Its D: A driver's foot on the accelerator and pressed down gradually more and more.
What happens to light when it changes speed?
A)It reflects
B)It polarizes
C)It refracts.
Answer:
The correct answer is C) Refracts
A car with mass m traveling at speed v has kinetic energy k. what is the kinetic energy of a second car that has the same mass m and twice the speed of the first car? k 1.4 k 2 k 4 k
Answer:
the kinetic energy of the second car would be D. 4k
Explanation:
took the test
A Geiger counter is like an electroscope that discharges whenever ions formed by a radioactive particle produce a conducting path. A typical Geiger counter consists of a thin conducting wire of radius 0.002 cm stretched along the axis of a conducting cylinder of radius 2.0 cm. The wire and the cylinder carry equal and opposites charges of 8.0 x 10-10 C all along their length of 10.0 cm. What is the magnitude of the electric field at the surface of the wire
The magnitude of the electric field at the surface of the wire in the Geiger counter can be found using the charge per unit length and the formula E = (1/(2πε0)) * (λ/r). Substituting the given values into the formula will yield the electric field at the wire's surface.
The student asked about calculating the magnitude of the electric field at the surface of the wire in a Geiger counter. To solve this, we will employ the formula for the electric field (E) generated by an infinitely long charged wire, which is given by E = (1/(2πε0)) * (λ/r), where λ is the charge per unit length and r is the radial distance from the wire. In this case, the charge per unit length λ is the total charge divided by the length of the wire, and r is the radius of the wire.
Given: Charge (Q) = 8.0 x 10-10 C, Length (L) = 10.0 cm = 0.1 m, and Wire radius (a) = 0.002 cm = 0.00002 m.
First, calculate the charge per unit length: λ = Q/L = (8.0 x 10-10 C) / (0.1 m) = 8.0 x 10-9 C/m.
Next, calculate the electric field using the radius of the wire (a) as radial distance (r): E = (1/(2πε0)) * (λ/a).
Using the value for the vacuum permittivity ε0 (approximately 8.854 x 10-12 C2/N·m2), the electric field on the surface of the wire is computed to be:
E = (1/(2π(8.854 x 10-12))) * (8.0 x 10-9 / 0.00002) N/C.
Simplifying this expression gives us the electric field at the surface of the wire.
Using the given values, the magnitude of the electric field at the surface of the wire (r = 0.002 cm or 0.00002 m) and L = 10 cm or 0.1 m is calculated as follows:
Convert the radius to meters: r = 0.002 cm = 0.00002 m.
Plugging in the values: E = (8.0 x 10-10 C) / (2π * 8.854 x 10-12 C2/Nm2 * 0.00002 m * 0.1 m).
Calculate the electric field: E = 2.87 x 106 N/C.
The magnitude of the electric field at the surface of the wire in the Geiger counter is therefore 2.87 x 106 Newtons per Coulomb (N/C).
The magnitude of the electric field at the surface of the wire is approximately [tex]\( 7.19 \times 10^{6} \, \text{N/C} \)[/tex].
The magnitude of the electric field at the surface of the wire is given by the formula:
[tex]\[ E = \frac{\sigma}{\varepsilon_0} \][/tex]
where [tex]\( \sigma \)[/tex] is the surface charge density of the wire, and [tex]\( \varepsilon_0 \)[/tex]is the vacuum permittivity [tex](\( 8.85 \times 10^{-12} \, \text{C}^2/\text{N} \cdot \text{m}^2 \)).[/tex]
The surface charge density [tex]\( \sigma \)[/tex] can be calculated by dividing the total charge [tex]\( Q \)[/tex] by the surface area [tex]\( A \)[/tex] of the wire. The surface area of a cylinder is given by [tex]\( A = 2\pi r h \)[/tex], where [tex]\( r \)[/tex] is the radius and [tex]\( h \)[/tex] is the height (or length in this case) of the cylinder.
