What is the wavelength of this wave
[tex]\mathfrak{\huge{\orange{\underline{\underline{AnSwEr:-}}}}}[/tex]
Actually Welcome to the Concept of the Waves.
Wavelength is the distance between two consecutive crest or trough.
hence, here the distance is 10cm
So the wavelength is 10cm
===> 10 cm
He-Ne Laser device emits photons of wave length 632.8 nm by rate 4.5 x 1020 photon/s, so
the power of the laser beam =
a. 3.14 W
b. 141.3 W
c. 314.1 w
d. 431.4 W
Answer: Option b.
Explanation:
We know:
Wavelength = 632.8 nm
Fluence = 4.5*10^20 photon/s
The energy of a single photon of wavelength λ is:
E = (h*c)/λ
where:
h = 6.6*10^(-34) J*s
c = 3*10^8 m/s
And we should rewrite the wave length in meters, so:
λ = 632.8 nm = 632.8*10^(-9) m
replacing these in the energy equation, we get:
E = (6.6*10^(-34)J*s)*(3*10^8 m/s)/(632.8*10^(-9) m) = 3.13*10^(-19) J
So each one of the 4.5x10^20 photon that flow each second have this energy, then the power is:
P = (3.13*10^(-19) J)*(4.5*10^20 /s) = 140.85 J/s
and 1 W = 140.85 J/S
Then the power is:
P = 140.85 W
Then the correct answer is the option b, where the units are a little bit different than mine because I used really simplified values for c and h.
Two teams of nine members each engage in a tug of war. Each of the first team's members has an average mass of 64 kg and exerts an average force of 1350 N horizontally. Each of the second team's members has an average mass of 69 kg and exerts an average force of 1367 N horizontally. (a) What is the acceleration (in m/s2 in the direction the heavy team is pulling) of the two teams
Answer:
[tex]a=0.13m/s^2[/tex]
Explanation:
From the question we are told that
Mass of first team man [tex]m_1=64kg[/tex]
Force of man first team man [tex]F_1=1350[/tex]
Mass of second team man [tex]m_2=69kg[/tex]
Force of man second team man [tex]F_2=1367N[/tex]
Generally the equation for net force F_n is mathematically given by
[tex]F_n=9(m_1+m_2)a[/tex]
[tex]9(m_1+m_2)a=9(f_2-f_1)[/tex]
[tex]9(64+69)a=9(1367-1350)[/tex]
[tex]a=\frac{9(1367-1350)}{9(64+69)}[/tex]
[tex]a=0.127819m/s^2[/tex]
Therefore the acceleration is given by
[tex]a=0.13m/s^2[/tex]
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Suppose that a uniform rope of length L resting on a frictionless horizontal surface, is accelerated along the direction of its length by means of a force F, pulling it at one end. A mass M is accelerated by the rope. Assuming the mass of the rope to be m and the acceleration is a. Stated in terms of the product ma, what is the tension in the rope at the position 0.3 L from the end where the force F is applied if the mass M is 1.5 times the mass of the rope m?
Answer:
2.2 ma
Explanation:
Given :
Length of the rope = L
Mass of the rope = m
Mass of the object pulled by the rope = M
M = 1.5 m
So, L [tex]$\rightarrow$[/tex] m
For unit length [tex]$\rightarrow \frac{m}{L}$[/tex]
∴ 0.3 L = [tex]$0.3 \ L \left(\frac{m}{L}\right)$[/tex]
= 0.3 m
And for remaining 0.7 L = [tex]$0.7 \ L \left(\frac{m}{L}\right)$[/tex]
= 0.7 m
By Newtons law of motion,
F - T = ( 0.3 m) a .........(1)
T = ( M + 0.7 m) a
T = ( 1.5 m + 0.7 m) a
T = ( 2.2 m ) a ..............(2)
So from equation (1) and (2), we have
Tension on the rope
T = 2.2 ma
The spacecraft was moved closer to the launcher by the same amount Wednesday as it was on Tuesday, and yet its speed went up much more. Claim 3 suggests that the magnetic force was much stronger on Wednesday than on Tuesday.
