Answer:
0.62 s
Explanation:
given,
maximum height of the flea = 0.49 m
velocity at maximum height = 0 m/s
now, calculating initial velocity
using equation of motion
[tex]v^2 = u^2 + 2 g h[/tex]
[tex]0^2 = u^2 - 2\times 9.8 \times 0.49[/tex]
[tex]u^2 = 9.604[/tex]
[tex]u = 3\ m/s[/tex]
now, calculating time he take to reach at the highest point
v = u + g t
0 = 3 - 9.8 x t
9.8 t = 3
t = 0.31 s
Time it will be in air will be twice the time it took to reach to the maximum height.
Time for which it was in air = 2 x 0.31 s = 0.62 s
A motorcycle with two riders weaves dangerously between parked cars in a crowded shopping center parking lot. As the motorcyclists dart between cars, they confront a moving car. Both the car and motorcycle veer to avoid a head-on collision. The motorcycle strikes the side of the oncoming car, throwing riders to the ground. The car stops abruptly, throwing the driver into the windshield. Nearby, Lisa and Paul (two college students) hear the sound of crunching metal and blaring horns and decide to join the small group that has gathered?
Answer:
a
Explanation:
Answer:
A
Explanation:
Which of the following are molecules?
Answer:
b c and e
Explanation:
Answer: C, D, and E are all molecules!
A certain spacecraft is x AU (Astronomical Units) from Earth. How long in seconds does it take for a signal to reach the Earth after it is transmitted from the spacecraft? Hint: An AU is about 149.9 Million Km, and light moves at 299,800 Km/s. Indicate your answer to the nearest whole second.
Answer:
The time taken for a signal to reach the Earth after it is transmitted from the spacecraft is (500x) seconds
Explanation:
Distance of spacecraft from Earth = x AU = x × 149.9×10^6 Km = (149.9×10^6x) Km
Speed of light = 299,800Km/s
Time taken for a signal to reach the Earth after it is transmitted from the spacecraft = distance of spacecraft from Earth ÷ speed of light = (149.9×10^6x)Km ÷ 299,800Km/s = (500x) seconds
If a substance can be separated by physical means and it is not the same throughout, what is it?
A: a homogeneous solution
B: a heterogeneous mixture
C: a pure substance
D: an element
E: a compound
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 Honda Civic travels in a straight line along a road. Its distance x from a stop sign is given as a function of time t by the equation x(t)= α t2− β t3, where α = 1.53 m/s2 and β = 4.80×10−2 m/s3. Calculate the average velocity of the car for each time interval:_______.
a) t = 0 to t = 2.00 s.
b) t = 0 to t = 4.00 s.
c) t = 2.00 s to t = 4.00 s.
Answer:
(a) [tex]V_{avg}=2.868m/s[/tex]
(b) [tex]V_{avg}=5.352m/s[/tex]
(c) [tex]V_{avg}=7.836m/s[/tex]
Explanation:
Given data
x(t)=αt²-βt³
α=1.53m/s²
β=0.0480m/s³
First we need to find distance x at these time so
x(t)=1.53t²-0.0480t³
at t=0
x(0)=1.53(0)²-0.0480(0)³=0m
at t=2
x(2)=1.53(2)²-0.0480(2)³=5.736m
at t=4s
x(4)=1.53(4)²-0.0480(4)³=21.408 m
For(a) Average velocity at t=0s to t=2s
The average velocity is given as
Vavg=Δx/Δt
[tex]V_{avg}=\frac{x_{f}-x_{i} }{t_{f}-t_{i}}\\ As\\x(0)=0m\\x(2)=5.736m\\V_{avg}=\frac{5.736m-0m }{2s-0s}\\V_{avg}=2.868m/s[/tex]
For(b) Average velocity at t=0s to t=4s
The average velocity is given as
Vavg=Δx/Δt
[tex]V_{avg}=\frac{x_{f}-x_{i} }{t_{f}-t_{i}}\\ As\\x(0)=0m\\x(4)=21.408m\\V_{avg}=\frac{21.408m-0m }{4s-0s}\\V_{avg}=5.352m/s[/tex]
For(c) Average velocity at t=2s to t=4s
The average velocity is given as
Vavg=Δx/Δt
[tex]V_{avg}=\frac{x_{f}-x_{i} }{t_{f}-t_{i}}\\ As\\x(2)=5.736m\\x(4)=21.408m\\V_{avg}=\frac{21.408m-5.736m }{4s-2s}\\V_{avg}=7.836m/s[/tex]
Final answer:
The average velocity of the Honda Civic over the time intervals from t=0 to t=2.00 s is 2.868 m/s, from t=0 to t=4.00 s is 5.352 m/s, and from t=2.00 s to t=4.00 s is 7.836 m/s, calculated using the given equation and the variables α and β.
