(a) 700 Hz
For standing waves on a string, the number of antinodes (n) corresponds to the order of the harmonic. So, three antinodes corresponds to the third harmonic. Also, the frequency of the nth-harmonic is the nth-integer multiple of the fundamental frequency, so we have:
[tex]f_3 = 3 f_1 = 420 Hz[/tex]
where [tex]f_1[/tex] is the fundamental frequency. Solving for f1, we find
[tex]f_1 = \frac{420 Hz}{3}=140 Hz[/tex]
And so now we can find the frequency of the 5th-harmonic:
[tex]f_5 = 5 f_1 = 5 (140 Hz)=700 Hz[/tex]
(b) 56.4 N
The fundamental frequency of a string is given by:
[tex]f_1 = \frac{1}{2L} \sqrt{\frac{T}{\mu}}[/tex]
where we have:
L = 60 cm = 0.60 m is the length of the string
[tex]\mu = 2.0 g/m = 0.002 kg/m[/tex] is the linear density
T = ? is the tension in the string
Solving the formula for T and using the fundamental frequency, f1=140 Hz, we find
[tex]T=\mu (2Lf_1)^2=(0.002 kg/m)(2(0.60 m)(140 Hz))^2=56.4 N[/tex]
The frequency of the fifth harmonic of the string is 700 Hz, and the tension in the string is 56.448 N.
Explanation:The question pertains to harmonic frequencies and tension in instruments, particularly a guitar string. Using the information given, we will apply the principles of standing waves on the string, wave speed derived from linear mass density and tension, and harmonic frequencies.
(a) If you're looking for the frequency of the fifth harmonic, note that the frequency of a particular harmonic is that harmonic number times the fundamental frequency. In this case, the string shows a standing wave with three antinodes, which refers to the third harmonic. So, the third harmonic frequency is 420 Hz; thus, the fundamental frequency is 420 Hz / 3 = 140 Hz. So, the frequency of the fifth harmonic is 5 times the fundamental frequency = 5*140 Hz = 700 Hz.
(b) To determine the tension in the string, you must first calculate the wave speed. The wave speed (Vw) can be derived from the formula for the frequency of a string, Vw=2Lf, where L is the length of the string in meters and f is the fundamental frequency. Here, L=0.6 m and f= 140 Hz, hence Vw = 2*(0.6m)*(140Hz) = 168 m/s. Given that wave speed is also calculated via the square root of tension (T) divided by linear mass density (μ), you can rearrange this formula to solve for T. This gives you T = (Vw)^2 * μ . The linear density is given as 2.0 g/m, but we need to convert it to kg/m to match the wave speed unit, so μ = 0.002 kg/m. Plugging in the values we find the tension: T = (168 m/s)^2 * 0.002 kg/m = 56.448 N.
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A 5.6 cm diameter parallel-plate capacitor has a 0.58 mm gap. What is the displacement current in the capacitor if the potential difference across the capacitor is increasing at 500,000 V/s?
Answer:
[tex]1.88\cdot 10^{-5} A[/tex]
Explanation:
The capacitance of a parallel plate capacitor is given by:
[tex]C=\frac{\epsilon_0 A}{d}[/tex] (1)
where
[tex]\epsilon_0[/tex] is the vacuum permittivity
A is the area of the plates
d is the separation between the plates
The charge stored on the capacitor is given by
[tex]Q=CV[/tex] (2)
where C is the capacitance and V is the voltage across the capacitor.
The displacement current in the capacitor is given by
[tex]J=\frac{Q}{t}[/tex] (3)
where t is the time elapsed
Substituting (1) and (2) into (3), we find an expression for the displacement current:
[tex]J=\frac{CV}{t}=\frac{\epsilon_0 A}{d} \frac{V}{t}[/tex]
where we have
[tex]A=\pi (\frac{d}{2})^2=\pi (\frac{0.056 m}{2})^2=2.46\cdot 10^{-3} m^2[/tex]
[tex]d = 0.58 mm = 5.8\cdot 10^{-4} m[/tex]
[tex]\frac{V}{t}=500,000 V/s[/tex]
Substituting into the equation, we find
[tex]J=\frac{(8.85\cdot 10^{-12} F/m)(2.46\cdot 10^{-3} m^2)}{5.8\cdot 10^{-4}m}(500,000 V/s)=1.88\cdot 10^{-5} A[/tex]
The displacement current in the parallel-plate capacitor is approximately 0.024 A.
Explanation:The displacement current in the capacitor can be calculated using the formula I = ε₀A(dV/dt), where ε₀ is the permittivity of free space, A is the area of the plates, and dV/dt is the rate of change of potential difference.
Given that the diameter of the capacitor is 5.6 cm, we can calculate the radius (r) by dividing the diameter by 2. The area (A) of the plates is then πr².
