Answer: lamda max= 16,5m lamda min= 0,0165m (air)
lamda max= 75m, lamda min=0,075m (water)
Explanatin
i really need help with this. cant anyone help me?
(A)
Explanation:
The number of atoms for each element on the left side must equal that on the right side. Only (A) satisfies that condition. The other other choices have mismatching number of atoms on either side.
what makes the morion mask expensive ?
Answer:
a lot of the masks are hand made a long time ago its really an art
Explanation:
Because it is a piece of art that is hard to craft
a force is something that ___ an object to ____
Answer:
which can cause an object with mass to change its volocity
what is the distance between two corresponding points of adjacent waves?
Answer:Wavelength
Explanation: The wavelength of a wave is the distance between any two corresponding points on adjacent waves.
Displacement (m) 0 20 40 40 80 80 60 400 Time (s) 10 10 20 30 40 50 60 70 80 a) Explain the different sections of the graph in as much detail as you can. b) Use the graph to determine the maximum velocity. c) Find the average velocity after 45 s. d) Find the instantaneous velocity at 45 s.
Answer:
because I dont know
Explanation:
first you add the multiply
The instantaneous velocity at 60 m is zero, at - 40 m is - 1 m / s, at 15 s is zero and at 25 s is 0.8 m / s if Displacement (m) 0 20 40 40 80 80 60 400 Time (s) 10 10 20 30 40 50 60 70 80.
What is Instantaneous velocity?Instantaneous velocity = Position with respect to time / Time
1 ) At d = 60 m, the instantaneous velocity is zero because the car is at rest.
2 ) At d = - 40 m, t = 40 s
Instantaneous velocity = - 40 / 40
Instantaneous velocity = - 1 m / s
3 ) At t = 15 s, d = 60 m
At 60 m, the car is at rest, so the Instantaneous velocity is zero
4 ) At t = 25 s, d = 20 m
Instantaneous velocity = 20 / 25
Instantaneous velocity = 0.8 m / s
Instantaneous velocity of a given curve in a position-time graph can be found by drawing a tangent to the curve and finding the slope of the tangent. Instantaneous velocity = 0
Instantaneous velocity = - 1 m / s
Instantaneous velocity = 0
Instantaneous velocity = 0.8 m / s
Therefore, The instantaneous velocity at 60 m is zero, at - 40 m is - 1 m / s, at 15 s is zero and at 25 s is 0.8 m / s if Displacement (m) 0 20 40 40 80 80 60 400 Time (s) 10 10 20 30 40 50 60 70 80.
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A steam pipe is covered with 1.50-cm thick insulating material of thermal conductivity of 0.200 cal/cm · °C · s. How much energy is lost every second when the steam is at 195°C and the surrounding air is at 20.0°C? The pipe has a circumference of 800 cm and a length of 40.0 m. Neglect losses through the ends of the pipe.
This question involves the concepts of Fourier's law of heat conduction and heat loss per unit time.
The heat energy loss per second is "3.12 x 10⁸ W".
Applying Fourier's Law of Heat Conduction here, we get:
[tex]Q = -kA\frac{\Delta T}{\Delta x}[/tex]
where,
Q = heat energy loss per unit time = ?
k = thermal conductivity = 0.2 cal/cm.°C.s [tex](\frac{4.184\ J}{1\ cal})(\frac{100\ cm}{1\ m})[/tex] = 83.68 J/m.°C.s
A = Surface area = (circumference)(length) = (8 m)(40 m) = 320 m²
ΔT = difference in temperature = 20°C - 195°C = -175°C
Δx =thickness of insulation = 1.5 cm = 0.015 m
Therefore,
[tex]Q = -(83.68\ J/m.^oC.s)(320\ m^2)(\frac{-175\ ^oC}{0.015\ m})\\\\[/tex]
Q = 3.12 x 10⁸ W = 312 MW
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The inclined plane in the figure above has two sections of equal length and different roughness. The dashed line shows where section 1 ends and section 2 begins. A block of M is placed at different locations on the incline. The coefficients of kinetic and static friction between the block and each section are shown in the table below.
If the block is at rest on section 1 of the incline, what is the magnitude of the force of static friction exerted on the block by the incline?
The static friction exerted on the block by the incline is [tex]\mu _ s _1 Mgcos \ \theta[/tex].