Explanation:
To find the average velocity of the Honda Civic over specific time intervals using the equation x(t) = α t^2 - β t^3, where α = 1.53 m/s^2 and β = 4.80×10^-2 m/s^3, we must calculate the change in position (Δx) over the change in time (Δt) for each interval.
a) t = 0 to t = 2.00 s
Initial position, x(0) = 0 (since any term with t = 0 results in 0).
Final position, x(2) = 1.53×2^2 - 4.80×10^-2×2^3 = 6.12 - 0.384 = 5.736 m.
Δx = 5.736 - 0 = 5.736 m, Δt = 2 s.
Average velocity = Δx/Δt = 5.736 m / 2 s = 2.868 m/s.
b) t = 0 to t = 4.00 s
Final position, x(4) = 1.53×4^2 - 4.80×10^-2×4^3 = 24.48 - 3.072 = 21.408 m.
Δx = 21.408 - 0 = 21.408 m, Δt = 4 s.
Average velocity = Δx/Δt = 21.408 m / 4 s = 5.352 m/s.
c) t = 2.00 s to t = 4.00 s
Using the previously calculated final positions for t = 2 and t = 4:
Δx = 21.408 - 5.736 = 15.672 m, Δt = 4 s - 2 s = 2 s.
Average velocity = Δx/Δt = 15.672 m / 2 s = 7.836 m/s.
Which of Galileo's observations directly disproved Ptolemy's epicycle model of the Solar System, showing that the Sun is at the center and not Earth?
Answer:
Venus observation.
Explanation:
Galileo had learned regarding the heliocentric (Sun-centered) idea of Copernicus, and acknowledged it. However, the theory was proven by Galileo's observations of Venus. Galileo concluded that Venus should travel round the Sun, sometimes passing behind and then beyond, instead of directly rotating around the Earth.
Three resistors are connected into the section of the circuit described by the diagram. Then the wire is cut at point x, and the two cut ends of the wire are separated. Through which of the three resistors, if any, does current still flow?
A. None of the resistors
B. All three resistors
C. R1 only
D. R1 and R2 only
Answer:
C. R1 only
Explanation:
As the wire is cut at x, there will be no current through the resistors R2 and R3. Then the current will only go from a to b through the R1 resistor.
Another way to think about this is that once the wire is cut at x, there is now infinite resistance at the point of cutting; therefore, the current can no longer flow through R2 and R3 resistors, but now it only flows through the R1 resistor.
Therefore, only choice C is correct.
A net force of 200 N acts on a 100-kg boulder, and a force of the same magnitude acts on a 130-g pebble during the same amount of time. Which object has the greatest value for each of these quantities?
Answer: pebble has the greatest value of acceleration due to its low mass.
Explanation:
According to newton's second law of motion,
Force = mass × acceleration
For boulder,
Acceleration = Force/mass
Force acting on boulder is 200N
Its mass is 100kg
Acceleration = 200/100
Acceleration = 2m/s²
Similarly for pebble,
Force on pebble = 200N (the same as boulder)
Its mass is 130g = 0.13kg(has to be converted to the standard unit which is kg)
Its acceleration = 200/0.13
Acceleration of pebble = 1538.5m/s²
Since the question doesn't specify what to compare, we will compare their accelerations.
Therefore, pebble has the greatest value of acceleration due to its low mass.
An electric drill starts from rest and rotates with a constant angular acceleration. After the drill has rotated through a certain angle, the magnitude of the centripetal acceleration of a point on the drill is twice the magnitude of the tangential acceleration. What is the angle?
Answer:
Explanation:
Given
magnitude of centripetal acceleration is twice the magnitude of tangential acceleration
Suppose [tex]\theta [/tex] is theta angle rotated by electric drill
it is given that it starts from rest i.e. [tex]\omega _0=0[/tex]
suppose [tex]\omega [/tex] and [tex]\alpha [/tex] is the final angular velocity and angular acceleration
using rotational motion equation
[tex]\omega ^2-\omega _0^2=2\times \alpha \times \theta [/tex]
where [tex]\theta [/tex]=angle turned by drill
[tex]\omega _0[/tex]=initial angular velocity
[tex]\omega [/tex]=final angular velocity
[tex]\alpha [/tex]=angular acceleration
[tex]\omega ^2-0=2\times \alpha \times \theta [/tex]
[tex]\omega ^2=2\alpha \theta ---1[/tex]
It is also given that centripetal acceleration is twice the magnitude of tangential i.e.
