Answer:
Option (B)
Explanation:
A heterogeneous mixture is usually defined as a combination of two or more chemical substances. It can be also elements as well as compounds. These contrasting components can be easily separated from one another by means of physical process. These are comprised of substances that are not even everywhere. There is variation in it.
Thus, the correct answer is option (B).
Answer:
B: a heterogeneous mixture
Explanation:
If the components of a substance can be separated by the physical means then the substance is a mixture and its component have not undergone any kind of chemical change with their original molecular structure.
The mixtures are of two types homogeneous and heterogeneous.
Homogeneous mixtures have a uniform composition of its components throughout the mixture whereas heterogeneous mixtures have a non-uniform composition of its constituents in the mixture.A library of e-books contains metadata for each book. The metadata is intended to help a search feature find
books that users are interested in. Which of the following is LEAST likely to be contained in the metadata of each e-book?
a. An archive containing previous versions of the e-book
b. The author and title of the e-book
c. The date the e-book was first published
d. The genre of the e-book (e.g., comedy, fantasy, romance, etc.)
The metadata is intended to help a search feature the least likely feature contained in the metadata is An archive containing previous versions of the e-book so, option A is correct.
What are e-books?The term "e-book" refers to a digital file with a body of text and graphics that is suited for electronic distribution and displays on-screen similarly to a printed book.
E-books can be made by converting a printer's source files into forms that are simple to download and read on a screen, or they can be taken from a database or a collection of text files that weren't made just for printing.
When businesses like Peanut Press started offering book content for reading on personal digital assistants (PDAs), handheld devices that were the forerunners of today's smartphones and tablet computers, the market for buying and selling e-books first became a mainstream industry in the late 1990s.
Thus, metadata does not contain a feature to find an archive containing previous versions of the e-book.
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On the earth, when an astronaut throws a 0.250-kg stone vertically upward, it returns to his hand a time T later. On planet X he finds that, under the same circumstances, the stone returns to his hand in 2T. In both cases, he throws the stone with the same initial velocity and it feels negligible air resistance. The acceleration due to gravity on planet X (in terms of g) is:_______
a) g/.
b) g/4.
c) 2g.
d) g/2.
e) g.
Answer:
d) g/2
Explanation:
We need to use one of Newton's equations of motion to find the position of the stone at any time t.
x(t) = x₀(t) + ut - ¹/₂at²
Where
x₀(t) = initial position of the stone.
x(t) - x₀(t) = distance traveled by the stone at any time.
u = initial velocity of the stone
a = acceleration of the stone
t = time taken
On both planets, before the stone was thrown by the astronaut, x = 0 and t = 0.
=> 0 = x₀(t)
=> x₀(t) = 0
On earth, when the stone returns into the hand of the astronaut at time T on earth, x = 0.
=> 0 = 0 + uT - ¹/₂gT² (a = g)
=> uT = ¹/₂gT²
=> g = 2u/T
On planet X, when the stone returns into the hand of the astronaut, time = 2T , x = 0.
=> 0 = 0 + u(2T) - ¹/₂a(2T)²
=> 2uT = 2aT²
=> a = u/T
By comparing we see that a = g/2.
Final answer:
The acceleration due to gravity on planet X is determined by comparing the time it takes for a stone thrown vertically upward to return to the thrower's hand. Using kinematic equations and given that it takes twice the time to return on planet X as it does on Earth, we deduce the acceleration on planet X is half that of Earth's gravity (g/2).
Explanation:
The question concerns finding the acceleration due to gravity on an unknown planet, which is an essential concept in physics. To determine the gravitational acceleration (g) on planet X, given that a stone thrown with the same initial upward velocity as on Earth returns in 2T instead of T, we can use the kinematic equations for motion under constant acceleration.
On Earth, the total time for the stone to return to the thrower's hand is T, which implies that the time to reach the maximum height is T/2. Applying the equation v = u + at, where v is the final velocity (0 m/s at maximum height), u is the initial upward velocity, a is the acceleration due to gravity (which is -g on Earth), and t is the time (T/2), we have 0 = u - g(T/2). Solving for u gives us u = g(T/2).
On Planet X, the total time for the stone to return to the thrower's hand is 2T, so the time to reach maximum height is T. Using the same equation with the acceleration on Planet X as aX, we get 0 = u - aX(T). Since u = g(T/2), we have g(T/2) = aX(T), leading to aX = g/2.
Therefore, the correct answer is (d) g/2.
normal hearing elsewhere. How much more intense is a 6000 Hz tone than a 80 Hz tone if they are both barely audible to the child?
Explanation:
There is no any direct relationship between the frequency of a sound relative the intensity of the sound.
Frequency is a perception of pitch of the sound whereas intensity determines the loudness of the sound. Intensity is proportional to amplitude of the sound.
[tex]\beta= log_{10}\frac{I}{I_o}[/tex]
So, it would be hard to determine how much more intense is a 6000 Hz tone than a 80 Hz tone.
