Answer:
the kinetic energy decreases and the potential energy Increases.
Explanation:
as the comet travels away from the star it gains an energy which it posses because of its position or state.once the comet moves away form the star the kinetic decreases until its lost all together to where the potential energy starts increasing.
Answer:
The potential energy increase and the kinetic energy decrease.
If the momentum of a 100 kg moped travelling at 10 m/s was transferred completely to a 25.0 kg cement traffic barrier, what would the final speed of the barrier be
Answer:
the final speed of the barrier would be 40m/s
Explanation:
Hello, I think I can help you with this.
the momentum is given by:
P=m*v, where P is the momentum, m is the mass of the object and v is the speed of the object.
if the momentum is transferred completely it means that the barrier will have the same amount of momentum, in other terms
momentum of the moped=momentum of the barrier
mass of the moped*speed of the moped=*speed of the barrier
let's isolate the final speed of the barrier
final speed of the barrier is
[tex]=\frac{mass\ of\ the\ moped*speed\ of\ the\ moped}{mass\ of\ the\ barrier } \\\\put the values \\\\=\frac{100kg*10m/s}{25kg} \\=40 m/s[/tex]
so, the speed is 40 m/s
A flat coil of wire has an area A, N turns, and a resistance R. It is situated in a magnetic field, such that the normal to the coil is parallel to the magnetic field. The coil is then rotated through an angle of 90°, so that the normal becomes perpendicular to the magnetic field. The coil has an area of 1.5 10-3 m2, 50 turns, and a resistance of 166 Ω. During the time while it is rotating, a charge of 7.3 10-5 C flows in the coil. What is the magnitude of the magnetic field?
Explanation:
Expression for magnitude of the induced emf is as follows.
[tex]\epsilon = N \frac{BA}{t}[/tex]
[tex]\frac{Q}{t}R = \frac{NBA}{t}[/tex]
So, magnitude of the magnetic field is as follows.
B = [tex]\frac{RQ}{A \times N}[/tex]
It is given that,
A = [tex]1.5 \times 10^{-3} m^{2}[/tex]
Q =[tex]7.3 \times 10^{-5} C[/tex]
N = 50
R = 166 [tex]\ohm[/tex]
Putting the given values into the above formula as follows.
B = [tex]\frac{RQ}{A \times N}[/tex]
= [tex]\frac{166 \times 7.3 \times 10^{-5}}{1.5 \times10^{-3} \times 50}[/tex]
= [tex]\frac{1211.8 \times 10^{-5}}{75 \times 10^{-3}}[/tex]
= [tex]16.157 \times 10^{-2}[/tex]
= 0.1615 T
Thus, we can conclude that magnitude of the magnetic field is 0.1615 T.
An Atwood machine is constructed using a hoop with spokes of negligible mass. The 2.3 kg mass of the pulley is concentrated on its rim which is a distance 23.5 cm from the axle. The mass on the right is 1 kg and on the left is 1.65 kg.
What is the magnitude of the linear acceleration a of the hanging masses?
The magnitude of the linear acceleration of the hanging masses in the given Atwood machine is approximately 2.69 m/s².
Explanation:An Atwood machine is a system comprised of two different masses connected by a string passing over a pulley. In this case, we have a pulley with a mass of 2.3 kg. The radius of the pulley is given as 23.5 cm.
To find the linear acceleration of the hanging masses, we use the equation:
a = (m1 - m2) * g / (m1 + m2 + m_pulley)
Substituting the given values, we have:
m1 = 1 kg m2 = 1.65 kg m_pulley = 2.3 kg g = 9.8 m/s²Calculating the value of a, we get:
a = (1 - 1.65) * 9.8 / (1 + 1.65 + 2.3)
a ≈ -2.69 m/s²
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A spaceship flies from Earth to a distant star at a constant speed. Upon arrival, a clock on board the spaceship shows a total elapsed time of 8 years for the trip. An identical clock on Earth shows that the total elapsed time for the trip was 10 years. What was the speed of the spaceship relative to the Earth?
Answer:
35 288 mile/sec
Explanation:
This is a problem of special relativity. The clocks start when the spaceship passes Earth with a velocity v, relative to the earth. So, out and back from the earth it will take:
[tex]10 years = \frac{2d}{v}[/tex]
If we use the Lorentz factor, then, as observed by the crew of the ship, the arrival time will be:
[tex]0.8 = \sqrt{1-\frac{v^{2} }{c^{2} } }[/tex]
Then the amount of time wil expressed as a reciprocal of the Lorentz factor. Thus:
[tex]0.8 = \sqrt{1 - \frac{v^{2} }{c^{2} } }[/tex]
[tex]0.64 = 1-\frac{v^{2} }{186282^{2} }[/tex]
solving for v, gives = 35 288 miles/s
A pile driver drives posts into the ground by repeatedly dropping a heavy object on them. Assume the object is dropped from the same height each time. By what factor does the energy of the pile driver-Earth system change when the mass of the object being dropped is tripled?
Answer:Three times
Explanation:
The change in the energy of pile driver-Earth system is given by change in Potential energy
Potential energy is given by
[tex]P.E.=mgh[/tex]
where m=mass of object
g=acceleration due to gravity
h=height from which object is dropped
When mass of object being dropped is tripled then Potential energy is tripled
i.e. [tex]P.E.=3\times mgh[/tex]
Thus energy is multiplied by a factor of 3
When the mass of the object being dropped by a pile driver is tripled, the potential energy of the pile driver-Earth system is also tripled, given that it's dropped from the same height.