[tex]\omega ^2r=\alpha \times r[/tex]
where r=radial distance of any point from axis of drill
i.e. [tex]\omega ^2=\alpha [/tex]
substitute this value to equation 1
we get
[tex]\theta =\frac{\omega ^2}{2\alpha }[/tex]
[tex]\theta =1\ rad[/tex]
Describe a situation that includes no less than four charges of any magnitude, but they combine so that another location, p, has no net electric field at that point. Describe where those charges could be or what magnitudes they could be
Answer:
Four charges of equal magnitude sitting at the vertices of a square
Explanation:
We can arrive at such a situation by thinking of a simple example first, a configuration of two charges. The force acting on the middle point of a straight line joining the two points(charges) will be zero. That is, the net Electric field will be zero as they cancel out being equal in magnitude and opposite in direction.
Now, we can extend this idea to a square having charge q at each vertex. If we put 'p' at the geometric center, we can see that the Electric fields along the diagonals cancel out due to the charges at the diagonally opposite vertices(refer to the figure attached). Actually, the only requirement is that the diagonally opposite charges are equal.
We can further take this to 3 dimensions. Consider a cube having charges of equal magnitude at each vertex. In this case, the point 'p' will yet again be the geometric center as the Electric field due to the diagonally opposite charges will cancel out.
A 3.0-kg block starts at rest at the top of a 37° incline, which is 5.0 m long. Its speed when it reaches the bottom is 2.0 m/s. What is the average friction force opposing its motion?
Answer: [tex]f_{r}[/tex] = 16.49N
Explanation: The object is placed on an inclined plane at an angle of 37° thus making it weight have two component,
[tex]W_{x}[/tex] = horizontal component of the weight = mgsinФ
[tex]W_{y}[/tex] = vertical component of weight = mgcosФ
Due to the way the object is positioned, the horizontal component of force will accelerate the object thus acting as an applied force.
by using newton's law of motion, we have that
mgsinФ - [tex]f_{r}[/tex] = ma
where m = mass of object=5kg
a = acceleration= unknown
Ф = angle of inclination = 37°
g = acceleration due to gravity = 9.8[tex]m/s^{2}[/tex]
[tex]f_{r}[/tex] = frictional force = unknown
we need to first get the acceleration before the frictional force which is gotten by using the equation below
[tex]v^{2} = u^{2} + 2aS[/tex]
where v = final velocity = 2m/s
u = initial velocity = 0m/s (because the object started from rest)
a= unknown
S= distance covered = length of plane = 5m
[tex]2^{2} = 0^{2} + 2*a*5\\\\4= 10 *a\\\\a = \frac{4}{10} \\a = 0.4m/s^{2}[/tex]
we slot in a into the equation below to get frictional force
mgsinФ - [tex]f_{r}[/tex] = ma
3 * 9.8 * sin 37 - [tex]f_{r}[/tex] = 3* 0.4
17.9633 - [tex]f_{r}[/tex] = 1.2
[tex]f_{r}[/tex] = 17.9633 - 1.2
[tex]f_{r}[/tex] = 16.49N
Xenon fluoride can be prepared by heating a mixture of Xe and F2 gases to a high temperature in a pressure-proof container. Assume that xenon gas was added to a 0.25 liter container until its pressure reached 0.12 atm at 0.0°C. Fluorine gas was then added until the total pressure reached 0.72 atm at 0.0°C. After the reaction was complete, the xenon was consumed completely, and the pressure of the F2 remaining in the container was 0.36 atm at 0.0°C. What is the empirical formula of the xenon fluoride?
Answer:
XeF₄
Explanation:
Dalton law's of partial pressure states that the total pressure equal to the sum of the partial pressure of the individual gases that make up the mixture.
Pt = Pxe + Pf₂
0.72 = 0.12 + Pf₂
0.72 - 0.12 = Pf₂
0.60 atm = Pf₂
after the the reaction was complete, the pressure of F₂ remaining = 0.36 atm
pressure of the consumed F₂ = 0.60 - 0.36 = 0.24 atm
Pxe / total pressure = number of mole / total number of mole of gas present
0.24 / 0.72 = nf / nt
0.12 / 0.72 = nxe / nt
comparing the two
(1/ 3) / ( 1/6) = (nf/ nt) / ( nxe/ nt)
nf / nxe = 2 / 1
the emperical formula = XeF₄
The empirical formula of the xenon fluoride is :
-XeF₄
"Dalton law's of partial pressure"It states that the Total pressure break even with to the whole of the fractional weight of the person gasses that make up the blend.