The distance between two successive crests of a certain transverse wave is 1.40 m. 25 crests pass a given point along the direction of travel every 12.4 s. Calculate the wave speed. Answer in units of m/s.
Answer: 2.82m/s
Explanation:
Wave speed(v) is the product of the frequency (f) of a wave and its wavelength (¶)
V = f¶
Since wavelength is the distance between two successive crests or trough, the wavelength given is 1.40m i.e ¶ = 1.40m
If the total crest is 25, this means that 25 oscillations occurs in 12.4s.
Since the number of oscillation made in one second is the frequency, frequency = 25/12.4 = 2.01cycle/seconds
Speed of the wave v = 2.01×1.4
V =
Wave speed is 2.82m/s
On the Moon, gravitational field is around 1.6 N/kg. How does this affect gravitational potential and kinetic energy of objects dropped from a height on the Moon as compared with Earth?
Answer:An object with mass in a gravitational field experiences a force called weight. This weight on the moon is about 1/6 but the mass of that object remains the same due to the moon's gravitational field which is less than the earth.
Explanation:There are factors that affect the gravitational potential and kinetic energy . They are:
1. The height of the object in the moon
2.The mass of the object
3. The strength of the gravitational field
4. The change in position of the object.
If a word problem gives the separation distance between two charges in centimeters (cm), you should ____. Question 4 options:
1) enter that value directly since cm is the standard unit
2) convert that distance to inches, the standard unit
3) convert that distance to meters, the standard unit
4) convert that distance to whatever units are given in the second distance
5) look to see whether the charges are like or opposite to know what to do
Answer:3
Explanation:
Convert the distance to meters,the standard units.
Final answer:
When a word problem provides the separation distance between two charges in centimeters, you should (3) convert that distance to meters, the standard SI unit for measurements in physics. Hence, (3) is the correct option.
Explanation:
If a word problem gives the separation distance between two charges in centimeters (cm), you should convert that distance to meters, the standard unit. Physicists use the International System of Units (SI units) for calculations to ensure consistency and comparison of data.
Since the SI unit for displacement is the meter (m), converting from centimeters to meters is necessary for performing electrostatic calculations involving Coulomb's law or any other physics formula.
For instance, when using Coulomb's law to find the force between two charges, the formula is F = k * (q₁ * q₂ / r²), where k is Coulomb's constant (8.987 x 10⁹ N m²/C²), q₁ and q₂ are the charges, and r is the separation distance in meters. If the problem states the distance in centimeters, you would convert the distance to meters by dividing by 100, since there are 100 centimeters in a meter.
A stuntman with a mass of 82.5 kg swings across a pool of water from a rope that is 12.0 m. At the bottom of the swing the stuntman's speed is 8.65 m/s. The rope's breaking strength is 1,000 N. Will the stuntman make it across the pool without falling in?
Answer:
The stuntman will not make it
Explanation:
At the bottom of the swing, the equation of the forces acting on the stuntman is:
[tex]T-mg = m\frac{v^2}{r}[/tex]
where:
T is the tension in the rope (upward)
mg is the weight of the man (downward), where
m = 82.5 kg is his mass
[tex]g=9.8 m/s^2[/tex] is the acceleration due to gravity
[tex]m\frac{v^2}{r}[/tex] is the centripetal force, where
v = 8.65 m/s is the speed of the man
r = 12.0 m is the radius of the circule (the length of the rope)
Solving for T, we find the tension in the rope:
[tex]T=mg+m\frac{v^2}{r}=(82.5)(9.8)+(82.5)\frac{8.65^2}{12.0}=1322 N[/tex]
Since the rope's breaking strength is 1000 N, the stuntman will not make it.
One of the many steps in the production of toothpaste is to screw the caps on the tubes, which is still a manual process, performed by on man, Mr. Bucket. Which statement about this situation is BEST?
A) This is most appropriate for an output measure of capacity.
B) In this case, the capacity of this step is not the maximum rate of output.
C) This is most appropriate for an input measure of capacity.
D) Utilization of the worker at this process step cannot be measures as it is a manual process.
Answer:A) This is most appropriate for an output measure of capacity.
Explanation: Toothpaste is a paste used to promote oral health and hygiene, Most toothpaste contain an active agent(floride) and it acts as an abrasive agent which helps to remove dental plaques or unwanted attachments to the teeth.
Among the options the Option(A) which says that THIS IS MOST APPROPRIATE FOR THE MEASURE OF CAPACITY is the statement that best Describes the situation. Toothpastes have a long history as it can be traced to about 4000years ago.
The partial pressure of ethanol in the air above a sample of liquid ethanol in a closed container is 19.5 torr, at a given experimental Temperature. Assuming the pressure inside the container is 758.6 torr at the experimental T, what is the molar percent of ethanol in the air?
