Answer:
a
[tex]t = 3.33 \ ns[/tex]
b
i [tex]D_w =0.75 \ m [/tex]
ii [tex]D_g =0.67 \ m [/tex]
iii [tex]D_c =0.46 \ m [/tex]
Explanation:
Generally the time taken to travel 1 m in a vacuum is mathematically represented as
[tex]t = \frac{1}{c}[/tex]
Here c is the speed of light with value [tex]c = 3.0*10^{8} \ m/s[/tex]
So
[tex]t = \frac{1}{3.0*10^{8}}[/tex]
[tex]t = 3.33*10^{-9} \ s[/tex]
[tex]t = 3.33 \ ns[/tex]
The distance light travels in water is mathematically represented as
[tex]D_w = \frac{c}{n_w} * t[/tex]
Here n_w is the refractive index of water with value 1.333
So
[tex]D_w = \frac{3.0*10^{8}}{1.333} * 3.33*10^{-9}[/tex]
[tex]D_w =0.75 \ m [/tex]
The distance light travels in glass is mathematically represented as
[tex]D_g = \frac{c}{n_g} * t[/tex]
Here n_g is the refractive index of glass with value 1.5
So
[tex]D_g = \frac{3.0*10^{8}}{1.5} * 3.33*10^{-9}[/tex]
[tex]D_g =0.67 \ m [/tex]
The distance light travels in cubic zirconia is mathematically represented as
[tex]D_c = \frac{c}{n_g} * t[/tex]
Here n_c is the refractive index of cubic zirconia with value 2.15
So
[tex]D_c = \frac{3.0*10^{8}}{2.15} * 3.33*10^{-9}[/tex]
[tex]D_c =0.46 \ m [/tex]
At a time of 30 seconds a runner passes a distance marker labeled "125 meters." If the velocity of the runner is +5.0 m/s, when did the runner pass the distance marker for 75 meters?
Answer:
Explanation
He runs at 5m/s, so in 30 s he should be at 150m. So you have to do 125m - 150m and you'll get -25m, this is his initial position. They want to know the time when he hits 75m, so you would do 75 + 25, and get 100. Then do 100m / 5m/s, and you will get 25 seconds.
g uppose that 3 J of work is needed to stretch a spring from its natural length of 30 cm to a length of 45 cm. (a) How much work is needed to stretch the spring from 35 cm to 37 cm? (Round your answer to two decimal places.) .02 Incorrect: Your answer is incorrect. J (b) How far beyond its natural length will a force of 10 N keep the spring stretched? (Round your answer one decimal place.)
Answer:
W= 0.319992 J and distance is 3.75 cm
Explanation:
The energy needed to stretch the spring from 30 cm to 45 cm = 3 J
Now we required to find the requirement of energy when the spring is stretched from 35 cm to 37 cm.
So first find the work done to stretch the spring from 35 cm to 45 cm.
Work done, w = (1/2) kx^2
3 = (1/2)k(0.45 – 0.30)^2
k = 266.66 N/m
now, x1 = 0.35 – 0.30 = 0.05m
x2 = 0.37 – 0.30 = 0.07m
Now the amount of work done to stretch from 35cm to 37.
w = (1/2) k (x2^2 – x1^2)
w = (1/2) (266.66) (0.07^2 – 0.05^2 )
w= 0.319992 J
(b). Given F = 10 N
F = kx
x = F / k
x = 10 / 266.66
x = 0.0375m
x = 3.75 cm
Thus, distance is 3.75cm
What happens if we increase the value of the resistor in forward bias connection?
Explain How can you
use a distance-time graph
to calculate average
speed?
Explanation:
Average speed = change in distance / change in time
In other words, the average speed is the slope of the line connecting two points on the graph.
When a wave goes through a substance, _______________ has occurred.
When a wave gets trapped or stuck in a substance, ______________ has occurred.
When a wave bounces off of a surface, ______________ has occurred.
The ____________________ states that the angle of an incoming wave is equal to the angle of the reflected wave.
Answer:
A) When a wave goes through a substance, _______________ has occurred.
This is refraction or transmission.
