Answer:
Convex lens and convex mirrors are similar that
1. They have the same image characteristics at various object positions
2. They possess a positive focal length
3. Both their ray lines converge to a particular focal point
Convex lens and concave mirror have certain similarities that are explained below.
The shape of concave mirror is spherical.
Convex lens is the combination of two concave mirrors.
When the ray lines are drawn at the convex lens, then the coverage of the rays leads parallel to its principal axis.
When the ray lines are drawn at the concave mirror, then the coverage of the rays leads parallel to its focal point.
The focal length of concave mirror and convex lens is positive.
Concave mirror and convex lens both project the real and inverted image of the object.
For more details, follow the link given below.
https://brainly.com/question/14295029.
A 11,000-kg train car moving due east at 21.0 m/s collides with and couples to a 23,000-kg train car that is initially at rest. What is the common velocty of the two-car train after the collisions?
Answer:
v = 6.79 m/s
Explanation:
It is given that,
Mass of a train car, m₁ = 11000 kg
Speed of train car, u₁ = 21 m/s
Mass of other train car, m₂ = 23000 kg
Initially, the other train car is at rest, u₂ = 0
It is a case based on inelastic collision as both car couples each other after the collision. The law of conservation of momentum satisied here. So,
[tex]m_1u_1+m_2u_2=(m_1+m_2)V[/tex]
V is the common velocity after the collisions
[tex]V=\dfrac{m_1u_1}{(m_1+m_2)}\\\\V=\dfrac{11000\times 21}{(11000+23000)}\\\\V=6.79\ m/s[/tex]
So, the two car train will move with a common velocity of 6.79 m/s.
1) A plane lands on a runway with a speed of 125 m/s, moving east, and it slows to a stop in 13.0 s. What is the magnitude (in m/s2) and direction of the plane's average acceleration during this time interval?2) Influenced by the gravitational pull of a distant star, the velocity of an asteroid changes from from +20.8 km/s to −17.1 km/s over a period of 2.01 years.(a) What is the total change in the asteroid's velocity? (Indicate the direction with the sign of your answer.)(b) What is the asteroid's average acceleration during this interval? (Indicate the direction with the sign of your answer.)
Answer: 1) a = 9.61m/s² pointing to west.
2) (a) Δv = - 37.9km/s
(b) a = - 6.10⁷km/years
Explanation: Aceleration is the change in velocity over change in time.
1) For the plane:
[tex]a=\frac{\Delta v}{\Delta t}[/tex]
[tex]a=\frac{125}{13}[/tex]
a = 9.61m/s²
The plane is moving east, so velocity points in that direction. However, it is stopping at the time of 13s, so acceleration's direction is in the opposite direction. Therefore, acceleration points towards west.
2) Total change of velocity:
[tex]\Delta v = v_{f}-v_{i}[/tex]
[tex]\Delta v = -17.1-(+20.8)[/tex]
[tex]\Delta v= -37.9[/tex]km/s
The interval is in years, so transforming seconds in years:
v = [tex]\frac{-37.9}{3.15.10^{-7}}[/tex]
[tex]v=-12.03.10^{7}[/tex]km/years
Calculating acceleration:
[tex]a=\frac{-12.03.10^{-7}}{2.01}[/tex]
[tex]a=-6.10^{7}[/tex]
Acceleration of an asteroid is a = -6.10⁷km/years .
Question 8 of 10
It takes a person 22 seconds to swim in a straight line from the south end of
a pool to the north end of the pool, a distance of 28 meters. What is the
swimmer's velocity?
A. 1.3 m/s south
B. 1.3 m/s north
C. 0.8 m/s south
D. 0.8 m/s north
The correct answer is B. 1.3 m/s north
Explanation:
Velocity is a factor that describes how fast or slow the motion of a body occurs and its direction. Moreover, this can be calculated by dividing the total displacement into the time of movement, and the final result is expressed in units such as meters per second followed by the direction, for example, 152 m/s south. The process to calculate the velocity of the swimmer is shown below.
[tex]v = \frac{d}{t}[/tex]
[tex]v = \frac{28 meters}{22 seconds}[/tex]
[tex]v = 1.27 m/s[/tex]
This means the velocity of the swimmer is 1.27 m/s, which can be rounded as 1.3 m/s. Additionally, if the direction is considered it would be 1.3 m/s north because the swimmer went from the south of the pool to its north.