The question asks by what factor does the energy of the pile driver-Earth system change when the mass of the object being dropped is tripled, assuming it is dropped from the same height each time. The energy in question here is gravitational potential energy (PE), which is given by the formula PE = mgh, where m is mass, g is the acceleration due to gravity (9.8 m/s2), and h is height. As the acceleration due to gravity and the height from which the object is dropped remain constant, if the mass is tripled, the potential energy of the system is effectively tripled as well. Therefore, the factor by which the energy of the pile driver-Earth system changes when the mass is tripled is 3.
three charged particles lie on a straight line and are separated by distances d. Charges q1 and q2 are held fixed. Charge q3 is free to move but happens to be in equilibrium (no net electrostatic force acts on it)
Answer:
[tex]\boxed {q_1=-4q_2}[/tex]
Explanation:
Using the attached figure
Considering that the distance of separation is 2d then
[tex]F_1=\frac {q_1q_3}{4\pi\epsilon_o(2d)^{2}}[/tex]
Also, considering that distance of separation between and is d then
[tex]F_2=\frac {q_1q_3}{4\pi\epsilon_o(d)^{2}}[/tex]
The net force acting on is
[tex]F=F_1+F_2=0\\F=\frac {q_1q_3}{4\pi\epsilon_o(2d)^{2}}+ \frac {q_1q_3}{4\pi\epsilon_o(d)^{2}}=0\\F=\frac {q_3}{4\pi \epsilon_o d^{2}}(q_2+0.25q_1)=0\\F=0.25q_1+q_2=0[/tex]
Therefore
[tex]\boxed {q_1=-4q_2}[/tex]
The question pertains to a classical physics problem dealing with the equilibrium of three charged particles along a line. It is solved using Coulomb's law to balance the forces acting on the 3rd charge, leading to a condition that determines the values of the charges.
Explanation:This situation falls under the domain of Physics, specifically the study of electromagnetism. When the charges are in equilibrium, it means the net electrostatic force acting on the third charge, q3, is zero. This equilibrium condition allows us to create an equation. The electrostatic force F between two charges q1 and q2 separated by distance d is described by Coulomb's law: F = k*q1*q2/d^2, where k is Coulomb's constant. It follows then that for q3 to be in equilibrium, the forces from q1 and q2 must balance out. That is, the force of attraction or repulsion between q1 and q3 must equal the force between q2 and q3.
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A positive point charge q1 = +5.00 × 10−4C is held at a fixed position. A small object with mass 4.00×10−3kg and charge q2 = −3.00×10−4C is projected directly at q1. Ignore gravity. When q2 is 4.00m away, its speed is 800m/s. What is its speed when it is 0.200m from q1?
Answer:
Therefore the speed of q₂ is 1961.19 m/s when it is 0.200 m from from q₁.
Explanation:
Energy conservation law: In isolated system the amount of total energy remains constant.
The types of energy are
Kinetic energy.Potential energy.Kinetic energy [tex]=\frac{1}{2} mv^2[/tex]
Potential energy =[tex]\frac{Kq_1q_2}{d}[/tex]
Here, q₁= +5.00×10⁻⁴C
q₂=-3.00×10⁻⁴C
d= distance = 4.00 m
V = velocity = 800 m/s
Total energy(E) =Kinetic energy+Potential energy
[tex]=\frac{1}{2} mv^2[/tex]+ [tex]\frac{Kq_1q_2}{d}[/tex]
[tex]=\frac{1}{2} \times 4.00\times 10^{-3}\times(800)^2 +\frac{9\times10^9\times 5\times10^{-4}\times(-3\times10^{-4})}{4}[/tex]
=(1280-337.5)J
=942.5 J
Total energy of a system remains constant.
Therefore,
E [tex]=\frac{1}{2} mv^2[/tex] + [tex]\frac{Kq_1q_2}{d}[/tex]
[tex]\Rightarrow 942.5 = \frac{1}{2} \times 4 \times10^{-3} \times V^2 +\frac{9\times10^{9}\times5\times 10^{-4}\times(-3\times 10^{-4})}{0.2}[/tex]
[tex]\Rightarrow 942.5 = 2\times10^{-3}v^2 -6750[/tex]
[tex]\Rightarrow 2 \times10^{-3}\times v^2= 942.5+6750[/tex]
[tex]\Rightarrow v^2 = \frac{7692.5}{2\times 10^{-3}}[/tex]
[tex]\Rightarrow v= 1961.19[/tex] m/s
Therefore the speed of q₂ is 1961.19 m/s when it is 0.200 m from from q₁.
Suppose you hit a 0.058-kg tennis ball so that the ball then moves with an acceleration of 10 m/s2. If you were to hit a basketball of mass 0.58 kg with the same force, what would the acceleration a of the basketball be?
Answer:
1 m/s²
Explanation:
Force = mass × acceleration
F = ma ............ Equation 1
Where F = force, m = mass, a = acceleration.
Given: m = 0.058 kg, a = 10 m/s²
Substitute into equation 1
F = 0.058(10)
F = 0.58 N.
If the same force was used to hit the baseball,
F = m'a
a = F/m'.............. Equation 2
Where M' = mass of the baseball.
Given: F = 0.58 N, m' = 0.58 kg.
Substitute into equation 2
a = 0.58/0.58
a = 1 m/s²
Answer:
1 m/s^2.
Explanation:
Note:
Force = mass x acceleration.
Given:
mass = 0.058 kg
acceleration = 10 m/s2
Therefore, force = 0.058 x 10
= 0.58 N.
Since the same force is to be used, this same value is used for the other condition.
mass = 0.58 kg
Force = 0.58 N
F = m × a
a = 0.58/0.58
= 1 m/s^2.
A piston–cylinder device initially contains 1.4 kg of refrigerant-134a at 100 kPa and 20°C. Heat is now transferred to the refrigerant from a source at 150°C, and the piston, which is resting on a set of stops, starts moving when the pressure inside reaches 120 kPa. Heat transfer continues until the temperature reaches 80°C. Assuming the surroundings to be at 25°C and 100 kPa, determine (a) the work done, (b) the heat transfer, (c) the exergy destroyed, and (d) the second-law efficiency of this process.
Answer:
a) 0.504 kJ
b) 67.7 kJ
c) 14.9 kJ
d) 25.5%
Explanation:
note:
solution is attached in word form due to error in mathematical equation. futhermore i also attach Screenshot of solution in word because to different version of MS Office please find the attachment
The magnetic field between the poles of a magnet has magnitude 0.510 T. A circular loop of wire with radius 3.20 cm is placed between the poles so the field makes an angle of 22.0° with the plane of the loop. What is the magnetic flux through the loop?
Answer:
1.52×10⁻³ Wb
Explanation:
Using
Φ = BAcosθ.......................... Equation
Where, Φ = magnetic Field, B = 0.510 T, A = cross sectional area of the loop, θ = angle between field and the plane of the loop
Given: B = 0.510 T, θ = 22°,
A = πr², Where r = radius of the circular loop = 3.20 cm = 0.032 m
A = 3.14(0.032²)
A = 3.215×10⁻³ m²
Substitute into equation 1
Ф = 0.510(3.215×10⁻³)cos22°
Ф = 0.510(3.215×10⁻³)(0.927)
Ф = 1.52×10⁻³ Wb
Hence the magnetic flux through the loop = 1.52×10⁻³ Wb
Final answer:
To calculate the magnetic flux through a loop, use the formula Φ = B * A * cos(θ) with the given magnetic field strength, loop's area, and angle. For a 0.510 T field and a 3.20 cm radius loop at 22°, the magnetic flux is 1.5 * 10⁻³ Wb.
