Given:
The potential of point A is
[tex]V_A=369.53\text{ V}[/tex]The potential of point B is
[tex]V_B=\text{ 217.85 V}[/tex]The work done is W = 0.55 milli Joule.
To find the value of charge in micro Coulomb.
Explanation:
The charge can be calculated as
[tex]\begin{gathered} Q=\frac{W}{V_B-V_A} \\ =\frac{0.55\times10^3\text{ J}}{217.85-369.53} \\ =-3.63\text{ C} \\ =\text{ -3.63 C}\times\frac{10^6\text{ }\mu C}{1\text{ C}} \\ =-3.63\times10^6\text{ }\mu C\text{ } \end{gathered}[/tex]A student pushes a baseball of m = 0.13 kg down onto the top of a vertical spring that has its lower end fixed to a table, compressing the spring a distance of d = 0.18 meters from its original equilibrium point. The spring constant of the spring is k = 970 N/m. Let the gravitational potential energy be zero at the position of the baseball in the compressed spring.
A. What is the maximum height, h, in meters, that the ball reaches above the equilibrium point?
B. What is the ball’s velocity, in meters per second, at half of the maximum height relative to the equilibrium point?
The maximum height reached by the baseball above the equilibrium point is 12.14 m.
The ball’s velocity, in meters per second, at half of the maximum height relative to the equilibrium point is 458.8 m/s.
What is the maximum height attained by the ball?The maximum height reached by the ball is determined as follows:
Data given:
compression of the spring is d = 0.18 m.mass of the baseball is m = 0.13 kgthe spring constant, K = 970 N/mThe gravitational potential energy at the compressed position is zero.
Based on the law of conservation of energy, the total energy of the system is conserved.
Let the final height from the bottom be h
m * g* h = ¹/₂ * K * d²
h = (k * d²) / (2 * m * g)
h = 970 * 0.18² / (2 * 0.13 * 9.81)
h = 12.32 m
Height above the equilibrium point = 12.32 - 0.18
Height above the equilibrium point = 12.14 m
Velocity is calculated1 as follows:
Half of the maximum height relative to the equilibrium point = 12.14/2 + (0.18) = 6.25 m
¹/₂ * m * v² = ¹/₂ * K * d²
m * v² = K * d²
v = √(K * d² / m)
v = √(970 * 6.25² / 0.13)
v = 458.8 m/s
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Jacob Grena raises a spoon 0.210 m above a table . If the spoon and its contents have a mass of 30.0 g, what is the gravitational potential energy associated with the spoon at that height relative to the table's surface?
Explanation:
The spoon is raised so it gains Gravitational potential energy. Formula to find Gravitational potential energy;
Gravitational potential energy = mass × Gravitational field strength × height of the body from the surface (table in this scenario)
In symbols; E = m×g×h
Substitute values:
m = 30g = 0.03kg (don't forget to convert grams to kg)
g = 10N/kg
h = 0.210 m
So it's;
0.03kg × 10N/kg × 0.210m = 0.063 Joules
SI Unit of energy is joules
Two traveling pulses on a rope move toward each other at a speed of 1.0 m/s. The waves have the same amplitude. The drawing shows the position of the waves at time t = 0 s. Which one of the following drawings depicts the waves on the rope at t = 4.0 s?
ANSWER:
STEP-BY-STEP EXPLANATION:
We know the graph when t = 0. Therefore, at t = 4 pulse at "A" will reach to "B" and pulse at "B" will reach to "A".
This means that the graph is contrary, the only one that fulfills this is the graph (e).
The velocity-time table represents the motion of a rightward-
moving motorcycle.
Magnitude =
Time (s)
0.0
0.5
1.0
1.5
2.0
What is the magnitude (i.e., value) and direction of the
acceleration?
Direction =
Velocity (m/s)
24.0, right
22.0, right
20.0, right
18.0, right
16.0, right
m/s/s
(No -
sign.)