Answer:
The molar percent of ethanol in the air is 2.6%
Explanation:
Molar percent of ethanol in the air = partial pressure of ethanol/total pressure × 100 = 19.5 torr/758.6 torr × 100 = 2.6%
in circuit A, two resistors R1 and R2 are connected in series. In circuit B, two identical resistors are connected in parallel. Both circuits are powered by identical energy sources and include no other resistors. In which circuit, if any, is the current greater across R1?
A. Circuit A
B. Neither circuit
C. Circuit B
The current across R1 in circuit A is the same as the current across the resistors in circuit B.
Explanation:In circuit A, two resistors R1 and R2 are connected in series. In circuit B, two identical resistors are connected in parallel. In a series circuit, the current is the same in each resistor. Therefore, the current across R1 in circuit A is the same as the current across the resistors in circuit B, since they are connected in parallel.
__________ refers to the spreading of the signal that different parts of the signal arrive at different times at the destination. a. turnaround time b. propagation delay c. dispersion d. insulation e. attenuation
Final answer:
Dispersion refers to the spreading of a signal where different parts of the signal reach the recipient at varying times, affecting the wave by causing signal broadening due to the speed differential of its frequency components.
Explanation:
The term that refers to the spreading of a signal where different parts arrive at the receiver at different times is known as dispersion. Dispersion affects waves by causing a broadening of the signal, as different frequency components of the wave travel at different speeds. This is especially important in optical fibers, where dispersion can cause pulse broadening, potentially limiting the bandwidth and impairing the integrity of the data being transmitted. Option C is correct .
Factors such as scattering and refraction directly relate to the phenomenon of dispersion. For instance, scattering causes the redirection of light within a medium and refraction is the bending of a wave when it enters a medium where its speed is different. In the context of signal transmission, residence time is the average time a molecule spends in its reservoir, though this concept is less directly related to the question at hand.
One of the moons of a planet is at an average distance of 0.7 AU from the planet and its period of revolution is 27.8 earth days. What is the period of revolution of another moon of the same planet that is at 1.2 AU from in?
a. 62.3 earth days
b. 128.4 earth days
c. 224 earth days
d. 27.8 earth days
Answer:
option A
Explanation:
given,
distance of the moon 1 from planet = 0.7 AU
Time period of moon 1 = 27.8 earth days
distance of the moon 2 from the planet = 1.2 AU
time period of the moon 2 = ?
using Kepler's formula
[tex]T^2 = \dfrac{4\pi^2}{GM}\ R^3[/tex]
now,
only variable in the formula is time period and distance.
so,
[tex]\dfrac{T_2^2}{T_1^2}=\dfrac{R_2^3}{R_1^3}[/tex]
[tex]\dfrac{T_2^2}{27.8^2}=\dfrac{1.2^3}{0.7^3}[/tex]
T₂² = 3893.49
T₂ = 62.3 earth days
Hence, the correct answer is option A
An automobile moves at a constant speed over the crest of a hill traveling at a speed of 88.5 km/h. At the top of the hill, a package on a seat in the rear of the car barely remains in contact with the seat. What is the radius of curvature (m) of the hill?
Answer:
[tex]r=61.65m[/tex]
Explanation:
Since the package remains in contact with the car's seat, the package's speed is equal to the car's speed. At the top on the mountain the package's centripetal force must be equal to its weight:
[tex]mg=F_c[/tex]
The centripetal force is defined as:
[tex]F_c=ma_c=\frac{mv^2}{r}[/tex]
Here v is the linear speed of the object and r is the radius of curvature. We need to convert the linear speed to [tex]\frac{m}{s}[/tex]:
[tex]88.5\frac{km}{h}*\frac{1000m}{1km}*\frac{1h}{3600s}=24.58\frac{m}{s}[/tex]
Now, we calculate r:
[tex]mg=\frac{mv^2}{r}\\r=\frac{v^2}{g}\\r=\frac{(24.58\frac{m}{s})^2}{9.8\frac{m}{s^2}}\\\\r=61.65m[/tex]
A bicyclist is moving toward a sheer wall while holding a tuning fork rated at 488 Hz. 1) If the bicyclist detects a beat frequency of 8 Hz (between the waves coming directly from the tuning fork and the echo waves coming from the sheer wall), calculate the speed of the bicycle. Assume the speed of sound is 343 m/s.
Answer:
speed of the bicycle = 2.81 m/s
Explanation:
given data
tuning fork rated f = 488 Hz
beat frequency fb = 8 Hz
speed of sound vs = 343 m/s
solution
we get here frequency that rider is listening that is
rider hearing frequency = 488 Hz + 8 Hz
rider hearing frequency = 496 Hz
and when it bouncing from sheer wall at frequency of related to stationary object
frequency = 488 Hz + 4 Hz
frequency = 492 Hz
so now we get speed of the bicycle that is
speed of the bicycle = ( [tex]\frac{492}{488}[/tex] × 343 ) - 343 Hz
speed of the bicycle = 2.81 m/s
The speed of the bicycle is 2.81 m/s.
Based on the given information,
• The tuning fork rated at frequency (f) 488 Hz.
• The beat frequency (fb) given is 8Hz.