Given:
- Radius of the wire, [tex]\( r = 0.002 \, \text{cm} = 2 \times 10^{-5} \, \text{m} \) (since 1 cm = 0.01 m)[/tex]
- Length of the wire, [tex]\( h = 10.0 \, \text{cm} = 0.1 \, \text{m} \)[/tex]
- Total charge on the wire, [tex]\( Q = 8.0 \times 10^{-10} \, \text{C} \)[/tex]
First, we convert the radius from centimeters to meters for consistency in units:
[tex]\[ r = 0.002 \, \text{cm} \times \frac{1 \, \text{m}}{100 \, \text{cm}} = 2 \times 10^{-5} \, \text{m} \][/tex]
Now, we calculate the surface area of the wire:
[tex]\[ A = 2\pi r h = 2\pi (2 \times 10^{-5} \, \text{m})(0.1 \, \text{m}) \][/tex]
[tex]\[ A = 4\pi \times 10^{-6} \, \text{m}^2 \][/tex]
Next, we calculate the surface charge density [tex]\( \sigma \)[/tex]:
[tex]\[ \sigma = \frac{Q}{A} = \frac{8.0 \times 10^{-10} \, \text{C}}{4\pi \times 10^{-6} \, \text{m}^2} \][/tex]
[tex]\[ \sigma = \frac{8.0 \times 10^{-10} \, \text{C}}{4\pi \times 10^{-6} \, \text{m}^2} \approx \frac{8.0 \times 10^{-10}}{12.56637 \times 10^{-6} \, \text{m}^2} \][/tex]
[tex]\[ \sigma \approx 6.3662 \times 10^{-5} \, \text{C/m}^2 \][/tex]
Finally, we calculate the magnitude of the electric field at the surface of the wire:
[tex]\[ E = \frac{\sigma}{\varepsilon_0} = \frac{6.3662 \times 10^{-5} \, \text{C/m}^2}{8.85 \times 10^{-12} \, \text{C}^2/\text{N} \cdot \text{m}^2} \][/tex]
[tex]\[ E \approx 7.19 \times 10^{6} \, \text{N/C} \][/tex]
Therefore, the magnitude of the electric field at the surface of the wire is approximately [tex]\( 7.19 \times 10^{6} \, \text{N/C} \)[/tex].
A water balloon weighing 19.6 N rests on a table. The balloon has an area of 0.015 m^2 in contact with the table. What pressure does the balloon exert on the table?
HURRY The compressed spring of a dart gun has potential energy of 50 J. If the spring constant is 200 N/m, what is the displacement of the spring?
A. 0.5 m
B. 0.2 m
C. 0.7 m
D. 0.4 m
Answer: its C on edge 2023. =0.7
Explanation: He's right just putting it in a quicker format.
What is a property of a transparent object? A. Almost all of the light rays that reach it are scattered. B. Almost all of the light rays that reach it are absorbed. C. Almost all of the light rays that reach it are reflected. D. Almost all of the light rays that reach it pass through it.
A rock of mass m is thrown horizontally off a building from a height h. the speed of the rock as it leaves the thrower's hand at the edge of the building is v0, as shown. m h v0 m what is the kinetic energy of the rock just before it hits the ground? 1. kf = 1 2 m v2 0 2. kf = 1 2 m v2 0 − m g h 3. kf = 1 2 m v2 0 + m g h 4. kf = m g h − 1 2 m v2 0 5. kf = m g h
[tex]K_f=mgh+\frac{1}{2}mv_o^2[/tex]
Given information:
A rock of mass m is thrown horizontally off a building from a height h
As, the total energy of rock at the time of leaving the thrower's hand is the sum of gravitational potential energy and the initial kinetic energy.
[tex]E=U_i+K_i\\E=mgh+(1/2)mv_0^2[/tex]
As, the height from the ground decreases, the potential energy before hitting the ground is zero and the total final energy is just kinetic energy:
[tex]E=K_f[/tex]
As, the law of conservation of energy states the total final energy must be equal to the initial energy.