Plugging in the given values and solving for I, we get:
I = (8.85×10⁻¹² F/m)(π(0.028 m)²)(500,000 V/s)
I ≈ 0.024 A
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When an object gains or loses electrons, it becomes A) electrically charged. B) electrically neutral. C) magnetically neutral. D) magnetically polarized.
The correct answer is electrically charged.
When an object gains or loses electrons, it becomes electrically charged because it acquires an imbalance of positive and negative charges.
Explanation:
- Atoms are made up of positively charged protons in the nucleus and negatively charged electrons orbiting the nucleus.
- In a neutral atom, the number of protons and electrons are equal, resulting in an overall neutral charge.
- If an atom gains or loses electrons, it disrupts the balance of positive and negative charges, resulting in an electrically charged state.
- If an atom loses electrons, it becomes positively charged because it has more protons than electrons.
- If an atom gains electrons, it becomes negatively charged because it has more electrons than protons.
The other options are incorrect:
B) Electrically neutral: This option is incorrect because gaining or losing electrons causes an object to become electrically charged, not neutral.
C) Magnetically neutral: This option is incorrect because it refers to magnetic properties, not electric charges.
D) Magnetically polarized: This option is incorrect because it refers to magnetic properties, not electric charges.
Therefore, when an object gains or loses electrons, it becomes electrically charged due to the imbalance of positive and negative charges.
Complete question:
When an object gains or loses electrons, it becomes
A) electrically charged.
B) electrically neutral.
C) magnetically neutral.
D) magnetically polarized.
In Burglar alarm LDR acts as a/an
a. off switch
b. on switch
c. AND gate
d. OR gate
In Burglar alarm, LDR acts an AND gate.
Answer: C
Explanation
The LDR is light dependent resistor. The principle used in the working of LDR is that the resistance is inversely proportional to the intensity of light falling on the diode.
In burglar alarm, LDR diode is combined with an IC 555.
Normally an LED source is made to be incident on the LDR diode with same intensity such that the resistance will be maintained constant.
As the LDR is connected with IC, the voltage will be high when light is falling on the diode.
The IC will give only two output states that is high and low. This confirms that LDR in burglar alarm act as AND gate.
As the thief enters and crosses the LED light, the intensity of the light falling on the diode will decrease leading to decrease in the voltage which will cause the alarm to beep.
In a Burglar alarm, LDR acts as an AND gate. Option C is correct. This is further explained below.
What is an alarm?Generally, an alarm is simply defined as a danger signal, generally in the form of loud noise or flashing light: If there's a problem, pull the safety cord to sound the alarm.
In conclusion, LDR serves as an AND gate in a security alarm.
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Newton believed that time and space are absolute, while Einstein believed that the ______ is absolute.
A. CURVE OF SPACE TIME
B. SPEED OF LIGHT
C. FOURTH DIMENSION
D. DILATION OF TIME
Curve of space time
Newton believed that time and space are absolute, while Einstein believed that the speed of light is absolute.
Color depends on what characteristic of light?
Answer:
On the wavelength
Explanation:
Visible light is just a small portion of the electromagnetic spectrum, which classifies all the electromagnetic waves from shortest wavelength (gamma rays) to longest wavelength (radio waves).
Visible light refers to the part of the spectrum which has wavelength between 380 nm and 750 nm. These are the only electromagnetic wave that our eyes can see, and depending on their wavelength, they appear as a different color. In particular, each color corresponds to a different range of wavelengths:
Violet: 380-450 nm
Blue: 450-495 nm
Green: 495-570 nm
Yellow: 570-590 nm
Orange: 590-620 nm
Red: 620-750 nm
Color perception relies on the wavelength of light, as each wavelength corresponds to a particular color. A white object reflects all visible wavelengths, while a black object absorbs them all. The perceived color of an object depends on which wavelengths its atoms absorb and reflect.
Explanation:The color of an object or light depends greatly on the characteristic of light known as wavelength. Wavelength defines the energy of a photon in the visual spectrum, and each wavelength corresponds to a specific color.
Let's take a white object as an example: when it is illuminated by white light, which comprises all visible wavelengths, it appears white as well. This is because it reflects all visible wavelengths equally. On the contrary, a black object absorbs all light and doesn't reflect any, which is why it appears black to our eyes.
Also characteristic is how the atoms of a material interact with different light wavelengths. For instance, a simple red object like a tomato absorbs all wavelengths except red - the energy levels in its atoms correspond to all visible photons except red, which is reflected back to our eyes leading to its color perception.
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A circuit has 3 resistors, all connected in parallel. Let R1 = 4.0 Ω, R2 = 3.0 Ω, and R3 = 6.0 Ω. What is the equivalent resistance of the circuit? Show all work and include units of measure.