• The speed of the sound is 343 m/s.
Now with the help of the given information, the rider is hearing a frequency of,
[tex]=(f+fb)\\=(488+8)\\=496 Hz[/tex]
However, it is bounding from sheer wall at the frequency of,
[tex]=(488+\frac{8}{2})\\= 492 Hz[/tex]
Now the speed of the bicycle will be,
[tex]=(\frac{492}{488} * 343)-343\\=2.811 m/s\\= 2.81 m/s[/tex]
Thus, the speed of the bicycle is 2.81 m/s.
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Which answer most completely describes the difference/s between the gravitational force and the electrostatic force? Question 6 options:
The gravitational force is much weaker than the electrostatic force
The gravitational force is weaker and can only attract objects but the electrostatic force is strong and can both attract and repel
The gravitational force occurs throughout the universe but the electrostatic force only occurs on Earth
The gravitational force can only attract but the electrostatic force can attract or repel
I believe the answer would be D
The gravitational force is weaker and can only attract objects but the electrostatic force is strong and can both attract and repel. - This is the most completely describes the differences between the gravitational force and the electrostatic force.
What are the differences between gravitational force and electrostatic force?The main differences are:
Electrostatic force acts on particles with charge, whereas gravitational force acts on particles with mass.While the electrostatic force can be either attracting or repulsive, the gravitational force is always attractive.While the electric potential can be either negative or positive, the gravitational potential is always negative.Gravitational force is weak but has low range whereas electrostatic force is strong bur short ranged.Learn more about electrostatic force here:
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You find a bag labeled "10 kg of lead-210". Its contents now weigh 1.25 kg. How many half-life periods have there been since the bag was weighed and labeled? Lead-210 is a radioisotope with a half-life of about 22 years.
Answer : The number of half-life periods will be, 3
Explanation : Given,
Initial amount of lead = 10 kg
Amount of lead after decay = 1.25 kg
Half-life = 22 years
Formula used :
[tex]a=\frac{a_o}{2^n}[/tex]
where,
a = amount of reactant left after n-half lives
[tex]a_o[/tex] = Initial amount of the reactant
n = number of half lives
[tex]t_{1/2}[/tex] = half-life
Now put all the given values in the above formula, we get:
[tex]1.25=\frac{10}{2^n}[/tex]
[tex]2^n=8[/tex]
[tex]2^n=2^3[/tex]
[tex]n=3[/tex]
Thus, the number of half-life periods will be, 3
A man (weighing 763 N) stands on a long railroad flatcar (weighing 3513 N) as it rolls at 19.8 m/s in the positive direction of an x axis, with negligible friction. Then the man runs along the flatcar in the negative x direction at 4.68 m/s relative to the flatcar. What is the resulting increase in the speed of the flatcar?
Answer:
0.8m/s
Explanation:
Weight of mas,F=763 N
Mass of man=[tex]\frac{F}{g}=\frac{763}{9.8}=77.86 kg[/tex]
By using [tex]g=9.8m/s^2[/tex]
Weight of flatcar=F'=3513 N
Mass of flatcar=[tex]\frac{3513}{9.8}=358.5 Kg[/tex]
Total mass of the system=Mass of man+mass of flatcar=77.86+358.5=436.36 kg
Velocity of system=19.8m/s
Let v be the velocity of flatcar with respect to ground
Velocity of man relative to the flatcar=[tex]-4.68m/s[/tex]
Final velocity of man with respect to ground=v-4.68
By using law of conservation of momentum
Initial momentum=Momentum of car+momentum of flatcar
[tex]436.36(19.8)=77.86(v-4.68)+358.5v[/tex]
[tex]8639.928=77.86v-364.3848+358.5v[/tex]
[tex]8639.928+364.3848=436.36 v[/tex]
[tex]9004.3128=436.36v[/tex]
[tex]v=\frac{9004.3128}{436.36}[/tex]
[tex]v=20.6 m/s[/tex]
Initial speed of flatcar=Speed of system
Increase in speed=Final speed-initial speed=20.6-19.8=0.8m/s
Henrietta is going off to her physics class, jogging down the sidewalk at a speed of 2.95 m/s . Her husband Bruce suddenly realizes that she left in such a hurry that she forgot her lunch of bagels, so he runs to the window of their apartment, which is a height 39.6 m above the street level and directly above the sidewalk, to throw them to her. Bruce throws them horizontally at a time 7.50 s after Henrietta has passed below the window, and she catches them on the run. You can ignore air resistance. With what initial speed must Bruce throw the bagels so Henrietta can catch them just before they hit the ground?
Answer:
Explanation:
Given
Speed of Henrietta is [tex]v=2.95\ m/s[/tex]
Height of tower [tex]h=39.6\ m[/tex]
Bruce throws the bangle after 7.5 s
During 7.5 s Henrietta travels
[tex]x=2.95\times 7.5=22.125\ m[/tex]
Suppose bangle hit the ground after t sec so bangle will has to cover a distance of x and distance traveled by Henrietta during this time t
Range of bangle when thrown with speed u
[tex]R=u\times t[/tex]
[tex]R=x+2.95\times t-----1[/tex]
bangle will also cover a vertical distance of 39.6 m
so using equation of motion
[tex]h=u_yt+\frac{1}{2}gt^2[/tex]
here initial vertical velocity is zero
[tex]39.6=0+\frac{1}{2}\cdot 9.81\cdot t^2[/tex]
[tex]t=\sqrt{8.0816}[/tex]
[tex]t=2.84\ s[/tex]
Substitute the value of t in
[tex]u\times 2.84=22.125+2.95\times 2.84[/tex]
[tex]u=10.743\ m/s[/tex]
A 6110-kg bus traveling at 20.0 m/s can be stopped in 24.0 s by gently applying the brakes. If the driver slams on the brakes, the bus stops in 3.90 s. What is the average force exerted on the bus in both these stops?
Answer:
For 24 seconds force exerted is 5092 N towards opposite direction of motion of bus.For 3.90 seconds force exerted is 31333 N towards opposite direction of motion of bus.Explanation:
We have equation of motion v = u + at
Initial velocity, u = 20 m/s
Final velocity, v = 0 m/s
Case 1:-
Time, t = 24 s
Substituting
v = u + at
0 = 20 + a x 24
a = -0.8333 m/s²
Force = Mass x Acceleration = 6110 x -0.8333 = -5092 N
Force exerted is 5092 N towards opposite direction of motion of bus.
Case 2:-
Time, t = 3.90 s
Substituting
v = u + at
0 = 20 + a x 3.90
a = -5.13 m/s²
Force = Mass x Acceleration = 6110 x -5.13 = -31333 N
Force exerted is 31333 N towards opposite direction of motion of bus.
The average force exerted on the bus with gentle braking is approximately -5092 N, and with slamming the brakes, it is approximately -31342 N. Negative values indicate the force direction is opposite to the motion.
Explanation:To calculate the average force exerted on the bus during both stops, we will use Newton's second law of motion, which states that the force applied to an object is equal to the mass of the object multiplied by the acceleration (F = ma). Here, the acceleration can be calculated using the formula for deceleration since the bus is coming to a stop:
For gentle braking: Deceleration = (Final velocity - Initial velocity) / Time = (0 - 20.0 m/s) / 24.0 s = -0.8333 m/s2.For slamming the brakes: Deceleration = (Final velocity - Initial velocity) / Time = (0 - 20.0 m/s) / 3.90 s = -5.1282 m/s2.Now, we can calculate the average force (F) exerted in both cases by multiplying the mass of the bus (m = 6110 kg) by the deceleration (a).
Gentle braking average force: F = m × a = 6110 kg × -0.8333 m/s2 = -5092 N (approximately).Slamming the brakes average force: F = m × a = 6110 kg × -5.1282 m/s2 = -31342 N (approximately).Since the bus is stopping, the negative sign indicates the direction of the force is opposite to the initial direction of motion.
A child in danger of drowning in a river is being carried down-stream by a current that flows uniformly with a speed of 2.20 m/s . The child is 500 m from the shore and 1100 m upstream of the boat dock from which the rescue team sets out.
If their boat speed is 7.30 m/s with respect to the water, at what angle from the shore must the boat travel in order to reach the child?
Final answer:
To reach a child being carried by a river current, the rescue boat, with a speed of 7.30 m/s versus water, must aim at a specific angle upstream accounting for the current's speed of 2.20 m/s. The calculation involves vector addition and trigonometry to counteract the current and ensure a direct path to the child.
Explanation:
The question involves calculating the angle at which a rescue boat must travel to reach a child being carried away by a river current. The river flows at a speed of 2.20 m/s, while the boat's speed in still water is 7.30 m/s. Given the distances from the shore to the child (500 m) and from the boat dock upstream to the child (1100 m), we need to determine the angle relative to the shore for the boat to make a direct rescue.
To solve this, we need to use the concept of vector addition, where the boat's velocity vector and the river's current vector combine to create a resultant vector that points directly to the child's location. The angle can be calculated using trigonometry, particularly the tangent function. However, since the specific calculations and steps to obtain the angle are not provided in the question or the previous examples, it is crucial to understand the principles of vectors and relative motion to approach this problem.
Essentially, the rescue team needs to aim the boat at an angle upstream to counteract the river's flow. The effect of the river's current will adjust the boat's path towards the child. It's a classic application of relative motion in two dimensions, requiring careful consideration of both the magnitude and direction of the vectors involved.
A wind turbine is initially spinning at a constant angular speed. As the wind's strength gradually increases, the turbine experiences a constant angular acceleration of 0.100 rad/s2. After making 2844 revolutions, its angular speed is 140 rad/s.
Complete answer
A wind turbine is initially spinning at a constant angular speed. As the wind's strength gradually increases, the turbine experiences a constant angular acceleration of 0.100rad/s2. After making 2844 revolutions, its angular speed is 140rad/s
(a) What is the initial angular velocity of the turbine? (b) How much time elapses while the turbine is speeding up?
Answer:
a) 126.59 radians per second
b) 134.1 seconds
Explanation:
We can use the rotational kinematic equations for constant angular acceleration.
a) For a) let’s use:
[tex]\omega^{2}=\omega_{0}^{2}+2\alpha\varDelta\theta [/tex] (1)
with [tex] \omega_{0}[/tex] the initial angular velocity, [tex] \omega[/tex] the final angular velocity, [tex] \alpha[/tex] the angular acceleration and [tex] \Delta \theta [/tex]the revolutions on radians (2844 revolutions = 17869.38 radians). Solving (1) for initial velocity:
[tex]\sqrt{\omega^{2}-2\alpha\varDelta\theta}=\omega_{0} [/tex]
[tex]\omega_{0}^2=\sqrt{(140)^2 -(2)(0.100)(17869.38)=126.59 \frac{rad}{s}}[/tex]
b) Knowing those values, we can use now the kinematic equation
[tex] \omega=\omega_{0}+\alpha t[/tex]
with t the time, solving for t:
[tex]t=\frac{\omega-\omega_0}{\alpha}=\frac{140-126.59}{0.1} [/tex]
[tex] t=134.1 s[/tex]
We divide the electromagnetic spectrum into six major categories of light, listed below. Rank these forms of light from left to right in order of increasing wavelength. To rank items as equivalent, overlap them.a.gamma raysb.X raysc.ultravioletd.visible lighte.infraredf.radio waves
Answer:
(1.) gamma rays. (2.) X-rays. (3.) ultraviolet. (4.) visible light. (5.) infrared.
(6.) radio waves
Explanation:
Notice that these wavelengths span an enormous range. The wavelengths of gamma rays can be smaller than the size of an atomic nucleus, while the wavelengths of radio waves can be many meters (or even milometers) long. Visible light spans only a very narrow range of wavelengths, from about 400 nano meters at the blue (violet) end to about 700 nano meters at the red end.
The types of light in the electromagnetic spectrum, ranked from shortest to longest wavelength, are gamma rays, X-rays, ultraviolet, visible light, infrared, and radio waves.
Explanation:The electromagnetic spectrum is made up of different types of waves, each characterized by specific wavelengths. The categories of light in the electromagnetic spectrum from shortest to longest wavelength are:
Gamma raysX-raysUltravioletVisible lightInfraredRadio wavesAs we move from gamma rays to radio waves, the wavelength increases. Gamma rays have the shortest wavelength, while radio waves have the longest.
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If the L-shaped rod has a moment of inertia I=9kgm2, F1=12N, F2=27N, and again F3=0, how long a time t would it take for the object to move through 45∘ ( π/4 radians)? Assume that as the object starts to move, each force moves with the object so as to retain its initial angle relative to the object. Express the time in seconds to two significant figures.
Answer:
Time, t = 2.80seconds
Explanation:
Calculating net Torque, T
Torque(net)= F2(0.08m) - F1(0.03)
Torque(net) = (27×0.08)-(12×0.03)
Torque(net)= 2.16 - 0.36 =1.8N/m
Pi/4= Torque(net)/ Inertia
3.142/4= 1.8/9 = 0.2rad/s^2=a
Pi/4= 1/2 at^2
3.142/4=1/2×0.2 ×t^2
0.7855= 0.1 t^2
t = sqrt(0.7856/0.1)
t= sqrt(7.855)
t=2.80 seconds
By Assuming that as the object starts to move, each force moves with the object so as to retain its initial angle relative to the object. then the time in seconds is 2.80seconds
What are the types of force ?Force can be denoted as the parameter obtained by pushing or pulling of any object which result in object’s interaction or movement, without force the objects can not move, it can be stopped or change the direction.
Force is a quantitative property and it is an interaction between two physical bodies, means an object and its environment, there are different types of forces in nature.
If an object is in its moving state then it will be either static or motion, if it is pushed or pulled the object will change its position and The external push or pull upon the object called as Force.
There are two types of forces such as contact force which are the force occurs when we apply some effort on the object such as Spring Force, Applied Force, Air Resistance Force, Normal Force, Tension Force, Frictional Force
Non-Contact forces can be defined as the force occur from a distance such as Electromagnetic Force, Gravitational Force, Nuclear Force
Calculating net Torque, T, Torque(net)= F2(0.08m) - F1(0.03)
Torque(net) = (27×0.08)-(12×0.03)
Torque(net)= 2.16 - 0.36 =1.8N/m
Pi/4= Torque(net)/ Inertia
3.142/4= 1.8/9 = 0.2rad/s^2=a
Pi/4= 1/2 at^2
3.142/4=1/2×0.2 ×t^2
0.7855= 0.1 t^2
t = sqrt(0.7856/0.1)
t= sqrt(7.855)
t=2.80 seconds
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You are operating a powerboat at night. You see green and white lights on another boat. What should you do?
