed Inversely proportionate to the mass of the objectsThe equation can be given as below,
[tex]Force =m \times acceleration[/tex]
[tex]\text { Acceleration }=\frac{F_{a}}{m}=\frac{4.784 \times 10^{6}}{2.92 \times 10^{6}}=1.638 \mathrm{m} / \mathrm{s}^{2}[/tex]
d) The pushing of rocket upward will happen as long as the engine gets fired. The propellant here is burned up, hence the mass of the rocket seems to be varied (total mass of all its parts). Thereby, the mass decreases (taotal mass) when the rocket moves upward.
e) The acceleration of the rocket increases when engines continue to fire
Let consider [tex]F_{a}[/tex] is constant, mass gets decreasing, then the acceleration would be increasing (as mass and acceleration are inversely proportionate to each other) .
If you were able to jog one mile in 10 minutes, how would you progressively increase your
performance by using each of the following? Be specific by using numbers in your answers.
1. (F) frequency variable?
2. (1) intensity variable?
3. (T) time variable?
Answer:
1. (F) Increasing the frequency variable reduces the time required.
2. (I) Intensity variable can be related the the energy expended.
3.(T) Time variable would affect the speed.
Explanation:
1. (F) increasing the frequency variable implies that the distance would be covered in lesser time. As a wave of 250Hz cover more distance than that of 200Hz.
2. (I) Intensity variable. An increase in this variable implies that more energy is expended.
3.(T) Time variable can either give the required out come or not. As increasing the time would mean that the speed has reduced, while decreasing the time means the distance would be covered quickly. If the required speed is 20m/s, then increasing it to 30m/s imples that lesser time would be recorded. If the reverse is the case, then more time would be recorded. time = distance/speed
1. reduces the time required.
2.energy expended.
3.affect the speed.
What is Frequency, time period and intensity ?Frequency, time period and intensity are quantitative dimensions which describe the amount of physical activity taking place during a given time.
1. (F) frequency variable - Frequency refers to the number of waves that pass a fixed point in unit time.
i.e. [tex]F =\frac{1}{T}[/tex]
When frequency is increased then distance would be covered in lesser time.
For example, a wave of 550Hz cover more distance than that of 500Hz.
2. (I) Intensity variable- The quantity of energy the wave conveys per unit time across a surface of unit area.
i.e. [tex]I = \frac{P}{4\pi r^{2} }[/tex]
When Intensity increases it implies that more energy is expended.
3.(T) Time variable -It is define as the ratio of distance and time.
i.e. [tex]time = \frac{distance}{speed}[/tex]
When time variable is increased than the quantity of energy the wave conveys per unit time across a surface of unit area, speed is reduced.
When time variable is decreased the distance would be covered quickly.
For example : If the speed of car is 80m/s, then by increasing the speed to 100m/s shows that lesser time would be required.
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A 12.0 V battery is attached to a circuit with a resistor of 5.5 Ohms. The internal resistance of the circuit is 1.75 Ohms. What is the terminal voltage of the circuit?
The end voltage is 9.01 volts.
Explanation:The emf of the battery is 12 volts.
The resistance of the circuit = 1.75 ohms.
The resistance of the applicant attached = 5.5 ohms.
So, total resistance of the circuit = [tex]1.75 + 5.5[/tex] ohms = 7.25 ohms.
According to the equation, the voltage applied is equal to the product of the current and the resistance.
V=iR.
So, 12 = [tex]i \times7.25[/tex].
So, i = 1.66.
Therefore, the voltage at the end of the circuit = [tex]12 - 1.66\times1.75[/tex]volts
= 9.01volts.
So the end voltage is 9.01 volts.
two different examples of forces that could cause a change in motion
Answer:
a)Kicking a ball; b) A car that gets hit
Explanation:
a)
A classic example is when a soccer player kicks a moving ball in another direction by modifying the original direction and speed that the ball brought.
b)
Another example is when a car goes on a highway and suddenly is hit by another car on its side modifying the original direction and speed.
1. Suppose you give a shopping cart a push and it begins t
give a shopping cart a push and it begins to roll across the parking
lot. Once you have stopped pushing it, it slows down and comes to a sto
observation consistent with Newton's First Law? Why or why not?
Newton's first law says the cart should keep going if there's no force acting on it to slow it down. But it slows down and stops. Is there any force acting on it ? You bet ! There's friction in the wheels, and a little bit of air resistance too.
An object is placed 12cm from a converging lens of focal length 18cm. Find the position of the image.
Answer:
The position of the image = 7.2 cm.