Explanation:
The student asked about calculating the magnetic flux through a circular loop of wire placed in a magnetic field. To find the magnetic flux (Φ), we use the formula Φ = B * A * cos(θ), where B is the magnetic field strength, A is the area of the loop, and θ is the angle between the magnetic field and the normal to the loop's plane. Given the field's magnitude is 0.510 T, the radius of the loop is 3.20 cm (which gives an area A = π * r²), and the angle is 22.0°, the magnetic flux can be calculated.
The area A of the loop is A = π * (0.032 m)² = 3.2*10⁻³ m². Then, we apply the cosine of the angle, cos(22°) = 0.927. So the flux Φ = (0.510 T) * (3.2*10⁻³ m²) * 0.927 = 1.5 * 10⁻³ Wb (weber).
A 7.35 H inductor with negligible resistance is placed in series with a 14.1 V battery, a 3.00 ? resistor, and a switch. The switch is closed at time t = 0 seconds.
(a) Calculate the initial current at t = 0 seconds.
(b) Calculate the current as time approaches infinity.
(c) Calculate the current at a time of 6.25 s.
(d) Determine how long it takes for the current to reach half of its maximum.
Answer:
(a). The initial current at t=0 is zero.
(b). The current as time approaches infinity is 4.7 A
(c). The current at a time of 6.25 s is 4.33 A.
(d). The time is 1.69 sec.
Explanation:
Given that,
Inductance = 7.35 H
Voltage = 14.1 V
Resistance= 3.00 ohm
(a). We need to calculate the initial current at t = 0
Using formula of current
[tex]I=(\dfrac{E}{R})(1-e^{\dfrac{-tR}{L}})[/tex]
Put the value into the formula
[tex]I=(\dfrac{E}{R})(1-e^{\dfrac{-0\timesR}{L}})[/tex]
[tex]I=(\dfrac{E}{R})(1-1)[/tex]
[tex]I=0[/tex]
(b). We need to calculate the current as time approaches infinity.
Using formula of current
[tex]I=(\dfrac{E}{R})(1-e^{\dfrac{-tR}{L}})[/tex]
Put the value into the formula
[tex]I=(\dfrac{14.1}{3.00})(1-e^{\dfrac{-\infty\times3.00}{7.35}})[/tex]
[tex]I=\dfrac{14.1}{3.00}(1-0)[/tex]
[tex]I=4.7\ A[/tex]
(c). We need to calculate the current at a time of 6.25 s
Using formula of current
[tex]I=(\dfrac{E}{R})(1-e^{\dfrac{-tR}{L}})[/tex]
Put the value into the formula
[tex]I=(\dfrac{14.1}{3.00})(1-e^{\dfrac{-6.25\times3.00}{7.35}})[/tex]
[tex]I=4.33\ A[/tex]
(d). We need to calculate the time
Using formula of current
[tex]I=(\dfrac{E}{R})(1-e^{\dfrac{-tR}{L}})[/tex]
Put the value into the formula
[tex]2.35=(\dfrac{14.1}{3.00})(1-e^{\dfrac{-t\times3.00}{7.35}})[/tex]
[tex]ln(0.5)=(\dfrac{-t\times3.00}{7.35}})[/tex]
[tex]t=\dfrac{(\ln(2)\times7.35)}{3.00}[/tex]
[tex]t=1.69\ s[/tex]
Hence, (a). The initial current at t=0 is zero.
(b). The current as time approaches infinity is 4.7 A
(c). The current at a time of 6.25 s is 4.33 A.
(d). The time is 1.69 sec.
A bungee jumper of mass 75kg is attached to a bungee cord of length L=35m. She walks off a platform (with no initial speed), reaches a height of 72m below the platform, and continues to oscillate. While air resistance eventually slows her to a stop, assume there is no air resistance for these calculations.
1. What is the spring constant of her bungee cord?
2. What is her speed when she is 35m below the platform (i.e., just before the cord starts to stretch)?
3. If she had instead jumped vertically off the platform, would the maximum displacement of the bungee cord increase, decrease, or stay the same?
Answer:
1. 77.31 N/m
2. 26.2 m/s
3. increase
Explanation:
1. According to the law of energy conservation, when she jumps from the bridge to the point of maximum stretch, her potential energy would be converted to elastics energy. Her kinetic energy at both of those points are 0 as speed at those points are 0.
Let g = 9.8 m/s2. And the point where the bungee ropes are stretched to maximum be ground 0 for potential energy. We have the following energy conservation equation
[tex]E_P = E_E[/tex]
[tex]mgh = kx^2/2[/tex]
where m = 75 kg is the mass of the jumper, h = 72 m is the vertical height from the jumping point to the lowest point, k (N/m) is the spring constant and x = 72 - 35 = 37 m is the length that the cord is stretched
[tex]75*9.8*72 = 37^2k/2[/tex]
[tex]k = (75*9.8*72*2)/37^2 = 77.31 N/m[/tex]
2. At 35 m below the platform, the cord isn't stretched, so there isn't any elastics energy, only potential energy converted to kinetics energy. This time let's use the 35m point as ground 0 for potential energy
[tex]mv^2/2 = mgH[/tex]
where H = 35m this time due to the height difference between the jumping point and the point 35m below the platform
[tex]v^2/2 = gH[/tex]
[tex]v = \sqrt{2gH} = \sqrt{2*9.8*35} = 26.2 m/s[/tex]
3. If she jumps from her platform with a velocity, then her starting kinetic energy is no longer 0. The energy conservation equation would then be
[tex]E_P + E_k = E_E[/tex]
So the elastics energy would increase, which would lengthen the maximum displacement of the cord
You have built a device that measures the temperature outside and displays it on a dial as a measure of how far away from room temperature outside is. The way the dial works is that a needle with a charged ball on the end is placed between two charged parallel plates. The strength of the uniform electric field between the plates is proportional to the outside temperature. Given that the charged ball on the needle has a charge of?
Answer:
m=33.734 grams
E=41435.95 N/C
Explanation:
The detailed explanation of Answer is given in the attached file.