(Tap field to change.)
NE
The direction of motion of a body or object depends on its velocity. Speed can be thought of as a scalar quantity in its simplest form. In essence, velocity is a vector quantity. It is the speed at which distance is changing.
What is velocity?Galileo Galilei, an Italian physicist, is credited with being the first to calculate speed by dividing it by the required amount of time and the distance traveled.
Speed, according to Galileo, is the distance traveled in a set amount of time.
The speed at which an object is moving is referred to as its velocity.
Examples of fast motion include an automobile driving north on a highway or a rocket blasting off.
The equation v = u + at, where v signifies the ultimate speed, can be used to express an object's final velocity, which is equal to its starting velocity plus acceleration times the distance traveled.
Therefore, the direction of motion of a body or object depends on its velocity. Speed can be thought of as a scalar quantity in its simplest form. In essence, velocity is a vector quantity. It is the speed at which distance is changing.
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Calculate the total capacitance of three capacitors 30µF, 20µF & 12µF connected in parallel across a d.c supply The answer is :Consider that the equivalent capacitance of three capacitors C1, C2 and C3 in parallel is given by:C=C1+C2+C3In this case:C1 = 30µFC2 = 20µFC3 = 12µFReplace the previous values into the formula for C and simplify:C=30μF+20μF+12μF=62μFHence, the total capacitance is 62µFQuestion 5 : Calculate the total charge on the capacitors connected in parallel if the supply voltage is 500V. Sketch a circuit diagram and label this to show how the charges are located
The circuit diagram is shown below:
From the diagram we notice that the same voltage will flow in every capacitor, this will be helpful later.
We know that this three capacitors are equivalent to a single equivalent capacitor with 62µF capacitance. The charge in this equivalent capacitor is:
[tex]Q_{eq}=(62\times10^{-6})(500)=0.031[/tex]Now, as we mentioned, the voltage is the same in each capacitor then the charge in each of them is:
[tex]\begin{gathered} Q_1=(30\times10^{-6})(500)=0.015 \\ Q_2=(20\times10^{-6})(500)=0.01 \\ Q_3=(12\times10^{-6})(500)=0.006 \end{gathered}[/tex]To check if this is correct we need to remember that the charge in the equivalent capacitor is equal to the sum of the charge in each capactior; for this case this conditon is fulfil; therefore we conclude that:
• The charge in the first capacitor is 0.015 C
,• The charge in the second capacitor is 0.01 C
,• The charge in the third capacitor is 0.006 C
The diagram with the labels is shown below:
A NASA probe is moving horizontally above the surface of the moon at a constant speed to the right, as depicted in the diagram below. It releases an instrument package when it is directly above Point P. As seen from the lunar surface, which path would the package likely follow after the release and why?
B, because the gravity of the moon will pull the instrument to the ground with constant acceleration and the lack of an atmosphere allows the package to fall straight down
D, because the gravity of the moon will pull the instrument to the ground with constant acceleration so that its vertical velocity increases as it falls, while the horizontal component of the velocity remains constant
C, because the reduced gravity of the moon pulls the package down vertically at a constant speed while the package travels horizontally at a constant speed resulting in a straight-line trajectory to the lunar surface.
A, because the gravity of the moon pulls the package down with constant acceleration, while the atmosphere of the moon creates horizontal drag on the package which reduces the horizontal component of the package’s velocity causing the package to be pulled backward as it falls
As seen from the lunar surface, the path followed by the package after the release is D, because the gravity of the moon will pull the instrument to the ground with constant a, so that its vertical velocity increases.
A is incorrect because there is no wind pushing the package backwards. B is incorrect because the package has an initial velocity. C is incorrect because vertical velocity is not constant due to the presence of gravity. E and F are incorrect because gravity acts immediately after the package is dropped.
D is correct because the horizontal component remains constant because there is no horizontal force acting on the package. This is because in outer space there is no atmosphere, so there will be no air resistance. The vertical component increases with respect to time because of constant acceleration due to gravitational pull on the package.