Explanation:
Given:
A Converging lens:
Object distance [tex](d_o)[/tex] = 12 cm = -12 cm (sign convention)
Focal length [tex](f)[/tex] = 18 cm
We have to find the position of the image.
Let the position of the image be [tex](d_i)[/tex] .
Sign convention:
The focal length of converging (convex) lens is always positive,while object distance is negative.
Using lens formula:
⇒ [tex]\frac{1}{object\ distance} + \frac{1}{image\ distance} =\frac{1}{focal\ length}[/tex]
⇒ [tex]\frac{1}{d_o} + \frac{1}{d_i} = \frac{1}{f}[/tex]
⇒ [tex]\frac{1}{d_i} =\frac{1}{f} - \frac{1}{d_o}[/tex]
⇒ [tex]\frac{1}{d_i} =\frac{1}{f} - \frac{1}{-d_o}[/tex]
⇒ [tex]\frac{1}{d_i} =\frac{1}{f} + \frac{1}{d_o}[/tex]
⇒ [tex]d_i=\frac{d_o\times f}{d_o+f}[/tex]
⇒ [tex]d_i=\frac{12\times 18}{12+18}[/tex]
⇒ [tex]d_i=\frac{216}{30}[/tex]
⇒ [tex]d_i=7.2[/tex] cm
So the position of the image = 7.2 cm.
At which point(s) is the net force the greatest?
At which point(s) is the net force zero?
Final answer:
The net force is the greatest when the forces are in the same direction and have the greatest magnitude. The net force is zero when the forces are balanced and cancel each other out.
Explanation:
The net force on an object is the vector sum of all the forces acting on it. The magnitude of the net force can be determined by considering the individual forces acting on the object. The net force is the greatest at points where the forces acting on the object are in the same direction and have the greatest magnitude.
The net force is zero when the forces acting on the object are balanced and cancel each other out. This occurs at points where the forces are equal in magnitude and opposite in direction.
For example, let's say we have two forces acting on an object: 10 N to the right and 5 N to the left. At the point where the forces are added, the net force would be 5 N to the right (10 N - 5 N). This is the greatest net force. At the point where the forces cancel each other out, the net force would be zero (10 N - 10 N).
Fiona claims that the diagram below shows simple machines, but Chad claims that it shows a compound machine.
Answer:
Simple machine
Explanation:
It is a simple machine because the person wants to raise the load by an inclined plane. Simple machines perform work with the mechanical advantage offered by the machine itself, such as using a bar as Lever, lifting a load by means of a pulley. simple machines multiply or change direction a force. While Composite machines are the Union of several simple machines that perform a given job, examples of combined machines are found on bicycles, a washing machine, a car, and others.
Answer:
b
Explanation:
took test, other answer said b and was verified
A 20-newton weight is attached to a spring.
causing it to stretch, as shown in the diagram.
What is the spring constant of this spring?
Answer:
40 N/m
Explanation:
Assuming that the missing diagram is as attached, then we can deduce from Hooke's law that the extension of a spring is directly proportional to the applied force. This is mathematically expressed as
F=kx
Here, F represent the applied force, x denote the extension of the spring while k is the spring constant.
From the attached diagram, extension of the spring x=1-0.5=0.5m
Substituting 20 N for F and 0.5 m for x then
20=0.5k
k=20/0.5=40 N/m
During a rising tide Ocean waves often become larger if the amplitude of a wave increases by a Factor of 1.1, by how much does the energy increase
Answer:
The energy of the wave will increase by a factor of 1.21.
Explanation:
Given:
The amplitude of the ocean wave is increased by a factor of 1.1.
The energy associated with a wave depends on the amplitude of the wave.
The energy (E) is directly proportional to the square of the amplitude (A) of the wave and is expressed as:
[tex]E=kA^2\\Where\ k\to constant\ of\ proportionality[/tex]
So, if the amplitude is doubled, then the energy will become 4 times and if the amplitude is halved, then the energy reduced by a factor of one by four.
Here, the amplitude is increased by a factor of 1.1. So, the energy associated with it will increase by a factor of square of 1.1 which is given as:
[tex]Increase\ in\ Energy=(1.1)^2=1.21\ of\ the\ initial\ value[/tex]
Therefore, the energy of the wave will increase by a factor of 1.21.
A short circuit is _____.
Answer:
in a device, an electrical circuit of lower resistance than that of a normal circuit, typically resulting from the unintended contact of components and consequent accidental diversion of the current.
Explanation:
Answer: A short circuit is simply a low resistance connection between the two conductors supplying electrical power to any circuit. This results in excessive current flow in the power source through the 'short,' and may even cause the power source to be destroyed.
Explanation: