Answer:
B. Community
Took science classes for 6 years now
In a physics lab, light with a wavelength of 560 nm travels in air from a laser to a photocell in a time of 17.2 ns . When a slab of glass with a thickness of 0.810 m is placed in the light beam, with the beam incident along the normal to the parallel faces of the slab, it takes the light a time of 20.8 ns to travel from the laser to the photocell.
Required:
What is the wavelength of the light in the glass?
Answer:
Distance traveled = 3 * 10E8 * 17.2 * 10E-9 = 5.16 m
.81 / 3 * 10E8 = 2.7 * 10E-9 normal time thru glass
(20.8 - 17.2) * E10-9 = 3.6 * 10E-9 additional time due to glass
c tg = c n ta where tg and ta are the times spent in glass and air
(Note you can also write Va = n Vg or D / ta = n D / tg)
n = tg / ta = 3.6 / 2.7 = 1.33 the index of refraction of the glass
Wavelength (air) = Wavelength (glass) * n
Wavelenght = 560 nm / 1.33 = 421 nm
please help me .finish this paper
Solution-1:-
[tex]\boxed{\sf \dfrac{10\times 1000}{60\times 60}}[/tex]
Solution:-2
[tex]\boxed{\sf Sodium\:and\:Potassium}[/tex]
Solution:-3
[tex]\boxed{\sf 320m}[/tex]
Solution:-4
[tex]\boxed{\sf Rough\:tiles\:are\:used\:in\:bathroom}[/tex]
Solution:-5
[tex]\boxed{\sf Mg_3N_2}[/tex]
Solution:-6
[tex]\boxed{\sf Grapes\:and\:Rambutan}[/tex]
Solution:-7
[tex]\boxed{\sf {}^{}_{}N}[/tex]
Solution:-8
[tex]\boxed{\sf Galactuse}[/tex]
Solution:-9
[tex]\boxed{\sf Y-X}[/tex]
Solution:-10
[tex]\boxed{\sf Cell\:wall}[/tex]
A barge is hauled along a straight-line section of canal by two horses harnessed to tow ropes and walking along the tow paths on
either side of the canal. Each horse pulls with a force of 839 N at an angle of 15° with the centerline of the canal. Find the sum of these
two forces on the barge.
answer in ___kN
Answer:
1.621 kN
Explanation:
Since each horse pulls with a force of 839 N at an angle of 15° with the centerline of the canal, the horizontal component of the force due to the first horse along the canal is F= 839cos15° N and its vertical component is F' = 839sin15° N(it is positive since it is perpendicular to the centerline of the canal and points upwards).
The horizontal component of the force due to the second horse along the canal is f = 839cos15° N and its vertical component is f' = -839sin15° N (it is negative since it is perpendicular to the centerline of the canal and points downwards).
So, the resultant horizontal component of force R = F + f = 839cos15° N + 839cos15° N = 2(839cos15°) N = 2(839 × 0.9659) = 2 × 810.412 = 1620.82 N
So, the resultant vertical component of force R' = F' + f' = 839sin15° N + (-839sin15° N) = 839sin15° N - 839sin15° N = 0 N
The magnitude of the resultant force which is the sum of the two forces is R" = √(R² + R'²)
= √(R² + 0²) (since R' = 0)
= √R²
= R
= 1620.82 N
= 1.62082 kN
≅ 1.621 kN
So, the sum of these two forces on the barge is 1.621 kN
Two wires are made of the same material and have the same length but different radii. They are joined end-to- end and a potential difference is maintained across the combination. Of the following quantities that is same for both wires is
A. Potential difference
B. Electric current
C. Current density
D. Electric field
Answer:
Current
I think The choose (B)
B. Electric current
During a particular thunderstorm, the electric potential difference between a cloud and the ground is Vcloud - Vground = 4.20 108 V, with the cloud being at the higher potential. What is the change in an electron's electric potential energy when the electron moves from the ground to the cloud?
Answer:
The electric potential energy is 6.72 x 10^-11 J.
Explanation:
Potential difference, V = 4.2 x 10^8 V
charge of electron, q = - 1.6 x 10^-19 C
Let the potential energy is U.