Hence, the final kinetic energy will be
[tex]K_f=mgh+\frac{1}{2}mv_o^2[/tex]
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How many electrons are in an electrically neutral atom of boron? A) 3 B) 5 C) 6 D) 8
Answer:
B. five electrons
Explanation:
A bag of sports equipment has a mass of 10.0 kilograms and a weight of
The weight of the bag of sports equipment is 98N
From the question,
We are to determine the weight of a bag of sports equipment that has a mass of 10.0 kg
The weight of an object with a given mass can be determined by using the formula
W = mg
Where
W is the weight
m is the mass
and g is the acceleration due to gravity (g = 9.8 m/s²)
From the given information,
m = 10.0 kg
∴ Weight of the bag of sports equipment,
W = 10.0 × 9.8
W = 98 N
Hence, the weight of the bag of sports equipment is 98N
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The distance from the sun to the Earth is 1.5 x 1011 m. How long does it take for light from the sun to reach the earth? Give your answer in seconds
it actually takes 500 seconds if you do the math properly but when converted to min 500 sec is 8 min and 20 sec
If we do the arithmetic correctly, it truly takes 500 seconds, but when we translate it to minutes, it takes 8 minutes and 20 seconds.
What is Time?Time is the ongoing progression of existence and things that happen in what seems to be an irrevocable order from the past, present, and forward into the future.
The four identical dimensions that make up the cosmos are put together to form a single, four-dimensional manifold that is properly called spacetime. Any physical item that is positioned at many points in time has a unique temporal component for each of those periods.
Time is calculated by dividing the distance which is travelled by the speed
For above given example,
the speed of the light is expressed as [tex]3*10^8 m/s[/tex]
Distance of earth from the sun is [tex]1.5 * 10^1^1[/tex] meters.
Thus, Time= Distance of the Earth/ Speed of the light
Time= [tex]1.5* 10^1^1 m[/tex]/ [tex]3*10^8m/s[/tex]
Time = [tex]0.5*10^1^1* 10^-^8\\0.5*10^3\\0.5*1000\\[/tex]
Time= 500seconds which is equals to 8 minutes and 20 seconds.
Thus, it takes 500 seconds for light to travel from the sun to the earth.
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The steel pipe is filled with concrete and subjected to a compressive force of 80 kn. determine the average normal stress in the concrete and the steel due to this loading. the pipe has an outer diameter of 100 mm and an inner diameter of 50 mm. est = 200 gpa, ec = 30 gpa.
The stress on the steel is 3.88mPa and the stress on the concrete is 1.94mPa
Data;
Compressive force = 80knouter diameter = 100m = 0.1minner diameter = 50mm = 0.05mSummation of Forces[tex]\sum f_y = 0 = Pst + Pc - \delta 0 = 0...equation(i)\\\delta st = \delta c\\[/tex]
We can solve for Pst and Pc
[tex]\frac{Pst * L}{\pi/4 * (0.1^2 - 0.05^2) * 200*10^9} = \frac{Pc * L}{\pi/4 * (0.05)^2 * 30*10^9}\\ 0.075Pst = 1.5Pc\\Pst = 20Pc...equation(ii)\\[/tex]
From equation(i) and (ii)
[tex]Pst + Pc + 0 = 80\\Pst = 20Pc\\20Pc + Pc = 80\\21Pc = 80\\Pc = \frac{80}{21} \\Pc = 3.81kN[/tex]
Let's solve for Pst
[tex]Pst = 20Pc\\Pst = 3.81 * 20\\Pst = 76.2kN[/tex]
Normal StressThe normal stress between the concrete and steel can be calculated as
The stress on the steel[tex]\sigma st = \frac{Pst}{A} \\\sigma st = \frac{76.2*10^3}{\frac{\pi }{4} * (0.1^2 * 0.05^2}\\\sigma st = 3.88mPa[/tex]
The stress on the concrete[tex]\sigma c=\frac{3.81*10^3}{\frac{\pi }{4}* (0.05)^2 } \\\sigma c = 1.94mPa[/tex]
The stress on the steel is 3.88mPa and the stress on the concrete is 1.94mPa
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A particular battery lasts 14.5 hours. what percentile is its lifetime on?
The battery with a lifetime of 14.5 hours is at the 40.13th percentile in terms of battery lifetimes, assuming a normal distribution with a mean of 15 hours and a standard deviation of 2 hours.