The given parameters;
mass of the block, = Mcoefficient of static friction in section 1, = [tex]\mu_s_1[/tex]angle of inclination of the plane, = θThe normal force on the block is calculated as follows;
Fₙ = Mgcosθ
The static friction exerted on the block by the incline is calculated as follows;
[tex]F_s = \mu_s F_n\\\\F_s = \mu _s_1(Mg cos\ \theta)\\\\F_s = \mu _s_1 Mgcos\ \theta[/tex]
Thus, the static friction exerted on the block by the incline is [tex]\mu _ s _1 Mgcos \ \theta[/tex]
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Hãy giải thích các khái niệm sau và lấy ví dụ cụ thể để minh họa:
Áp suất chân không tuyệt đối, áp suất tuyệt đối, áp suất khí trời, áp suất dư và áp suất chân
không?
Hãy giải thích các khái niệm sau và lấy ví dụ cụ thể để minh họa:
A car driving at 32.6 miles/hours slowed to a stop in 2.4 seconds
The driver's acceleration was
miles/hour second. Record your answer to two places past the decimal point
Explanation:
using v=u+at
0=32.6 + (a×(2.4/3600))
-32.6=(a×(2.4/3600))
a=-32.6×3600/2.4
a= -48900miles/hr^2
What is the value of the magnitude of the difference of vectors A and B, |A - B|?
Answer:
Let, θ be the angle between vector a and vector b.
So, the angle between vector a and vector (-b) must be (180° - θ).
Magnitude of the difference of vectors a and b
= |a - b|
= Magnitude of the resultant of the vector sum of the vectors a and (-b)
= |{√|a|^2 + |b|^2 + 2 * |a| * |b| * cos (180° - θ)}|
= |{√|a|^2 + |b|^2 - 2 * |a| * |b| * (cos θ)}|.
Explanation:
If a car has a suspension system with a force constant of 5.00x104 N/m, how much energy must the car's shocks remove to dampen an oscillation starting with a maximum displacement of 0.0750 m
Answer: 140.625
Explanation: Because energy gained due to damped motion and needs to be absorbed
d. The soccer player accelerates from rest at a rate of 3.1 m/s2. How far does he
run in 1.5 seconds? (1 point)
Answer:
= 6.975 m
Explanation:
3.1 m = 1 s2
? = 2.25 s2
(3.1 * 2.25)/1
= 6.975m
Please explain what the passage mean in your own words.
Remember the Sabbath day, to keep it holy. Six days shalt thou labor, and do all thy work: But the seventh day is the Sabbath of the Lord thy God: in it thou shalt not do any work, thou, nor thy son, nor thy daughter, thy manservant, nor thy maidservant, nor thy cattle, nor thy stranger that is within thy gates: For in six days the Lord made heaven and earth, the sea, and all that in them is, and rested the seventh day: wherefore the Lord blessed the Sabbath day, and hallowed it.
A race car traveling at 10 meters per second accelerates at 1.5 meters per second squared while moving a distance of 600 meters. Which of the following best represents the final speed of the race car?
Answer:
Explanation:
Givens
vi = 10 m/s
a = 1.5 m/s^2
d = 600 m
vf = ?
Formula
vf^2 = vi^2 + 2*a*d
Solution
vf^2 = 10^2 + 2*1.5 * 600
vf^2 = 100 + 1800
vf^2 = 1900
sqrt(vf^2) = sqrt(1900)
vf = 43.59 m/s
A car equipped with a massless spring in front collides with a stationary car as shown in the diagram. The mass of the incoming car is three times the mass of the target car. While the spring is being compressed, the cars are moving closer together, and while it is expanding they are moving farther apart, but at the instant that the spring is fully compressed, they have no relative motion, which means they are moving at the same speed. At this instant, the cars are moving with a speed of 2.4 m/s.
(a) Find the speed of the incoming car before the collision.
(b) Find the fraction of the system’s energy that is stored in the spring when
it is fully compressed.
(c) Find the final speed and direfction for both cars after the spring expands and the cars separate.
Answer:
Explanation:
Let m be target car mass
a)
conservation of momentum
(3m)u + m(0) = (3m + m)(2.4)
3u = 4(2.4)
u = 3.2 m/s
b)
system energy is originally in the striking car as kinetic energy
KE = ½(3m)3.2² = 15.36m J
at full compression, kinetic energy is
½(3m + m)2.4² = 11.52m J
so spring potential energy fraction of total energy is
(15.36 - 11.52)/15.36 = 0.25 or 25%
c)
The center of mass (CoM) is moving at 2.4 m/s
The CoM sees the larger mass car approach at
3.2 - 2.4 = 0.8 m/s and will see it depart at - 0.8 m/s
so a ground based observer would see it moving at
v₁ = 2.4 + -0.8 = 1.6 m/s in its original direction
The CoM sees the smaller mass car approach at
0 - 2.4 = -2.4 m/s and will see it depart at + 2.4 m/s
so a ground based observer would see it moving at
v₂ = 2.4 + 2.4 = 4.8 m/s in its original direction of the larger car
so the relative velocity of approach at 3.2 - 0 = 3.2 m/s
equals the relative velocity of departure at 4.8 - 1.6 = 3.2 m/s
checks.
