Answer:
Explanation:
Position of n the order fringe = n λ D / d
for n = 2
position = 2 λ D / d
λ = 710 nm , D = 1.1 m
d = .65 x 10⁻³
position 1 = 2 x 710 x 10⁻⁹ x 1.1 / .65 x 10⁻³
= 2403.07 x 10⁻⁶ m
= 2.403 x 10⁻³ m
= 2.403 mm .
For λ = 655 nm
position = 2 λ D / d
λ = 655 nm , D = 1.1 m
d = .65 x 10⁻³
position 2 = 2 x 655 x 10⁻⁹ x 1.1 / .65 x 10⁻³
= 2216.91 x 10⁻⁶ m
= 2.217 x 10⁻³ m
= 2.217 mm .
Difference between their position
= 2.403 - 2.217 = .186 mm .
HELPP physics final will give brainliest
A sled is pulled with a force of 540 N at an angle of 40° with the horizontal. What are the horizontal and vertical components of this force?
Answer:
Fx = 467.65N
Fy = 270N
Explanation:
Given
Force = 540N
angle of inclination = 40 degree
Horizontal component Fx = Fcos 30
Fx = 540cos30
Fx = 540(0.8660)
Fx = 467.65N
Hence the horizontal component is 467.65N
Vertical component Fy = Fsin 30
Fy = 540sin30
Fy = 540(0.5)
Fy = 270N
Hence the vertical component is 270N
Where do magnetic fields occur?
Answer:
The Magnetosphere and MagnetsExplanation:
[tex]--------------------------------------------[/tex]
According to the National Geographic, "Earth’s magnetic field dominates a region called the magnetosphere, which wraps around the planet and its atmosphere. Solar wind, charged particles from the sun, presses the magnetosphere against the Earth on the side facing the sun and stretches it into a teardrop shape on the shadow side. The magnetosphere protects the Earth from most of the particles, but some leak through it and become trapped. When particles from the solar wind hit atoms of gas in the upper atmosphere around the geomagnetic poles, they produce light displays called auroras. These auroras appear over places like Alaska, Canada and Scandinavia, where they are sometimes called “Northern Lights.” The “Southern Lights” can be seen in Antarctica and New Zealand. The magnetic field is the area around a magnet that has magnetic force. All magnets have north and south poles."
[tex]--------------------------------------------[/tex]
Hope this helps! <3
[tex]--------------------------------------------[/tex]
Your electric drill rotates initially at 5.21 rad/s. You slide the speed control and cause the drill to undergo constant angular acceleration of 0.311 rad/s2 for 4.13 s. What is the drill's angular displacement during that time interval?
Answer:
The drill's angular displacement during that time interval is 24.17 rad.
Explanation:
Given;
initial angular velocity of the electric drill, [tex]\omega _i[/tex] = 5.21 rad/s
angular acceleration of the electric drill, α = 0.311 rad/s²
time of motion of the electric drill, t = 4.13 s
The angular displacement of the electric drill at the given time interval is calculated as;
[tex]\theta = \omega _i t \ + \ \frac{1}{2}\alpha t^2\\\\\theta = (5.21 \ \times \ 4.13) \ + \ \frac{1}{2}(0.311)(4.13)^2\\\\\theta = (21.5173 ) \ + \ (2.6524)\\\\\theta =24.17 \ rad[/tex]
Therefore, the drill's angular displacement during that time interval is 24.17 rad.
in the case shown below, the 1 kg rock rides on a horizontal disk that rotates at constant speed 5m/s about its vertical axis. the radius of the disk is 1 meter. What is the magnitude of the friction?
Answer:
25
Explanation:
3.00 kg block moving 2.09 m/s right hits a 2.22 kg block moving 3.92 m/s left. afterwards, the 3.00 kg block moves 1.71 m/s left. find the velocity of the 2.22 kg block afterwards
Momentum is conserved, so the total momentum before collision is equal to the total momentum after collision. Take the right direction to be positive. Then
(3.00 kg) (2.09 m/s) + (2.22 kg) (-3.92 m/s) = (3.00 kg) (-1.71 m/s) + (2.22 kg) v
where v is the velocity of the 2.22 kg block after collision. Solve for v :
6.27 kg•m/s - 8.70 kg•m/s = -5.13 kg•m/s + (2.22 kg) v
(2.22 kg) v = 2.70 kg•m/s
v = (2.70 kg•m/s) / (2.22 kg)
v ≈ 1.22 m/s
i.e. a velocity of about 1.22 m/s to the right.