Pt = Pxe + Pf₂0.72 = 0.12 + Pf₂0.72 - 0.12 = Pf₂0.60 atm = Pf₂Pressure of F₂ remaining = 0.36 atm
Pressure of the consumed F₂ = 0.60 - 0.36 = 0.24 atm
Pxe / total pressure = number of mole / total number of mole of gas present
0.24 / 0.72 = nf / nt
0.12 / 0.72 = nxe / nt
comparing the two:
(1/ 3) / ( 1/6) = (nf/ nt) / ( nxe/ nt)
nf / nxe = 2 / 1
The emperical formula = XeF₄
Learn more about "Emperical Formula" :
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A water break at the entrance to a harbor consists of a rock barrier with a 50.0-m-wide opening. Ocean waves of 20.0-m wavelength approach the opening straight on. At what angles to the incident direction are the boats inside the harbor most protected against wave action?
The angles to the incident direction are the boats inside the harbor most protected against wave action will be 23.57 °
What is diffraction ?Diffraction is the phenomenon that occur when a wave of light encounter an obstacle or a slit generally.
considering wide opening of harbor as thickness d
the ocean wave as light source (coherent )
boats inside the harbor as screen where diffraction pattern is going to happen
so , destructive interference should happen (to minimize the amplitude of wave ) in order to save the boats from its effect
n * lambda = d sin (theta )
n=1 ( first order minima )
sin(theta ) = lambda / d
sin( theta ) = 20 /50
sin(theta) = 2/5
theta = sin inverse (2/5)
theta = 23.57 °
The angles to the incident direction are the boats inside the harbor most protected against wave action will be 23.57 °
learn more about diffraction
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A car is traveling at 33.0 m/s when the driver uses the brakes to slow down to 29.0 m/s in 2.50 seconds. How many meters will it travel during that time?
Explanation:
We have equation of motion v = u + at
Initial velocity, u = 33 m/s
Final velocity, v = 29 m/s
Time, t = 2.50 s
Substituting
v = u + at
29 = 33 + a x 2.50
a = -1.6 m/s²
We have equation of motion v² = u² + 2as
Initial velocity, u = 33 m/s
Acceleration, a = -1.6 m/s²
Final velocity, v = 29 m/s
Substituting
v² = u² + 2as
29² = 33² + 2 x -1.6 x s
s = 77.5 m
Distance traveled during that time is 77.5 m
Final answer:
The car will travel 77.5 meters while decelerating from 33.0 m/s to 29.0 m/s over a period of 2.50 seconds by using the average speed during the deceleration period.
Explanation:
The question deals with calculating the distance covered by a car while decelerating from 33.0 m/s to 29.0 m/s over 2.50 seconds. We can solve this by finding the car's average speed during this period and multiplying it by the time duration. Here's how we can solve it:
Step-by-Step Calculation
Find the average speed: (Initial speed + Final speed) / 2 = (33.0 m/s + 29.0 m/s) / 2 = 31.0 m/s.
Multiply the average speed by the time taken to decelerate to find the distance covered: Distance = Average speed * Time = 31.0 m/s * 2.50 s = 77.5 meters.
Therefore, the car will travel 77.5 meters during that time.
A uniform plank of length 6.1 m and mass 33 kg rests horizontally on a scaffold, with 1.6 m of the plank hanging over one end of the scaffold. L l x How far can a painter of mass 60 kg walk on the overhanging part of the plank x before it tips? The acceleration of gravity is 9.8 m/s 2 . Answer in units of m.
Answer:
Explanation:
consider the principle of moment
when a system is in equilibrium, the clockwise moment (torque) about the pivot is equal to the counterclockwise moment ( torque). Since the plank is uniform the weight of the plank act at the middle which = 6.1 m / 2 = 3.05 m
the distance that can support the weight of the man = d
mass of the man = 70
70 × d = 33 × ( 3.05 - 1.6)
d = 47.85 / 60 = 0.798 m, if the man work beyond this point he will fall.
Recall the observed behavior of the compass during the mapping of the magnetic field. Which of the following descriptions best matches the behavior you observed? Two correct answers required for full credit.
A. Where the field is strong the compass needle readily and quickly aligns with rapid oscillations around the equilibrium direction.
B. Where the field is strong the compass needle slowly oscillates around the final equilibrium position.
C. Where the field is weak the compass needle rapidly oscillates around the final equilibrium position.
D. Where the field is weak the compass needle slowly oscillates around the final equilibrium position.
E. Where the field is weak the compass needle readily and quickly aligns with rapid oscillations around the equilibrium direction.