Imagine you and a friend are trying to rearrange the furniture in your classroom. You push on a desk with a force of 50 N to the right. Your friend pushes on the same desk with a force of 50 N to the left. What is the net force on the desk?
Answer:
The net force on the desk is zero.
Why did the physicist conclude that cats can be either liquid or solid?
Answer:
Explanation:
https://www.science.org/content/article/cats-behave-liquids-tampons-play-music-and-other-advances-honored-ig-nobel-prizes
Answer:
Liquid or Both
Explanation:
Cats exist in their own three states of matter: loaf, cinnamon roll, and liquid. The last category is certainly the most confusing because cats should be solids — and of course, they can be.
A girl sitting on a beach counts six waves passing a buoy in 3.0 s. She measured the distance between the wave crests to be 1.5 meters. What is the speed of the waves the girl observed?
[tex]\\ \sf\longmapsto Frequency=\dfrac{1}{Time}=\dfrac{1}{3}=0.33s^{-1}[/tex]
Velocity be v
[tex]\\ \sf\longmapsto v=f\lambda[/tex]
[tex]\\ \sf\longmapsto V=0.33(1.5)[/tex]
[tex]\\ \sf\longmapsto V\approx 0.5ms^{-1}[/tex]
Given,
Wavelength = 1.5 metres
T = 3 seconds OR 3.0 seconds
We know that the formula of Frequency–
1/Time
According to the question we have seen that Time is 3 seconds
[tex] \sf \longmapsto \: Frequency = \frac{1}{3 .0 \: seconds} [/tex]
[tex] \sf \longmapsto \: Frequency = 0.33 seconds^{-1} [/tex]
We Know,
Velocity = Frequency
Solving Further[tex]\sf \longmapsto Velocity =0.33 \times 1.5 [/tex]
[tex]\sf \longmapsto \: 0.5 \: m/s ^{-1}[/tex]
Therefore the speed is[tex]\sf \: 0.5 \: m/s ^ {-1}[/tex]A worker pushes 1500N crate with a horizontal force of 345N a distance of 24m.
How much work is done by the Floor on a crate?
Answer:
8280J
Explanation:
W = f x d
345 x 24 = 8280J
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Answer:
.
Explanation:
Could a car drive on a frictionless surface? Explain using the terms action
force and reaction force. *
Answer:
No, it cannot. The car needs the friction of the surface to drive because the car pushes the surface backwards, and the surfaces makes a reaction force pushing the car forward, and that works because of the friction. In a frictionless surface the tires would rotate in the same place
Please help on question e and F
E- what will be the total charge of the compound formed?
F what type of band will form?
Answer:
E-MgO(Magnesium Oxide)
Draw an energy pie chart above each of the cart’s four positions.
i. Choose your reference point for determining height.
ii. Create a pie chart that accurately represents the ratio of the three forms of energy and label the sections of the pie Eg for gravitational potential energy, Ek for kinetic energy, and Eth for thermal energy.
Answer:
Attached below! Ignore the name please. Thanks!
Explanation:
100 POINTS PLEASE answer ASAP
Answer:
uhmmm
Explanation:
uhmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmm
Including reaction time, the stopping distance is more than 20 feet at 10 miles per hour, at 20 miles per hour it will be about:
30 feet
63 feet
40 feet
75 feet
Answer:
A. 30 feet
Explanation:
Including reaction time, the stopping distance is more than 20 feet at 10 miles per hour, at 20 miles per hour it will be about 63 feet. Option B is correct.
What is speed?Speed is defined as the rate of change of the distance or the height attained. it is a time-based quantity.
Given conditions;
10 miles per hour ⇒ 20 feet
1 mile per hour ⇒ 2 feet
20 miles per hour ⇒ 20 × 2 feet
20 miles per hour ⇒ 20 × 2 feet
20 miles per hour ⇒40 feet
If the stopping distance is more than 20 feet at 10 miles per hour at 20 miles per hour it must be more than 40 feet.So that the correct answer is 63 feet.
Hence option B is correct.
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How are the parts of birds and fish similar in structure and function?