Certainly, to find the percentile of the battery's lifetime, we'll use the Z-score formula and then refer to the standard normal distribution table. Let's assume a normal distribution with a mean (μ) of 15 hours and a standard deviation (σ) of 2 hours. The battery's lifetime is X = 14.5 hours.
The Z-score is calculated as:
Z = (X - μ) / σ
Substitute the values:
Z = (14.5 - 15) / 2 = -0.25
Now, we need to find the percentile associated with this Z-score. Looking up -0.25 in the standard normal distribution table, we find that the corresponding percentile is approximately 40.13%.
The question probable may be:
What is the percentile of a battery's lifetime with a mean (μ) of 15 hours and a standard deviation (σ) of 2 hours when its actual lifetime is X = 14.5 hours? Use the Z-score formula and the standard normal distribution table to determine the percentile associated with the Z-score of -0.25.
This battery's lifetime falls in the 22.66th percentile.
The lifetime of a battery is normally distributed with a mean (μ) of 16 hours and a standard deviation (σ) of 2 hours. To find what percentile a particular battery lasting 14.5 hours falls into, we need to calculate its Z-score.
The Z-score is given by:
[tex]Z = \frac{X- \mu}{\sigma}[/tex]
where X is the observed value.
The Z-score of the given battery is:
Z = [tex]\frac{14.5 - 16}{2}[/tex] = [tex]\frac{-1.5}{2}[/tex] = - 0.75
According to the Z-table, a Z-score of - 0.75 corresponds to a cumulative probability of approximately 0.2266.
This means that 14.5 hours is at the 22.66th percentile.
Thus, the lifetime of a particular battery lasting 14.5 hours falls at the 22.66th percentile of the lifetime distribution.
The complete question is:
The lifetime of a battery in a certain application is normally distributed with mean μ = 16 hours and standard deviation σ = 2 hours.
Activities:
What is the probability that a battery will last more than 19 hours?Find the 10th percentile of the lifetimes.A particular battery lasts 14.5 hours. What percentile is its lifetime on?What is the probability that the lifetime of a battery is between 14.5 and 17 hours?Almost all of the electricity that people use is produced by
It takes a person one half hour to run 6 kilometers at a constant rate along a straight-line path. What is the velocity of the person?
A train travels 67 kilometers in 1 hours, and then 81 kilometers in 5 hours. What is its average speed?
Graph the velocity of a car accelerating at a uniform rate from 7.0 m/s to 12.0 m/s in 2.0 s. Calculate the accerleration
Final answer:
Velocity vs. time graph for uniform acceleration is a straight line, acceleration calculated using the formula a = (Vf - Vi) / t. The acceleration during the first 8 seconds is 1 m/s², and during the last 6 seconds it is -1 m/s².
Explanation:
To graph the velocity of a car accelerating at a uniform rate from 7.0 m/s to 12.0 m/s in 2.0 s, plot a velocity vs. time graph where the time axis (x-axis) spans at least 2 seconds and the velocity axis (y-axis) spans from 7.0 m/s to 12.0 m/s. The graph will be a straight line starting at the point (0, 7.0) and ending at the point (2.0, 12.0) since the acceleration is uniform.
To calculate the acceleration (a), use the formula a = (Vf - Vi) / t, where Vf is the final velocity, Vi is the initial velocity, and t is the time taken for the change in velocity. Substituting the given values, we get a = (12.0 m/s - 7.0 m/s) / 2.0 s = 2.5 m/s².
The acceleration of the car during the first 8 seconds (from 2 m/s to 10 m/s) is calculated by a = (Vf - Vi) / t = (10 m/s - 2 m/s) / 8 s = 1 m/s². During the last 6 seconds, when the car slows down from 10 m/s to 4 m/s, the acceleration is a = (4 m/s - 10 m/s) / 6 s = -1 m/s². Note the negative acceleration indicates a deceleration or slowing down.
During one year, eight moose in a population died and two moose were born. Three moose immigrated from another population and five emigrated to find mates.
What was the population growth during this year?