Answer:
Option A & B
Explanation:
This is cause the compass needle is magnetized anointed in such a way that it responds to magnetic field strength.
A rectangular block floats in pure water with 0.400 in. above the surface and 1.60 in. below the surface. When placed in an aqueous solution, the block of material floats with 0.800 in. below the surface. Estimate the specific gravities of the block and the solution.
Answer:
specific gravity = 0.8
specific gravity of solution = 2
Explanation:
given data
rectangular block above water = 0.400 in
rectangular block below water = 1.60 in
material floats below water = 0.800 in
solution
first we get here specific gravity of block that is
specific gravity = block vol below ÷ total block vol × specific gravity water ..............1
put here value we get
specific gravity = [tex]\frac{1.60}{1.60+0.400}[/tex] × 1
specific gravity = 0.8
and now we get here specific gravity of solution that is express as
specific gravity of solution = total block vol ÷ block vol below × specific gravity block ........................2
put here value we get
specific gravity of solution = [tex]\frac{1.60+0.400}{0.800}[/tex] × 0.8
specific gravity of solution = 2
A 170 g air-track glider is attached to a spring. The glider is pushed in 11.2 cm against the spring, then released. A student with a stopwatch finds that 14 oscillations take 11.0 s. What is the spring constant?
Answer:k = 10.83 N/m²
Explanation: The angular frequency (ω), spring constant (k) and mass is related by the formulae below
ω = √k/m
But ω = 2πf, where f = frequency.
f = number of oscillations /time taken
Number of oscillations = 14, time taken = 11s
f = 14/11 = 1.27Hz.
ω = 2×22/7×1.27
ω = 7.98 rad/s.
By substituting this parameters into ω = √k/m
Where ω = 7.98rad/s, m = 170g = 170/1000 = 0.17kg.
7.98 = √k/0.17
By squaring both sides
(7.98)² = k/ 0.17
k = (7.98)² × 0.17
k = 10.83 N/m²
Final answer:
To find the spring constant (k) of the spring system, we use the period of oscillation formula. After calculating the period of one oscillation (T = 0.7857 s), we solve for k = m/(T/2π)^2 and find that k is approximately 44.1 N/m.
Explanation:
The student asked how to find the spring constant of a spring if a 170 g air-track glider is attached to it, pushed in 11.2 cm, released, and makes 14 oscillations in 11.0 seconds. The spring constant (k) can be found using the formula for the period of a mass-spring system (T = 2π√(m/k)) where m is mass and k is the spring constant.
First, calculate the period of one oscillation by dividing the total time by the number of oscillations: T = 11.0 s / 14 = 0.7857 s. Then rearrange the period formula to solve for k: k = m/(T/2π)2. Convert the mass to kilograms (m = 0.170 kg) and substitute the values to calculate k.
Doing the calculation: k = 0.170 kg / (0.7857 s / 2π)2 gives us the spring constant k. Upon solving, we find that the spring constant is approximately 44.1 N/m.
u are holding the axle of a bicycle wheel with radius30 cm and mass 1.05 kg. You get the wheel spinning at arate of 77 rpm and then stop it by pressing the tire againstthe pavement. You notice that it takes 1.37 s for the wheelto come to a complete stop. What is the angular accelerationof the wheel
Answer:
-5.9 rad/s^{2}
Explanation:
radius (r) = 30 cm = 0.3 m
mass (m) = 1.05 kg
initial speed (u) = 77 rpm
final speed (v) = 0 rpm
time (t) = 1.37 s
angular acceleration =[tex]\frac{(final speed-initial speed)rad/s}{time}[/tex]
therefore
initial speed (U) = 77 rpm = 77 x (2π/60) = 8.06 rad/s
final speed (v) = 0 rpm = 0 rad/s
angular acceleration = [tex]\frac{0-8.06}{1.37}[/tex] = -5.9 rad/s^{2}
Answer:
-5.886 rad/s^2.
Explanation:
radius, r = 30 cm
= 0.3 m
mass, m
= 1.05 kg
initial speed, wo = 77 rpm
Converting from rpm to rad/s,
= 77 rpm * 2pi rad * 1 min/60 s
= 8.063 rad/s
final speed, wi = 0 rad/s
time, t = 1.37 s
angular acceleration = Δw/Δt
= (wi - wo)/t
= 8.063/1.37
= -5.886 rad/s^2.
Suppose that the voltage of the battery in the circuit is 3.3 V, the magnitude of the magnetic field (directed perpendicularly into the plane of the screen) is 0.66 T, and the length of the rod between the rails is 0.19 m. Assuming that the rails are very long and have negligible resistance, find the maximum speed attained by the rod after the switch is closed.