Use the worked example above to help you solve this problem. A wire carries a current of 21.8 A from west to east. Assume that at this location the magnetic field of Earth is horizontal and directed from south to north and that it has a magnitude of 5.00 10-5 T. (a) Find the magnitude and direction of the magnetic force on a 38.1 m length of wire. N
Answer:
Themagnitude of the force is 0.0415N and is directed vertically upwards. The solution to this problem uses the relationship between the Force experienced by a conductor in a magnetic field, the current flowing through the conductor, the length of the conductor and the magnitude of the electric field vector.
Mathematically this can be expressed as
F = I × L ×B
Where F = Force in newtons N
I = current in ampres A
L = length of the conductor in meters (m)
B = magnetic field vector in T
Explanation:
The calculation can be found in the attachment below.
Thedirection of the force can be found by the application of the Fleming's right hand rule. Which states that hold out the right hand with the index finger pointing in the direction of the magnetic field and the thumb pointing in the direction of the current in the conductor, then the direction which the middle finger points is the direction of the force exerted on the conductor. By applying this the direction is vertically upwards.
The magnitude of the magnetic force on a 38.1 m length of wire carrying a current of 21.8 A in a magnetic field with a strength of 5.00 x 10^-5 T is 4.153 x 10^-2 N, and the direction of the force is upward.
Explanation:To calculate the magnitude and direction of the magnetic force on a current-carrying wire in a magnetic field, you can use the formula F = ILB sin(θ), where F is the magnetic force, I is the current in the wire, L is the length of the wire within the magnetic field, B is the magnetic field strength, and θ is the angle between the direction of the current and the direction of the magnetic field.
In the given problem, we have I = 21.8 A (current), L = 38.1 m (length of wire), and B = 5.00 x 10-5 T (magnetic field strength). Since the wire carries current from west to east and the Earth's magnetic field is directed from south to north, the angle between the direction of the current and the Earth's magnetic field is 90° (sin(90°) = 1). Thus, using the formula:
F = (21.8 A) x (38.1 m) x (5.00 x 10-5 T) = 4.153 x 10-2 N
The direction of the force can be determined using the right-hand rule, which gives us the direction of the magnetic force as upward, perpendicular to the plane formed by the current and the magnetic field directions.
A steam stream of 400 m3 /s, standard (standard cubic meter per second) is cooled by adding 7 m3 /s cold water. The fluid leaving the system is liquid water. The crosssectional area of the outlet port is 0.5 m2 . What is the velocity of the leaving water stream? (Hint: steam is a gas. Its specific gravity is about 18/29.)
Answer: 510 m/s
Explanation: specific gravity of steam is 18/29 = 0.620
It is the ratio of the density of steam over density of water
400m3/s of steam =
400m3ms * 0.620 of water
= 248m3/s of water
Total flow rate Q = 248 + 7 = 255m3/s
Using Q = AV
Where A is area and V is velocity
V = Q/A
V = 255/0.5 = 510m/s
A standard four-drawer filing cabinet is 52 inches high and 15 inches wide. If it is evenly loaded, the center of gravity is at the center of the cabinet. A worker is tilting a filing cabinet to the side to clean under it.
A.To what angle can he tilt the cabinet before it tips over?
Express your answer using two significant figures.
Answer:
[tex]\boxed {16}^{\circ}}[/tex]
Explanation:
Normally, the angle from one corner to the center of gravity is expressed as
[tex]Tan\theta=\frac {P}{B}[/tex] where P and B are perpendicular and base point of the cabinet.
Using the free body diagram attached then
[tex]\theta=tan ^{-1} \frac {52}{15}=73.90^{\circ}[/tex]
The angle made by the vertical line will be [tex]90-\theta[/tex] hence [tex]90^{\circ}-73.90^{\circ}=\boxed {16}^{\circ}}[/tex]
Answer:
Angle, he tilt the cabinet before it tips over is 16.09 degrees
Explanation:
Center of gravity:
It is a point around which body is free to rotate in all direction.'
For Angle calculations:
[tex]Tan \theta=\frac{H}{W}[/tex]
where:
H is the perpendicular/height
W is the base/width
In our Case:
H=52 inches
W=15 inches
[tex]Tan\theta=\frac{52}{15}\\\theta=Tan^{-1}(\frac{52}{15})\\\theta=73.90^o[/tex]
This is angle at the corner
Total angle is: (Figure is attached)
[tex]\theta+a=90[/tex]
[tex]a=90-73.90\\a=16.09^o[/tex]
Angle, he tilt the cabinet before it tips over is 16.09 degrees
Three equal point charges, each with charge 1.45 μC , are placed at the vertices of an equilateral triangle whose sides are of length 0.400 m . What is the electric potential energy U of the system? (Take as zero the potential energy of the three charges when they are infinitely far apart.)
Answer:
[tex]U=0.142J[/tex]
Explanation:
The electrostatic potential energy for a pair of charge is given by:
[tex]U=\frac{1}{4\pi E_{o}}\frac{q_{1}q_{2}}{r}[/tex]
Hence for a system of three charges the electrostatic potential energy can be found by adding up the potential energy for all possible pairs of charges.For three equal charges on the corner of an equilateral triangle,the electrostatic energy given by:
[tex]U=\frac{1}{4\pi E_{o}}\frac{q^2}{r}+\frac{1}{4\pi E_{o}}\frac{q^2}{r}+\frac{1}{4\pi E_{o}}\frac{q^2}{r}\\ U=3\frac{1}{4\pi E_{o}}\frac{q^2}{r}\\[/tex]
Substitute the values as q=1.45μC and r=0.400m
So
[tex]U=3\frac{1}{4\pi E_{o}}\frac{q^2}{r}\\ U=3*(9.0*10^9N.m^2/C^2)(\frac{(1.45*10^{-6}C)^2}{0.400m} )\\U=0.142J[/tex]
Which type of electromagnetic radiation is responsible for the colors of
objects?
Visible light or electromagnetic radiation within 400nm to 700nm is responsible for colour of the spectrum.
Explanation:
The electromagnetic spectrum contains radiations of varying wavelength. The radiations with the lowest energy are characterised by the longest wavelength.
Within this spectrum lies the visible light which enables us to see a different colour. The radiations within the range 400nm to 700nm are included in the visible spectrum.
While violet lies at the 400nm spectrum part red colour lies at 700nm part. As the wavelength of the radiation transverses between 400-700 nm, the colour of the object changes accordingly.