Answer:
Just maintain the course you are on and your speed.
Explanation:
When you see a green and a white light, then from the port (left) side you will start coming across another craft. You are the stand-alone craft in this scenario and therefore should stay consistent with your pace and direction. The other craft must take important and early measures to keep the craft away from you.
A beam of light is traveling through a medium at 200,000 km/s. It enters a different medium and speeds up to almost 250,000 km/s. Finally, it enters a third medium and halts. What were the media the light traveled through, in order? Hurry!!!!!!!!!!!
Answer
From the question we can see that the in medium 1 speed of light is 200,000 Km/s where as in Medium 2 speed of light is 250,000 km/s.
We can conclude that medium 1 is denser than Medium 2.
In third medium Light speed halts so, the third medium is opaque.
We can say that
Medium 1 can be either water, glass.
Medium 2 will be gas
Medium 3 will be an opaque material.
Answer:
D
Explanation:
Which one of the following statements is true?
1. Burning coal contributes to sulfur aerosols in the stratosphere; these sulfur aerosols have moderated global warming.
2. Sulfur compounds released into the atmosphere have relatively short lifetimes of only an hour or two.
3. Volcanoes are an anthropogenic source of sulfur compounds.
4. The widespread use of nuclear energy has released significant amounts of sulfur dioxides that have made their way into the stratosphere.
Answer:
1. Burning coal contributes to sulfur aerosols in the stratosphere; these sulfur aerosols have moderated global warming.
Explanation:
Sulfur aerosols both absorb as well as scatter solar radiation coming from the sun , thus saving the earth from getting wormed up . In this way it has moderating effect on the global warming . This is also called global dimming effect. Less of the solar radiation is able to reach earth surface due to these aerosol.
A car is traveling at 22 m/s and slows to 4.9 m/s in 4.0 seconds. What is the acceleration? Give the answer to one decimal place.
Answer:
a= - 4.2 m/s².
Explanation:
Given that
u = 22 m/s
v= 4.9 m/s
t= 4 s
The average acceleration = a
We know v = u +at
v=final velocity
u=initial velocity
Now by putting the values in the above equation
4.9= 22 + a x 4
[tex]a=\dfrac{4.9-22}{4}\ m/s^2[/tex]
a= - 4.2 m/s².
Therefore the acceleration will be - 4.2 m/s².
a= - 4.2 m/s².
Negative indicates that velocity and acceleration is is opposite direction.
Why are biogeochemical cycles important? Earth is a closed system. That means the amount of rocks, metals, carbon, nitrogen, oxygen, and water on Earth.... So, it is essential that biogeochemical cycles..... these materials as they move through Earth's subsystems.
Answer:
The earth is considered to be a closed system, where the concentration of rocks, metals, C, N, O and H₂O on earth remains constant.
So, it is essential that the biogeochemical cycle renews these materials while moving through the earth subsystems.
Explanation:
The biogeochemical cycle is usually defined as the circulation of chemical components such as Carbon, Nitrogen, Phosphorous, Oxygen, rocks, and metals within the different spheres and interaction of biotic and non-biotic factors.
These components are continuously in motion from one sphere to another. For example, the organism takes up the elements through food and some of these components such as carbon is released into the atmosphere in the form of CO₂. After the death of the organisms, the decomposers feed on these elements, where some of it is consumed by them and the remaining is eliminated into the atmosphere.
So, this is how the elements are renewed, and they are conserved. The total concentration of these elements remains constant considering the earth to be a closed system.
Thus, the correct answers are given above.
Answer:
remains constant and renews
Explanation:
Force is defined as mass times acceleration. Starting with SI base units, derive a unit for force. Using SI prefixes suggest a convenient unit for the force resulting from a collision with a 10-ton trailer truck moving at 55 miles per hour. (Assume 1 second deceleration time)
Force is derived from the equation F = ma, resulting in units of Newtons (N). For a 10-ton truck decelerating over 1 second from 55 mph, we convert mass to kg and speed to m/s to calculate force, likely resulting in a force measured in kilonewtons (kN) or meganewtons (MN) due to the large mass and deceleration involved.
Explanation:The subject of this question is how to derive a unit for force starting with SI base units, and finding a convenient unit for the force resulting from a specific collision scenario. Starting from the base, force (F) can be defined using Newton's second law, F = ma (force equals mass times acceleration), where mass has units of kilograms (kg), and acceleration is measured in meters per second squared (m/s²). Thus, the unit of force in the SI system is the Newton (N), defined as the force needed to accelerate a 1-kg object by 1 m/s².
To find a convenient unit for the force resulting from a collision with a 10-ton trailer truck moving at 55 miles per hour and assuming a 1 second deceleration time, we first convert the truck's mass to kilograms (10 tons = 9,071.85 kg, assuming 1 ton = 907.185 kg) and its speed to meters per second (55 mph ≈ 24.5872 m/s, assuming 1 mile = 1.60934 km and 1 hour = 3600 seconds). The deceleration rate (a) can be calculated assuming the final velocity (v) is zero and the time (τ) is 1 second, giving us a = -24.5872 m/s² (negative sign denotes deceleration). Thus, applying F = ma, the force of the collision can be calculated, and due to the large numbers involved, units such as kilonewtons (kN) or meganewtons (MN) may be more convenient depending on the exact outcome of the calculation.