Answer:
Maximum velocity of rod will be equal to 26.315 m/sec
Explanation:
We have given voltage of the battery in the circuit V = 3.3 volt
Magnetic field B = 0.66 Tesla
Length of the rod l = 0.19 m
We know that emf is given by e = BVl
We have to find the maximum velocity of the rod
Here velocity is maximum
So [tex]e=v_{max}Bl[/tex]
So [tex]3.3=v_{max}\times 0.66\times 0.19[/tex]
[tex]v_{max}=26.315m/sec[/tex]
So maximum velocity of rod will be equal to 26.315 m/sec
From the following list of words, choose the correct one to complete the statement: absorption, reflection, emission, transmission.
a. Light coming from your computer screen is an example of _________________.
b. Yellow light hitting a yellow banana is an example of _______________.
c. Blue light hitting a yellow banana is an example of _______________.
d. Sunlight passing through a glass door is an example of ______________.
Answer:
a) Emission
b) Reflection
c) Absorption
d) Transmission
Explanation:
a) The screen of computers (most of them LED or LCD) are a bunch of tiny lights that take power of our pc to produce light so they EMMIT light.
b) Sun's light contains all the visible colors of electromagnetic spectrum, when sun light hits an object, due their chemical characteristics the object absorbs a bunch of wavelengths and reflects others, the reflected light is the color we perceive with our eyes. So, for a banana, it absorbs almost all the other colors and reflects only yellow light.
c) As explained on b) a banana absorbs all the other colors except yellow so blue is absorbed on it.
d) When a light hits transparent medium, some rays are reflected and other passes through the medium so that rays are called transmitted and that is the case of Sunlight passing through a glass door.
Final answer:
Correct words are emission for light from a computer screen, reflection for yellow light on a banana, absorption for blue light on a banana, and transmission for sunlight through a glass door. These terms describe how light interacts with different materials through emission, reflection, absorption, and transmission.
Explanation:
From the list of terms provided, the correct words to complete the statements about the behavior of light are:
a. Light coming from your computer screen is an example of emission.
b. Yellow light hitting a yellow banana is an example of reflection.
c. Blue light hitting a yellow banana is an example of absorption.
d. Sunlight passing through a glass door is an example of transmission.
When we talk about light waves interacting with materials, we discuss concepts such as reflection, absorption, and transmission. Reflection occurs when light bounces off an object, such as a yellow light reflecting off a yellow banana. Absorption occurs when an object takes in the light energy, which might later be transformed into other forms of energy like heat, as is the case when a yellow banana absorbs blue light. Lastly, transmission refers to the passage of light through a transparent or semi-transparent material, such as sunlight passing through a glass door.
A model airplane with mass 0.750-kg is tethered by a wire so that it flies in a circle of radius 30.0-m. The airplane engine provides a force of 0.800-N perpendicular to the tethering wire. (Consider the airplane to be a point mass) (a) Find the torque that the net thrust produces about the center of the circle. (b) Find the angular acceleration of the airplane. (c) Find the linear acceleration of the airplane tangent to its flight path.
Answer:
a) 24.0 N.m b) 3.6*10⁻² rad/s² c) 1.07 m/s²
Explanation:
a) If the force that produces the torque is perpendicular to the tethering wire, we can determine its magnitude just as follows:
τ = F*r = 0.800 N * 30.0 m = 24.0 N*m (1)
b) We can express the torque we found above, using the rotational form of Newton´s 2nd Law, as follows:
τ = I* α (2)
where I is the rotational inertia regarding an axis passing through the center of the circle and α is the angular acceleration of the airplane.
If we consider the airplane as a point mass, the rotational inertia I can be calculated as follows:
I = m*r² = 0.750 Kg * (30.0)² m² = 675 Kg*m²
From (1) and (2), we can solve for α, as follows:
[tex]\alpha = \frac{T}{I} = \frac{24.0 N*m}{675.0 kg*m2} = (3.6e-2) rad/s2[/tex]
⇒α = 3.6*10⁻² rad/s²
c) Applying the definition of the angular velocity, and the definition of an angle, we can find the following realtionship between the linear and angular velocity:
v = ω*r
Dividing both sides by Δt, we can extend this relationship to the linear and angular acceleration, as follows:
a = α*r
a = 3.6*10⁻² rad/s²* 30.0 m = 1.07 m/s²
a. The torque that the net thrust produces about the center of the circle is 24 Newton.