U = q V
U = 1.6 x 10^-19 x 4.2 x 10^8
U = 6.72 x 10^-11 J
At a distance of 14,000 km from the center of Planet Z-99, the acceleration due to gravity is 32 m/s2. What is the acceleration due to gravity at a point 28,000 km from the center of this planet
A body of mass m feels a gravitational force due to the planet of
F = GmM/R ² = ma
where
• G = 6.67 × 10⁻¹¹ N•m²/kg² is the universal gravitational constant
• M is the mass of the planet
• R is the distance between the body and the planet's center
• a is the acceleration due to gravity
Solving for a gives
a = GM/R ²
Notice that 28,000 km is twice 14,000 km. The equation says that the acceleration varies inversely with the square of the distance. So if R is changed to 2R, we have a new acceleration of
GM/(2R)² = 1/4 × GM/R ² = a/4
so the acceleration of the body at 28,000 km from the planet's center would be (32 m/s²)/4 = 8 m/s².
Which two factors does the power of a machine depend on? А. work and distance B.. force and distance C. work and time D. time and distance?
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[tex]AnimeVines[/tex] [tex]is[/tex] [tex]here![/tex]
The answer is...
C. Work and time.
[tex]HopeThisHelps!![/tex]
[tex]AnimeVines[/tex]
A small object with mass 0.20 kg swings as a pendulum on the end of a long light rope. For small amplitude of swing, the period of the motion is 3.0 s. If the object is replaced by one with mass 0.400 kg, what is the period for small amplitude of swing? (a) 1.5 s (b) 3.05 (c) 6.0 s (d) 12.0 s (e) none of the above answers
Answer:
The correct option is (e) "none of the above".
Explanation:
Given that,
A small object with mass 0.20 kg swings as a pendulum on the end of a long light rope. For small amplitude of swing, the period of the motion is 3.0 s.
If the object is replaced by one with mass 0.400 kg, then we have to find the period for small amplitude of the swing.
We know that the time period can be calculated as :
[tex]T=2\pi \sqrt{\dfrac{l}{g}}[/tex]
Where
l is the length
g is acceleration due to gravity
It means the time period is independent of the mass. So, if the mass is replaced by one with mass 0.400 kg, there is no effect on the time period.
What are the major sources of energy utilized during a 100 meter race, a 1000 meter race, and a marathon
Answer:
The energy from food and then from plants and then from sun.
As sun is the ultimate source of energy.
Explanation:
Distance = 100 m, 1000m, marathon
As the distance is covered by the person, so the muscular energy is used and thus the energy comes form out food.
As we know that the energy can neither be created nor be destroyed it can transform from one form to another.
So, the energy form the food which we consume is converted into the kinetic energy as we run.
Which quantities below of a solid object on this planet are NOT the same as on Earth?
Choose all
possible answers.
Weight
Mass
Volume
Density
Acceleration when it falls vertically.
Color
Answer:
Weight, acceleration when it falls vertically, are not same as that of earth.
Explanation:
Weight of the object is given by the product of mass of the object and the acceleration due to gravity of the planet.
So, the weight of object is not same as that on earth.
The mass is defined as the amount of matter contained in the object.
So, the mass of the object is same as that of earth.
The volume of the object is defined as the space occupied by the object.
So, the volume of the object is same as that of earth.
The density is defined as the ratio of mass of the object to its volume.
So, the density of the object is same as that of earth.
The acceleration due to gravity on a planet depends on the mass of planet and radius of planet.
So, the acceleration is not same as that of earth.
The color of the object is its characteristic.
It is same as that of earth.
This percentage of all water on the planet is salt water . 97 % 95% 93% 91%
hurry please !
Answer:
none of those are right, its technically 96.5%. so i would say 97% is your best bet because thats closest and it just rounds up :)
Explanation:
A car starting at rest accelerates at 3m/seconds square How far has the car travelled after 4s?
Answer:
24 meters
Explanation:
Find the final velocity. 12m/s
d=[final-initial]/2×time
D=(6m/s)×4=24 m/s
A cylindrical water tank has a height of 20cm and a radius of 14cm. If it is filled to 2/5 of its capacity, calculate.
I. Quantity of water in the tank
II. Quantity of water left to fill the tank to its capacity.
Answer:
4.926 L Y 7.389 L
Explanation:
first you calculate the tank volume
V = π[tex](14 cm)^{2}[/tex](10 cm = [tex]12315 cm^{3}[/tex]
then you convert to liters
[tex]12315 cm^{3}[/tex] = 12.315 l
then you calculate the liters of water
2/5(12.35 l) = 4.926 l
finally we calculate the amount without water
12.315 l - 4.926 l = 7.389 l
HERE IS MORE INFORMATION ON THE SUBJECT. THEY REMOVED THE
ENGLISH SITE BUT YOU CAN USE TRANSLATOR
LINK: https://gscourses.thinkific.com/courses/fisicai
In a large chemical factory, a feed pipe carries a liquid at a speed of 5.5 m/s. A pump pushes the liquid along at a gauge pressure of 140,000 Pa. The liquid travels upward 6.0 m and enters a tank at a gauge pressure of 2,000 Pa. The diameter of the pipe remains constant. At what speed does the liquid enter the tank
Answer:
v₂ = 15.24 m / s
Explanation:
This is an exercise in fluid mechanics
Let's write Bernoulli's equation, where the subscript 1 is for the factory pipe and the subscript 2 is for the tank.