-8 hope it helps i just took the test
What is the frequency of a wave that has a wave speed of 20 m/s and a wavelength of 0.20 m
If the above two waveforms were sound waves, we would hear the ___________ wave louder. If the above two waveforms were light waves, we would see the ________ wave dimmer.
Answer:
the answer is green, red
The charge per unit length on a long, straight filament is -94.5 µc/m. (a) find the electric field 10.0 cm from the filament, where distances are measured perpendicular to the length of the filament. (take radially inward toward the filament as the positive direction.)
The electric field at 10.0 cm from a long, straight filament with a charge per unit length of -94.5 µC/m can be calculated using the formula E = λ / (2πε₀r), producing a value directed radially inward.
Explanation:Finding the Electric Field from a Charged Filament
The problem is asking us to calculate the electric field generated by a long, straight filament with a given charge per unit length. By using Gauss's law, we can find the electric field at a distance from the filament. Since the electric field of a uniformly charged infinite line is symmetrical, it only depends on the distance from the line, and it is directed radially.
We can use the formula for the electric field created by an infinitely long straight filament, which is:
E = λ / (2πε0r)
Where:
ε0 is the permittivity of free space (approximately 8.85 x 10-12 C2/N·m2).
r is the distance from the filament, which is 10.0 cm (or 0.10 meters).
By substituting the known values into the equation:
E = (-94.5 x 10-6 C/m) / (2π x 8.85 x 10-12 C2/N·m2 x 0.10 m)
We perform the calculation to find the magnitude and direction of the electric field. The negative sign of λ indicates that the electric field is directed radially inward, towards the filament.
A 0.55-kg ball, attached to the end of a horizontal cord, is revolved in a circle of radius 1.3 m on a frictionless horizontal surface. if the cord will break when the tension in it exceeds 75 n, what is the maximum speed the ball can have?
The force act on the body tries to attract the body inward towards the circle when the body is executing circular motion. The maximum value of speed will be 13.3 m/sec.
What is centripetal force?The force operating on an object in curvilinear motion directed toward the axis of rotation or center of curvature is known as centripetal force.
Newton is the unit of centripetal force.
The centripetal force is always perpendicular to the direction of displacement of the item. The centripetal force of an item traveling on a circular route always works towards the center of the circle.
Due to rotation, the tension is act in the chord which is equal to the centripetal force act the ball.
[tex]\rm T = F_c= \frac{mv^2}{r}[/tex]
Tension in the cord = T= 75 N
Centripetal force = [tex]F_c[/tex]
Mass of the ball= m= 0.55 Kg
Linear speed =v= ?
The radius of the circle.= r = 1.3 m
T is the maximum tension before the cord breaks. So according to the condition the obtained velocity will be ;
[tex]\rm v = \sqrt{\frac{Tr}{m} } \\\\ \rm v = \sqrt{\frac{75\times1.3}{0.55} }\\\\ \rm v = 13.3 \;m/sec[/tex]
Hence the maximum value of speed will be 13.3 m/sec.
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a boat travels in a straight path that is 25 . west of north which describes the values of the west and north components of the boat's displacement ?
Both components are positive numbers.
Both components are negative numbers.
The west component is a negative number, and the north component is a positive number.
The west component is a positive number, and the north component is a negative number.
Answer:
C on edge
Explanation:
The surface gravity on jupiter is about three times as much as the surface gravity on earth. this means
A 0.200 kg block of a substance requires 3.59 kJ of heat to raise its temperature by 20.0 K. What is the specific heat of the substance? A. 2,020 J/(kg * K) B. 383 J/(kg*K) C. 130 J/(kg * K) D. 897 J/(kg*K)
One mole of ideal gas is slowly compressed to one-third of its original volume. in this compression, the work done on the gas has magnitude 672 j . for the gas, cp=7r/2.
Myth: An organism's kingdom only describes physical characteristics.
Fact:
Evidence:
The kingdom of an organism does not only describe physical characteristics. It is evident by that plant and animal kingdom are classified on the basis of photosynthetic ability.
Kingdom:
In classification kingdom is the highest in the hierarchy. A kingdom includes the organism of similar classes.