Consider the two subclaims for Claim 3 and answer the question below.
Claim 3.A: The magnetic force was much stronger on Wednesday because the magnet was stronger.
Claim 3.B: The magnetic force was much stronger on Wednesday because the magnetic force is stronger closer to the magnets.
Which claim do you think is more convincing, and why?
Answer:
The magnetic force was much stronger on Wednesday because the magnetic force is stronger closer to the magnets.
Explanation:
The magnetic force was much stronger on Wednesday than on Tuesday because the magnetic force is stronger closer to the magnets. Therefore claim 3B is more convincing.
What is magnetic force?Magnetic force can be described as a consequence of electromagnetic force which is caused due to the motion of charges. A moving charge surrounds itself with a magnetic field and the force that arises due to interacting magnetic fields.
The magnetic force between two moving charges is the effect exerted upon either charge by a magnetic field generated by the other. The magnetic force depends on the charge, the motion of each of the objects, and the separation between them.
The magnitude of the force is determined by the cross product of velocity and the magnetic field is equal to q.[v × B]. The resultant force can be described as perpendicular to the direction of the velocity and the magnetic field.
Therefore, the magnetic force was much stronger closer to the magnets.
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How did Einstein’s and Newton’s theories differ in terms of explaining the cause of gravity?
thank you
Answer:
Newton's theory identified mass as the factor that causes gravity. On the other hand, Einstein's theory identified the curvature of space-time as the factor that causes gravity.
Answer:
Hey mate...
Explanation:
This is ur answer....
In the 17th century Newton concluded that objects fall because they are pulled by Earth's gravity. Einstein's interpretation was that these objects do not fall. According to Einstein, these objects and Earth just freely move in a curved spacetime and this curvature is induced by mass and energy of these objects.
Hope it helps you,
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a car moved 120km to the north. what is its displacement?
Consider a large truck carrying a heavy load, such as steel beams. A significant hazard for the driver is that the load may slide forward, crushing the cab, if the truck stops suddenly in an accident or even in braking. Assume, for example, that a 10 000-kg load sits on the flatbed of a 20 000-kg truck moving at 12.0 m/s. Assume that the load is not tied down to the truck, but has a coefficient of friction of 0.500 with the flatbed of the truck.
A) Calculate the minimum stopping distance for which the load will not slide forward relative to the truck.
B) Is any piece of data unnecessary for the solution?
a) mass of the load.
b) mass of the truck.
c) velocity.
d) coefficient of static friction.
e) all are necessary.
Answer:
A)
the minimum stopping distance for which the load will not slide forward relative to the truck is 14 m
B)
data that were not necessary to the solution are;
a) mass of truck and b) mass of load
Explanation:
Given that;
mass of load [tex]m_{LS}[/tex] = 10000 kg
mass of flat bed [tex]m_{FB}[/tex] = 20000 kg
initial speed of truck [tex]v_{0}[/tex] = 12 m/s
coefficient of friction between the load sits and flat bed μs = 0.5
A) the minimum stopping distance for which the load will not slide forward relative to the truck.
Now, using the expression
Fs,max = μs [tex]F_{N}[/tex] -------------let this be equation 1
where [tex]F_{N}[/tex] = normal force = mg
so
Fs,max = μs mg
ma[tex]_{max}[/tex] = μs mg
divide through by mass
a[tex]_{max}[/tex] = μs g ---------- let this be equation 2
in equation 2, we substitute in our values
a[tex]_{max}[/tex] = 0.5 × 9.8 m/s²
a[tex]_{max}[/tex] = 4.9 m/s²
now, from the third equation of motion
v² = u² + 2as
[tex]v_{f}[/tex]² = [tex]v_{0}[/tex]² + 2aΔx
where [tex]v_{f}[/tex] is final velocity ( 0 m/s )
a is acceleration( - 4.9 m/s² )
so we substitute
(0)² = (12 m/s)² + 2(- 4.9 m/s² )Δx
0 = 144 m²/s² - 9.8 m/s²Δx
9.8 m/s²Δx = 144 m²/s²
Δx = 144 m²/s² / 9.8 m/s²
Δx = 14 m
Therefore, the minimum stopping distance for which the load will not slide forward relative to the truck is 14 m
B) data that were not necessary to the solution are;
a) mass of truck and b) mass of load
While sitting in a boat, a fisherman observes that 2 complete waves pass by his position every 4 seconds. What is the period of these waves?A)2s B)8s C) 0.5 s D) 4s
Answer:
2
Explanation:
Because 2 waves 4 secs means 1 in 2s
The period of these waves is 2s. and The right option is A)2s.