Answer:
Birds and fish both have a backbone structure, which makes them structurally and functionally similar.
Explanation:
A remote-control car has batteries that store 1.8 kJ of energy. When the car is
driven across a sports field there is friction dragging on the car, producing a force
of 2 N.
Complete the following sentence: If the car is 100% efficient, it could travel a
distance of metres on a single battery charge.
Answer:
Explanation:
Complete the following sentence: If the car is 100% efficient, it could travel a distance of 900 meters on a single battery charge.
W = Fd
1800 = 2d
d = 900
Assuming 100 % efficient means turning battery power into motion.
A car totally 100% efficient would have zero friction and could travel an infinite distance on a full battery charge.
name a suitable thermometer for a winter temperature at noth pole?
following the new year and it will not have a great day and service offering the best way to our attention that you are interested please send the following URL Adobe Flash player to by the following link unsubscribe the new year and the new Sprint network from the new one for the new one and the following URL to the new year is available for remote playback is most convenient time ago the new one is available for the new one is available using to our u to the followings to our to our attention the information contained IN the product and the product and it will take a look the following URL Adobe Flash memory of a great day and it was great to our to by the the following link to the new year and the following link unsubscribe if possible I would like a few minutes late and I have been working in lowest terms if I can be in touch and it was great meeting with the new year is available for remote playback not allowed because the new year is available using this email and the following URL Adobe
Helium gas is confined within a chamber that has a moveable piston. The mass of the piston is 8.7 kg; and its radius is 0.013 m. If the system is in equilibrium, what is the pressure exerted on the piston by the gas
Helium gas is in a chamber with a moveable piston, whose mass is 8.7 kg and its radius is 0.013 m. The pressure exerted on the piston by the gas is 1.6 × 10⁵ Pa.
Helium is confined in a chamber with a moveable 8.7-kg piston (m). We can calculate the weight of the piston (w) using Newton's second law of motion.
[tex]w = m \times g = 8.7kg \times 9.8 m/s^{2} = 85 N[/tex]
The radius of the piston (r) is 0.013 m. We can calculate the area of the piston (A) using the following expression.
[tex]A = \pi \times r^{2} = \pi \times (0.013)^{2} = 5.3 \times 10^{-4} m^{2}[/tex]
We can calculate the pressure exerted by the piston (P) using the following expression.
[tex]P = \frac{w}{A} = \frac{85N}{5.3 \times 10^{-4}m^{2} } = 1.6 \times 10^{5} Pa[/tex]
Since the system is in equilibrium, the pressure exerted on the piston by the gas is also 1.6 × 10⁵ Pa.
Helium gas is in a chamber with a moveable piston, whose mass is 8.7 kg and its radius is 0.013 m. The pressure exerted on the piston by the gas is 1.6 × 10⁵ Pa.
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Multiple-Concept Example 10 provides one model for solving this type of problem. Two wheels have the same mass and radius of 4.0 kg and 0.35 m, respectively. One has the shape of a hoop and the other the shape of a solid disk. The wheels start from rest and have a constant angular acceleration with respect to a rotational axis that is perpendicular to the plane of the wheel at its center. Each turns through an angle of 13 rad in 8.0 s. Find the net external torque that acts on each wheel.
Newton's second law for rotational motion allows finding the net torque for each body is:
Ring torque is: τ = 0.199 Nm Solid disc the torque is: τ = 0.995 N m
Newton's second law for rotational motion establishes a relationship between the torque, the moment of inertia, and the angular acceleration of the body.
τ = I α
Where τ is the torque, I the moment of inertia and α the angular acceleration.
They indicate that the rotated angle is θ = 13 rad in a time of 8.0 s, let's use the rotational kinematics relations.
θ = w₀ t + ½ α t²
The body starts from rest, therefore its inertial velocity is zero.
θ = ½ α t²
[tex]\alpha =\frac{2 \theta }{t^2}[/tex]
Let's calculate
α = [tex]\frac{2 \ 13^2}{8.0^2 }[/tex]
α = 0.406 rad / s²
The moments of inertia of symmetrical bodies are tabulated:
Ring I = m R² Solid disc I = ½ m R²
Let's look for every torque.
Ring
τ = m R² α
τ = 4.0 0.35² 0.406
τ = 0.199 N m
τ = ½ m R² α
τ = ½ 4.0 0.35² 0.406
τ = 0.0995 N m
In conclusion using Newton's second law for rotational motion we can find the net torque for each body is:
Ring torque is: τ = 0.199 Nm Solid disc the torque is: τ = 0.995 N m
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