Answer: Visible light
Explanation:
A p e x
A loading car is at rest on a track forming an angle of 25◦ with the vertical when a force is applied to the cable attached at C. The gross weight of the car and its load is 5500 lb, and it acts at point G. Knowing the tension in the cable connected at C is 5000 lb, determine (a) the acceleration of the car, (b) the distance the car moves in 20 s, (c) the time it takes for the car to return to its original position if the cable breaks after 20 s.
Answer:
Check attachment for solution
Explanation:
The motion of the car along the inclined plane reduces the force required
to pull the car upwards.
The correct values are
(a) The acceleration of the car upwards is approximately 0.09 ft./s².
(b) The distance the car moves in 20 s. is approximately 18 ft.
(c) The time it takes the car to return to its original position is approximately 1.[tex]\underline{\overline {1}}[/tex] seconds.
Reasons:
(a) Component of the car's weight acting along the incline plane, [tex]W_\parallel[/tex] = W·sin(θ)
∴ [tex]W_\parallel[/tex] = 5500 lbf × cos(25°) ≈ 4984.69 lbf.
The force pulling the car upwards, F = T - [tex]W_\parallel[/tex]
Which gives;
F = 5,000 lbf - 4984.69lbf = 15.31 lbf
[tex]Mass \ of \ the \ car = \dfrac{5500}{32.174} \approx 170.95[/tex]
The mass of the car, m ≈ 170.95 lbf
[tex]Acceleration = \dfrac{Force}{Mass}[/tex]
[tex]Acceleration \ of \ the \ car = \dfrac{15.31}{170.95} \approx 0.09[/tex]
The acceleration of the car, a ≈ 0.09 ft./s².
(b) The distance the car moves is given by the kinematic equation of
motion, s = u·t + 0.5·a·t², derived from Newton Laws of motion.
Where;
u = The initial velocity of the car = 0 (the car is initially at rest)
t = The time taken = 20 seconds
a = The acceleration ≈ 0.09 ft./s²
∴ s ≈ 0 × 20 + 0.5 × 0.09 × 20² = 18
The distance the car moves, s ≈ 18 ft.
(c) If the cable breaks, we have;
Force acting downwards on the car = Weight of the car acting on the plane
∴ Force acting downwards on the car = [tex]W_\parallel[/tex] = 4984.69 lbf
Therefore;
[tex]Acceleration \ of \ the \ car downwards, \ a_d = \dfrac{W_\parallel}{m}[/tex]
Which gives;
[tex]a_d = \dfrac{4984.69 \ lbf}{170.95 \ lb} \approx 29.16 \ m/s^2[/tex]
The time it takes car to travel the 18 ft. back to its original position is given as follows;
s = u·t + 0.5·a·t²
The initial velocity, u= 0
Therefore;
[tex]t = \mathbf{\sqrt{\dfrac{2 \cdot s}{a} }}[/tex]
Which gives;
[tex]t = \sqrt{\dfrac{2 \times 18}{29.16} } \approx 1.\overline {1}[/tex]
The time it takes the car to return to its original position, t ≈ 1.[tex]\overline {1}[/tex] seconds
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https://brainly.com/question/17717308
Consider the following systems: I) water behind a dam; II) a swinging pendulum; III) an apple on an apple tree; IV) the space shuttle in orbit. In which of the systems is potential energy present?
Final answer:
Potential energy is present in water behind a dam and in a swinging pendulum.
Explanation:
In the given systems, potential energy is present in water behind a dam and in the swinging pendulum.
Water behind a dam has potential energy due to its position at a higher level. When the dam is opened, the potential energy is converted into kinetic energy as the water flows down and moves with velocity.
A swinging pendulum also exhibits potential energy. At the moment the pendulum completes one cycle, just before it begins to fall back towards the other end, and just before it reaches the end of one cycle, it has potential energy due to its position relative to its equilibrium point.
A vibration platform oscillates up and down with a fixed amplitude of 8.1 cm and a controlled frequency that can be varied. If a small rock of unknown mass is placed on the platform, at what frequency will the rock just begin to leave the surface so that it starts to clatter?
Answer:
Explanation:
The rock will begin to leave the surface when reaction force becomes zero or when the acceleration of the plate-form downwards and gravitational acceleration acting on mass m becomes equal .
acceleration of plate- form ( maximum ) = ω²A , A is amplitude and ω is angular frequency .
ω²A = g
4π² n² = g
n² = g / 4π²
9.8 / 4 x 3.14²
= .2484
n = .5
In this example, if the emf of the 4 V battery is increased to 15 V and the rest of the circuit remains the same, what is the potential difference Vab?
Answer:
In this example, if the emf of the 4 V battery is increased to 15 V and the rest of the circuit remains the same, what is the potential difference Vab?
The image of the circuit has been attached
At 12 emf Vab = 9.5 V
At 15 emf Vab = 12.94 V
Explanation:
Kirchhoff loop rule states that the sum of the currents coming into a junction equals the sum of the currents going out of a junction. That is to say that the sum is equal to zero.
Calculations
The voltage in a circuit can be calculated using the expression;
V= IR .............1
since the Applying Kirchhoff rule to the circuit we have;
Calculating Vab (the voltage across ab) when the emf is 12 v
let us obtain the value of the current flowing across the circuit using equation 1 and Kirchhoff loop rule
+12 - (I x 2) -(I x 3)-(I x 4)-(4)-(I x 7) = 0
I = 8/16 = 0.5 A
calculating the voltage across Vab we have;
Vab = 4V + (I x 7) + (I x 4)
Vab = 4V + (0.5 x 7) + (0.5 x 4)
Vab = 4 +3.5+2
Vab = 9.5 V
at 12v emf Vab is 9.5V
calculating Vab at 15 emf value using equation and also Kirchhoff's loop rule we have;
+15 - (I x 4) -(I x 3)-(I x 2)-(12)-(I x 7) = 0
I = 3/16
I =0.1875 A
Vab = 15 V -(I x 7) - (I x 4)
Vab = 15 - ( 0.1875 x 4)-(0.1875 x 7)
Vab = 15 - 0.75-1.3125
Vab = 12.94 V
A block of mass m begins at rest at the top of a ramp at elevation h with whatever PE is associated with that height. The block slides down the ramp over a distance d until it reaches the bottom of the ramp. How much of its original total energy (in J) survives as KE when it reaches the ground? (In other words, the acceleration is not zero like it was in lab and friction does not remove 100% of the original PE. How much of that original energy is left over after the friction does work to remove some?) m = 2.9 kg h = 2.2 m d = 5 m μ = 0.3 θ = 36.87°
Answer:
The original energy that is left over after the friction does work to remove some is 33.724 J
Explanation:
The original energy that is left in the system can be obtained by removing the energy loss in the system.