The period of a wave is the time taken by a wave to complete one cycle.
The formula of period from the question is given below.
⇒ Formula:
T = t/2.................... Equation 1⇒ Where:
T = Period of the wavet = time taken for two complete oscillationFrom the question,
⇒ Given:
t = 4 seconds.⇒ Substitute these value above into equation 1
T = 4/2T = 2 seconds.Hence, The period of these waves is 2s.
The right option is A)2s
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If the hoop has speed vo at the bottom of
the pipe, what is its speed when it has rolled
halfway up the side of the pipe?
The speed of the hoop when it has rolled halfway up the side of the pipe is √(v₀² - gR).
Conservation of energyThe speed of the hoop when it has rolled halfway up the side of the pipe is calculated as follows;
K.E = P.E
- ¹/₂mv₀² + ¹/₂Iω² = (mgh₀ - mghf)
- ¹/₂mv₀² + ¹/₂Iω² = (0 - 0.5mgh) (hf = 0.5h) (half way up)
¹/₂Iω² = ¹/₂mv₀² - 0.5mgh
where;
I is moment of inertia of the hoop = mr²ω is angular speed = v/r¹/₂(mr²)(vf/r)² = ¹/₂mv₀² - 0.5mgh
¹/₂vf² = ¹/₂v₀² - ¹/₂gh
vf² = v₀² - gh
vf = √(v₀² - gh)
where;
h is the distance traveled half-way up the pipe = Rvf = √(v₀² - gR)
[tex]v_f = \sqrt{v_0^2 - gR}[/tex]
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A vertical piston cylinder assembly contains 10.0kg of a saturated liquid-vapor water mixture with initial quality of 0.85 The water receives energy by heat transfer until the temperature reaches 320*C. The piston has a mass of 204kg and area of 0.005m2. Atmospheric pressure of 100kPa acts on the top side of the piston. Local gravitational acceleration is 9.81m/s2 Calculate the amount of heat transfer between the water and the surroundings in kJ. Enter a numeric value only. 6735.66
Answer:
Explanation:
From the given information:
At state 1:
Initial Quality [tex]= x_1 = 0.85[/tex]
mass = 10.0 kg
At state 2:
Temperature [tex]T_2 = 320^0[/tex]
mass of the piston [tex]m_p = 204 \ kg[/tex]
area of the piston [tex]A_p = 0.00 5 \ m^2[/tex]
Atmospheric pressure [tex]P_{atm}= 100 \ kPa = 100 \times 10^3 \ Pa[/tex]
Gravitational acceleration = 9.81 m/s²
[tex]\mathbf{P= P_1=P_2}[/tex], This is because there exists no restriction to the movement of the piston and provided the process is frictionless. So, the process 1-2 is regarded as constant.