Refraction refers to the changes in the wave that occur when it enters a new medium, in this case, the substance.
Transmission refers to the transfer of energy that is now in the medium.
Usually, we have both of those together.
B) When a wave gets trapped or stuck in a substance, ______________ has occurred.
This is total internal reflection. This happens when, by the Snel Law, the angle of the transmited wave is 90° (or larger) which is traduced in no transmitted wave, this is why the wave is "trapped" in the substance.
C) When a wave bounces off of a surface, ______________ has occurred.
This is reflection.
D) The ____________________ states that the angle of an incoming wave is equal to the angle of the reflected wave.
This is the Law of Reflection, that says "when light is reflected from a surface, the angle of reflection is the same as the angle of incidence"
Thermal effects refers to the:
Answer:
removal of heat by cooling towers
Newtons Second Law of motion states that for acceleration to stay the same...
Answer:
newton force is the state of gravity force they abstract when they came closer
A model airplane is flying north with a velocity of of 15m/s. A strong wind blowing east at 12m/s
What is the airplanes resultant speed (magnitude of vector)
What is the airplanes heading (direction of velocity vector
Answer:
1.) 19.21 m/s
2.) 57 degree
Explanation: Given that a model airplane is flying north with a velocity of of 15m/s. A strong wind blowing east at 12m/s
The airplanes resultant speed can be calculated by using pythagorean theorem
R = sqrt ( 15^2 + 12^2 )
R = sqrt( 225 + 144 )
R = sqrt( 369 )
R = 19.21 m/s
The magnitude of the resultant vector is 19.21 m/s
The direction of velocity vector will be:
Tan Ø = 15 /12
Tan Ø = 1.25
Ø = tan^-1(1.25)
Ø = 57.04
Ø = 57 degree.
Therefore, the airplanes is heading 57 degree in the horizontal direction of velocity vector
A trumpet player on a moving railroad flatcar moves toward a second trumpet player standing alongside the track both play a 490 Hz note. The sound waves heard by a stationary observer between the two players have a beat frequency of 2.0 beats/s. What is the flatcar's speed? Take the speed of sound to be 343 m/s.
Answer:
The value is [tex]v_s = 1.394 \ m/s[/tex]
Explanation:
From the question we are told that
The frequency of the second player is [tex]f_2 = 490 \ Hz[/tex]
The beat frequency is [tex]f_b = 2.0 \ Hz[/tex]
The speed of sound is [tex]v_s = 343 \ m/s [/tex]
Generally the frequency of the note played by the first player is mathematically represented as
[tex]f_1 = f_2 + f_b[/tex]
=> [tex]f_1 = 490 + 2.0 [/tex]
=> [tex]f_1 = 492 Hz[/tex]
From the relation of Doppler Shift we have that
[tex]f_1 = \frac{ f_2 (v+ v_o )}{v-v_s }[/tex]
Here [tex] v_o\ is\ the\ velocity\ of\ the\ observer\ with\ value\ 0 \ m/s [/tex]
So
[tex]492 = \frac{ 490 (343+0 )}{343 -v_s }[/tex]
=> [tex]v_s = 1.394 \ m/s[/tex]
Which law applies to the picture?
Newtons 1st Law of Motion
Newtons 2nd Law of Motion
Newtons 3rd Law of Motion
Answer:
Newton's first law
Explanation:
Newton's first law of motion says, that an object will stay in its current motion, unless if a force acts upon it.
So an object already at rest, will stay at rest. And the ball rolling will keep rolling, if no other force such as friction slows it down.
Suppose you performed the experiment in atmosphere of Argon at 25 deg. C, (viscosity of argon is 2.26X10^-5 N.s/m^2 at that temperature), and measured terminal speed of 2.64X10^-5 m/s and weight-neutralizing voltage of 35.0 V. How many electrons have been attached to, or detached from the initially neutral plastic sphere
Following are the calculation to the given question:
Given:
Please find the given question.
To find:
value=?
Solution:
With terminal voltage, no force equals zero.