Answer:
[tex]1.33\Omega[/tex]
Explanation:
The equivalent resistance of three resistors connected in parallel is given by:
[tex]\frac{1}{R_T} = \frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}[/tex]
where in this problem we have
[tex]R_1 = 4.0\Omega[/tex]
[tex]R_2 = 3.0 \Omega[/tex]
[tex]R_3 = 6.0 \Omega[/tex]
Substituting into the equation, we find
[tex]\frac{1}{R_T} = \frac{1}{4.0\Omega}+\frac{1}{3.0\Omega}+\frac{1}{6.0\Omega}=0.75\Omega^{-1}[/tex]
And so, the equivalent resistance is
[tex]R_T=\frac{1}{0.75 \Omega^{-1}}=1.33\Omega[/tex]
A particle moves along the x axis. It is initially at the position 0.250 m, moving with velocity 0.050 m/s and acceleration -0.240 m/s2. Suppose it moves with constant acceleration for 3.70 s. (a) Find the position of the particle after this time. m (b) Find its velocity at the end of this time interval. m/s
(a) -1.208 m
The position of the particle at time t is given by
[tex]x(t) = x_0 + v_0 t + \frac{1}{2}at^2[/tex]
where:
[tex]x_0 = 0.250 m[/tex] is the initial position
[tex]v_0 = 0.050 m/s[/tex] is the initial velocity
[tex]a=-0.240 m/s^2[/tex] is the acceleration
Substituting into the equation t=3.70 s, we find the position after 3.70 seconds:
[tex]x(3.70 s) = 0.250 m + (0.050 m/s)(3.70 s) + \frac{1}{2}(-0.240 m/s^2)(3.70 s)^2=-1.208 m[/tex]
(b) -0.838 m/s
The velocity of the particle at time t is given by:
[tex]v(t) = v_0 + at[/tex]
where
[tex]v_0 = 0.050 m/s[/tex] is the initial velocity
[tex]a=-0.240 m/s^2[/tex] is the acceleration
Substituting t = 3.70 s, we find the velocity after 3.70 seconds:
[tex]v(3.70 s) = 0.050 m/s + (-0.240 m/s^2)(3.70 s)=-0.838 m/s[/tex]
Which best describes what happens when light traveling through air enters water at an angle? It moves along straight lines in air and changes direction when it enters water. It moves in a curve in air and moves in straight lines when it enters water. It moves along straight lines in air and continues along the same lines when it enters water. It moves in a curve in air and continues moving in the same curve when it enters water.
Answer:
It moves along straight lines in air and changes direction when it enters water.
Explanation:
When light travels from one medium to another, it undergoes a phenomenon called refraction: the ray of light changes speed and also direction, but it continues to move in a straight line.
There is a relationship between the direction of the incident ray and the direction of the refracted ray: (Snell's Law)
[tex]n_1 sin \theta_1 = n_2 sin \theta_2[/tex]
where
n1 is the index of refraction of the first medium
[tex]\theta_1[/tex] is the angle that the incident ray forms with the normal to the surface
n2 is the index of refraction of the second medium
[tex]\theta_2[/tex] is the angle that the refracted ray forms with the normal to the surface
Answer:
A: It moves along straight lines in air and changes direction when it enters water.
Explanation:
Metals are both good heat conductors and good electrical conductors because of the
Metals are both good heat conductors and good electrical conductors because of the looseness of outer electrons in metal atoms
_________________________
Unlike nonmetals that are insulators, most metals are both good heat conductors and good electrical conductors. This is because within a solid metal there is one or more outer electrons in each atom that become detached and can move freely through the solid metal, but what about the other electrons? Well, the y remain bound to the positively charged nuclei, and bound within the material in almost fixed positions. On the other hand, insulators are material having few, very few or no electrons that are allowed to move freely through the material.
Answer:
looseness of outer electrons in metal atoms
Explanation:
Unlike in insulators the valance electrons in metals are free to roam inside the boundaries of the metal slab. These electrons are called free electrons.
These electrons are free to move because there is no gab between their valance band and conduction band.
Since they are free to move they can transfer charges easily and be a good conductor.
Also heat exchange takes place when one electron collides with another moving electron or a stationary atom.
This is why metals are both good heat conductors and good electrical conductors.
What kind of waves do earthquakes produce
body and surface waves is the answer
Andy rubs a rubber rod with a piece of animal fur. The total charge of the system before rubbing is zero. The system is made up of the rubber rod and the fur. Initially, the charge on the rod is zero and it is electrically neutral. The same is true for the fur. When Sandy rubs the rubber rod with the fur, the rubber rod acquires negative (-) charges. How does the system change, overall, based on the law of conservation of charge?