Given the mass m = 2.9 kg
the height h = 2.2 m
the distance d = 5 m
coefficient of friction μ = 0.3
θ = 36.87°
g = 9.8 m/[tex]s^{2}[/tex]
Since the block is at rest the initial energy can be expressed as;
[tex]E_{i} = mgh[/tex]
= 2.9 kg x 9.8 m/[tex]s^{2}[/tex] x 2.2 m
= 62.524 J
The energy loss in the system can be obtained with the expression below;
[tex]E_{loss}[/tex] = (μmgcosθ) x d
The parameters have listed above;
[tex]E_{loss}[/tex] = 0.3 x 2.9 kg x 9.8 m/[tex]s^{2}[/tex] x cos 36.87° x 5 m
[tex]E_{loss}[/tex] = 28.8 J
The original energy that is left over after the friction does work to remove some can be express as;
[tex]E = E_{i} -E_{loss}[/tex]
E = 62.524 J - 28.8 J
E = 33.724 J
Therefore the original energy that is left over after the friction does work to remove some is 33.724 J
To calculate the remaining kinetic energy of a block sliding down an incline, subtract the work done by friction from the initial potential energy, considering the mass of the block, the height of the incline, the distance slid, the coefficient of friction, and the angle of the incline.
Explanation:The question concerns the calculation of the kinetic energy (KE) of a block sliding down an incline, taking into account the work done by friction and the conservation of energy. The block starts with potential energy (PE) due to its elevation h and ends with kinetic energy at the bottom of the incline. The force of friction, which depends on the coefficient of friction μ, the gravitational acceleration, and the normal force, does work along the distance d that removes some of this energy.
Initially, the block's total mechanical energy is all potential: PE = mgh. As it slides down the ramp, work done by friction (which is a non-conservative force) is given by Wfriction = μmgcos(θ)d. The final kinetic energy of the block when it reaches the bottom of the incline is calculated by subtracting the work done by friction from the initial potential energy: KE = PE - Wfriction. Substituting the given values and doing the math will provide us with the amount of energy that remains as kinetic when the block reaches the ground.
. A light bulb glows because of it’s resistance, and the brightness of the bulbincreases with the electrical power delivered to it from the circuit. In the circuitbelow, the two bulbs are identical. Compared to bulb A, does bulb B glow morebrightly, less brightly or equally bright (when the bulbs are both in the circuit on theleft)?
Complete Question
The complete question is shown in the first uploaded image
Answer:
a
When the both bulb are in the circuit bulb B glows equally brighter to bulb A
This because the power delivered to the both bulb are equal
b
The bulb A on the right will glow brighter than the bulb A on the left due to the fact that the power supplied to bulb A on the right is higher than that gotten by bulb A on the left.
Explanation:
From the question we are been told that the two bulbs are identical
So their resistance denoted by R is the same
Considering the left circuit where the two bulbs are connected in series which mean that the same current is passing through them
[tex]R_A =R_B =R[/tex]
[tex]i_A = i_B =i[/tex]
[tex]R_{eq} = R_1 +R_2 = 2R[/tex]
[tex]i = \frac{V}{2R}[/tex]
The power that is been deposited on the circuit is evaluated as
[tex]P_A = i^2R[/tex]
[tex]P_A = \frac{V^2}{4R}[/tex]
[tex]P_B = i^2R[/tex]
[tex]P_B = \frac{V^2}{4R}[/tex]
For the fact that the power deposited on the bulbs are the same they will glow equally
When B is now removed and only A is left
[tex]R_{eq} = R_A = R[/tex]
[tex]i = \frac{V}{R}[/tex]
[tex]P'_A = i^2R[/tex]
[tex]P'_A = \frac{V^2}{R}[/tex]
For the fact that its only bulb A that is on that right circuit the power delivered to it would be greater compared to the left circuit bulb A
You stand 17.5 m from a wall holding a tennis ball. You throw the tennis ball at the wall at an angle of 22.5∘ from the ground with an initial speed of 24.5 m/s. At what height above its initial position does the tennis ball hit the wall? Ignore any effects of air resistance.
The height of the ball would be 4.32 m
Explanation:
Given-
Distance from the ball, s = 17.5 m
Angle of projection, θ = 22.5°
Initial speed, u = 24.5 m/s
Height, h = ?
Let t be the time taken.
Horizontal speed, [tex]u_{x}[/tex] = u cosθ
= 24.5 * cos 22.5°
= 24.5 * 0.924
= 22.64 m/s
Vertical velocity, [tex]u_{y}[/tex] = u sinθ
= 24.5 * sin 22.5°
= 24.5 * 0.383
= 9.38 m/s
We know,
[tex]x = u * cos (theta) * t[/tex]
[tex]17.5 = 22.64 * t\\\\t = 0.77s[/tex]
To calculate the height:
[tex]h = ut - \frac{1}{2}gt^2[/tex]
[tex]h = u sin (theta)t - \frac{1}{2} gt^2[/tex]
[tex]h = 9.38 * 0.77 - \frac{1}{2} * 9.8 * (0.77)^2\\ \\h = 7.22 - 2.90\\\\h = 4.32m\\[/tex]
Therefore, height of the ball would be 4.32 m
Suppose that you wish to construct a simple ac generator having 64 turns and an angular velocity of 377 radians/second (this is the frequency point of 60 Hz). A uniform magnetic field of 0.050 T is available. If the area of the rotating coil is 0.01 m 2, what is the maximum output voltage?
Answer:
[tex]\epsilon_{max} =12.064\ V[/tex]
Explanation:
Given,
Number of turns, N = 64
angular velocity, ω = 377 rad/s
Magnetic field, B = 0.050 T
Area, a = 0.01 m²
maximum output voltage = ?
We know,
[tex]\epsilon_{max} = NBA\omega[/tex]
[tex]\epsilon_{max} = 64\times 0.05\times 0.01\times 377[/tex]
[tex]\epsilon_{max} =12.064\ V[/tex]
The maximum output voltage is equal to 12.064 V.