Therefore, as seen from the lunar surface, the path followed by the package after the release is D, because the gravity of the moon will pull the instrument to the ground with constant acceleration so that its vertical velocity increases as it falls, while the horizontal component of the velocity remains constant
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Below is a diagram of a 5.0 kg block being dragged to the right, along a horizontal surface. The
coefficient of dynamic friction, is 0.40. Take acceleration due to gravity, g, as 9.81 ms ².
5 kg
30. N
What is the acceleration of the block? Give your answer correct to two significant figures, in
m-s-2, without units.
The acceleration of the block when the coefficient of friction is 0.40 is 3.924 m/s².
The ratio of the frictional resistive force to the perpendicular force pushing the objects together is known as the coefficient of friction.
Acceleration is the rate at which an object's velocity with respect to time changes.
The block of mass 5 kg is dragged to the right on a horizontal surface.
The coefficient of friction is 0.40.
The acceleration due to gravity is 9.81 m/s².
The force on the block is 30 N.
Now, the force is defined as the product of the mass and acceleration of the object.
Now, using the conservation of force:
F = μmg
ma = μmg
a = μg
a = 0.40 × 9.81
a = 3.924 m/s²
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Hafthor bjornson broke the deadlift record in April 2020, lifting 501kg. A) How much weight (in N) did he lift?B) How hard was the floor pushing up on the weights when they were on the floor?
Given, the mass that Hafthor Bjornson lifted, m=501 kg
A)
The weight is given by the product of the mass and the acceleration due to gravity.
Thus the weight lifted by him is,
[tex]\begin{gathered} W=mg \\ =501\times9.8 \\ =4909.8\text{ N} \end{gathered}[/tex]Thus the weight he lifted is 4909.8 N
B)
When the weight is on the floor the force applied by the floor on the weights is equal to the weight itself. This force is called the normal force.
Thus the force applied by the floor on the weights is 4909.8 N
Calculate the height of an image in a converging lens with a height of 6.1 cm, an image location of 6.2 cm, and a distance of 3.6 cm for the object's placement from the lens. Final answer will have 2 decimal places and might not follow sig fig rules.
the formula we will use is
1/f = 1/p + 1/q
where f is = focal length
p is the distance away (distance from the lens to the object)
and q is the image distance
The highest frequencies humans can hear isabout 20000 Hz.What is the wavelength of sound in airat this frequency?The speed of sound is310 m/s.
Given
The highest frequency human can hear is f=20000 Hz
Speed of the sound,v=310 m/s
To find
The wavelength
Explanation
We know,
The wavelength is given by
[tex]\begin{gathered} \lambda=\frac{v}{f} \\ \Rightarrow\lambda=\frac{310}{20000} \\ \Rightarrow\lambda=0.0155\text{ m} \end{gathered}[/tex]Conclusion
The wavelength is 0.0155 m
An Olympic long jumper is capable of jumping 8.0 m. Assuming his horizontal speed is 9.1 m/s as he leaves the ground, how long is he in the air and how high does he go?
Given data:
* The initial velocity of the jumper is u = 9.1 m/s.
* The horizontal range in the given case is 8 m.
Solution:
(a). By the kinematic equation, the time taken by the jumper in terms of the initial velocity and the horizontal range is,
[tex]R=ut+\frac{1}{2}at^2[/tex]where a is the acceleration of the jumper in the horizontal direction,
As there is no force acting on the jumper in the horizontal direction, thus, the value of acceleration is zero.
Substituting the known values,
[tex]\begin{gathered} 8=9.1\times t \\ t=\frac{8}{9.1} \\ t=0.88\text{ s} \end{gathered}[/tex]Thus, the time for which the jumper remains in the air is 0.88 seconds.