The net charge in the system is 0. All electrons that were sent to the rod are missing from the fur. Hence, the rod is a (-) while the fur is a (+).
Answer:
In order to satisfied the law of conservation of charge, If the rubber rod becomes negatively charged, then the fur becomes positively charged.
Explanation:
This law states that charge is neither created nor destroyed, it can only be transferred from one system to another. In this case the rubber rod obtains electrons from the fur, in consequence the rod, which was originally neutral becomes negatively charged (the number of electrons on it increase), and the fur which was originally neutral becomes positively charged (the number of electrons on it decrease).
A policeman in a stationary car measures the speed of approaching cars by means of an ultrasonic device that emits a sound with a frequency of 41.2 khz. A car is approaching him at a speed of 33.0 m/s. The wave is reflected by the car and interferes with the emitted sound producing beats. What is the frequency of the beats? The speed of sound in air is 330 m/s.
Answer:
4.4 kHz
Explanation:
The frequency of the beats is given by the frequency of the original ultrasound, [tex]f=41.2 kHz[/tex], and the frequency of the ultrasound reflected back from the car, [tex]f'[/tex]:
[tex]f_B = f'-f[/tex] (1)
The frequency of the reflected wave can be found by using the Doppler effect formula:
[tex]f'=\frac{v}{v-v_s}f[/tex]
where
v = 340 m/s is the speed of sound
[tex]v_s =33.0 m/s[/tex] is the speed of the car
[tex]f=41.2 kHz[/tex] is the frequency of the original sound
Substituting,
[tex]f'=\frac{340 m/s}{340 m/s-33.0 m/s}(41.2 kHz)=45.6 kHz[/tex]
So, the beat frequency (1) is
[tex]f_B = 45.6 kHz - 41.2 kHz=4.4 kHz[/tex]
The frequency of the beats is about 9.2 kHz
[tex]\texttt{ }[/tex]
Further explanationLet's recall the Doppler Effect formula as follows:
[tex]\boxed {f' = \frac{v + v_o}{v - v_s} f}[/tex]
f' = observed frequency
f = actual frequency
v = speed of sound waves
v_o = velocity of the observer
v_s = velocity of the source
Let's tackle the problem!
[tex]\texttt{ }[/tex]
Given:
actual frequency = f = 41.2 kHz
velocity of the car = v_c = 33.0 m/s
speed of sound in air = v = 330 m/s
Asked:
frequency of the beats = Δf = ?
Solution:
Firstly , we will calculate the observed frequency by using the formula of Doppler Effect as follows:
[tex]f' = \frac{v + v_c}{v - v_c} \times f[/tex]
[tex]f' = \frac{330 + 33}{330 - 33} \times 41.2[/tex]
[tex]f' = \frac{363}{297} \times 41.2[/tex]
[tex]f' = \frac{11}{9} \times 41.2[/tex]
[tex]f' = 50 \frac{16}{45} \texttt{ kHz}[/tex]
[tex]f' \approx 50.4 \texttt{ kHz}[/tex]
[tex]\texttt{ }[/tex]
Next , we could calculate the frequency of the beats as follows:
[tex]\Delta f = f' - f[/tex]
[tex]\Delta f \approx 50.4 - 41.2[/tex]
[tex]\Delta f \approx 9.2 \texttt{ kHz}[/tex]
[tex]\texttt{ }[/tex]
Conclusion:The frequency of the beats is about 9.2 kHz
[tex]\texttt{ }[/tex]
Learn moreDoppler Effect : https://brainly.com/question/3841958Example of Doppler Effect : https://brainly.com/question/810552[tex]\texttt{ }[/tex]
Answer detailsGrade: College
Subject: Physics
Chapter: Sound Waves
Part A - Whale communication.
Blue whales apparently communicate with each other using sound of frequency 17.0Hz , which can be heard nearly 1000km away in the ocean. What is the wavelength of such a sound in seawater, where the speed of sound is 1521m/s ?
?1 =
m
Part B - Dolphin clicks.
One type of sound that dolphins emit is a sharp click of wavelength 1.50cm in the ocean. What is the frequency of such clicks?
f =
kHz
Part C - Dog whistles.
One brand of dog whistles claims a frequency of 26.0kHz for its product. What is the wavelength of this sound?
? =
cm
Part D - Bats.
While bats emit a wide variety of sounds, one type emits pulses of sound having a frequency between 39.0kHz and 78.0kHz . What is the range of wavelengths of this sound?
?min,?max =
mm
Part E - Sonograms.