b. The angular acceleration of the airplane is equal to [tex]0.036 \;rad/s^2[/tex]
c. The linear acceleration of the airplane tangent to its flight path is[tex]1.08 \;m/s^2[/tex]
Given the following data:
Mass of airplane = 0.750 kgRadius = 30.0 mForce = 0.800 Newtona. To find the torque that the net thrust produces about the center of the circle:
Mathematically, the torque produced by a perpendicular force is given by:
[tex]T = Fr[/tex]
Where:
T is the torque.F is the perpendicular force.r is the radius.Substituting the given parameters into the formula, we have;
[tex]T = 0.8 \times 30[/tex]
Torque, T = 24 Newton
b. To find the angular acceleration of the airplane, we would use Newton's Second Law of rotational motion:
[tex]T = I\alpha[/tex]
But, [tex]I = mr^2[/tex]
[tex]I = 0.750 \times 30^2\\\\I = 0.750 \times 900[/tex]
Moment of inertia, I = [tex]675\;Kgm^2[/tex]
[tex]\alpha = \frac{T}{I} \\\\\alpha = \frac{24}{675}\\\\\alpha = 0.036 \;rad/s^2[/tex]
c. To find the linear acceleration of the airplane tangent to its flight path:
[tex]a = r\alpha \\\\a = 30 \times 0.036[/tex]
Linear acceleration, a = [tex]1.08 \;m/s^2[/tex]
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Austin left the park traveling 4 mph. Then, 3 hours later, Wyatt left traveling the same direction at 10 mph. How long until Wyatt catches up with Austin?
ANSWER: 2 hours
EXPLANATION: After three hours Austin has traveled 12 miles. Wyatt will start up and in an hour will be at 10 miles, but by that time Austin will be at 16 miles. One more mile and Wyatt will be at 20 miles and so will Austin. So that would make the answer 2 hours.
In The Funeral of St. Bonaventure, Francisco de Zurbarán used the principle of __________ to create emphasis and focal point.
Answer:
In The Funeral of St. Bonaventure, Francisco de Zurbarán used the principle of _contrast_ to create emphasis and focal point
Explanation:
Contrast relates to the combination of opposing components and effects as an art theory. Lighter and darker colours, polished and rough materials, large and small shapes. Contrast can be used for the production of diversity visual interest and excitement.
answers If visible light and radio waves are both examples of electromagnetic waves, why can’t we see radio waves?
Answer:
Radio Waves:
Radio waves being the lowest-energy form of light and are known to be produced by electrons spiraling around magnetic fields. These Magnetic fields are generated by stars, including our sun, and many weird celestial objects like black holes and neutron stars.
Explanation:
All electromagnetic radiation is light, but we can only see a small portion of this radiation that is the portion we call visible light. Cone-shaped cells in our eyes act as receivers tuned to the wavelengths in this narrow band of the spectrum
Visible light and radio waves are both examples of electromagnetic waves, but visible light has shorter wavelengths that our eyes can detect, while radio waves have longer wavelengths that are outside the range of what our eyes can see.
Explanation:Visible light and radio waves are both examples of electromagnetic waves, but they have different wavelengths. Visible light has a shorter wavelength than radio waves, which is why we can see it. Our eyes are sensitive to the wavelengths of visible light, but not to the longer wavelengths of radio waves.
The human eye can detect wavelengths within a certain range, known as the visible spectrum, which includes the colors of the rainbow. Radio waves have much longer wavelengths, ranging from meters to kilometers. These longer wavelengths are outside the range of what our eyes can detect, so we cannot see radio waves.
However, even though we can't see radio waves, we can still use them for various purposes, such as wireless communication, radar, and satellite transmission.
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To have a negative ion, you must have: A. added a positron to the outer electron shell. B. taken away a proton from the nucleus. C. added an electron to the outer electron shell. D. added a positron to the nucleus. E. None of these; only positive ions can exist in nature.
C. added an electron to the outer electron shell.
Explanation:
Atoms consist of three particles:
- Protons: they are located in the nucleus, they have positive charge of [tex]+e[/tex], and mass of [tex]1.67\cdot 10^{-27}kg[/tex]
- Neutrons: they are also located in the nucleus, they have no electric charge, and mass similar to that of the proton
- Electrons: they orbit around the nucleus, they have negative charge of [tex]-e[/tex], and mass around 1800 smaller than the proton
Normally, atoms are neutral (no electric charge), because they have an equal number of protons and electrons.
However, sometimes atoms can give off or take electrons from other atoms. We have two cases:
If an atom gives off an electron, it remains with an excess of positive charge, so it becomes a positive ionIf an atom takes an electron from another atom, it remains with an excess of negative charge, so it becomes a negative ionTherefore, in order to form a negative ion, the atom must have
C. added an electron to the outer electron shell
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A train travels due south at 25 m/s (relative to the ground) in a rain that is blown toward the south by the wind. The path of each raindrop makes an angle of 66° with the vertical, as measured by an observer stationary on the ground. An observer on the train, however, sees the drops fall perfectly vertically. Determine the speed of the raindrops relative to the ground.