Answer:
the answer is B
Explanation:
confirmed
The force of gravity pulls down on your school with a total force of 400,000 newtons. The force of gravity pulling down on your school would be exactly twice as much if your school: a Had twice as much mass b Was twice as tall c Had twice as much volume d Covered twice as much area
Answer: a Had twice as much mass
Explanation:
The data that we have is:
"The force of gravity pulls down on your school with a total force of 400,000 newtons. "
First, remember that, by the second Newton's law that:
F = a*m
F = force
a = acceleration
m = mass
In the case of the gravitational force, the gravitational acceleration is a constant: a = 9.8m/s^2
Then, if we want to have twice as much force the only thing that we can change in the equation is the mass:
Then if the initial force is written as:
F = a*m
twice as much that force is written as:
2*F = a*x
x is a variable that represents the new mass.
We know that F = a*m
2*F = 2*a*m
2*a*m = a*x
2*m = x
Then, if we want to have twice as much force, we should have twice as much mass.
does anyone know the answer to these 3 questions?
Answer:
independent
dependent
dependent
A merry-go-round is rotating at constant angular speed. Two children are riding the merry-go-round: Ana is riding at point A and Bobby is riding at point B.
Part A Which child moves with greater magnitude of velocity? Ana has the greater magnitude of velocity. Bobby has the greater magnitude of velocity. Both Ana and Bobby have the same magnitude of velocity.
Part B Who moves with greater magnitude of angular velocity? Ana has the greater magnitude of angular velocity. Bobby has the greater magnitude of angular velocity. Both Ana and Bobby have the same magnitude of angular velocity.
Part C Who moves with greater magnitude of tangential acceleration? Ana has the greater magnitude of tangential acceleration. Bobby has the greater magnitude of tangential acceleration. Both Ana and Bobby have the same magnitude of tangential acceleration.
Part D Who has the greater magnitude of centripetal acceleration? Ana has the greater magnitude of centripetal acceleration. Bobby has the greater magnitude of centripetal acceleration. Both Ana and Bobby have the same magnitude of centripetal acceleration.
Part E Who moves with greater magnitude of angular acceleration? Ana has the greater magnitude of angular acceleration. Bobby has the greater magnitude of angular acceleration. Both Ana and Bobby have the same magnitude of angular acceleration.
Answer:
bobby has a greater magnitude of velocity because because when angular speed is constant linear velocity is proportional to radius of the circular path
B. They both have same magnitude of angular velocity since the angular speed of the merrygoround is constant
C. Also they both have the same tangential acceleration because the angular speed is constant and tangential is zero for both of them
D. Centripetal acceleration of Bobby is greater
E.they both have the same angular acceleration because angular Speed I constant so angular acceleration is zero for both
(a) A standard sheet of paper measures 8 1/2 by 13 inches. Find the area of one such sheet of paper in m2.8.5(!meter/39.37in) — 0.2159 13 (!meter/39.37in ) — 0.3302 0.2159 X 0.3302 = 0.0713
= 0.0713 m2 (b) A second sheet of paper is three-quarters as long and three-quarters as wide as the one described in part (a). By what factor is its area less than the area found in part(a)?= _____________ times less
Answer:
a) A = 0.07129 m²
b) A / A ’= 1.77
Explanation:
In this exercise we are asked to find the area in SI units, so let's start by reducing the dimensions to SI units.
width a = 8.5 inch (2.54 cm 1 inch) (1 m / 100 cm)
a = 0.2159 m
length l = 13 inch (2.54 cm / 1 inch) (1 m / 100 cm)
l = 0.3302 m
The area of a rectangle is
A = l a
A = 0.3302 0.2159
A = 0.07129 m²
b) we have a second sheet with reduced dimensions
a ’= 3/4 a
l ’= ¾ l
Let's find the area of this glossy sheet
A ’= l’ a ’
A ’= ¾ l ¾ a
A ’= 9/16 l a
To find the factor we divide the two quantities
A / A ’= l a 16 / (9 l a
A / A ’= 1.77
Find p1, the gauge pressure at the bottom of tube 1. (Gauge pressure is the pressure in excess of outside atmospheric pressure.) Express your answer in terms of quantities given in the problem introduction and g, the magnitude of the acceleration due to gravity.
Answer:
(a). The gauge pressure at the bottom of tube 1 is [tex]P_{1}=\rho g h_{1}[/tex]
(b). The speed of the fluid in the left end of the main pipe [tex]\sqrt{\dfrac{2g(h_{2}-h_{1})}{(1-(\gamma)^2)}}[/tex]
Explanation:
Given that,
Gauge pressure at bottom = p₁
Suppose, an arrangement with a horizontal pipe carrying fluid of density p . The fluid rises to heights h1 and h2 in the two open-ended tubes (see figure). The cross-sectional area of the pipe is A1 at the position of tube 1, and A2 at the position of tube 2.