Ultrasound is used to view the interior of the body, much as x rays are utilized. For sharp imagery, the wavelength of the sound should be around one-fourth (or less) the size of the objects to be viewed. Approximately what frequency of sound is needed to produce a clear image of a tumor that is 1.00mm across if the speed of sound in the tissue is 1550m/s ?
f =
MHz
A) 89.5 m
The wavelength of a wave is given by:
[tex]\lambda=\frac{v}{f}[/tex]
where v is the wave speed and f the frequency. For the sound emitted by the whales,
[tex]f=17.0 Hz[/tex]
[tex]v=1521 m/s[/tex]
Therefore, the wavelength is
[tex]\lambda=\frac{1521 m/s}{17.0 Hz}=89.5 m[/tex]
B) 101.4 kHz
The speed of the sound emitted by the dolphins in the water is still
v = 1521 m/s
While the wavelength is
[tex]\lambda=1.50 cm=0.015 m[/tex]
So, re-arranging the previous equation we find the frequency:
[tex]f=\frac{v}{\lambda}=\frac{1521 m/s}{0.015 m}=101,400 Hz=101.4 kHz[/tex]
C) 1.31 cm
The frequency of the dog whistle is
f = 26.0 kHz = 26,000 Hz
While the speed of sound in air is
v = 340 m/s
Therefore, the wavelength is
[tex]\lambda=\frac{v}{f}=\frac{340 m/s}{26,000 Hz}=0.0131 m=1.31 cm[/tex]
D) 4.4 - 8.7 mm
The speed of the sound waves emitted by the bats in air is
v = 340 m/s
The minimum frequency is
f = 39.0 kHz = 39,000 Hz
So the corresponding wavelength is
[tex]\lambda=\frac{v}{f}=\frac{340 m/s}{39,000 Hz}=0.0087 m=8.7 mm[/tex]
The maximum frequency is
f = 78.0 kHz = 78,000 Hz
So the corresponding wavelength is
[tex]\lambda=\frac{v}{f}=\frac{340 m/s}{78,000 Hz}=0.0044 m=4.4 mm[/tex]
E) 6.2 MHz
The wavelength of the sound must be 1/4 the size of the tumor, so
[tex]\lambda=\frac{1}{4}(1.00 mm)=0.25 mm=2.5\cdot 10^{-4}m[/tex]
while the speed of sound across the tissue is
v = 1550 m/s
So the frequency must be
[tex]f=\frac{v}{\lambda}=\frac{1550 m/s}{2.5\cdot 10^{-4} m}=6.2\cdot 10^6 Hz=6.2 MHz[/tex]
A 50 kg skydiver is falling downwards and accelerating 6 m/s2 down. What is the net force on the skydiver?
300 N, up
8.3 N, down
300 N, down
500 N, up
Net Force = (mass) x (acceleration) (Newton #2)
Net Force = (50 kg) x (6 m/s² down)
Net Force = (50 * 6) (kg-m/s² down)
Net Force = 300 Newtons down
A 50 kg skydiver is falling downwards and accelerating 6 m/s² down. The net force on the skydiver is 300 N down.
What is force?A force is an effect that can alter an object's motion according to physics. An object with mass can change its velocity, or accelerate, as a result of a force. An obvious way to describe force is as a push or a pull. A force is a vector quantity since it has both magnitude and direction.
Net Force = (50 kg) x (6 m/s² down)
Net Force = 300 kg-m/s² down
A 50 kg skydiver is falling downwards and accelerating 6 m/s² down. The net force on the skydiver is 300 N down.
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Nuclei with too many neutrons undergo _____ emission as neutrons are converted to protons.
A bike is pushed with a net force of 2.5 N, and it accelerates 0.5 m/s2. If you pushed the bike with twice that amount of net force, the acceleration would be __________.
0.25 m/s2 because net force and acceleration have an inverse relationship.
0.5 m/s2 because the net force and acceleration do not affect each other.
More than 0.5 m/s2, but less than 1 m/s2 because net force and acceleration do not have a proportional relationship.
1.0 m/s2 because net force and acceleration have a proportional relationship.
1.0m/s2 because net force and acceleration have a proprtional relationship
Consider motion in one dimension. (The sign of the vector quantities is their direction indicator.) An object moves in the positive x-direction with speed 0.9 m/s for 3.6 s. It stops for 4.7 s and then moves in the negative x direction with speed 1 m/s for 2.9 s. What is the total distance traveled by the object in units of meter? Enter a number with two digits behind the decimal point.
Answer:
6.14 m
Explanation:
The motion of the object is divided into three parts:
- Part 1: the object moves with speed
v = +0.9 m/s (the positive sign means positive x-direction)
for t = 3.6 s
Therefore, the distance covered is
[tex]d_1 =vt=(0.9 m/s)(3.6 s)=3.24 m[/tex]
- Part 2: the object stops for 4.7 s, so during this time the distance travelled by the object is zero.
- Part 3: the object moves with speed
v = -1 m/s (negative sign means negative x-direction)
for a time of
t = 2.9 s
We are only concerned in the distance travelled, so we can ignore the negative sign, and the distance covered in this part is
[tex]d_3 = vt=(1 m/s)(2.9 s)=2.90 m[/tex]
So, the total distance covered is
[tex]d=d_1+d_2+d_3=3.24 m+0 +2.90 m=6.14 m[/tex]
A moving electron enters an electric field. The initial velocity of the electron is in the same direction as the electric field. How will the electric field affect the motion of the electron?
The electron will accelerate in the opposite direction of its motion.
The electron will accelerate in the same direction as its motion.
The electron's speed will remain constant as it changes direction.
The electron's speed will increase as it changes direction
Answer:
The electron will accelerate in the opposite direction of its motion.
WILL GIVE BRAINLIEST!!!
Which describes how a simple machine can make work easier?
A.
It magnifies the input force so that force is greater.
B.
It magnifies the potential energy so that the kinetic energy is greater.
C.
It magnifies the output force so that the input force is greater.
D.
It magnifies the kinetic energy so that the potential energy is greater.
A simple machine can make work easier by reduce the amount of energy needed to perform a task, therefore, B. it magnifies the potential energy so that the kinetic energy is greater is the correct answer:)
Study the scenario.
As a spacecraft moves beyond Pluto, its thrusters are turned on for a short time to boost its speed to approximately 32,500 mi/h. The thrusters are then turned off.
If no other forces ever act on the spacecraft what will it do?
it will continue moving in the same direction, but it will speed up because there are no forces to slow it down
it will continue moving at 32,500 mi/h the same direction because forces are not required for an object to continue its motion
it will stop immediately because a force is needed constantly for it to move at all
it will continue moving in the same direction, but it will gradually slow down because a force is needed to maintain its speed.
If no other forces act on the object, according to Newton’s first law, the spacecraft will continue moving at a constant velocity, assuming that a planet or something with large mass doesn’t cross its path. Forces are not required to continue the motion of an object on a frictionless plane at a constant rate.
If no other forces ever act on the spacecraft, it will continue moving at 32,500 mi/h in the same direction, because forces are not required for an object to continue its motion.
The spacecraft will eventually curve away from a straight line, and its speed will change, as it comes within the gravitational influence of one or more of the objects out there in the Kuiper Belt and the Oort Cloud.
An object moves in a circular path with constant speed v. Which of the following statements is true concerning the object?
Its acceleration is constant, but its velocity is changing.
Both its velocity and acceleration is changing.
Its velocity is constant, but its acceleration is changing.
Its velocity and acceleration remain constant.
Answer:
Both its velocity and acceleration is changing.
Explanation:
Before answering, we must remind that both velocity and acceleration are vectors, so they both consist of a magnitude and a direction.
We can easily answer the question by looking at the direction of the two vectors only. In a circular motion, in fact:
- The velocity is always tangential to the circular path --> this means that its direction changes at every instant, so velocity is not constant
- The acceleration always points towards the centre of the circular path --> this means that its direction changes at every instand, so acceleration is changing as well
The object moving in a circular path at a constant speed has a changing velocity due to the alteration in direction. Also, its acceleration is continuously changing, directed towards the center. So, both velocity and acceleration of the object are changing.
Explanation:In this scenario, the object is moving in a circular path at a constant speed v. The term speed refers to the magnitude of velocity, which is a scalar quantity. Velocity, on the other hand, is a vector quantity including both magnitude (speed) and direction. Even if the speed is constant, the changing direction as the object moves along the circular path means the velocity of the object is continually changing.
The acceleration of the object is also non-zero and is directed towards the center of the circular path. This is known as centripetal or radial acceleration. Therefore, although the speed of the object remains constant, its velocity (due to changes in direction) and acceleration (towards the center) are continuously changing. Hence, the statement 'Both its velocity and acceleration are changing' is accurate.
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Which of the following statements are true for images formed by thin lenses?Check all that apply.A diverging lens always produces a virtual image of an object placed in front of it.A diverging lens always produces an upright image of an object placed in front of it.A converging lens always produces an inverted image of an object placed in front of it.A converging lens always produces an image that is larger than the object.A diverging lens always produces an image that is larger than the object.A converging lens always produces a real image of an object placed in front of it.
Answer:b
Explanation:
A diverging lens always produces a virtual and upright image of an object placed in front of it.
To determine the correct statement among all the options, we need to know about the properties of the image formed by converging and diverging lens.
What are the properties of images formed by converging lens?The properties of image formed by converging lens depends on the position of the object.The image for a far away object is real, inverted, diminished and is produced opposite side of the lens.As the object approaches the lens, the image remains real and inverted, but eventually becomes magnified.When the object passes the focal point of the lens, the image becomes virtual, erect, magnified and produced on the same side of the lens.What are the properties of image formed by diverging lens?The image formed by a diverging lens is always erect (upright), virtual and diminished.