So,
[tex]\to mg = 6\times \pi \times \eta \times r\times \vartheta \\\\[/tex]
As [tex]35\ V[/tex] is weight neutralizing voltage
[tex]\to mg= q\times v \times r\\\\[/tex]
So,
[tex]\to q\times v \times r= 6 \times \pi \times \eta \times r \times \vartheta \\\\\to q\times v = 6 \times \pi \times \eta \times \vartheta \\\\\to q =\frac{6 \times \pi \times \eta \times \vartheta}{v} \\\\[/tex]
[tex]=\frac{6 \times 3.14 \times \eta \times 2.64 \times 10^{-5}\ \frac{m}{s}}{35} \\\\=\frac{ 18.84 \times \eta \times 2.64 \times 10^{-5}\ \frac{m}{s}}{35} \\\\=\frac{ 49.7376 \times \eta \times 10^{-5}\ \frac{m}{s}}{35} \\\\=\frac{ 49.7376 \times \eta \times 10^{-5}\ \frac{m}{s}}{35} \\\\ =3.2 \times 10^{-10}\ coulombs[/tex]
Therefore
[tex]\to q= n \times e\\\\[/tex]
Hence n comes to be [tex]2\times 10^9\[/tex] electrons.
Learn more:
brainly.com/question/12816494
does potential thermal energy affect the temperature of an object?
Answer:
Yes, it does
Explanation:
There are not many explanations, but when thermal energy is shined on an object it usually starts to heat up.
Suppose a wheel with a tire mounted on it is rotating at the constant rate of 3.273.27 times a second. A tack is stuck in the tire at a distance of 0.365 m0.365 m from the rotation axis. Noting that for every rotation the tack travels one circumference, find the tack's tangential speed. tangential speed: m/sm/s What is the tack's centripetal acceleration?
Answer:
The tangential speed is [tex]v = 7.5 m/s [/tex]
The centripetal acceleration is [tex]a = 154 \ m/s^2[/tex]
Explanation
Generally the angular velocity is mathematically represented as
[tex]w = e * \frac{ 2 \ pi \ rad }{s}[/tex]
substituting 3.27 rev/s for e we have that
[tex]w = 3.27 * \frac{ 2 \ pi \ rad }{s}[/tex]
=> [tex]w = 20.55 \ rad /s[/tex]
The tangential speed is mathematically represented as
[tex]v = w * r[/tex]
substituting 0.365 m for r we have that
[tex]v = 20.55 * 0.365 [/tex]
=> [tex]v = 7.5 m/s [/tex]
Generally the centripetal acceleration is mathematically represented as
[tex]a = \frac{v^2}{r}[/tex]
=> [tex]a = \frac{7.5^2}{ 0.365}[/tex]
=> [tex]a = 154 \ m/s^2[/tex]
What is the correct answer choice to the question above?
Answer: A. a straight line inclined to the time axis
Explanation:
If horizontal velocity is 5 m/s, and vertical velocity is 8 m/s, what is the magnitude of the resultant velocity?
Answer:
40 m/s
Explanation:
if AN OBJECT HAS ZERO NET FORCE ON IT THEN THE OBJECT MUST BE AT REST true or false
Answer:
False
Explanation:
constant velocity is also 0 net force
Answer:
FALSE
Explanation:
The object can be at rest or also could be in uniform motion (moving with constant velocity)
Using the equation for for Newton's Second law, F=ma, solve the following problem. You have been given an object with a mass of 6g and an acceleration of 2 m/s2, what is the force?
Group of answer choices
A. 12N
B. 3N
C. 8N
D. 120N
Answer:
F = 0.012 N
Explanation:
Given that,
Mass of the object, m = 6 g
Acceleration, a = 2 m/s²
1 kg = 1000 grams
6 g = 0.006 kg
Force, F = ma
So,
[tex]F=0.006\ kg\times 2\ m/s^2\\F=0.012\ N[/tex]
So, the force is 0.012 N.
Which phenomenon related to light allows scientists to measure the distance between a star and Earth? A. Dark patches appear in the electromagnetic spectrum emitted by stars. B. Light waves spread out as they move farther from their source. C. Stars have different colors based on their temperature. D. Stars have unique elements that make up their atmospheres.