Find the speed of the fluid in the left end of the main pipe.
(a). We need to calculate the gauge pressure at the bottom of tube 1
Using bernoulli equation
[tex]P_{1}=\rho g h_{1}[/tex]
(b). We need to calculate the speed of the fluid in the left end of the main pipe
Using bernoulli equation
Pressure for first pipe,
[tex]P_{1}=\rho gh_{1}[/tex].....(I)
Pressure for second pipe,
[tex]P_{2}=\rho gh_{2}[/tex].....(II)
From equation (I) and (II)
[tex]P_{2}-P_{1}=\dfrac{1}{2}\rho(v_{1}^2-v_{2}^2)[/tex]
Put the value of P₁ and P₂
[tex]\rho g h_{2}-\rho g h_{1}=\dfrac{1}{2}\rho(v_{1}^2-v_{2}^2)[/tex]
[tex]gh_{2}-gh_{1}=\dfrac{1}{2}(v_{1}^2-v_{2}^2)[/tex]
[tex]2g(h_{2}-h_{1})=v_{1}^2-v_{2}^2[/tex]....(III)
We know that,
The continuity equation
[tex]v_{1}A_{1}=v_{2}A_{2}[/tex]
[tex]v_{2}=v_{1}(\dfrac{A_{1}}{A_{2}})[/tex]
Put the value of v₂ in equation (III)
[tex]2g(h_{2}-h_{1})=v_{1}^2-(v_{1}(\dfrac{A_{1}}{A_{2}}))^2[/tex]
[tex]2g(h_{2}-h_{1})=v_{1}^2(1-(\dfrac{A_{1}}{A_{2}}))^2[/tex]
Here, [tex]\dfrac{A_{1}}{A_{2}}=\gamma[/tex]
So, [tex]2g(h_{2}-h_{1})=v_{1}^2(1-(\gamma)^2)[/tex]
[tex]v_{1}=\sqrt{\dfrac{2g(h_{2}-h_{1})}{(1-(\gamma)^2)}}[/tex]
Hence, (a). The gauge pressure at the bottom of tube 1 is [tex]P_{1}=\rho g h_{1}[/tex]
(b). The speed of the fluid in the left end of the main pipe [tex]\sqrt{\dfrac{2g(h_{2}-h_{1})}{(1-(\gamma)^2)}}[/tex]
scott and jafar are ready to work a problem. in this simulation, assume the coordinates of the points are as follows. a (0 s, 30 m); b (10 s, 50 m); c (20 s, 38 m); d (30 s, 0 m); e (40 s, −38 m); and f (50 s, −50 m) recall the definitions of average speed, vavg ≡ d δt , and average velocity in the x direction, vx ,avg ≡ δx δt . find the average velocity from circled a to circled b.
Answer:
Explanation:
point a represents time 0 and position coordinate 30
point b represents time 10 s , and position coordinate 50 m .
time elapsed = 10 - 0 = 10 s .
displacement = 50 m - 30 m
= 20 m
average velocity = displacement / time elapsed
= 20 / 10
= 2 m /s .
In lifting a heavy weight from the floor, one should use the power of the __________ in order to avoid straining the lower back.
Answer:
Hip and knee extensors
Explanation:
These are gluteus muscles and hamstring muscles the there are the major movers for your body and are very important in pelvic alignment and lower back support during weight lifting
If the position versus time graph of an object is a horizontal line, the object is
In a position-time graph, the velocity of the moving object is represented by the slope, or steepness, of the graph line. If the graph line is horizontal, like the line after time = 5 seconds in Graph 2 in the Figure below, then the slope is zero and so is the velocity. The position of the object is not changing.
Explanation:
The slope of a position time graph gives the velocity of an object. It can be calculated as follows:
[tex]v=\dfrac{dx}{dt}[/tex]
Where
[tex]dx[/tex] is the change in position
[tex]dt[/tex] is the change in time
If the position vs time graph of an object is a horizontal line, it shows that the object is at rest. At this point the object is not moving.
Answer:
If the position versus time graph of an object is a horizontal line, the object is at rest.
Reference:
https://brainly.com/question/12025472
Convert 252 cL into uL
Answer:
Explanation:
252,000
List two scientific questions that would be best explained using a model.