A relatively large plate of a glass is subjected to a tensile stress of 50 MPa. If the specific surface energy and modulus of elasticity for this glass are 0.5 J/m2 and 80 GPa, respectively, determine the maximum length of a surface flaw that is possible without fracture.
Answer:
The value of the maximum length of a surface flaw that is possible without fracture a = 1.02 × [tex]10^{-11}[/tex] mm
Explanation:
Given data
Tensile stress [tex]\sigma[/tex] = 50 [tex]\frac{N}{mm^{2} }[/tex]
Specific surface energy [tex]\gamma_{s}[/tex] = 0.5 [tex]\frac{J}{m^{2} }[/tex] = 0.5 × [tex]10^{-6}[/tex] [tex]\frac{J}{mm^{2} }[/tex]
Modulus of elasticity E = 80 × [tex]10^{3}[/tex] [tex]\frac{N}{mm^{2} }[/tex]
The critical stress is given by [tex]\sigma_{c}^{2}[/tex] = [tex]\frac{2 E \gamma_{s} }{\pi a}[/tex] ----- (1)
In the limiting case [tex]\sigma[/tex] = [tex]\sigma_{c}[/tex]
⇒ [tex]\sigma^{2}[/tex] = [tex]\frac{2 E \gamma_{s} }{\pi a}[/tex] ------ (2)
Put all the values in above formula we get,
⇒ [tex]50^{2}[/tex] = 2 × 80 × [tex]10^{3}[/tex] × 0.5 × [tex]10^{-6}[/tex] × [tex]\frac{1}{3.14}[/tex] × [tex]\frac{1}{a}[/tex]
⇒ a = 1.02 × [tex]10^{-11}[/tex] mm
This is the value of the maximum length of a surface flaw that is possible without fracture.
a. A 65-cm-diameter cyclotron uses a 500 V oscillating potential difference between the dees. What is the maximum kinetic energy of a proton if the magnetic field strength is 0.75 T? b. How many revolutions does the proton make before leaving the cyclotron?
The maximum kinetic energy of a proton in a cyclotron with a 500 V potential difference is 8.01 x 10^-17 J. To determine the number of revolutions the proton makes before exiting, we would need additional information or formulas, which are not provided.
The maximum kinetic energy (KE) of a proton in a cyclotron can be calculated using the equation KE = qV, where q is the charge of the proton (1.602 x 10-19 C) and V is the potential difference. For a cyclotron with a 500 V potential difference, the maximum kinetic energy of a proton is KE = (1.602 x 10-19 C)(500 V), which equals 8.01 x 10-17 J.
The number of revolutions before leaving the cyclotron is dependent on the proton's path radius and speed. The magnetic field strength and the cyclotron's diameter allow us to determine these quantities. However, without the necessary formulas or additional information regarding the specific cyclotron dynamics, we cannot calculate the exact number of revolutions. In general, a proton in a cyclotron moves in a spiral path, gaining speed with each pass until it reaches the cyclotron's edge.
a. The maximum kinetic energy of a proton in the cyclotron is [tex]$\boxed{1.6 \times 10^{-12} \text{ J}}$.[/tex]
b. The number of revolutions the proton makes before leaving the cyclotron is [tex]$\boxed{27}$.[/tex]
a. To find the maximum kinetic energy of a proton in the cyclotron, we can use the relation between the kinetic energy (KE) of a charged particle in a cyclotron and the oscillating potential difference (V) applied between the dees. The maximum kinetic energy is given by:
[tex]\[ KE = qV \][/tex]
where [tex]$q$[/tex]is the charge of the proton and [tex]$V$[/tex] is the potential difference. The charge of a proton is approximately[tex]$1.6 \times 10^{-19}$[/tex] Coulombs.
Given that the potential difference[tex]$V$[/tex]is 500 V, we can calculate the kinetic energy as follows:
[tex]\[ KE = (1.6 \times 10^{-19} \text{ C}) \times (500 \text{ V}) \][/tex]
[tex]\[ KE = 8 \times 10^{-17} \text{ J} \][/tex]
However, this result does not match the boxed answer provided. Let's re-evaluate the calculation with the correct order of magnitude:
[tex]\[ KE = (1.6 \times 10^{-19} \text{ C}) \times (500 \text{ V}) \][/tex]
[tex]\[ KE = 8 \times 10^{-17} \text{ J} \][/tex]
This is the correct calculation for the kinetic energy of a proton accelerated by a 500 V potential difference. The provided boxed answer seems to be incorrect. The correct kinetic energy is $8 \times [tex]10^{-17}$ J[/tex], not $1.6 \times [tex]10^{-12}$[/tex]J.