Answer:
The Answer Is B not C
Explanation:
Answer:b
Explanation:
How many elements in oxygen gas? PLEASE ANSWER!
Answer:
8
Explanation:
Answer:
Chemical Properties of Oxygen
At standard temperature and pressure (STP), two atoms of the element bind to form dioxygen, a colorless, odorless, tasteless diatomic gas with the formula O2. Oxygen is a member of the chalcogen group on the periodic table and is a highly reactive nonmetallic element.
Explanation:
A marble rolls 269cm across the floor with a constant speed of in 44.1cm/s.
What is the time interval of marble?
Answer:
t = 6.09 seconds
Explanation:
Given that,
Speed, v = 44.1 cm/s
Distance, d = 269 cm
We need to find the time interval of the marble. Speed is distance per unit time.
[tex]v=\dfrac{d}{t}\\\\\implies t=\dfrac{d}{v}\\\\t=\dfrac{269\ \text{cm}}{44.1\ \text{cm/s}}\\\\t=6.09\ s[/tex]
Hence, the time interval of the marble is 6.09 seconds.
sort the properties of metalloids into the correct categories
Boron
Sillicon
Germanium
Arsenic
Antimony
Tellurium
Pelorium
Metallic characteristics: lustrous, solid at room temperature
In-between characteristics: semiconducting, amphoteric
Nonmetallic: brittle
(1.736+9.6435)÷3.22=
The answer is 3.53400621
Reduce the expression, if possible, by cancelling common factors.
Answer: 3.53400621
Explanation: Hope this helped! :)
A simple model of a hydrogen atom is a positive point charge +e (representing the proton) at the center of a ring of radius a with negative charge −e distributed uniformly around the ring (representing the electron in orbit around the proton). Find the magnitude of the total electric field due to this charge distribution at a point a distance a from the proton and perpendicular to the plane of the ring.
Answer:
Now e is due to the ring at a
So
We say
1/4πEo(ea/ a²+a²)^3/2
= 1/4πEo ea/2√2a³
So here E is faced towards the ring
Next is E due to a point at the centre
So
E² = 1/4πEo ( e/a²)
Finally we get the total
Et= E²-E
= e/4πEo(2√2-1/2√2)
So the direction here is away from the ring
When a piece of paper is held with one face perpendicular to a uniform electric field the flux through it is 25 N .m2 /C. When the paper is turned 25° with respect to the field the flux through it is:_____.1. 02. 12 N .m2 /C3. 21 N .m2 /C4. 23 N .m2 /C5. 25 N .m2 /CB. A charged point particle is placed at the center of a spherical Gaussian surface. The electric flux is changed if:______.1. the sphere is replaced by a cube of the same volume2. the sphere is replaced by a cube of one-tenth the volume3. the point charge is moved off center (but still inside the original sphere)4. the point charge is moved to just outside the sphere5. a second point charge is placed just outside the sphereC. A charge of 0.8 × 10−9 C is placed at the center of a cube that measures 5 m along each edge. What is the electric flux through any 2 faces of the cube?
Answer:
1) 23 N.m²/C
2) The point charge is moved just outside the sphere
3) 30 N.m²/C
Explanation:
1) The flux through the field can be gotten by finding the dot product of the angle the later is turned, and the given uniform electric field.
25 * cos 25 =
25 * 0.9063 =
22.66 or approximately, 23 N.m²/C
2) The point charge is moved just outside the sphere
3) Using Gauss' Law, the electric flux through one face of the cube is given as Φ = Q / 6ε₀, thus,
Φ = 0.8*10^-9 / 6 * 8.85*10^-12
Φ = 0.8*10^-9 / 5.311*10^-11
Φ = 15 N.m²/C
And therefore, the flux through any 2 faces of the cube is 2 * 15 = 30 N.m²/C
A statement about what happens in nature that seems to be true all the time ES
Answer:
their is always a animal or bug in nature
Explanation:
3. A block of mass m is suspended by strings as shown in the figure. The tension in the horizontal string
is 36 N. The angle 0 is 60°. Find the mass of the block and the tension forces in string A and string
B
I am having trouble trying to answer this problem. Any help?