A 100kg couch is being pushed with 196N of force. As it slides along the ground it experiences a coefficient of friction of 0.1. What is the net force in this situation?
A 300N
B 202N
C 398N
D 98N
Answer:98
Explanation:hope this helps!
A skier is moving 8.33 m/s when
he starts to slide UP a 8.44°
frictionless slope. How much time
does it take him to come to a stop?
(Unit = s)
Answer:
5.79
Explanation:
Vf=Vi+at
0=8.33+(-9.8)sin8.44t
t=8.33/(9.8sin8.44)=0.83/sin 8.44
=5.79
What is the original source of the energy stored in fossil fuels?
plants
the Sun
air
O water
The sun
Answer:
Plants the chloroplasts
(1-dimension) A fish has a mass of 6 kg and is moving at a speed of 4m/s to the right. What is its momentum?
Answer:
24 kg m/sExplanation:
The momentum of an object can be found by using the formula
momentum = mass × velocity
From the question we have
momentum = 6 × 4
We have the final answer as
24 kg m/sHope this helps you
Which scenario is an example of the transfer of thermal energy by radiation?
A. Water boils in a pan.
B. Hot air circulates in an oven.
C. An ice cube melts in a person's hand.
D. A frozen lake melts under the Sun.
Correct answer is D
Answer:
its D: A frozen lake melts under the sun.
Explanation:
Radiation is the transfer of heat energy through space by electromagnetic radiation. Most of the electromagnetic radiation that comes to the earth from the sun is invisible. Only a small portion comes as visible light. Light is made of waves of different frequencies.
Find the state of Georgia in the southeastern U.S. What can you say about the weather in this state?
What is the frequency of this wave?
can someone pls explain and answer
Answer:
f = 1 Hz
Explanation:
From the attached figure, we find that the time period of the wave is 1 second.
It is a longitudinal wave. It travel in the form of compression and rarefaction. The relation between time period T and frequency f is given by :
[tex]f=\dfrac{1}{T}\\\\\text{As T = 1 s}\\\\f=1\ s^{-1}\\\\=1\ Hz[/tex]
Hence, the frequency of this wave is 1 Hz.
what is the initial position of the object?
a. 2m b. 4m c. 6m d. 8m e. 10 m
what is the velocity of the object?
a. -10 m/s b. -5 m/s c. 0 m/s d. 5 m/s e. 10 m/s
which of the following is true?
a. the object increase its velocity.
b. the object decrease its velocity.
c. the objects velocity stays unchanged.
d. the object stays at rest.
e. more information is required.
Answer:
e:10 m b: -5 m/s b.The object decreases its velocity.
Explanation:
Ask For
6.
A horizontal 100 N force is applied to a 50 kg classmate resting on a level tile
floor. The coefficient of kinetic friction is 0.15.
a. Draw a force diagram to represent this situation.
b. What is the acceleration of the classmate?
c. Suppose the classmate was resting on a carpet where the coefficient of static
friction is 0.25. Is the horizontal 100 N force sufficient to cause the classmate to
accelerate? Draw a force diagram, and then explain why or why not.
d. If a dog has a mass of 12 kg, what is its weight on Neptune?
11.7N/kg
Answer:
133.8 N
Explanation:
Recall that the acceleration of gravity in Neptune is estimated as 11.15 m/s^2
Therefore, the weight of the dog on this planet would be:
Weight = mass x acceleration of gravity = 12 kg x 11.15 m/s^2 = 133.8 N
A. A piece of paper near a magnet
B. An aluminum nail near a magnet
C. An iron nail, not near a magnet
D. An iron nail near a magnet
Answer:
it’s c not d
Explanation:
took the test
Answer: D!!!
Explanation: jus got it wrong from the other answer.
The Slingshot is a ride for two people. It consists of a single passenger cage, two towers, and two elastic bands. Potential energy is stored in the elastic bands and the passenger cage is released. On the way up, this potential energy in the elastic bands is converted into the kinetic energy of the cage. At the maximum height of the ride, the energy has been converted into gravitational potential energy of the cage. The slingshot has two towers of height h = 76 m. The towers are a distance d = 31 m apart. Each elastic band has an unstretched length of L0 = 41 m and a spring constant of k = 310 N/m. The total mass of the passengers and cage is m = 410 kg. The car is pulled down to the ground in the middle of the two towers.