The force resulting from the collision with a 10-ton trailer truck moving at 55 miles per hour is approximately 245.87 kN.
Starting with the equation for force, F = ma, where:
- F represents force,
- m represents mass,
- a represents acceleration.
In the International System of Units (SI), the base units are:
- Mass ( m ) is measured in kilograms (kg).
- Acceleration ( a ) is measured in meters per second squared [tex](\( \mathrm{m/s^2} \)).[/tex]
Therefore, the unit of force ( F ) in the SI system is:
[tex]\[ \mathrm{kg \cdot m/s^2} \][/tex]
This unit is commonly known as the Newton (N).
Now, let's calculate the force resulting from a collision with a 10-ton trailer truck moving at 55 miles per hour with a deceleration time of 1 second.
First, we need to convert the mass of the trailer truck from tons to kilograms:
[tex]\[ 1 \text{ ton} = 1000 \text{ kg} \][/tex]
So, [tex]\( 10 \text{ tons} = 10,000 \text{ kg} \).[/tex]
Next, let's convert the speed from miles per hour to meters per second:
[tex]\[ 1 \text{ mile} = 1609.34 \text{ meters} \][/tex]
[tex]\[ 1 \text{ hour} = 3600 \text{ seconds} \][/tex]
So, [tex]\( 55 \text{ miles per hour} \)[/tex] is equivalent to:
[tex]\[ \frac{55 \times 1609.34}{3600} \text{ meters per second} \approx 24.587 \text{ m/s} \][/tex]
Given that deceleration time ( t ) is [tex]\( 1 \text{ second} \)[/tex], we can use the formula for acceleration:
[tex]\[ a = \frac{v}{t} \][/tex]
where v is the change in velocity.
Since the truck is decelerating, the change in velocity is from [tex]\( 24.587 \text{ m/s} \) to \( 0 \text{ m/s} \).[/tex]
[tex]\[ a = \frac{0 - 24.587}{1} \text{ m/s}^2 = -24.587 \text{ m/s}^2 \][/tex]
Now, we can calculate the force:
[tex]\[ F = m \times a = 10,000 \text{ kg} \times (-24.587 \text{ m/s}^2) \][/tex]
[tex]\[ F \approx -245,870 \text{ N} \][/tex]
Since force is a vector quantity and its direction is opposite to the direction of motion, the negative sign indicates that the force is acting in the opposite direction of the truck's motion.
For convenience, we can express this force in kilonewtons ( kN ):
[tex]\[ -245,870 \text{ N} = -245.87 \text{ kN} \][/tex]
So, the suggested convenient unit for the force resulting from the collision is [tex]$\mathrm{kN}$[/tex].
Complete Question:
Force is defined as mass times acceleration. Starting with SI base units, derive a unit for force. Using SI prefixes suggest a convenient unit for the force resulting from a collision with a 10-ton trailer truck moving at 55 miles per hour. (Assume 1 second deceleration time)
a. MN
b. GN
c. mN
d. kN
e. TN
If a trapeze artist rotates once each second while sailing through the air, and contracts to reduce her rotational inertia to one fourth of what it was, how many rotations per second will result?
Answer:
There are finally 4 rotations per second.
Explanation:
If a trapeze artist rotates once each second while sailing through the air, and contracts to reduce her rotational inertia to one fourth of what it was. We need to find the final angular velocity. It is a case of conservation of angular momentum such that :
[tex]I_1\omega_1=I_2\omega_2[/tex]
Let [tex]I_1=I[/tex] , [tex]I_2=\dfrac{I}{4}[/tex] and [tex]\omega_1=1[/tex]
So,
[tex]\omega_2=\dfrac{I_1\omega_1}{I_2}[/tex]
[tex]\omega_2=\dfrac{I\times 1}{(I/4)}[/tex]
[tex]\omega_2=4\ rev/sec[/tex]
So, there are finally 4 rotations per second. Hence, this is the required solution.
Final answer:
When a trapeze artist reduces their rotational inertia to one fourth while rotating, their rotational speed increases to four times its original rate to conserve angular momentum, resulting in 4 rotations per second.
Explanation:
The question pertains to the conservation of angular momentum, which is a principle in physics stating that if no external torque acts on an object, the total angular momentum will remain constant. For a trapeze artist or any rotating body, if the moment of inertia decreases, the rotational speed (number of rotations per second) must increase proportionally to conserve angular momentum.
If a trapeze artist is rotating once each second and contracts to reduce the rotational inertia to one fourth, her rotational speed must increase to four times what it was to conserve angular momentum. Therefore, the new rotation rate will be 4 rotations per second.