Thus we can conclude the statements (a) and (b) are correct.
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What range of sound wavelengths in air at room temperature are audible to a human with ideal hearing (f= 20 Hz - 20 kHz)?
Answer:
0.017 m - 17 m
Explanation:
The relationship between wavelength and frequency of a wave is:
[tex]\lambda=\frac{v}{f}[/tex]
where v is the speed of the wave, [tex]\lambda[/tex] the wavelength and f the frequency.
The speed of sound waves in air at room temperature is
v = 340 m/s
For the lowest frequency, f = 20 Hz, the wavelength is
[tex]\lambda=\frac{340 m/s}{20 Hz}=17 m[/tex]
For the highest frequency, f = 20 kHz = 20,000 Hz, the wavelength is
[tex]\lambda=\frac{340 m/s}{20,000 Hz}=0.017 m[/tex]
Two charged spheres are 20 cm apart and exert an attractive force of 8 x 10-9 n on each other. What will the force of attraction be when the spheres are moved to 10 cm apart?
Answer:
[tex]3.2\cdot 10^{-8} N[/tex]
Explanation:
The inital electrostatic force between the two spheres is given by:
[tex]F=k\frac{q_1 q_2}{r^2}[/tex]
where
[tex]F=8\cdot 10^{-9} N[/tex] is the initial force
k is the Coulomb's constant
q1 and q2 are the charges on the two spheres
r is the distance between the two spheres
The problem tells us that the two spheres are moved from a distance of r=20 cm to a distance of r'=10 cm. So we have
[tex]r'=\frac{r}{2}[/tex]
Therefore, the new electrostatic force will be
[tex]F'=k\frac{q_1 q_2}{(r')^2}=k\frac{q_1 q_2}{(r/2)^2}=4k\frac{q_1 q_2}{r^2}=4F[/tex]
So the force has increased by a factor 4. By using [tex]F=8\cdot 10^{-9} N[/tex], we find
[tex]F'=4(8\cdot 10^{-9} N)=3.2\cdot 10^{-8} N[/tex]
The force of attraction between the soheres when they are 10cm apart is 3.2 * 10^-8 N
what is force of attraction?This is the force pulling bodies together given by
the force of attration, F = ( G q1 q2 ) / d^2
G refera to constant of attraction
q1 and q2 is the charges of the objects
d is the distance between the objects
G q1 q2 are all constant with respect to this question
let K = G * q1 * q2
F = k / d^2
for d = 20 cm = 0.2 m
F20 = k / 0.2^2
8*10^-9 = k / 0.2^2
K = 8*10^-9 *0.2^2
K = 3.2*10^-10
for d = 10 cm = 0.1 m
F10 = K / d^2
= (3.2*10^-10) * 0.1^2
= 3.2*10^-8 N
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Which disease is caused when the body tissues cannot get enough oxygen?
I believe it's hypoxia for your tissues and hypoxemia for your blood.
Answer:
Anemia
Explanation:
Edge 2021
A switch that connects a battery to a 10 µF capacitor is closed.Several seconds later you find that the capacitor plates are charged to 30 µC.What is the emf of the battery?
Answer:
3 V
Explanation:
The relationship between capacity (C), charge stored (Q) and potential difference across a capacitor (V) is
[tex]C=\frac{Q}{V}[/tex] (1)
The problem asks us to find the emf of the battery several seconds after the switch has been closed: this means that the capacitor had enough time to fully charge, so the potential difference across the capacitor (V) is equal to the emf of the battery.
Since we have:
[tex]C=10 \mu F=10\cdot 10^{-6} F[/tex]
[tex]Q=30 \mu C=30\cdot 10^{-6} C[/tex]
We can solve the equation for V, and we find
[tex]V=\frac{Q}{C}=\frac{30 \cdot 10^{-6}C}{10 \cdot 10^{-6} F}=3 V[/tex]
A switch that connects a battery to a 10 µF capacitor is closed for several seconds and the capacitor plates are charged to 30 µC then the voltage across the capacitor will be 3 V.
The emf of the battery will be equal to the voltage across the capacitor which is 3 V.
How do you calculate the emf of the battery?
The EMF or electromotive force can be defined as the energy supplied by a battery or a cell per coulomb (Q) of charge passing through it. The magnitude of emf is equal to V (potential difference) across the cell terminals when there is no current flowing through the circuit.
Hence the emf of the battery will be equal to the voltage across the capacitor.
The voltage across a capacitor can be calculated as given below.
[tex]V = \dfrac {Q}{C}[/tex]
Where Q is the amount of charge stored on each plate and C is the capacitance.