Calculate the maximum height, in meters, of the ride.
my bad I was in a herie last time can you please answer my question , I am going to give you the 5 points back for this question and extra 30 points ,
Explanation:
first its going to say 10 points for my question but after that I well make answer a small question and give you 30points. like whats your favorite color . stay tuned .
Answer:
The maximum height reached by the ride after it was pulled to the ground is 51.6 m.
The given parameters;
Distance between the two towers, d = 31 mHeight of the tower, h = 76 mUnstretched length of the band, L₀ = 41 mElastic constant of the band, k = 310 N/mThe distance half-way between the bands;
[tex]\frac{d}{2} = \frac{31}{2} = 15.5 \ m[/tex]
The maximum length of the band when stretched is calculated as;
[tex]c^2 = 15.5^2 + 76^2\\\\c^2 = 6016.25\\\\c = \sqrt{6016.25} \\\\c = 77.57 \ m[/tex]
The extension of the elastic band;
x = 77.57 m - 41 m
x = 36.37 m
The elastic potential energy stored in the band;
[tex]E = \frac{1}{2} kx^2\\\\E = \frac{1}{2} \times 310 \times (36.57)^2\\\\E = 207,291.56 \ J[/tex]
The elastic potential energy of the elastic band will be converted into kinetic energy of the ride and the speed of the ride is calculated as;
[tex]E = \frac{1}{2} mv^2\\\\207,291.56 = \frac{1}{2} \times 410 \times v^2\\\\v^2 = \frac{207,291.56}{(0.5\times 410)} \\\\v^2 = 1011.178\\\\v = \sqrt{1011.178} \\\\v = 31.8 \ m/s[/tex]
The maximum height reached by the ride is calculated as;
[tex]P.E = K.E\\\\mgh = \frac{1}{2} mv^2\\\\h = \frac{v^2}{2g} \\\\h = \frac{(31.8)^2}{(2\times 9.8)} \\\\h = 51.6 \ m[/tex]
Thus, the maximum height reached by the ride is 51.6 m.
Learn more here: https://brainly.com/question/156316
What is the distance between a 900 kg compact car and a 1600 kg pickup truck if the gravitational force between them is about 0.0001 N?
Answer:
The distance is 0.96m
Explanation:
Given
m1= 900kg
m2= 1600kg
Force F= 0.0001nN
G=6.67430*10^-11 Nm^2/kg^2
Required
The distance r
Step two:
the formula for the force is given as
F = Gm1m2/r2
make r subject of the formula
[tex]r= \sqrt{\frac{Gm1m2}{F} }[/tex]
[tex]r= \sqrt{\frac{6.67430*10^-11*900*1600}{0.0001} }\\\\r= 0.00009610992/0.0001`}\\\\r= 0.96m[/tex]
Answer:
The distance is 0.96m
Explanation:
Given
m1= 900kg
m2= 1600kg
Force F= 0.0001nN
G=6.67430*10^-11 Nm^2/kg^2
Required:
The distance r
Step two:
the formula for the force is given as
F = Gm1m2/r2
make r subject of the formula
[tex]r= \sqrt{\frac{Gm1m2}{F} }[/tex]
[tex]r= \sqrt{\frac{6.67430*10^-11*900*1600}{0.0001} }\\\\r= 0.00009610992/0.0001`}\\\\r= 0.96m[/tex]
Answer:
The distance between the compact car and pickup truck is 0.96048 m
Explanation:
The gravitational force is directly proportional to the product of the masses of the interacting object, it is also inversely proportional to the square of the distance between them. This is shown in equation 1;
[tex]F =G \frac{m_{1} X m_{2} }{d^{2} }[/tex]............ 1
Where F is the gravitational force = 0.0001 N
G is the gravitational constant = 6.673 x [tex]10^{-11} Nm^{2} kg^{-2}[/tex]
[tex]m_{1}[/tex] is the mass of the compact car = 900kg
[tex]m_{2}[/tex] is the mass of the pickup truck = 1600kg
d is the distance and its unknown ?