For Example: plant and animal kingdom
Plant Kingdom:
It includes the multicellular organisms who have cell wall and produce glucose using Photosynthesis. Such as Bryophytes, Pteridophytes, and gymnosperms.Animal Kingdom:
This includes the the organisms with cell membrane only. They rely o others for their nutrition. Such as mammals and fishesTherefore, the fact is that the organism's kingdom does not only describes physical characteristics.
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Mutations are avoided during replication because DNA polymerase is able to _____?
A.) repair itself
B.) repair the DNA
C.) encode a message
D.) add more codes
Meanwhile DNA construction, maximum DNA polymerases "review their performance," getting the bulk of mis-paired bases in a method called proofreading. Shortly after DNA synthesis, any left mis-paired bases can be identified and substituted in a procedure called mismatch repair. If DNA gets destroyed, it can be fixed by different mechanisms, including chemical repeal, excision restoration, and double-stranded break repair.
A ball is thrown at a 60.0° angle above the horizontal across level ground. it is thrown from a height of 2.00 m above the ground with a speed of 23.7 m/s and experiences no appreciable air resistance. the time the ball remains in the air before striking the ground is closest to
The time the ball remains in the air before striking the ground is closest to 4.20 seconds.
To find the time the ball remains in the air before striking the ground, we can use the equations of motion. Since the ball is thrown at an angle above the horizontal, we'll need to analyze both the horizontal and vertical components of its motion separately.
Given:
- Initial velocity [tex](\(v_0\))[/tex] = 23.7 m/s
- Launch angle[tex](\(\theta\))[/tex] = 60.0°
- Initial height (h) = 2.00 m
First, we'll find the horizontal and vertical components of the initial velocity:
Horizontal component: [tex]\(v_{0x} = v_0 \cdot \cos(\theta)\)[/tex]
Vertical component: [tex]\(v_{0y} = v_0 \cdot \sin(\theta)\)[/tex]
[tex]\[v_{0x} = 23.7 \, \text{m/s} \cdot \cos(60.0^\circ)\]\\v_{0x} \approx 23.7 \, \text{m/s} \cdot \frac{1}{2}\]\\v_{0x} \approx 11.85 \, \text{m/s}\]\\v_{0y} = 23.7 \, \text{m/s} \cdot \sin(60.0^\circ)\]\\v_{0y} \approx 23.7 \, \text{m/s} \cdot \frac{\sqrt{3}}{2}\]\\v_{0y} \approx 20.57 \, \text{m/s}\][/tex]
Now, let's find the time it takes for the ball to reach the maximum height using the vertical motion equation:
[tex]\[v_y = v_{0y} - g \cdot t\][/tex]
At the maximum height, the vertical velocity[tex](\(v_y\))[/tex] becomes zero. So:
[tex]\[0 = v_{0y} - g \cdot t_{\text{max}}\][/tex]
Solving for [tex]\(t_{\text{max}}\):[/tex]
[tex]\[t_{\text{max}} = \frac{v_{0y}}{g}\]\\t_{\text{max}} = \frac{20.57 \, \text{m/s}}{9.8 \, \text{m/s}^2}\]\\t_{\text{max}} \approx 2.10 \, \text{s}\][/tex]
Now, let's find the total time the ball is in the air. We know that the time to reach maximum height and the time to fall from maximum height to the ground are equal due to symmetry (neglecting air resistance). Therefore, the total time in the air is twice the time to reach maximum height:
Total time in the air [tex]\( = 2 \times t_{\text{max}}\)[/tex]
[tex]\[ \text{Total time in the air} \approx 2 \times 2.10 \, \text{s} \]\\ \text{Total time in the air} \approx 4.20 \, \text{s} \][/tex]
So, the time the ball remains in the air before striking the ground is closest to 4.20 seconds.
An engine performs 2700 J of work on a scooter. The scooter and rider have a combined mass of 150 kg. If the scooter started at rest, what is the speed of the bike after the work is performed?
If you mix equal amounts of cyan pigments and magenta pigments on a sheet of white paper, what color will you see on the paper?
A. Red
B. Blue
C. Black
D. Yellow
E. Cyan
F. Magenta