Answer:
The wavelength is approximately 611 nm
Explanation:
We can use the formula for the condition of maximum of interference given by:
[tex]d\,sin(\theta)=m\,\lambda\\(0.000250\,\,m)\,\,sin(1.12^o)=8\,\lambda\\\lambda=\frac{1}{8} \,(0.000250\,\,m)\,\,sin(1.12^o)\\\lambda \approx 610.8\,\,nm[/tex]
We can also use the formula for the distance from the central maximum to the 5th minimum by first finding the tangent of the angle to that fifth minimum:
[tex]tan(\theta)=\frac{y}{D}\\ tan(\theta)=\frac{0.0333}{3.02} =0.011026[/tex]
and now using it in the general formula for minimum:
[tex]d\,sin(\theta)\approx d\,tan(\theta)=(m-\frac{1}{2} )\,\lambda\\\lambda\approx 0.00025\,(0.011026)/(4.5)\,\,m\\\lambda\approx 612.55\,\,nm[/tex]
Answer:
The correct answer is [tex]6.1\times10^{-7}\:m[/tex]
Explanation:
The distance from the central maxima to 5th minimum is:
[tex]x_{5n}-x_{0} =3.33\:cm=0.033\:m[/tex]
The distance between the slits and the screen:
[tex]L = 302\:cm = 3.02\:m[/tex]
Distance between 2 slits: [tex]d = 0.00025\:m[/tex]
[tex](n-\frac{1}{2})\lambda=\frac{d(x_n)}{L}[/tex]
For fifth minima, n = 5... so we have:
[tex]x_{5n}=\frac{9\lambda L}{2d}[/tex]
For central maxima, n = 0... so we have:
[tex]x_{0}=\frac{n\lambda L}{d}=0[/tex]
So the distance from central maxima to 5th minimum is:
[tex]\frac{9\lambda \:L}{2d}-0=0.033[/tex] (Putting the values, we get):
[tex]\Rightarrow \lambda = 6.1\times 10^{-7}\:m[/tex]
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Two teams are playing tug-of-war. Team A, on the left, is pulling on the rope with an effort of 5000 N. If the rope is moving at a constant velocity, how hard and in which direction is team B pulling?
A. 2500 N to the left
B. 5000 N to the right
C. 2500 N to the right
D. 5000 N to the left
Explanation:
If Team A is on the left, B is on the right
if the force is constant, it means that the effort applied is equal.
So Team B is pulling 5000N to the right.
A student throws a 120 g snowball at 7.5 m/s at the side of the schoolhouse, where it hits and sticks. What is the magnitude of the average force on the wall if the duration of the collision is 0.15 s
Answer:
The magnitude of the average force on the wall during the collision is 6 N.
Explanation:
Given;
mass of snowball, m = 120 g = 0.12 kg
velocity of the snowball, v = 7.5 m/s
duration of the collision between the snowball and the wall, t = 0.15 s
Magnitude of the average force can be calculated by applying Newton's second law of motion;
F = ma
where;
a is acceleration = v / t
a = 7.5 / 0.15
a = 50 m/s²
F = ma
F = 0.12 x 50
F = 6 N
Therefore, the magnitude of the average force on the wall during the collision is 6 N.
As a motorcycle takes a sharp turn, the type of motion that occurs is called _______________ motion.
Answer:
circular motion
Explanation:
As a motorcycle takes a sharp turn, the type of motion that occurs is called circular motion.
Circular motion is a movement of an object along a circular path. As this motorcycle makes the sharp turn, it is acted upon by a centripetal force which directs the motorcycle towards the center.
Therefore, circular motion is the correct answer to the question.
Answer:
Circular Motion
Oil at 150 C flows slowly through a long, thin-walled pipe of 30-mm inner diameter. The pipe is suspended in a room for which the air temperature is 20 C and the convection coefficient at the outer tube surface is 11 W/m2 K. Estimate the heat loss per unit length of tube.