Answer:
Is a flame hottest when it is blue? Is it cold today?
Explanation:
Verification questions. These are basic data collecting questions. They are useful in building knowledge.
What are two examples of healthcare fraud ?
you travel 4.0km east, 4.0km north, then 5.0km at 53.1 degrees north of west in a total of 5 hours. What is the magnitude and direction of your average velocity
Answer:
Explanation:
We shall convert all the displacement in vector form .
i and j represents east and north respectively .
D₁ = 4 i
D₂ = 4 j
D₃ = - 5 cos 53.1 i + 5 sin 53.1 j
= -3i + 4 j
Total displacement = D₁ + D₂ + D₃
= 4i + 4 j - 3i + 4 j
= i + 8j
magnitude of displacement = √( 1² + 8² )
= 8.06 km
velocity = 8.06 / 5
= 1.61 km / h
Direction from x axis in anticlockwise direction .
Tanθ = 8 / 1 = 8
θ = 83° north of east .
Using the lensmaker's formula (equation (5) of your lab manual), calculate the index of refraction of the acrylic lens. You should use the f_are you calculated in part (1) above instead of the value for the focal length of the concave lens that you measured. Remember that the focal length of a concave lens is negative so in this case, f = -f_are.
Answer:
n = 1 + R / f
Explanation:
The equation of the constructor is optical is
1 / f = 1 / p + 1 / q
where f is the focal length, p and q are the distance to the object and image, respectively
The exercise tells us that it is a concave lens with focal length fo, in these lenses the focal length is negative. The relationship to calculate the focal length is
1 / f = (n -n₀) (1 /R₁ - 1 /R₂)
where is n₀ the refractive index of the medium that surrounds the lens in this case it is air with n₀ = 1, you do not indicate the type of lens, but the most used lens is the concave plane, in this case R₂ = ∞, so which 1 / R₂ = 0, let's substitute
1 / f = (n-1) / R₁
n - 1 = R₁ / f
let's calculate
n = 1- R₁ / f
remember that the radius of curvature is negative, so the equation is
n = 1 + R / f
What is the magnitude of the emf induced in the secondary winding at the instant that the current in the solenoid is 3.2 A
Complete Question
A very long, straight solenoid with a cross-sectional area of 2.39cm2 is wound with 85.7 turns of wire per centimeter. Starting at t= 0, the current in the solenoid is increasing according to i(t)=( 0.162A/s2) t2. A secondary winding of 5 turns encircles the solenoid at its center, such that the secondary winding has the same cross-sectional area as the solenoid. What is the magnitude of the emf induced in the secondary winding at the instant that the current in the solenoid is 3.2A ?
Answer:
The value is [tex]\epsilon = 1.83 *10^{-5} \ V[/tex]
Explanation:
From the question we are told that
The cross-sectional area is [tex]A = 2.39 \ cm^2 = \frac{2.39}{10000} = 0.000239 \ m^2[/tex]
The number of turns is [tex]N = 85.7 \ turns/cm = 8570 \ turns / m[/tex]
The initial time is t = 0s
The current on the solenoid is [tex]I(t) = (0.162 \ A/s^2) t^2[/tex]
The number of turns of the secondary winding is [tex]n = 5 \ turns[/tex]
Generally At I = 3.2 A
[tex]3.2 = (0.162 )t^2[/tex]
=> [tex]t^2 = 19.8[/tex]
=> [tex]t = 4.4 \ s[/tex]
Generally induced emf is mathematically represented as
[tex]\epsilon = A * \mu_o * n * N \frac{d(I)}{dt}[/tex]
[tex]\epsilon = 0.000239 * 4\pi * 10^{-7} * 8570 * 5 * (0.162) * 2t[/tex]
[tex]\epsilon = 0.000239 * 4\pi * 10^{-7} * 8570 * 5 * (0.162) * 2 * 4.4[/tex]
[tex]\epsilon = 1.83 *10^{-5} \ V[/tex]
Which of the following statements is/are correct?
A. The instantaneous speed is always equal to the magnitude of instantaneous velocity.
B. The average speed is always equal to the magnitude of average velocity.
C. Neither of these statements is correct.
D. Both of these statements are correct.
Answer:
The correct answer is:
The instantaneous speed is always equal to the magnitude of instantaneous velocity. (A)
Explanation:
Speed is the ratio of distance travelled to time. It is a scalar quantity.