To calculate the applying force balance over the piston by using force balance in the vertical direction:
[tex]\mathbf{P_{AP} = P_{atmA_p} + m_pg}[/tex]
∴
(100 × 10³)×0.005 + 204 × 9.31 = P × 0.05
P = 500248 Pa
P = 500.25 kPa
At state 1:
[tex]\mathbf{P_1 = P = 500.25 \ kPa}[/tex]
[tex]x_1 = 0.85[/tex]
Hence, this is a saturated mixture of liquid and vapor
Using the steam tables at 500.25 kPa
[tex]V_f = 1.093 \times 10^{-3} \ m^3/kg \\ \\ V_g = 0.375 \ m^3/kg \\ \\ U_f = 639.72 \ kJ/kg \\ \\ U_g = 2560.72 \ kJ/kg[/tex]
∴
Specific volume at state 1 is given as:
[tex]V_1 = [ V_f +x_1(v_g -v_f) ] \ at \ 500.25 \ kPa \\ \\ V_1 = 0.319 \ m^3/kg[/tex]
volume at state 1 is given by:
[tex]V_1 = mV_1 = 10 \times 0.319 \\ \\ V_1 = 3.19 \ m^3[/tex]
Similarly, the specific internal energy is:
[tex]U_1 = [U_f +x_1 (U_o-Uf)] \ at \ 500.25 \ kPa[/tex]
[tex]U_1 = 639.72 +0.82 (2560.72 -639.72)[/tex]
[tex]U_1 = 2272.57 \ kJ/kg[/tex]
At state 2:
[tex]P = P_1 = P_2 = 500.25 \ kPa \\ \\ T_2 = 320^0 \ C[/tex]
Using steam tables at P = 500.25 kPa and T = 320° C
[tex]V_2 = 0.541 \ m^3/kg \\ \\ U_2 = 2835.08 \ kJ/kg[/tex]
∴
[tex]V_2 = mV-2 = 10 \times V_2 = 5.41 \ m^3[/tex]
[tex]\text{Now; Applying the 1st law of thermodynamics to the system}[/tex]
[tex]_1Q_2 -_1W_2 = \Delta V =m(u_2-u_1) \\ \\ where;\ _1W_2 = P(V_2-V_1) \\ \\ _1Q_2 -P(V_2-V_1) = m(u_2-u_1) \\ \\ _1Q_2 - 500.25(5.91 -3.19) = 10( 2835.08 -2272.57) \\ \\ \mathbf{ _1Q_2 = 6735.66 \ kJ}[/tex]
Which of the following is an example of an object with kinetic energy?
a. a plane lifting off of the runway
b. a bobsled perched at the top of a run
c. a snowball tumbling down a hill
d. both a and c
Answer:
A and C
Explanation:
Both have mass and are in motion
HELP ME HOW ARE BLACK HOLES FROMED
Answer:
Stellar black holes form when the center of a very massive star collapses in upon itself.
Answer:
A black hole can be formed by the death of a massive star. When such a star has exhausted the internal thermonuclear fuels in its core at the end of its life, the core becomes unstable and gravitationally collapses inward upon itself, and the star's outer layers are blown away.
Explanation:
What is the earliest time at which the oscillator shown below is stationary?
Answer:
0.0s
Explanation:
I got it right in acellus
Answer: 0.0
Explanation:
when is an object considered to be in motion
Answer:
An object is considered to be in motion only when its position changes over time with reference to a point which will be known as orgin
Explain why the same side of the moon is always facing Earth.
Answer:
The moon keeps the same face pointing towards the Earth because its rate of spin is tidally locked so that it is synchronized with its rate of revolution (the time needed to complete one orbit). In other words, the moon rotates exactly once every time it circles the Earth.
When you are driving on the freeway and following the car in front of you, how close is too close? Let's do an estimation.
1. Pick a car model (preferably the one you drive, but can also be any car of your dream), and find its stopping distance at highway speeds (you can usually find this type of data online).
2. Assuming that the car in front of you suddenly does a hard brake. For simplicity, assume that its braking performance is about the same as yours. Then also assume a reasonable amount of reaction time on your part (the time delay between seeing the brake lights lit up and applying your own brake). In order for you not to run into the car your are following, what's the closest distance you need to keep between the two cars?
3. Redo the same calculation if the vehicle in front of you is a typical big-rig truck. Find its braking data online.
4. There is a rule of thumb which says that you must stay one car length behind the car in front of you for every 10 mi/h of driving speed. From your calculation, does this rule make sense?