P₁ + ½ ρ v₁² + ρ g y₁ = P₂ + ½ ρ v₂² + ρ g y₂
They indicate the pressure in the factory P₁ = 140000 Pa, the velocity
v₁ = 5.5 m / s and the initial height is zero y₁ = 0
the tank is at a pressure of P2 = 2000 Pa and a height of y₂ = 6.0 m
P₁ -P₂ + ρ g (y₁ -y₂) + ½ ρ v₁² = ½ ρ v₂²
let's calculate
140,000 - 2000 + ρ 9.8 (0- 6) + ½ ρ 5.5² = ½ ρ v₂²
138000 - ρ 58.8 + ρ 15.125 = ½ ρ v2²
v₂² = 2 (138000 /ρ - 58.8 + 15.125)
v₂ = [tex]\sqrt{\frac{276000}{\rho } - 43.675 }[/tex]
In the exercise they do not indicate what type of liquid is being used, suppose it is water with
ρ = 1000 kg / m³
v₂ = [tex]\sqrt{\frac{276000}{1000} - 43.675}[/tex]
v₂ = 15.24 m / s
write a note on unity of ant
Answer: When a pathogen enters their colony, ants change their behavior to avoid the outbreak of disease. In this way, they protect the queen, brood and young workers from becoming ill. These results, from a study carried out in collaboration between the groups of Sylvia Cremer at the Institute of Science and Technology Austria (IST Austria) and of Laurent Keller at the University of Lausanne, are published today in the journal Science.
Explanation: search for it.
Can a conductor be given limitless charge
Answer:
No
Explanation:
You could try to give it enough to fill all valence electrons in all of the atoms in the conductor, but practically this could not be achieved.
Consider two oppositely charged, parallel metal plates. The plates are square with sides L and carry charges Q and -Q. What is the magnitude of the electric field in the region between the plates
Answer:
E = [tex]\frac{Q}{L^2 \epsilon_o}[/tex]
Explanation:
For this exercise we use that the electric field is a vector, so the resulting field is
E_total = E₁ + E₂ (1)
since the field has the same direction in the space between the planes
Let's use Gauss's law for the electric field of each plate
Let's use a Gaussian surface that is a cylinder with the base parallel to the plate, therefore the normal to the surface and the field lines are parallel and the angle is zero so cos 0 = 1
Ф = ∫ .dA = [tex]q_{int}[/tex] /ε₀
if we assume that the charge is uniformly distributed on the plate we can define a charge density
σ = q_{int} A
as the field exists on both sides of the plate on the inside
E A = A σ / 2ε₀
E = σ / 2ε₀
we substitute in equation 1
E = σ /ε₀
for the complete plate
σ = Q / A = Q / L²
we substitute
E = [tex]\frac{Q}{L^2 \epsilon_o}[/tex]
A solid conducting sphere of radius ra is placed concentrically inside a conducting spherical shell of inner radius rb1 and outer radius rb2. The inner sphere carries a charge Q while the outer sphere does not carry any net charge. The potential for rb1 < r < rb2 is:________
Answer:
The right answer is "[tex]\frac{KQ}{r_b_2}[/tex]".
Explanation:
As the outer spherical shell is conducting, so there is no electric field in side from
⇒ [tex]r_b_1 < r < r_b_2[/tex].
So the electric potential at all points inside the conducting shell that from
⇒ [tex]r_b_1<r<r_b_2[/tex]
and will be similar as well as equivalent to the potential on the outer surface of the shell that will be:
⇒ [tex]v=\frac{KQ}{r_b_2}[/tex]
Thus the above is the right solution.
A 59.0 kg bungee jumper jumps off a bridge and undergoes simple harmonic motion. If the period of oscillation is 0.250 mins, what is the spring constant (in N/m) of the bungee cord, assuming it has negligible mass compared to that of the jumper
Answer:
The spring constant of the spring is 10.3 N/m.