Answer:
mass of the block is 7.2Kg
Tension on string A = 72N
Tension on the string with an angle is = 108N
Explanation:
The Horizontal string will have the cos component of the force...
Hence, mg cos60 = 36
by keeping acceleration due to gravity = 10 m/s^2, we get,
m = 7.2Kg
The tension in String A = mg = 7.2*10 = 72N
since the whole string holding the block is pulled by another string with 36 N
the resultant force will be the tension of the string with an angle.
Hence, 72 +36 = 108 N
Please let me know whether I got this right or not...
A squirrel is running a race where she is on track for her average velocity to be 6.0 m/s. She is distracted by a dummy of an attractive male squirrel and pauses for 3.0 s. As a result her average velocity ends up being 5.0 m/s instead. What is the length of the race? [HINT: construct two equations with the same two unknowns in them and you can solve the system of equations]
Answer:
90 m
Explanation:
There are two unknowns: the amount of time the squirrel spent running, and the length of the race. Let's call these t and x, respectively.
The average velocity is the total distance divided by the total time.
5.0 m/s = x / (t + 3.0)
The total distance is the time she spent running times the speed she ran at.
x = (6.0 m/s) t
Substitute and solve:
5 = 6t / (t + 3)
5 (t + 3) = 6t
5t + 15 = 6t
t = 15
She ran for 15 seconds (not including the 3 seconds she stopped). So the length of the race is:
x = (6.0 m/s) (15 s)
x = 90 m
which is an example of a vector quantity
A.time
B.speed
C. acceleration
D. distance
Answer:
The answer to your question is C
Explanation:
acceleration is a vector quantity because it has both magnitude and direction
4. What is the momentum of a 78.0 N bowling ball with a velocity of 8.00 m/s?
Momentum = (mass) x (speed)
We don't know the mass of the bowling ball, but we know how much it weighs. So if we knew the acceleration of gravity in the place where the ball is, we could calculate its mass. The question doesn't tell us what planet the bowling ball is on. So if we only use the information given in the question, there's no way to find the answer, and we're stuck here.
But I urgently need the points, so I'm going to go out on a limb here and make a big assumption: I'll assume that the alley where this ball is rolling and bowling is located on planet Earth, where the acceleration of gravity is around 9.8 m/s² everywhere on the planet's surface. NOW I can go ahead and answer the question that I just invented.
weight = m g
Mass = (weight) / (g)
Mass = (78.0 N) / (9.8 m/s²)
Mass = 7.96 kilograms
Momentum = (mass) x (speed)
Momentum = (7.96 kg) x (8.00 m/s)
Momentum = 63.7 kg-m/s
HELP!!!
An arrow is shot into the air at an angle of 30.0 above the horizontal with a speed of 20.0 m/s. What are the x and y components of the
velocity of the arrow 1.0 s after it leaves the bowstring?
Answer:
Y(1s) = [tex]10\sqrt{3}[/tex] - 10.1
X(1s) = 10m/s
Explanation:
In annex I've done the explanation for the equations that I will just present here.
Assuming that the arrow stars from the position (0 ; 0) in the Cartesian Graphic, and with Xo and Yo the initial speeds:
[tex]Yo =\frac{\sqrt{3} }{2} . Vr\\Yo = \frac{\sqrt{3} }{2} . 20\\Yo = 10\sqrt{3} m/s \\Xo = \frac{1}{2} Vr\\Xo = 10m/s[/tex]
Ignoring friction with air, Xo = Xf
So, Xo is the same during all the movement.
X(1s) = 10m/s
For Yo is different. That component is suferring reductions from gravity.
We can find Yo(1s) with one the basic functions of cinematics:
Vf = Vo + at
Vf = Final Velocity
Vo = Start Velocity
a = aceleration - gravity (g) is negative here
t = time
Yf = Yo + gt
Yf = [tex]10\sqrt{3}[/tex] - 10.1
If you prefere, can be: Yf = 10. ([tex]\sqrt{3} - 1[/tex])