Answer:
1.01 W/m
Explanation:
diameter of the pipe d = 30 mm = 0.03 m
radius of the pipe r = d/2 = 0.015 m
external air temperature Ta = 20 °C
temperature of pipe wall Tw = 150 °C
convection coefficient at outer tube surface h = 11 W/m^2-K
From the above, we assumed that the pipe wall and the oil are in thermal equilibrium.
area of the pipe per unit length A = [tex]\pi r ^{2}[/tex] = [tex]7.069*10^{-4}[/tex] m^2/m
convectional heat loss Q = Ah(Tw - Ta)
Q = 7.069 x 10^-4 x 11 x (150 - 20)
Q = 7.069 x 10^-4 x 11 x 130 = 1.01 W/m
The heat loss per unit length of tube should be considered as the 1.01 W/m.
Calculation of the heat loss:Since
diameter of the pipe d = 30 mm = 0.03 m
radius of the pipe r = d/2 = 0.015 m
external air temperature Ta = 20 °C
temperature of pipe wall Tw = 150 °C
convection coefficient at outer tube surface h = 11 W/m^2-K
Now
area of the pipe per unit length A should be
= πr^2
= 7.069*10^-4 m^2/m
Now
convectional heat loss Q = Ah(Tw - Ta)
Q = 7.069 x 10^-4 x 11 x (150 - 20)
Q = 7.069 x 10^-4 x 11 x 130
= 1.01 W/m
hence, The heat loss per unit length of tube should be considered as the 1.01 W/m.
Learn more about heat here: https://brainly.com/question/15170783
Choose the friction which opposes the relative motion between surfaces in motion a.Static friction b.Kinetic friction c.Sliding friction d.Both kinetic and sliding friction
Answer:
d. Both kinetic and sliding friction
Explanation:
Kinetic friction, commonly known as sliding friction, happens when a body with its surfaces in contact is in relative motion with another. It's the frictional force slowing it down, and finally stopping a moving body. One can describe sliding friction as the resistance any two objects create while sliding against each other. It is often documented as the force required to hold a surface moving along another surface. It is determined by two variables- one is material of the object and another is its weight.
A pendulum at position A is released and swings through position B to position Con the other side.
B
1. Describe the total mechanical energy at each of the following positions. (3)
A.
B.
C
Explanation:
Given the conditions A,B and C when the pendulum is released, at point A the initial velocity of the pendulum is zero(0), the potential energy stored is maximum(P.E= max),
the conditions can be summarized bellow
point A
initial velocity= 0
final velocity=0
P.E= Max
K.E= 0
point B
initial velocity= maximum
final velocity=maximum
P.E=K.E
point C
initial velocity= min
final velocity=min
P.E= 0
K.E= max
when a 0.622kg basketbll hits the floor its velocit changes from 4.23m/s down to 3.85m/s up. if the averge force was 72.9N how much time was it in contact with the floor?
Answer:
Time, t = 3.2 ms
Explanation:
It is given that,
Mass of basketball, m = 0.622 kg
Initial velocity, u = 4.23 m/s
Final velocity, v = 3.85 m/s
Average force acting on the ball, F = 72.9 N
We need to find the time of contact of the ball with the floor. Let t is the time of contact. So,
[tex]F=ma\\\\F=\dfrac{m(v-u)}{t}\\\\t=\dfrac{m(v-u)}{F}\\\\t=\dfrac{0.622\times (3.85-4.23)}{72.9}\\\\t=0.0032\ s\\\\\text{or}\\\\t=3.2\ ms[/tex]
So, the ball is in contact with the floor for 3.2 ms.
1. Find the energy required to melt 255g of ice at 0°C into water at 0°C
Answer:
E = 85170 J (/ 85.2 kJ)
Explanation:
Take the latent heat of fusion of water be 334J / g.
From the equation E = ml,
E = energy required (unknown),
mass m = 255g,
latent heat of fusion l = 334J / g,
E = 255 x 334
E = 85170 J (/ 85.2 kJ)