Let us make d the subject formula in equation 1
[tex]d = \sqrt{G\frac{m_{1} m_{2} }{F } }[/tex] .... 2
Substituting into equation 2 we have
[tex]d = \sqrt{\frac{6.673x10^{-11} x 900 x 1600}{0.0001N} }[/tex]
d = 0.96048m
Therefore the distance between the compact car and pickup truck is 0.96048 m
a mass of 2.00 kg rest on a rough horizontal table. The coefficient of static friction between the block and the table is 0.60. The block is attached to a hanging mass by a string that goes over a smooth pulley,as shown in the diagram. Determine the largest mass that can hang in this way without forcing the block to slide.
Answer:
1.2 kg
__________________________________________________________
We are given:
Mass of the block = 2 kg
Coefficient of Static Friction = 0.6
__________________________________________________________
Friction Force on the Block:
Finding the Normal Force:
We know that the normal force will be equal and opposite to the weight of the 2 kg block
So, Normal Force = mg
replacing the variables with the given values
Normal Force = (2)(9.8) [Taking g = 9.8]
Normal Force = 19.6 N
Friction force on the Block:
We know that:
Coefficient of Static Friction = Static Friction Force/Normal Force
replacing the variables
0.6 = Static Friction force / 19.6
Static Friction force = 0.6*19.6 N [Multiplying both sides by 19.6]
Static Friction force = 11.76 N
__________________________________________________________
Largest Mass that can Hang:
We know that the Static Friction force is 11.76 N, this means that a force of 11.76 N will be applied to keep the object at rest
So, if the weight of the second block is less than the static friction force, it will hang
Weight of the second block ≤ 11.76
We know that weight = mg
mg ≤ 11.76
m(9.8) ≤ 11.76 [since g = 9.8]
m ≤ 1.2 kg [dividing both sides by 9.8]
From this, we can say that the maximum mass of the second block is 1.2 Kg
A 100 kg. football player and a 20 kg. child sit on rolling carts and push off
from each other. Which person applies more force?*20 kg. child applies more force
100 kg. football player applies more force
Same force is applied
Answer:
the forces are the same
Explanation:
Please help
1. Who was Ptolemy and Copernicus?
2. Describe their ideas about the planets and the sun
Answer:
1 +2=3 so thats it
Explanation:
Because i need points
Two objects of equal mass are a distance of 5.0 m apart and attract each other with a gravitational force of 3.0 x 10^-7 N find their mass.
A) 150 kg
B) 9.8 kg
C) 11.000 kg
D) 340 kg
Answer
I Think Its 150
What is the wavelength of a wave whose velocity is 12m/s and has a frequency of .75 Hz?
Answer:
16 meters
Explanation:
Use the formula that relates frequency velocity and wavelength:
velocity = wave-length x frequency
in our case:
12 m/s = wave-length * 0.75 Hz
wave-length= 12/0.75 m = 16 meters
Calculate the RMS speed of helium atoms near the surface of the Sun at a temperature of about 5300 K.
Answer:
RMS velocity, [tex]v_{rms}=5748.75\ m/s[/tex]
Explanation:
We need to find the RMS speed of helium atoms near the surface of the Sun at a temperature of about 5300 K.
The formula for RMS speed of a gas is given by :
[tex]v_{rms}=\sqrt{\dfrac{3RT}{m}}[/tex]
Where
R is ideal gas constant, R = 8.314 J /mol K
T = 5300 K
m is molar mass of Helium, [tex]m = 4\times 10^{-3}\ Kg/mol[/tex]
Substituting all the values in above formula :
[tex]v_{rms}=\sqrt{\dfrac{3\times 8.314\times 5300}{4\times 10^{-3}}}\\\\=5748.75\ m/s[/tex]
So, the RMS speed Helium atoms 5748.75 m/s.
Carter's favorite ride at playland amusement park is the rollercoaster. The roller coaster
car and passengers have a combined mass of 1620kg and they descend the first hill at
an angle of 45.0 degrees to the horizontal. With what force is the rollercoaster pulled
down the hill?
Answer:
F = 11226.02 N
Explanation:
The roller coaster car and passengers have a combined mass of 1620kg.
They descend the first hill at an angle of 45.0 degrees to the horizontal.
We need to find force is the rollercoaster pulled down the hill.