Velocity is the rate of change of displacement with time. It is a scalar quantity, having both magnitude and direction
Instantaneous velocity is the rate of change of position for a very small time interval (at a particular point in time), while the instantaneous speed at any given time, is the magnitude of instantaneous velocity. Note that the velocity has both magnitude and direction, so at a particular point in time, the magnitude part of velocity is the same as the speed.
The formula of instantaneous velocity is: [tex]V_{i}=\lim_{\Delta t\rightarrow 0}\frac{ds}{dt}[/tex]
The formula of instantaneous speed is: [tex]Speed_{(i)}=\frac{ds}{dt}[/tex]
On the other hand, average speed does not equal the magnitude of average velocity, because velocity depends on displacement while speed depends on distance, and if an object in motion changes direction at any point in the motion, the velocity reduces because the angle of change in the direction will be considered, hence the speed will be greater than the velocity, therefore, the average speed is not the same as the magnitude of average velocity.
When sunlight interacts with a raindrop to produce a rainbow, how many refractive events occur for each drop
Answer:
The correct answer will be "2".
Explanation:
The mechanism of rainbow formation starts whenever sunlight is shone on a raindrop. When the sun's light waves hit and go through a droplet of water, their pace changes a little. This induces bending or "refracting" of the direction of light.Light moves in such a single direction, unless someone represents something, refracts, or diffracts everything that. When some of these things have happened, the balanced information of color is isolated, in which each could still be seen.When the Apollo 11 lunar module Eagle lands on the moon it comes to a stop 10m above the surface of the moon.The last 10m it freely falls to the surface of the Moon. i) How long does it take for the Eagle to touch down
Answer:
3.5s
Explanation:
Using equation of motion number 2
S= ut + 1/2 gt²
Substituting
10= 0.5+1.62t²
t= 3.5seconds
We will use the same coordinate system for a number of the questions on this exam.
Place a sheet of paper flat on a table in front of you as you sit facing it.
In the horizontal plane toward the right side of the paper is the positive x direction.
In the horizontal plane toward the top of the paper ( away from you in the direction you are facing ) is the positive y direction.
In the vertical direction toward the ceiling is the positive z direction
A spatially uniform magnetic field is steadily changing its size. When a clock reads 2.00s the field is 1.75T in the positive z direction. In the x y plane there is a coil of wire with 55 loops and an area of 0.12m%. The EMF ( voltage change) from one end of the coil to the other is a steady 12.7 Volts. One of these is the magnetic field when the clock reads 2.37s. Which is it?
Answer:
Explanation:
Let the magnetic field after .37 s is B
magnetic flux associated with coil = n BA , where B is magnetic field and A is area of coil and n is no of turns
= 55 x .12 x 1.75
= 11.55 Tm²
magnetic flux after .37 s
= 55 x B x .12
= 6.6 B
rate of change of flux = (66B - 11.55) / .37 = EMF induced
(6.6B - 11.55) / .37 = 12.7
6.6 B - 11.55 = 4.7
6.6 B = 16.25
B = 2.46 T .
A clock battery wears out after moving 10,900 C of charge through the clock at a rate of 0.450 mA. (a) How long (in s) did the clock run
Answer:
2.42×10⁷ s
Explanation:
From the question above,
Applying,
Q = It.................... Equation 1
Where Q = quantity of charge in coulombs, I = electric current in Ampere, t = time in seconds
make t the subject of the equation
t = Q/I................ Equation 2
Given: Q = 10900 C, I = 0.450 mA = 4.5×10⁻⁴ A
Substitute these values into equation 2
t = 10900/(4.5×10⁻⁴)
t = 2.42×10⁷ s
Hence the clock runs for 2.42×10⁷ s
i need help Mr or ms tutor
Explanation:
Height is the x-axis, and gravitational potential energy is the y-axis. As the height increases, the gravitational potential energy increases linearly.
Determine the discharge of a river that is 10 m wide, has a semicircular cross-sectional area, and has a velocity of 3 m/s.
Answer:
117.8 m³/s
Explanation:
given that
the diameter of the river, d = 10 m wide
Velocity of flow of water, v, = 3 m/s
To get the discharge of a river, we have to find the area of the said river first.
Area of the river, a = area of semi circle = 1/2 * area of a circle
Area = 1/2 * πr²
Area = 1/2 * 3.142 * 5²
Area = 39.275 m²
Discharge of the river therefore is
Discharge, q = A * V, where
A is area of the river, and
V is the velocity of flow of the river.