Explanation:
Given that,
Mass of a bungee jumper, m = 59 kg
The period of oscillation, T = 0.25 min = 15 sec
We need to find the spring constant of the bungee cord. We know that the period of oscillation is given by :
[tex]T=2\pi\sqrt{\dfrac{m}{k}}[/tex]
Where
k is the spring constant
[tex]T^2=4\pi^2\times \dfrac{m}{k}\\\\k=4\pi^2\times \dfrac{m}{T^2}\\\\k=4\pi^2\times \dfrac{59}{(15)^2}\\\\k=10.3\ N/m[/tex]
So, the spring constant of the spring is 10.3 N/m.
1. Consider a 1000 kg car rounding a curve on a flat road of radius 50 m at a speed of
50 km/h (14 m/s).
a. Will the car make the turn if the pavement is dry and the coefficient of static
friction is 0.60?
Answer:
The car will make the turn perfectly
Explanation:
Given that the centripetal force= mv^2/r
M= mass of the car
v = speed of the car
r= radius
Hence;
F = 1000 × (14)^2/50
F= 3920 N
The frictional force = μmg
μ = coefficient of static friction
m= mass
g = acceleration due to gravity
Frictional force= 0.6 × 1000× 10
Frictional force = 6000 N
The car will not skid off the curve because the frictional force is greater than the centripetal force.
A compact disk with a 12 cm diameter is rotating at 5.24 rad/s.
a. What is the linear speed _______m/s
b. What is the centripetal acceleration of a point on its outer rim _______
c. Consider a point on the CD that is halfway between its center and its outer rim. Without repeating all of the calculations required for parts (a) and (b), determine the linear speed of this point. _______m/s
d. Determine the centripetal acceleration of this point. _______
Answer:
(a) 31.44 m/s (b) 164.74 m/s²
Explanation:
Given that,
The diameter of a disk, d = 12 cm
Radius, r = 6 cm
Angular speed = 5.24 rad/s
(a) Linear speed,
[tex]v=r\omega\\\\v=6\times 5.24\\\\v=31.44\ m/s[/tex]
(b) Centripetal acceleration,
[tex]a=\dfrac{v^2}{r}\\\\a=\dfrac{31.44^2}{6}\\\\a=164.74\ m/s^2[/tex]
Object A has a mass m and a speed v, object B has a mass m/2 and a speed 4v, and object C has a mass 3m and a speed v/3. Rank the objects according to the magnitude of their momentum.
Required:
Rank from smallest to largest.
Answer:
Momentum of object A = Momentum of object C < momentum of B.
Explanation:
The momentum of an object is equal to the product of mass and velocity.
Object A has a mass m and a speed v. Its momentum is :
p = mv
Object B has a mass m/2 and a speed 4v. Its momentum is :
p = (m/2)×4v = 2mv
Object C has a mass 3m and a speed v/3. Its momentum is :
p = (3m)×(v/3) = mv
So,
Momentum of object A = Momentum of object C < momentum of B.
An electric field E⃗ =5.00×105ı^N/C causes the point charge in the figure to hang at an angle. What is θ?
We have that the angle is
[tex]\theta=32.53[/tex]
From the Question we are told that
E⃗ =5.00×105ı^N/C
Generally the equation for Tension is mathematically given
[tex]W=Tcos\theta[/tex]
Where
[tex]tan\theta=\frac{2.5*10^{-9}(5*10{5})}{2*10^{-3}(9.8)}[/tex]
[tex]\theta=32.53[/tex]
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A 12.0 g sample of gas occupies 19.2 L at STP. what is the of moles and molecular weight of this gas?
At STP, 1 mole of an ideal gas occupies a volume of about 22.4 L. So if n is the number of moles of this gas, then
n / (19.2 L) = (1 mole) / (22.4 L) ==> n = (19.2 L•mole) / (22.4 L) ≈ 0.857 mol
If the sample has a mass of 12.0 g, then its molecular weight is
(12.0 g) / n ≈ 14.0 g/mol
Wind instruments like trumpets and saxophones work on the same principle as the "tube closed on one end" that we examined in our last experiment. What effect would it have on the pitch of a saxophone if you take it from inside your house (at 76 degrees F) to the outside on a cold day when the outside temperature is 45 degrees F ?
Answer:
f = v / 4L
the frequency of the instruments is reduced by the decrease in the speed of the wave with the temperature.