(b). By the kinematics equation, the initial velocity of the jumper in the upward direction is,
[tex]v_y-u_y=gt^{\prime}_{}\ldots\ldots\ldots(1)[/tex]where u_y is the initial velocity, v_y is the final velocity of the jumper at the top of vertical displacement, g is the acceleration due to gravity, and t' is the time taken by the jumper to reach the top of vertical displacement,
The jumper will come to rest at the higher position, thus, the final velocity of the jumper at the highest position is zero.
The time taken by the jumper to reach the maximum height is equal to the time taken by the jumper to reach the ground from the maximum height.
As the total time for the complete motion of the jumper is t, thus, the time taken by the jumper to reach the maximum height from the ground is,
[tex]\begin{gathered} t^{\prime}=\frac{t}{2} \\ t^{\prime}=\frac{0.88}{2} \\ t^{\prime}=0.44\text{ s} \end{gathered}[/tex]Substituting the known values in the equation (1),
[tex]\begin{gathered} 0-u_y=-9.8\times0.44_{} \\ u_y=4.312\text{ m/s} \end{gathered}[/tex]By the kinematics equation, the maximum height reached by the jumper is,
[tex]h=u_yt^{\prime}+\frac{1}{2}gt^{\prime}^2[/tex]Substituting the known values,
[tex]\begin{gathered} h=4.312\times0.44+\frac{1}{2}\times(-9.8)\times(0.44)^2 \\ h=1.9-0.95 \\ h=0.95\text{ m} \end{gathered}[/tex]Thus, the maximum height reached by the jumper is 0.95 meters.
The couple required to hold a triple turn of 1.5cm² area in equilibrium when carrying a current 2A at 70° to a field with 0.15T is?
The couple or torque required to hold the triple turn is 1.27 x 10⁻⁴ Nm.
What is the couple or torque required?
The couple required to hold the triple turn is calculated as follows;
τ = M x Bsinθ
where;
M is the magnetic moment B is the magnetic field strengthThe magnetic moment is calculated as follows;
M = NIA
where;
N is number of turns = 3I is current = 2 AA is the area of the loop = 1.5 cm² = 0.00015 m²M = (3) x (2) x (0.00015)
M = 0.0009 m²A
The torque or couple required is calculated as;
τ = (0.0009) x (0.15 x sin70)
τ = 1.27 x 10⁻⁴ Nm
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An unwary football player collides head-on with a padded goalpost while running at 7.5 m/s and comes to a full stop after compressing the padding and his body by 0.27 m. Take the direction of the player’s initial velocity as positive.1.assuming constant acceleration calculate the his acceleration during the collision in meters per second squared.2 how long does the collision last in seconds.
Answers:
1. a = -104.16 m/s²
2. t = 0.072
Explanation:
To find the acceleration, we will use the following equation:
[tex]v^2_f=v^2_i+2ax[/tex]where vf is the final velocity, vi is the initial velocity, a is the acceleration and x is the distance. So, replacing vf by 0 m/s, vi by 7.5 m/s, and x by 0.27m, we get:
[tex]\begin{gathered} 0^2=7.5^2+2a(0.27) \\ 0=56.25+0.54a \end{gathered}[/tex]Then, solving for a, we get:
[tex]\begin{gathered} 0-56.25=56.25+0.54a-56.25 \\ -56.25=0.54a \\ \frac{-56.25}{0.54}=\frac{0.54a}{0.54} \\ -104.16m/s^2=a \end{gathered}[/tex]Therefore, the acceleration during the collision is -104.16 m/s²
Then, to calculate how long the collision last, we will use the following equation:
[tex]v_f=v_i+at[/tex]So, replacing the values and solving for t, we get:
[tex]\begin{gathered} 0=7.5-104.16t \\ 104.15t=7.5 \\ t=\frac{7.5}{104.15}=0.072s \end{gathered}[/tex]Therefore, the collision last 0.072 seconds
Shown here are astronomical objects located at different distances from earth. rank the objects based on their distances from earth, from farthest to nearest.