We firstly find the rectangular component of the downward. The force acting in the downward direction is mgsinθ such that,
F = mgsinθ
= 1620 × 9.8 × sin(45)
= 11226.02 N
So, the roller coaster is pulled down the hill with a force of 11226.02 N.
Hey guys this is Ap physics please help I need this to pass i will mark brainliest for a good attempt
Split up the forces into components acting parallel to and perpendicular to the slope. See the attached picture for the reference axes.
The box stays on the surface of the plane, so that the net force acting perpendicular to it is 0, and the only acceleration is applied in the parallel direction.
Let m be the mass of the box, θ the angle the plane makes with the ground, and a the acceleration of the box. By Newton's second law, we have
• net parallel force
∑ Force (//) = W (//) - F = m a
(that is, the net force in the parallel direction is the sum of the parallel component of the weight W and the friction F which acts in the negative direction)
• net perpendicular force
∑ Force (⟂) = W (⟂) + N = 0
Notice that
W (//) = W sin(θ) … … … which is positive since it points down the plane
W (⟂) = -W cos(θ) … … … which is negative since it points opposite the normal force N
So the equations become
W sin(θ) - F = m a
-W cos(θ) + N = 0
Solving for a gives
a = (W sin(θ) - F ) / m
which is good enough if you know the magnitude of the friction force.
If you don't, you can write F in terms of the coefficient of kinetic friction between the box and plane, µ, as
F = µ N
so that
a = (W sin(θ) - µ N ) / m
and the normal force itself has a magnitude of
N = W cos(θ)
so that
a = (W sin(θ) - µ W cos(θ) ) / m
The weight W has magnitude m g, where g is the magnitude of the acceleration due to gravity, so
a = (m g sin(θ) - µ m g cos(θ) ) / m
a = g (sin(θ) - µ cos(θ))
8x = -6. What does x equal?
Answer:
x=-3/4
Explanation:
Among the elements potassium, lithium, and iron, the metallic bonds are likely to be strongest in
彼が好きな男のような話は妻と関係があったのでそう
Explanation:これが難しい場合はコメントで教えてくださいが何でも
lol okay
from: a random person :)
Answer: Iron
Explanation: Gradpoint
How high does a rocket have to go above the earth's surface to be subject to a gravitational field from the earth that is 50.0 percent of its value at the earth's surface?
A) 2.650 km
B) 3,190 km
C) 9.020 km
D) 12.700 km
Answer:
A) 2.650 km
Explanation:
The relationship between acceleration of gravity and gravitational constant is:
[tex]g = \frac{Gm}{R^2}[/tex] ---- (1)
Where
[tex]R = 6,400 km[/tex] -- Radius of the earth.
From the question, we understand that the gravitational field of the rocket is 50% of its original value.
This means that:
[tex]g_{rocket} = 50\% * g[/tex]
[tex]g_{rocket} = 0.50 * g[/tex]
[tex]g_{rocket} = 0.5g[/tex]
For the rocket, we have:
[tex]g_{rocket} = \frac{Gm}{r^2}[/tex]
Where r represent the distance between the rocket and the center of the earth.
Substitute 0.5g for g rocket
[tex]0.5g = \frac{Gm}{r^2}[/tex] --- (2)
Divide (1) by (2)
[tex]\frac{g}{0.5g} = \frac{Gm}{R^2}/\frac{Gm}{r^2}[/tex]
[tex]\frac{g}{0.5g} = \frac{Gm}{R^2}*\frac{r^2}{Gm}[/tex]
[tex]\frac{1}{0.5} = \frac{1}{R^2}*\frac{r^2}{1}[/tex]
[tex]2 = \frac{r^2}{R^2}[/tex]
Take square root of both sides
[tex]\sqrt 2 = \frac{r}{R}[/tex]
Make r the subject
[tex]r = R * \sqrt 2[/tex]
Substitute [tex]R = 6,400 km[/tex]
[tex]r = 6400km * \sqrt 2[/tex]
[tex]r = 6400km * 1.414[/tex]
[tex]r = 9 049.6\ km[/tex]
The distance (d) from the earth surface is calculated as thus;
[tex]d = r - R[/tex]
[tex]d = 9049.6\ km - 6400\ km[/tex]
[tex]d = 2649.6\ km[/tex]
[tex]d = 2650\ km[/tex] --- approximated