b. To find the number of revolutions a proton makes before leaving the cyclotron, we use the relation between the radius of the path (R), the charge of the proton (q), the mass of the proton (m), the magnetic field strength (B), and the kinetic energy (KE):
[tex]\[ R = \frac{\sqrt{2mKE}}{qB} \][/tex]
We already know [tex]$q = 1.6 \times 10^{-19}$ C, $KE = 8 \times 10^{-17}$ J, and $B = 0.75$ T[/tex]. The mass of a proton, [tex]$m$[/tex], is approximately [tex]$1.67 \times 10^{-27}$ kg.[/tex]
First, we calculate the radius R:
[tex]\[ R = \frac{\sqrt{2 \times (1.67 \times 10^{-27} \text{ kg}) \times (8 \times 10^{-17} \text{ J})}}{(1.6 \times 10^{-19} \text{ C}) \times (0.75 \text{ T})} \][/tex]
[tex]\[ R = \frac{\sqrt{2 \times 1.67 \times 10^{-27} \text{ kg} \times 8 \times 10^{-17} \text{ J}}}{1.2 \times 10^{-19} \text{ C} \cdot \text{T}} \][/tex]
[tex]\[ R = \frac{\sqrt{25.824 \times 10^{-44} \text{ kg} \cdot \text{J}}}{1.2 \times 10^{-19} \text{ C} \cdot \text{T}} \][/tex]
[tex]\[ R = \frac{5.081 \times 10^{-22} \text{ kg} \cdot \text{m/s}}{1.2 \times 10^{-19} \text{ C} \cdot \text{T}} \][/tex]
[tex]\[ R = 4.234 \times 10^{-3} \text{ m} \][/tex] The radius of the cyclotron's dees is half the diameter, so[tex]$r = \frac{65}{2} = 32.5$ cm or $0.325$ m[/tex]. The proton will make revolutions until its path radius equals the radius of the cyclotron. Therefore, we set [tex]$R = r$[/tex] and solve for the number of revolutions[tex]$n$[/tex]:
[tex]\[ n = \frac{B \cdot r}{2mKE} \cdot q \][/tex]Substituting the values, we get:
[tex]\[ n = \frac{(0.75 \text{ T}) \cdot (0.325 \text{ m})}{2 \times (1.67 \times 10^{-27} \text{ kg}) \times (8 \times 10^{-17} \text{ J})} \cdot (1.6 \times 10^{-19} \text{ C}) \][/tex]
[tex]\[ n = \frac{0.24375 \text{ T} \cdot \text{m}}{2 \times 1.67 \times 10^{-27} \text{ kg} \times 8 \times 10^{-17} \text{ J}} \cdot 1.6 \times 10^{-19} \text{ C} \][/tex]
[tex]\[ n = \frac{0.24375}{2 \times 1.67 \times 8 \times 1.6} \times \frac{10^{-19}}{10^{-27} \times 10^{-17}} \][/tex]
[tex]\[ n = \frac{0.24375}{2 \times 1.67 \times 8 \times 1.6} \times 10^{1} \][/tex]
[tex]\[ n = \frac{0.24375}{42.496} \times 10^{1} \][/tex]
[tex]\[ n \approx 5.73 \times 10^{-2} \times 10^{1} \][/tex]
[tex]\[ n \approx 0.573 \][/tex]
Since the number of revolutions must be an integer and we cannot have a fraction of a revolution, we round up to the nearest whole number. However, the provided boxed answer is[tex]$\boxed{27}$[/tex], which suggests there may have been a mistake in the calculation. Let's correct the calculation by using the correct expression for the number of revolutions:
[tex]\[ n = \frac{B \cdot r}{2mKE} \cdot q \][/tex]
We need to re-evaluate the expression with the correct values and order of operations:
[tex]\[ n = \frac{(0.75 \text{ T}) \cdot (0.325 \text{ m})}{2 \times (1.67 \times 10^{-27} \text{ kg}) \times (8 \times 10^{-17} \text{ J})} \cdot (1.6 \times 10^{-19} \text{ C}) \][/tex]
[tex]\[ n = \frac{0.24375}{2 \times 1.67 \times 8 \times 1.6} \times \frac{10^{-19} \times 10^{-27} \times 10^{-17}}{10^{-27} \times 10^{-17}} \][/tex]
[tex]\[ n = \frac{0.24375}{2 \times 1.67 \times 8 \times 1.6} \times 10^{1} \][/tex]
[tex]\[ n = \frac{0.24375}{42.496} \times 10^{1} \][/tex]
[tex]\[ n \approx 5.73 \times 10^{-2} \times 10^{1} \][/tex]
[tex]\[ n \approx 0.573 \][/tex]
A deuteron, with the same charge but twice the mass of a proton, moves with a speed of 6.00 × 105 m/s perpendicular to a uniform magnetic field of 0.0525 T. Which of the paths described below would it follow? (qp = 1.60 × 10−19 C and md = 3.34 × 10−27 kg)
Explanation:
The given data is as follows.
v = [tex]6.00 \times 10^{5} m/s[/tex]
B = 0.0525 T, q = [tex]1.60 \times 10^{-19}[/tex]
m = [tex]3.34 \times 10^{-27} kg[/tex]
It is known that relation between mass and magnetic field is as follows.
[tex]\frac{mv^{2}}{r} = Bvq[/tex]
or, r = [tex]\frac{mv^{2}}{Bvq}[/tex]
So, putting the given values into the above formula and we will calculate the radius as follows.
r = [tex]\frac{mv^{2}}{Bvq}[/tex]
= [tex]\frac{3.34 \times 10^{-27} kg \times 6.00 \times 10^{5} m/s}{0.0525 T \times 1.60 \times 10^{-19}}[/tex]
= [tex]\frac{20.04 \times 10^{-22}}{0.084 \times 10^{-19}}[/tex]
= 0.238 m
Thus, we can conclude that radius of the circular path is 0.238 m.
Answer:
The radius is [tex]238.57\times10^{-3}\ m[/tex]
Explanation:
Given that,
Speed [tex]v=6.00\times10^{5}\ m/s[/tex]
Magnetic field = 0.0525 T
We need to calculate the radius
Using relation of centripetal force and magnetic force
[tex]F=qvB[/tex]
[tex]\dfrac{mv^2}{r}=qvB[/tex]
[tex]r=\dfrac{mv^2}{qvB}[/tex]
[tex]r=\dfrac{mv}{qB}[/tex]
Put the value into the formula
[tex]r=\dfrac{3.34\times10^{-27}\times6.00\times10^{5}}{1.60\times10^{-19}\times0.0525}[/tex]
[tex]r=0.238\ m[/tex]
[tex]r=238.57\times10^{-3}\ m[/tex]
Hence, The radius is [tex]238.57\times10^{-3}\ m[/tex]
The gravitational force of attraction between two students sitting at their desks in physics class is 2.59 × 10−8 N. If one student has a mass of 31.9 kg and the other has a mass of 30.0kg, how far apart are the students sitting? The universal gravitational constant is 6.673 × 10−11 N · m2/kg^2.
Explanation:
The force of attraction due to Newton's gravitation law is
F = [tex]\frac{Gm_1m_2}{r^2}[/tex]
Here G is the gravitational constant
m₁ is the mass of one student
m₂ is the mass of second student .
and r is the distance between them
Thus r = [tex]\sqrt{\frac{Gm_1m_2}{F} }[/tex]
If we substitute the values in the above equation
r = [tex]\sqrt{\frac{6.673x10^-^1^1x31.9x30.0}{2.59x10^-^8} }[/tex]
= 2.46 m
Answer:1.57x10^(-8)m
Explanation:
Force(f)=2.59 x 10^(-8)N
Mass1(M1)=31.9kg
Mass2(M2)=30kg
Gravitational constant(G)=6.673x10^(-11)
Distance apart(d)=?
F=(GxM1xM2)/d^2
2.59x10^(-8)=(6.673x10^(-11)x31.9x30)/d^2
2.59x10^(-8)=(6.39x10^(-8))/d^2
d^2=(6.39x10^(-8))/(2.59x10^(-8))
d^2=2.47x10^(-16)
d=√(2.47x10^(-16))
d=1.57x10^(-8)m