- star on far side of Andromeda Galaxy
- star on near side of Andromeda Galaxy
- star on far side of Milky Way Galaxy
- star near center of Milky Way Galaxy
- Orion Nebula
- Alpha Centauri
- Pluto- The Sun
The distance of astronomical objects is measure very carefully. These are having a different unit. This is the astronomical unit. The distances are very huge.
The distance between objects in space is vast and very difficult to calculate. These are learned under solar system mathematics. The values for these distances are cumbersome for astronomers and scientists to manipulate. Therefore, scientists use a unit of measurement called an astronomical unit.
Let us understand the distances first.
To know the distance of stars in Andromeda Galaxy, we should first know the distance of Andromeda Galaxy. The distance of Andromeda Galaxy from Earth is 2.5 million light years away. The astronomical unit used is the light years.
Thus, from this we can conclude that,
The star near to Andromeda Galaxy must be at a distance of 2.5 million light years away.The star far side from Andromeda Galaxy will be more than 2.5 million light years away.Now to know about the stars in Milky Way Galaxy, the distance of milky way galaxy from Earth is approximately 9 light years away.
So,
The star far from Milky Way Galaxy should be more than 9 light years away from Earth.The star near to the Milky Way Galaxy should be close to 9 light years away from Earth.Orion Nebula is 1,344 light years away.Alpha Centauri is approximately 4.3 light years away from the Earth.The Pluto is approximately 5 billion km away from the Earth.The Sun is approximately 148 million km away from the Earth.Thus, from this we can conclude that,
The farthest is the star on far side of Andromeda Galaxy, then the star on near side of Andromeda Galaxy, then comes the star on far side of Milky Way Galaxy, then the star near center of Milky Way Galaxy, then it is the Alpha Centauri, then the Orion Nebula and then it is the Pluto with the Sun being the nearest one.
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a 298 kg boat is being propelled forward with a force of 2,365 N. What is the acceleration of the boat if it has a resistance force (rewarded) due to wind and water of 878 N? (Write answer as a 2 digit number)
The acceleration of the boat is 4.9m/s²
Mass of boat= 298 kg
Forward force= 2365 N
Resistance force= 878N
We need to apply the concept of laws of motion
Net force= Forward force- Resistance force
Net force= 2365-878 N
= 1487 N
Net force= mass x acceleration
2365= acceleration x 298
acceleration = 4.9 m/s²
Therefore, the acceleration of the boat is 4.9 m/s²
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What is the force of gravity between two 50.0kg masses that are separated by 0.300m?3.71x10-8N5.59x10-7N2.78x104N1.85x10-6N
We will have the following:
[tex]\begin{gathered} F=G\frac{m_1m_2}{r^2}\Rightarrow F=\frac{(6.67\ast10^{-11}m^3\ast kg^{-1}\ast s^{-2})(50kg)(50kg)}{(0.3m)^2} \\ \\ \Rightarrow F=1.852777778...\ast10^{-6}N\Rightarrow F\approx1.85\ast10^{-6}N \end{gathered}[/tex]So, the force is approximately 1.85*10^-6 N.
Your engineering team has created a 46.9 kg spider robot that moves along a strand of web for
Halloween. The spider begins at rest and moves straight down the strand increasing its speed
at a constant rate. It covers 3.04 m in a time of 6.08 s. What is the Tension, in Newtons, in
the strand of web?
The fabric of the spiderwebs is so thin that it adheres easily to the hook side of a Velcro strip. Push Velcro onto the object you want to attach your webs to, then remove the adhesive backing. Grab a few of the strands and put them onto the Velcro to keep the webs in place.
Explain about the web for Halloween?A black spider spread between two slices of buttered bread is said to offer a witch great power. Spiders were rumoured to help witches cast charms. Old customs claim that if you see a spider on Halloween, the ghost of a departed loved one is purportedly watching over you.