Explanation:
In wind instruments the wave speed must meet
v = λ f
λ = v / f
from v is the speed of sound that depends on the temperature
v = v₀ [tex]\sqrt{1+ \frac{T [C]}{273} }[/tex]
where I saw the speed of sound at 0ºC v₀ = 331 m/s the temperature is in degrees centigrade, we can take the degrees Fahrenheit to centigrade with the relation
(F -32) 5/9 = C
76ºF = 24.4ºC
45ºF = 7.2ºC
With this relationship we can see that the speed of sound is significantly reduced when leaving the house to the outside
at T₁ = 24ºC v₁ = 342.9 m / s
at T₂ = 7ºC v₂ = 339.7 m / s
To satisfy this speed the wavelength of the sound must be reduced, so the resonant frequencies change
λ / 4 = L
λ= 4L
v / f = 4L
f = v / 4L
Therefore, the frequency of the instruments is reduced by the decrease in the speed of the wave with the temperature.
A T-shirt cannon launches a shirt at 5.30 m/s from a platform height of 4.00 m from ground level. How fast (in m/s) will the shirt be traveling if it is caught by someone whose hands are at 5.20 m from ground level (b) 4.00 m from ground level?
Answer:
(a) the velocity of the shirt is 2.14 m/s
(b) the velocity of the shirt is 5.3 m/s
Explanation:
Given;
initial velocity of the shirt, u = 5.3 m/s
height of the platform above the ground, h = 4.00 m
(a) When the shirt is caught by someone whose hand is 5.20 m from the ground level, the height traveled by the shirt = 5.2 m - 4.0 m = 1.2 m
The velocity at this position is calculated as;
[tex]v^2 = u^2 + 2(-g)h\\\\v^2 = u^2 - 2gh\\\\v^2 = 5.3^2 - (2\times 9.8 \times 1.2)\\\\v^2 = 4.57\\\\v= \sqrt{4.57} \\\\v = 2.14 \ m/s[/tex]
(b) When the shirt is caught by someone whose hand is 4.00 m from the ground level, the height traveled by the shirt = 4.00 m - 4.00 m = 0 m
The velocity at this position is calculated as;
[tex]v^2 = u^2 + 2(-g)h\\\\v^2 = u^2 - 2gh\\\\v^2 = 5.3^2 - (2\times 9.8 \times 0)\\\\v^2 = 28.09\\\\v= \sqrt{28.09} \\\\v = 5.3 \ m/s[/tex]
A baseball pitcher throws a fastball by spinning his arm at 27.7m/s. The ball has a mass of 0.700kg and experiences a net centripetal force of 625N. How long is the pitchers arm (the radius of the curve)?
In the historical sense, postmodern society is simply a society that occurs after the modern society. ... Many of the elements of a society like this are reactions to what the modern society stood for: industrialism, rapid urban expansion, and rejection of many past principles.
A ten loop coil of area 0.23 m2 is in a 0.047 T uniform magnetic field oriented so that the maximum flux goes through the coil. The average emf induced in the coil is
Answer:
Explanation:
From the question we are told that:
Number of turns [tex]N=10[/tex]
Area [tex]a=0.23m^2[/tex]
Magnetic field [tex]B=0.947T[/tex]
Generally the equation for maximum flux is mathematically given by
[tex]\phi=NBa[/tex]
[tex]\phi=10*0.047*0.23[/tex]
[tex]\phi=0.1081wbi[/tex]
Therefore induced emf
[tex]e= \frac{d\phi}{dt}[/tex]
Since
[tex]t=0[/tex]
Therefore
[tex]e=0[/tex]
A penny of mass 3.10 g rests on a small 20.0 g block supported by a spinning disk with radius of 12.0 cm. The coefficients of friction between block and disk are 0.850 (static) and 0.575 (kinetic) while those for the penny and block are 0.395 (kinetic) and 0.495 (static). What is the maximum rate of rotation in revolutions per minute that the disk can have, without the block or penny sliding on the disk
Answer:
do this Q yourself because i havent read the chapter
The maximum rate of rotation in revolutions per minute that the disk can have, without the block or penny sliding on the disk is 63 rpm.
How to solveThis is calculated using the coefficient of static friction between the penny and block, which is 0.495.
The maximum angular velocity of the disk is when the force of static friction is just sufficient to prevent the penny from sliding.
This force is equal to the mass of the penny multiplied by the acceleration due to gravity, multiplied by the coefficient of static friction.
The angular velocity of the disk is then calculated from this force and the radius of the disk.
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A force of 15 N toward the WEST is applied to a 4.0 kg box. Another force of 42 N toward the EAST is also applied to the 4.0 kg box. The net force on the 4.0 kg box
is
[tex]\implies F_1 < F_2[/tex]
[tex] \implies F_{net} = F_2 - F1[/tex]
[tex]\implies F_{net} = 42 -15[/tex]
[tex]\implies \underline{ \boxed{ F_{net} = 27 \: N}}[/tex]
The net force on the 4.0 kg box is 27 N towards EAST.