Answer:
Explanation:
Given
Train travels towards south with a velocity if [tex]v_t=25\ m/s[/tex]
Rain makes an angle of [tex]\theta =66^{o}[/tex] with vertical
If an observer sees the drop fall perfectly vertical i.e. horizontal component of rain velocity is equal to train velocity
suppose [tex]v_r[/tex] is the velocity of rain with respect to ground then
[tex]v_r\sin\theta =v_t[/tex]
[tex]v_r\times \sin (66)=25[/tex]
[tex]v_r=27.36\ m/s[/tex]
Therefore velocity of rain drops is 27.36 m/s
If you fire a projectile from the ground, it hits the ground some distance R away (called "the range"). If you keep the launch angle fixed, but double the initial launch speed, what happens to the range?
Answer:
range becomes 4 times
Explanation:
We know that the range of a projectile is given as:
[tex]R=\frac{u^2.\sin(2\theta)}{g}[/tex]
where:
[tex]R=[/tex] range of the projectile
[tex]u=[/tex] initial velocity of projectile
[tex]\theta=[/tex] initial angle of projection form the horizontal
g = acceleration due to gravity
When the initial velocity of launch is doubled:
[tex]R'=\frac{(2u)^2.\sin(2\theta)}{g}[/tex]
[tex]R'=\frac{4u^2.\sin(2\theta)}{g}[/tex]
[tex]R'=4R[/tex]
range becomes 4 times
Doubling the initial launch speed of a projectile, while keeping the angle of launch fixed, results in the range being quadrupled.
Explanation:The range R of a projectile motion is given by the formula R = ((v^2)*sin(2*theta))/g, where v is the initial launch speed, theta is the launch angle, and g is the acceleration due to gravity. If the initial launch speed v is doubled, the new range R' would be R' = ((2v)^2)*sin(2*theta))/g, which simplifies to R' = 4R.
Thus, if you double the initial launch speed, the range of the projectile is quadrupled, assuming the launch angle is fixed.
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Vector A is 3.00 units in length and points along the positive x-axis. Vector B is 4.00 units in length and points along the negative y-axis. Use graphical methods to find the magnitude and direction of the following vectors:
To find the magnitude and direction of the resultant vector R, we can use graphical methods. First, find the components of vectors A and B along the x and y axes. Then, use the Pythagorean theorem to find the magnitude of R and the inverse tangent function to find the direction of R.
Explanation:To find the magnitude and direction of the resultant Vector B, with a magnitude of 4 units, points along the negative y-axis, so its x-component is 0 and its y-component is -4.
To find the components of R, we can simply add the corresponding components of A and B: Rx = Ax + Bx
= 3 + 0 = 3, Ry = Ay + By = 0 + (-4) = -4.
Using the Pythagorean theorem, we can find the magnitude of R: R = sqrt(Rx^2 + Ry^2)
= sqrt(3^2 + (-4)^2) = sqrt(9 + 16) = sqrt(25) = 5 units.
To find the direction of R, we can use the inverse tangent function: Rtheta = atan(Ry/Rx)
= atan((-4)/3)
= atan(-4/3) = -53.13 degrees.
However, since vector B points along the negative y-axis, the direction of R is 90 degrees minus the calculated angle: Rtheta = 90 - 53.13 = 36.87 degrees.
Therefore, the magnitude of R is 5 units and it points at an angle of 36.87 degrees north of the x-axis.
A train, traveling at a constant speed of 25.8 m/s, comes to an incline with a constant slope. While going up the incline, the train slows down with a constant acceleration of magnitude 1.66 m/s². What is the speed of the train after 8.30 s on the incline?
The final velocity of the train after 8.3 s on the incline will be 12.022 m/s.
Answer:
Explanation:
So in this problem, the initial speed of the train is at 25.8 m/s before it comes to incline with constant slope. So the acceleration or the rate of change in velocity while moving on the incline is given as 1.66 m/s². So the final velocity need to be found after a time period of 8.3 s. According to the first equation of motion, v = u +at.
So we know the values for parameters u,a and t. Since, the train slows down on the slope, so the acceleration value will have negative sign with the magnitude of acceleration. Then
v = 25.8 + (-1.66×8.3)
v =12.022 m/s.
So the final velocity of the train after 8.3 s on the incline will be 12.022 m/s.