The spiderwebs stick in the hook side of a Velcro strip with ease because they are constructed of a very thin fabric. When you wish to attach your webs to something, push Velcro onto the surface and peel off the sticky backing. To ensure that the webs stay in place, grab a handful of the strands and push them onto the Velcro.
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A) A recipe calls for 5.0 qt of milk. What is this quantity in cubic centimeters?Express your answer in cubic centimeters.B) A gas can holds 2.0 gal of gasoline. What is this quantity in cubic centimeters?Express your answer in cubic centimeters.
A.
[tex]\begin{gathered} 1qt=946.353cm^3 \\ 5qt=\text{?} \\ \text{quantity}=946.353\times5 \\ \text{quantity}=4731.765cm^3 \end{gathered}[/tex]B.
[tex]\begin{gathered} 1\text{ gal=}3785.41\text{ }cm^3 \\ 2\text{ gal}=\text{?} \\ \text{quantity}=2\times3785.41 \\ \text{quantity}=7570.82cm^3 \end{gathered}[/tex]A/An _____ is described as a device that is used to measure potential difference across any part of a circuit.ammeterfuseground fault interruptervoltmeter
Answer:
[tex]\text{Voltmeter}[/tex]Explanation: We need to find an instrument that measures potential difference across any part of the circuit, a potential difference is basically voltage difference across a circuit difference and the device used to measure this difference is known as:
[tex]\text{ Voltmeter}[/tex]Voltmeter used in a circuit
find the pressure increase in a fluid when a force of 25 N is exerted on a closed syringe where the piston radius is 2 cm.
Given:
• Force, F = 25 N
,• Radius, r = 2 cm
Let's find the pressure increase.
To find the pressure increase, apply the formula:
[tex]\Delta P=\frac{Force}{Area}=\frac{F}{\pi r^2}[/tex]Where:
F = 25 N
r is the radius in meters = 2 cm
P is the pressure
Convert the radius from cm to meters.
Where:
100 cm = 1 m
2 cm = 2 /100 = 0.02 m
Hence, we have:
[tex]\begin{gathered} \Delta P=\frac{25}{\pi(0.02)^2} \\ \\ \Delta P=\frac{25}{\pi(0.0004)} \\ \\ \Delta P=\frac{25}{0.00125664} \\ \\ \Delta P=19894.4\text{ Pa}\approx1.99\times10^4\text{ Pa} \end{gathered}[/tex]Therefore, the pressure increase is 1.99 x 10⁴ Pa.
ANSWER:
1.99 x 10⁴ Pa
Using an electric current, you can split liquid water to form two new substances, hydrogen and oxygen gases. Is this a change in state. Explain your answer.
This process (called electrolysis) is not have a change of state. This comes from the fact that the substance (in this case water) does not keep its identity during the change, this means that we don't end up with the same substance; rather we have two new ones, hydrogen and oxygen.
How many cubic inches are there in 3.25 yd3?Express the volume in cubic inches to three significant figures.What is the mass in grams of 16.86 mL of acetone?Express your answer to four significant figures and include the appropriate units.What is the volume in milliliters of 7.06 g of acetone?Express your answer to three significant figures and include the appropriate units.
Answer:
[tex]\text{ 3.25 yd}^3=151,632in^3[/tex]Explanation: We need to convert cubic-yards into cubic inches, this can be simply done as follows:
[tex]\frac{46656\text{ Cubic inches}}{1\text{ Cubic Yard}}^{}[/tex]Therefore we have:
[tex]\begin{gathered} 3.25\text{ Cubic yards }\times\text{ }\frac{46656\text{ Cubic Inches}}{1\text{ Cubic Yard}}=151,632\text{ Cubic Inches} \\ \therefore\rightarrow \\ \text{ 3.25 yd}^3=151,632in^3 \end{gathered}[/tex]Rtz coordinate system
Due to the car goes around the trajectory without sliding, then. it is necessary that centripetal force equals the friction force.