Discharge, q = 39.275 * 3
Discharge, q = 117.825
Distance is the length of a path followed by a particle. The displacement of a particle is defined as its change in position in some time interval.
a. True
b. False
Answer:
a. True
Explanation:
Distance is described with only magnitude. It is defined as the total path covered by an object, in other words it is the length of a path followed by a particle.
Displacement is described with both magnitude and direction. It is distance traveled in a specified direction or change in position in some time interval.
Therefore, the correct option is " a. True"
The peak value of an alternating current in a 1500-W device is 6.4 A. What is the rms voltage across it
Answer:
6787.5 V
Explanation:
From the question,
P = IV..................... Equation 1
Where P = Power, I = rms current, V = rms voltage.
make V the subject of the equation
V = P/I................. Equation 2
Given: P = 1500 W, I = 6.4/√2 = 4.525 A
Substitute these values into equation 2
V = 1500(4.525)
V = 6787.5 V
Hence the rms voltage = 6787.5 V
hor and Hulk are fighting eachother, back and forth in a straight line in the master's colosseum. Hulk punches thor into a wall 30 meters away from the start. Thor then hit's hulk with a hammer, hurling Hulk 45 meters back from the wall. This all happens in the course of 3 seconds. What is the speed of their fight across the arena?
a. 25 m/s
b. 10 m/s forward from Hulk and Thor's starting point.
c. 5 m/s backwards from hulk's starting point
d. 5 m/s
Answer:
A
Explanation:
cant be c because its asking for speed and speed doesnt have direction
The effective spring constant describing the potential energy of the HBr molecule is 410 N/m and that for the NO molecule is 1530 N/m.A) Calculate the minimum amplitude of vibration for the NO molecule.
B) Calculate the minimum amplitude of vibration for the HCl molecule.
Answer:
a. the minimum amplitude of vibration for the NO molecule A [tex]\simeq[/tex] 4.9378 pm
b. the minimum amplitude of vibration for the HCl molecule A [tex]\simeq[/tex] 10.9336 pm
Explanation:
Given that:
The effective spring constant describing the potential energy of the HBr molecule is 410 N/m
The effective spring constant describing the potential energy of the NO molecule is 1530 N/m
To calculate the minimum amplitude of vibration for the NO molecule, we use the formula:
[tex]\dfrac{1}{2}kA^2= \dfrac{1}{2}hf[/tex]
[tex]kA^2= hf[/tex]
[tex]A^2= \dfrac{hf}{k}[/tex]
[tex]A = \sqrt{\dfrac{hf}{k}}[/tex]
[tex]A = \sqrt{\dfrac{(6.626 \times 10^{-34} \ J. s ) ( 5.63 \times 10^{13} \ s^{-1})}{1530 \ N/m}}[/tex]
[tex]A = \sqrt{\dfrac{3.730438 \times 10^{-20} \ m}{1530 }[/tex]
[tex]A = \sqrt{2.43819477 \times 10^{-23}\ m[/tex]
[tex]A =4.93780799 \times 10^{-12} \ m[/tex]
A [tex]\simeq[/tex] 4.9378 pm
The effective spring constant describing the potential energy of the HCl molecule is 480 N/m
To calculate the minimum amplitude using the same formula above, we have:
[tex]A = \sqrt{\dfrac{hf}{k}}[/tex]
[tex]A = \sqrt{\dfrac{(6.626 \times 10^{-34} \ J. s ) ( 8.66 \times 10^{13} \ s^{-1})}{480 \ N/m}}[/tex]
[tex]A = \sqrt{\dfrac{5.738116\times 10^{-20} \ m}{480 }[/tex]
[tex]A = \sqrt{1.19544083\times 10^{-22}\ m[/tex]
[tex]A = 1.09336217 \times 10^{-11} \ m[/tex]
[tex]A = 10.9336217 \times 10^{-12} \ m[/tex]
A [tex]\simeq[/tex] 10.9336 pm
Why do we perceive the sky as blue instead of red?
The blue sky we observe depends upon two factors: how sunlight interacts with Earth's atmosphere, and how our eyes perceive that light.
If this helps please mark me the brainiest
how sunlight interacts with Earth's atmosphere, and how our eyes perceive that light.nation:
which of the following is not a method of caculating percentage of body fat
Body mass index, known as BMI, is not a method of calculating body fat percentage, but it is a technique used to check nutritional status and to see if a person is within normal range with respect to weight and height.. This technique is measured by the formula: BMI = Weight (Kg) / (Height (m)) ².
Ergo, the answer is BMI!!!!!!!!!!!
Hope this helped you!