[tex]V = \dfrac { 30 }{10}[/tex]
[tex]V = 3\;\rm V[/tex]
Hence we can conclude that the emf of the battery is 3 V.
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Using our understanding of orbits, which orbits (near or far) are moving more slowly?
near
far
The closer the planet is to the sun's gravitational pull, the planet will move faster because it takes less time to orbit the sun. The planets that are farther away have less gravitational pull, which means they will move slower and take more time to orbit the sun.
Using our understanding of orbits, objects in lower, near-in orbits move slower than objects in higher, far-out orbits.
Here's a set of examples:
International Space Station ...
Orbital height -- 254 miles
Speed in Orbit -- 76 mi/sec
Geostationary TV satellite ...
Orbital height -- 22,200 miles
Speed in orbit -- about 1.9 mi/sec
Moon ...
Orbital height -- 238,000 miles
Speed in orbit -- about 0.6 mi/sec
Most binary systems with an invisible companion contain a large, bright star and a small, dim star hidden by the light of its larger companion. However, in some binary systems, the unseen companion may be a black hole instead of a small star. One of the most likely black hole candidates in a binary system is an object called Cygnus X-1. If you were an astronomer trying to prove that Cygnus X-1 is a black hole, which of the following observations would provide evidence in support of your claim?CHECK ALL THAT APPLYA. Cygnus X-1 has a mass one-tenth that of the Sun.B. Cygnus X-1 produces a large amount of X-ray emissions.C. Cygnus X-1 produces no X-ray emissions.D. Cygnus X-1 has a diameter less than 300 kilometers.E. Cygnus X-1 has a mass of 10 solar masses.F. Cygnus X-1 has a diameter much greater than 300 kilometers.
Answer:
Brain signals are converted
Two identical cars are traveling at the same speed. One is heading due east and the other due north, as the drawing shows. Which statement is true regarding the kinetic energies and momenta of the cars?A) They have the same kinetic energies, but different momenta.B) They have different kinetic energies, but the same momenta.C) They have the same kinetic energies and the same momenta.D) They have different kinetic energies and different momenta.
Answer:
A) They have the same kinetic energies, but different momenta.
Explanation:
Before answering, let's remind that:
- Kinetic energy is a scalar quantity, which is given by:
[tex]K=\frac{1}{2}mv^2[/tex]
where m is the mass of the object and v is the speed.
- Momentum is a vector quantity, whose magnitude is given by:
[tex]p=mv[/tex]
where m is the mass of the object and v is the velocity, and whose direction is the same as the velocity.
By looking at the two definitions, we can say that:
- the two cars have same kinetic energy, because they have same mass and same speed
- The two cars have different momenta, because they do not have same velocity (in fact, their directions are different, since one is heading east and the other one is heading north)
The identical cars traveling at the same speed will have the same kinetic energies, but different momenta because momentum is a vector quantity dependent on direction.
Explanation:The correct answer would be A) They have the same kinetic energies, but different momenta.
The two identical cars traveling at the same speed will have identical kinetic energies since kinetic energy depends only on mass and speed. The kinetic energy (KE) of an object can be calculated using the formula KE=1/2mv², where m is the mass and v is the speed of the object. Therefore, since the cars are identical and moving at the same speed, their kinetic energies will be the same.
However, their momenta will be different because momentum is a vector quantity, meaning direction matters. The formula for momentum is p=mv, where m is mass and v is velocity (speed in a given direction). Since one car is going east and the other is going north, their momenta are in different directions, hence, their momenta are different.
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I traveled 3.8 miles in 12 minutes. Calculate my average velocity in miles per hour. 2) A sprinter increased his/her speed from 18 mph to 28 mph over 3.5 seconds. Calculate his/her acceleration.
1) Average velocity: 19 mi/h
The average velocity is given by:
[tex]v=\frac{d}{t}[/tex]
where
d is the displacement
t is the time taken
In this problem,
d = 3.8 mi is the displacement
[tex]t=12 min \cdot \frac{1}{60 min/h}=0.2 h[/tex] is the time taken
Substituting,
[tex]v=\frac{3.8 mi}{0.2 h}=19 mi/h[/tex]
Note that this is actually the average speed, not the average velocity, since there is no information about the direction of the motion.
2) [tex]10,309 mi/h^2[/tex]
The acceleration is given by
[tex]a=\frac{v-u}{t}[/tex]
where we have
v = 28 mph is the final velocity
u = 18 mph is the initial velocity
t = 3.5 s is the time taken
Conerting the time from seconds to hours,
[tex]t=3.5 s \cdot \frac{1}{3600 s/h}=9.7\cdot 10^{-4} h[/tex]
So the acceleration is
[tex]a=\frac{28 mph-18 mph}{9.7\cdot 10^{-4} h}=10,309 mi/h^2[/tex]