Then, we have:
Fc = Fr
where,
Fc = m(v^2/R) centripetal force
Fr = μN = μ*m*g friction force
Then, by replacing the precious expressions we obtain:
m(v^2/R) = μ*m*g
And by solving for v:
[tex]v=\sqrt[]{\mu\cdot g\cdot R}[/tex]where,
μ = 1.00
g = 9.8 m/s^2
R = 70.0 m
By replacing we get:
[tex]undefined[/tex]
nd in Atlanta you decide to drive around the city. You turn a corner and are driving up a steep hill. Suddenly, a small boy runs out on the street chasing a ball. You slam on the brakes and skid to a stop leaving a 50-foot-long skid mark on the street. The boy calmly walks away but a policemen watching from the sidewalk walks over and gives you a speeding ticket. He points out that the speed limit on this street is 25mph. After you recover your wits, you begin to examine the situation. You determine that the street makes an angle of 25◦with the horizontal and that the coefficient of static friction between your tires and the street is0.80. You also find that the coefficient of kinetic friction between your tires and the street is 0.60. Your car’s information book tells you that the mass of your car is 1600 kg. You weigh 140 lbs. Will you fight the ticket
skid = 50 ft
Speed limit = 25 mph
angle = 25°
friction coefficient = 0.80
mass = 1600 kg
weight = 140 lbs
A Hydrogen atom is a low density hot gas will give out what type of spectrum?A. A uniform spectrum containing all colorsB. A series of emission lines with equal spaces between the colorsC. A series of emission lines spaced in mathematical sequenceD.a uniform spectrum crossed by numerous dark absorption lines
The emission of photons takes place when an electron from higher energy orbitals jumps to a lower energy orbital.
Therefore the light emitted will correspond to the energy difference between the orbitals.
When the atom emits the photons, they will have energy equal to the energy difference between the orbitals of the Hydrogen. Therefore the spectrum obtained by the hydrogen gas will contain only those lines which correspond to the energy difference of the orbitals.
Therefore the hydrogen will emit a spectrum that contains a series of emission lines spaced in a mathematical sequence.
Therefore the correct answer is option C.
7.0 J of work is done to draw a bowstring back. The bow launches an arrow with a mass of 0.09 kg straight upward.
(a) What is the arrow's kinetic energy as it leaves the bow? (Round your answer to one decimal place.)
J
(b) What is the arrow's speed? (Round your answer to one decimal place.)
m/s
(c) What maximum height does the arrow reach? (Round your answer to one decimal place.)
m
The final answer is
(a) The KE = 3.88 * 10
(b) The speed of the arrow is 2.93 * 10
(c) Height which the arrow attains is 4.38 * 10
When the arrow is drawn from a bowstring back then there will be some work done. The pulling of the bowstring will have a distance moved and the energy used to do this.
Given,
Work done = 7 J
Mass of the arrow = 0.09 kg
The computations for a and b are the following:
To calculate the Kinetic energy,Work done = force x distance
7.0 J = force x 0.09
force = 7/0.09=77.77 = 7.77 x 10
force = m. a = 0.09 x a
77.77 / 0.09 = a = 864.1 = 8.64*10²
KE = 0.5 x m x v²
= 0.5 x 0.09 x 864.1
= 3.88 * 10
The speed of the arrow can be calculated asspeed 38.8/(0.5 x 0.09) = v² = 862 sq-root
Hence, v = 2.93*10¹ m/s is the answer
Height that the arrow reaches29.3² = 2gh
= 29.3²/(2 x 9.8) = 858.49 / (2 x 9.8) = 43.8 m
h = 43.8 m or 4.38 * 10¹ is the answer
Therefore, we can conclude that the kinetic energy of the arrow that leaves bowstring is 3.88*10 with the speed of 29.3 m/s and it reaches the height of 43.8 m.