Answer:
1) v= 90km/h d = 70 m, 2) x₁ = v t_r, x₁ = 6.25 m, 3) x₁=6.25 no change
4) x = 22 m
Explanation:
1) for the first part, you are asked to find the minimum safety distance with the vehicle in front
The internet is searched for the stopping distance for two typical speeds on the highway
v (km/ h) v (m/s) d (m)
90 25 70
100 27.78 84
the safe distance is this distance plus the distance traveled during the person's reaction time, which can be calculated with infirm movement
v = x / t_r
x₁ = v t_r
the average reaction time is t_r = 0.25s for a visual stimulus and t_r 0.17 for an auditory stimulus
therefore the safe distance is
x_total = x₁ + d
2) The distance is the sum of the distance traveled in the reaction
x₁ = v t_r
for v = 90 km / h
x₁ = 25 0.25
x₁ = 6.25 m
for v = 100 km / h
x₁ = 27.78 0.25
x₁ = 6.95 m
the total distance is
x_total = x₁ + d
for v = 90 km / h
x_total = 25 0.25 + 70
x_total = 76.25 m
this is the distance until the cars stop and do not collide
3) the stopping distance of a truck is
v = 90 km / h d = 100 m
in this case we see that the braking distance is much higher,
the safe distance is given by the distance traveled during the reaction, as the truck brakes slower than the car this distance does not change
4) let's analyze the empirical rule: maintain the length of a car for each increase in speed of v = 10 m / h = 4.47 m / s
for the car case at v = 90km / h = 25 m / s
according to this rule we must this to
x = 25 / 4.47 = 5.6 cars
each modern car is about 4 m long so the distance is
x = 22 m
we see that this distance is much greater than the reaction distance so it does not make much sense
examples of buildup of static energy?
Answer:
Materials that can lose or gain electrons in this way are called triboelectric, according to Northwestern University. One common example of this would be shuffling your feet across carpet, particularly in low humidity which makes the air less conductive and increases the effect.
Explanation:
Hope that helps :)
HELP - What are some good topics to add to a Pipeline presentation? So yes, a presentation about pipelines, like for example in the US. Have any ideas??
An amusement park ride called the Rotor debuted in 1955 in Germany. Passengers stand in the cylindrical drum of the Rotor as it rotates around its axis. Once the Rotor reaches its operating speed, the floor drops but the riders remain pinned against the wall of the cylinder. Suppose the cylinder makes 26.0 rev/min and has a radius of 3.70 m. 1) What is the coefficient of static friction between the wall of the cylinder and the backs of the riders
Answer:
μs = 0.36
Explanation:
While the drum is rotating, the riders, in order to keep in a circular movement, are accelerated towards the center of the drum.This acceleration is produced by the centripetal force.Now, this force is not a different type of force, is the net force acting on the riders in this direction.Since the riders have their backs against the wall, and the normal force between the riders and the wall is perpendicular to the wall and aiming out of it, it is easily seen that this normal force is the same centripetal force.In the vertical direction, we have two forces acting on the riders: the force of gravity (which we call weight) downward, and the friction force, that will oppose to the relative movement between the riders and the wall, going upward.When this force be equal to the weight, it will have the maximum possible value, which can be written as follows:[tex]F_{frmax} = \mu_{s}* F_{n} = m * g (1)[/tex]
where μs= coefficient of static friction (our unknown)As we have already said Fn = Fc.The value of the centripetal force, is related with the angular velocity ω and the radius of the drum r, as follows:[tex]F_{n} = m* \omega^{2} * r (2)[/tex]
Replacing (2) in (1), simplifying and rearranging terms, we can solve for μs, as follows:[tex]\mu_{s} = \frac{g}{\omega^{2} r} (3)[/tex]
Prior to replace ω for its value, is convenient to convert it from rev/min to rad/sec, as follows:[tex]\omega = 26.0 \frac{rev}{min} * \frac{1min}{60 sec} *\frac{2*\pi rad }{1 rev} = 2.72 rad/sec (4)[/tex]
Replacing g, ω and r in (3):[tex]\mu_{s} = \frac{g}{\omega^{2} r} = \frac{9.8m/s2}{(2.72rad/sec)^{2} *3.7 m} = 0.36 (5)[/tex]The part of the eye that gives its shape is the:
a race car goes around a circular track of radius 150 m at speed of 10.0 m/s. How long does it take to complete one lap?