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A 5 cm spring is suspended with a mass of 1 g attached to it which extends the spring by 3.2 cm. The same spring is placed on a frictionless flat surface and charged beads are attached to each end of the spring. With the charged beads attached to the spring, the spring's extension is 0.01 m. What are the charges, in micro-Coulombs, of the beads?
First, we need to find the spring constant
Force of gravity = force of tension from spring
mg = kx
(1x10^-3)(9.8) = k (3.2x10^-2)
k = .30625
Now we can look at the other situation
Since the spring was moved 0.01 meters, we can find the force
F = kx = (.01)(.30625) = .0030625 N
Now we can set the electric force equal to the force of the beads
kq^2/r^2 = .0030625 N
q = 292 microCoulombs
I need help with the following question:
USING THE FOLLOWING CONVERSION FACTOR:
Conversions Longitude:
1 degree= 52.505 miles
1 minute= 4620.5 feet
Conversions for Latitude:
1 degree= 69.005 miles
1 minute= 6072.5 feet
How Do I convert these coordinates into feet?
The length of side A (40° 51.485' N, 74° 12.080' W & 40° 51.485' N, 74° 11.883' W) = ------------- ft
The length of side B (40° 51.485' N, 74° 12.080' W & 40° 51.800' N, 74° 11.883' W) = ----------- ft
The length of the side A is 1,13,25,866.44 ft N, 2,70,34,989.4 ft W
The length of the side B is 1,13,25,866.44 ft N, 2,70,34,989.4 ft W
Length is the measurement of anything from end to end.
Conversions Longitude:
Longitude is the angular distance of a location east or west of the meridian in Greenwich, England, or west of the standard meridian of a celestial object.
1 degree= 52.505 miles
1 minute= 4620.5 feet
Conversions for Latitude:
Latitude is the angular distance of a location north or south of the Earth's equator.
1 degree= 69.005 miles
1 minute= 6072.5 feet
Also 1 mile = 5280 feet
The length of side A (40° 51.485' N, 74° 12.080' W)
40° 51.485' N = 40 * 52.505 * 5280 + 51.485 * 4620.5
40° 51.485' N = 1,10,88,000 + 2,37,866.44
40° 51.485' N = 1,13,25,866.44 feet.
74° 12.080' W = 74 * 69.005 * 5280 + 12.080 * 6072.5
74° 12.080' W = 2,69,61,633.6 + 73,355.8
74° 12.080' W = 2,70,34,989.4 feet
The length of side B (40° 51.485' N, 74° 12.080' W)
40° 51.485' N = 1,13,25,866.44 feet.
74° 12.080' W = 2,70,34,989.4 feet.
Side A = Side B = 1,13,25,866.44 ft N, 2,70,34,989.4 ft W.
To know more about length,
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On a fishing trip Justin rides in a boat 12 km south. The fish aren't biting so they go 4 km west. They then follow a school of fish 1 km north. What is his total displacement?
ANSWER:
The displacement is 11.7 km
STEP-BY-STEP EXPLANATION:
The displacement is the vector sum from the start point to the end point. Vector addition is applying the formula of the distance between two points.. We will use this formula:
[tex]d=\sqrt[]{(x_2-x^{^{}}_1)^2+(y_2-y_1)^2^{}}[/tex]To better understand the exercise, we will draw a picture of the situation, it would look like this:
Replacing the points (-4, -11) and (0, 0)
[tex]\begin{gathered} d=\sqrt[]{(-4-0^{}^{}_{})^2+(-11_{}-0_{})^2} \\ d=\sqrt[]{16+121^{}} \\ d=\sqrt[]{137} \\ d=11.7 \end{gathered}[/tex]Solve: What work is done when 3.0 C is moved through an electric potential difference of 1.5 V?1) 0.5 J2) 2.0 J3) 4.0 J4) 4.5 J
We know that the work done in an electric potential difference is
[tex]\begin{gathered} W=V\cdot Q \\ W=1.5V\cdot3.0C \\ W=4.5J \end{gathered}[/tex]Therefore, the work done is 4.5 J.