Answer:
94.25 seconds
Explanation:
Solve for period (T) using: v=(2*pi*r)/T
rearrange: vT=2*pi*r
rearrange: T=(2*pi*r)/v
Plug in values.
T=(2*pi*150)/10
T=94.25 seconds
If a race car goes around a circular track of a radius of 150 m at speed of 10.0 m/s ,then the time taken to complete the one lap would be 94.25 seconds.
What is speed?The total distance covered by any object per unit of time is known as speed. It depends only on the magnitude of the moving object. The unit of speed is a meter/second. The generally considered unit for speed is a meter per second.
As given in the problem a race car goes around a circular track of radius 150 m at speed of 10.0 m/s.
vT = 2 × π × r
T = (2 × π × r)/ v
T = (2 × π× 150)/10
T = 94.25 seconds
Thus, the time taken to complete the one lap would be 94.25 seconds.
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A box with mass 1.10 kg is located on a horizontal tabletop with friction. The coefficient of kinetic friction is 0.500. The tabletop is square and measures 1.00 m on its side. The box starts at one corner and finishes at the diagonal edge. The path it follows is by first traveling one edge, turning and traveling to the final location. You push the box by exerting a force on it that makes an angle of 30.0o with the horizontal. How much work does the friction force do
Answer:
W = 6.5 W
Explanation:
Work is defined by
W = F . d
W = f d cos tea
where the point represents the scaled product and the bold letters indicate vectors
they ask us the work of the friction force
we write the translational equilibrium equation
y Axis
N -W = 0
N = mg
x axis
F - fr = 0
F = fr
the formula for the friction force is
fr = μ N
we substitute
fr = μ m g
we substitute in the equation of work
W = fr d cos θ
W = μ m g d cos θ
let's calculate
W = 0.500 1.10 9.8 Σ d_i cos θ_i
W = 5.39 d cos tea
we have two displacement
the first on one side of the box, suppose that side is on the y-axis, therefore the angle between the displacement and the friction force is 70º
and there is a second displacement in the x axis, in this case the angle between the friction force and the displacement is 30º
therefore the total workload is the sum of those work
W = 5.39 (1 cos 70 + 1 cos 30)
W = 5.39 (0.342 + 0.866)
W = 6.5 W
If you increase the frequency of a sound wave four times, what will happen to its speed?
А
The speed will increase four times.
B.
The speed will decrease four times,
C. The speed will remain the same.
D.
The speed will increase twice.
E.
The speed will decrease twice.
You start biking at the top of a steep hill. As you travel downhill, you apply
your brakes to control your speed. What are the energy transformations
taking place in this system?
A. Kinetic energy to mechanical energy to chemical energy
B. Potential energy to kinetic energy to heat energy
C. Thermal energy to mechanical energy to potential energy
D. Electric energy to kinetic energy to chemical energy
If your water heat has an efficiency of 95 percent, how much energy would it take to heat 45kg of water from 23 C to 60 C. (The specific heat of water is 4.18 J/g/C.) Please show your work
Answer: 6611.715 joules
Explanation:
Q = MxCxdeltaT = 6959.7 which is 100%
95% = 6611.715
In a thunderstorm at 32.0°C, Reginald sees a bolt of lightning and hears the thunderclap 2.00s later. How far from Reginald did the lightning strike? The speed of sound through air at 32.0°C is 350.2 m/s. Show your work.
PLEASE HELP, THANKS!
Answer:
d = 700.4 m
Explanation:
Given that,
The speed of sound through air at 32.0°C is 350.2 m/s.
Reginald sees a bolt of lightning and hears the thunderclap 2.00s later.
We need to find how far from Reginald did the lightning strike. Let the distance be d. So,
Speed = distance/time
d = vt
So,
d = 350.2× 2
d = 700.4 m
So, the required distance is 700.4 m.
Help please :)
Infrared waves are transmitted by ....... ?
Netwtons 1st law of motion
Answer:
Newton's first law states that, if a body is at rest or moving at a constant speed in a straight line, it will remain at rest or keep moving in a straight line at constant speed unless it is acted upon by a force. This postulate is known as the law of inertia.
Explanation: