In an experiment in space, one proton is held desk bound and every other proton is released from relaxation at a distance of 3.50 mm. After the proton is launched its initial acceleration is four.14*10^{five}
Given data:
Electrostatic force = kq^{2} /r^{2}
F = ma
ma = [tex]= kq^{2} /r^{2}[/tex]
[tex]a = = kq^{2} /r^{2}m[/tex] ....(1)
Here,
k = [tex]9*10^{-19} C[/tex]
r = 3.50 mm
m = [tex]1.67 * 10^{-27}[/tex]
Putting the details given above into equation 1
[tex]a = \frac{9*10^{9}(1.6*10^{-19})^{2} }{(2.5*10^{-3})^{2}*1.67*10^{-27} }[/tex]
[tex]a = 4.14*10^{5}[/tex]
Electrostatic forces, as an instance, pull or push on objects without genuinely getting into contact with them. while sure substances are rubbed collectively, a phenomenon known as "rate" may be transferred from one surface to some other. Uncharged gadgets are pulled by way of charged objects, which in turn may additionally push or pull towards different charged ones.
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The stemplot below shows the heights (in inches) of students in a clas
5
25569
2477
6
7 01
Which of the following is a height of a student in the class?
OA. 69 inches
OB. 47 inches
O C. 77 inches
OD. 70 inches
D. The height of a student in the class from the given data is 70.
What is stem plot?A stem plot is a way to plot data where the data is split into stems (the largest digit) and leaves (the smallest digits).
Heights of the students in the class5, 25, 56, 92, 47, 76, 70, 1
Thus, the height of a student in the class from the given data is 70.
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How strong is the electric field between two parallel plates 4.4 mm apart if the potential difference between them is 220 v?
The electric field between the plates is 50,000 N/C.
CalculationIt is provided that the separation d between the plates is 4.4 mm and the potential difference V between the plates is 220 V.
The electric field E between the plates can be formulated as follows.
E=[tex]\frac{V}{d}[/tex]
substituting the values in the equation,
E= [tex]\frac{220}{0.0044}[/tex]
E= 50,000 N/C
Hence, we can say that the electric field between the plates is 45833.33 N/C.
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Find online the annual 10-k report for costco wholesale corporation (cost) for sep. 2, . answer the following questions from the financial statements:
The annual 10-k report costco wholesale corporation is
a. 4.28 billion dollars.
b. 1.128 billion.
c. $2,393 billion
d. $481 million less.
Costco Wholesale Corporation: What is it?
The fee is a multibillion-dollar large container retail organisation that operates a network of comrades for the large container shops. it's miles the fifth-biggest retailer inside the globe as of the year 20220. It offers wine and natural goods for sale. From these operations, price receives $4,285 in coins. The coins spent on the brand new assets become system for sales of $2,393 million USD, and the depreciation expenditure changed into $1,127 USD. The value increases its holdings to roughly 481 million USD.
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A car starts from rest and accelerates uniformly for a distance of
204 m over an 6.5-second time interval. The car's acceleration
is _____ m/s².
Answer: 8.10m/s
Explanation:
using second eq of motion s=ut+1/2at
2
as u=0, a=
t
2
2s
=
5.21
2
2∗110
=8.10m/s
2
When a dilute gas expands quasi-statically from 0.50 to 4.0 l, it does 250 j of work. assuming that the gas temperature remains constant at 300 k, what is the change in the internal energy of the gas?
The change in the internal energy of the gas is ΔE int =0 J.
Q=250J heat is absorbed by the gas in this process.
What is isothermal process?
A thermodynamic process known as an isothermal process keeps the system's temperature constant. Because heat is transferred into or out of the system so slowly, thermal equilibrium is preserved.
Givens:
Vi =0.5 L=0.5×10^ −3 m^3
Vf =4.0 L=4.0×10^−3 m^3
W= 250 J
T= 300 K
Because the gas temperature is constant, the process is ISOTHERMAL. As a result, the system's internal energy is constant.
ΔE int =0 J
We know that, ΔE int =Q−W
And we are aware that there has been no change in the system's inherent energy.
Therefore,0=Q−W
Q=W=250J
Q=250J
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Question:
When a dilute gas expands quasi-statically from 0.50 to 4.0 L, it does 250 J of work. Assuming that the gas temperature remains constant at 300 K, (a) what is the change in the internal energy of the gas? (b) How much heat is absorbed by the gas in this process?
If you drove 90km/hr for 2 hours, and 35km/hr for 0.5 hours, what would your average speed be
Answer:
1,25m.s^-1
Explanation:
Speed=distance /time
=90000/72000
you walk 9.6km in 4.5h , what the average speed
How many walking paces are there approximately as you walk down Main Street (0.25 miles)?
1 mile = 5280 feet;
1 foot = 12 inches;
22 inches = 1 walking pace
The number of walking paces as you walk down main street (0.25 miles) is 720 walking paces
How to convert 0.25 mile to feet1 mile = 5280 feet
Therefore,
0.25 mile = (0.25 mile × 5280 feet) / 1 mile
0.25 mile = 1320 feet
How to convert 1320 feet to inches1 feet = 12 inches
Therefore,
1320 feet = (1320 feet × 12 inches) / 1 feet
1320 feet = 15840 inches
How to convert 15840 inches to walking pace22 inches = 1 walking pace
Therefore,
15840 inches = (15840 inches × 1 walking pace) / 22 inches
15840 inches = 720 walking paces
Thus, we can conclude that as you walk down main street (0.25 miles), there are 720 walking paces
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5. A ball is dropped from rest at a height of 25.0 m above the ground. A. How fast is the ball moving when it is 10.0 m above the ground? B. How much time is required for it to reach the ground level? Ignore the effects of air resistance.
A. The ball is moving 17.15 m/s when it was 10 m above the ground
B. The time required for the ball to reach the ground is 2.26 s
How to determine the time to fall till 10 m above the ground Initial velocity (u) = 0 m/sAcceleration due to gravity (g) = 9.8 m/s²Height (h) = 25 - 10 = 15 mTime (t) = ?h = ½gt²
15 = ½ × 9.8 × t²
15 = 4.9 × t²
Divide both side by 4.9
t² = 15 / 4.9
Take the square root of both side
t = √(15 / 4.9)
t = 1.75 s
A. How to determine the velocity when the ball is 10 m above the ground Time till 10 m above the ground (t) = 1.75 sAcceleration due to gravity (g) = 9.8 m/s²Velocity (v) = ?v = gt
v = 9.8 × 1.75
v = 17.15 m/s
B. How to determine the time taken to reach the groundInitial velocity (u) = 0 m/sAcceleration due to gravity (g) = 9.8 m/s²Height (h) = 25 mTime (t) = ?h = ½gt²
25 = ½ × 9.8 × t²
25 = 4.9 × t²
Divide both side by 4.9
t² = 25 / 4.9
Take the square root of both side
t = √(25 / 4.9)
t = 2.26 s
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an uncharged capacitor is connected to the terminals of a 4.0 v battery, and 12 μc flows to the positive plate. the 4.0 v battery is then disconnected and replaced with a 8.0 v battery, with the positive and negative terminals connected in the same manner as before.
The additional charge that flows to the positive plate from the 8 Volts battery is 12 x 10⁻³ Coulomb's.
We have an uncharged capacitor which is charged in two different ways using battery of two different voltages. Firstly, it is connected to the terminals of a 4.0 V battery, and 12 μc of charge flows to the positive plate. Then, the 4.0 V battery is disconnected and replaced with a 8.0 V battery
We have to determine the Additional charge that flows to the Positive plate from this 8 V battery.
What is Capacitance ?Capacitance is the ability of an object to collect and store energy in the form of an electrical charge. Mathematically -
C = Q/V
According to the question, we have -
Electric potential of battery 1 = V[1] = 4 V
Charge flow from battery 1 to the positive plate = Q[1] = 12 x 10⁻³ C
Electric potential of battery 2 = V[2] = 8 V
Now, using the following relation, we will find Q[2] {Charge flow from battery 2 to the positive plate} →
V[1] / V[2] = Q[1] / Q[2]
Substituting the values -
Q[2] = (12 x 10⁻³ x 8)/4 = 24 x 10⁻³ C
Additional Charge that flowed to the positive plate →
Q[2] - Q[1] = 12 x 10⁻³ Coulomb's
Hence, the additional charge that flows to the positive plate is [tex]12\times 10^{-3}[/tex] Coulomb's.
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[ The question given is incomplete, the complete question is -
"An uncharged capacitor is connected to the terminals of a 4.0 v battery, and 12 μc flows to the positive plate. the 4.0 v battery is then disconnected and replaced with a 8.0 v battery, with the positive and negative terminals connected in the same manner as before. How much additional charge flows to the positive plate" ]
Ok someone help me brainlest will be given
1.With what will a car hit a tree if the car has a mass of 3,000 kg acceleration of m/s^2
2.What force would be needed to make 10kg bowling ball acceleration of 3m/s^2
3. What would be the mass of a truck if it is accelerating at a rate of 5m/s^2 and hits a parked car with a force of 14000 N
4.What is the acceleration of a softball if it has a mass of 0.50 kg and hits the catcher's glove with a force of 25 N
5.How much does a suitcase weigh if has a mass of 22.5 kg
6. What would the mass of a rock falling from the sky if it hits the ground with the force of 147N
7.Find the acceleration of the stacks of books pictured if the stack has a total mass 1.5 kg
8. A kg model airplane is traveling at a speed 33m/s.
The operator than increases the speed up to 45 m/s in seconds
How much force did the engine need in order to make this change.
Answer:
According to Newton's second law of motion, the acceleration of a body is directly proportional to the force acting on the body and inversely proportional to its mass. The formula for this law is
a
=
F
m
, from which we get the formula
F
=
m
a
. When mass is in kg and acceleration is in
m/s/s
or
m/s
2
, the unit of force is
kgm/s
2
, which is read as kiligram-meter per second squared. This unit is replaced with an N in honor of Isaac Newton. Your problem can be solved as follows:
Known/Unknown:
m
=
3000kg
a
=
2m/s
2
Equation:
F
=
m
a
Solution:
F
=
m
a
=
3000kg
x
2m/s
2
=
6000kgm/s
2
=
6000N
Explanation:
find the net force σf⃗ 3 acting on particle 3 due to the presence of the other two particles. report you answer as a magnitude σf3 and a direction θ measured from the positive x axis.
Therefore, the net force acting on particle 3 is 139.68 × 10—5 N, and the angle made it by with at the +x axis is 19.1°.
The charge of particle 1 is q1.
q1 = -6.6nC
The charge of particle 2 is q2.
q2 = -13.2 nC
Sides of the equilateral triangle = 3 cm = 3 × 10 —² m.
The force acting on particle 1 is,
[tex]F _{1} = \frac{kq_{1} q_{3}}{a ^{2} } [/tex]
[tex] = \frac{9 \times 10 ^{9} \times 6.6 \times 10 ^{ - 9} \times 8 \times 10 ^{ - 9} }{9 \times 10 ^{ - 4} } [/tex]
[tex]F _{1} = 52.8 \times 10 ^{ - 5} \: N[/tex]
[tex]F _{1} = F _{1} \: cos(60°)i + F _{1} \: sin(60°)j[/tex]
[tex]= ( - 26.4 \: i \: + 45.7 \: j \: ) ×10 ^{ - 5} [/tex]
The force acting on particle 2 is,
[tex] = \frac{9 \times 10 ^{9} \times 3.2 \times 10 ^{ - 9} \times 8 \times 10 ^{ - 9} }{9 \times 10 ^{ - 4} } [/tex]
[tex] = - 105.6 \times 10 ^{ - 5} \: j[/tex]
The net force acting on particle 3 is,
[tex]F_{Net} =F _{1} + F _{2}[/tex]
[tex] = ( - 132 \times 10 ^{ - 5} \: i + 45.7 \times 10 ^{ - 5} \: j)[/tex]
[tex]F_{Net} =139.68 \times \: 10 ^{ - 5} \: N[/tex]
Thus, the net force acting on particle 3 is 139.68 × 10—5 N.
Let the angle made by it with at the +x axis be θ.
[tex]cos \: θ = \frac{ - 132 \times 10 ^{ - 5} }{139.6 \times 10 ^{ - 5} } [/tex]
θ = 19.1°
Therefore, the net force acting on particle 3 is 139.68 ×
10—5 N and the angle made it by with at the +x axis is 19.1°.
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Limits of Measurement (page 19)
12. Circle the letter of each expression that has four significant figures.
a. 1.25 x 10¹
12.51
c. 0.0125
d. 0.1255
13 Is the following sentence true or false? The precision of a
Option B has four significant figures.
The number of significant single digits (0 through 9 inclusive) in the coefficient of an expression in scientific notation is referred to as significant figures. The amount of precision or assurance an engineer or scientist uses to state a quantity is indicated by the number of significant figures in a statement. Rounding off an expression once a calculation is completed yields significant figures. The number of significant figures in the result of any computation must be equal to or less than the number of significant figures in the expression or component with the least accuracy. The digits in a number that have significance in terms of accuracy or precision are known as significant figures. Any digit that is not zero is among them. As in 3003 or 45.60009, there are zeros in between non-zero digits.
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9. The first system of measurement that was based on units of ten was the ____ system
Answer:
metric
Explanation:
In the lewis formula for hydrazinium ion, n2h5 , the total number of lone electron pairs around the two nitrogen atoms is:________
In the Lewis formula for hydrazinium ion, N 2H 5 +, the total number of lone electron pairs around the two nitrogen atoms is 1.
What is Lewis structure?
Lewis structures, often referred to as Lewis dot formulas, Lewis dot structures, electron dot structures, or Lewis electron dot structures (LEDS), are diagrams that depict the interactions of atoms inside molecules as well as any lone pairs of electrons that may be present. Any molecule with a covalent link, as well as coordination compounds, can have a Lewis structure. Gilbert N. Lewis, who described it in his 1916 article, The Atom and the Molecule, gave the Lewis structure its name. Lewis structures add lines between atoms to represent shared pairs in a chemical bond, extending the idea of the electron dot diagram.
The two nitrogen atoms are surrounded by just one lone electron pair in the hydrazinium ion's Lewis formula, n2h5+.
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the total time it takes a projectile fired straight up at 10 m/s to reach the top of its path and return to its starting point is about
The total time it takes a project fired straight up at 10m/s and Return to its starting point is about time, t= 2s.
At top speed v=0,
V= urate
0=10+ -g×t
t=1 sec
Top to Bottom it will take one more sec
t= 1+1
t=2 sec/s.
What is total time duration?Unless there has been a break in service of less than twelve (12) months, in which case the employee will get an adjusted seniority date upon restoration to a Unit 6 classification, Total Time is calculated as the amount of time since the hire date.
During a gap in employment, an employee is not permitted to gain seniority. When an employee returns to a Unit 6 classification after more than twelve (12) months of absence, their starting seniority date is reset entirely.
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1. DCI Forces and Motion A ball is rolling down a flat, frictionless ramp with
a constant velocity of 13 m/s. What is the acceleration of the ball over three
seconds? Over an infinite number of seconds? Explain your answer.
The acceleration of the ball is zero.
What is the acceleration?The term acceleration refers to the rate of change of velocity with time thus we can write;
a = d(v)/dt. This simply means the change in velocity with respect to time in the calculus notation.
This implies that before we can talk of acceleration the velocity must be changing. We can not have acceleration when the velocity is constant. It violates the very definition of acceleration. As such, the acceleration of the ball is zero.
Thus whether over a period of three seconds or over an infinite period of time, the acceleration would remain zero because the velocity is constant and does not change over the time interval as given.
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how does the current motion of hnlc on oahu compare to the direction of pacific plate motion relative to the hawaiian hotspot over the past million years?
Now the islands are moving more north compared to before compare to the direction of pacific plate motion relative to the hawaiian hotspot over the past million years.
The Pacific Ocean's ocean floor is made up of multiple plates. In relation to the plate holding North America and to hot areas rising through the mantle from below the plates, the largest one, the Pacific Plate, is moving north-west (they generate islands like Hawaii). The plate is travelling around Hawaii at a rate of 7 cm per year.
A hot spot that provided magma produced deep under the mantle of the earth, which was used to create the long volcanic chain, did so over the course of about 70 million years. Over the course of around 70 million years, a hot spot supplied magma that was formed deep under the earth's interior (mantle), pushing its way through the earth's surface and ocean floor to construct volcanic islands.
The hot point continuously created new volcanoes on the Pacific Plate, creating the volcanic chain, while the Pacific Plate was shifted by tectonic forces inside the Earth.The Pacific Plate will move in the direction and at the rate determined by the distances between and the approximate ages of several of the Hawaiian volcanoes.
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a rock is thrown vertically upward at an initial velocity of 70 m/s. what is the approximate acceleration of the rock when it reaches its maximum height?
A rock is thrown vertically upward at an initial velocity of 70 m/s.
The approximate acceleration of the rock when it reaches its maximum heigh is- zero
0 is the approximate acceleration of the rock whilst it reaches its most peak.given that rock is hurled vertically up into the air at a speed of 70 meters according to second. what's the rock's angular acceleration when it reaches its maximum point?(count on that the positive y direction is upward)?given that has reached its maximum factor, very last velocity V will identical 0.calculating the time the usage of the first regulation of movement.additionally, because the orientation is vertical instead of horizontal, the acceleration can be identical to zero.To learn more about acceleration here:-
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If the final pressure is 0.500 times the initial pressure, what is the ratio of the final density to the initial density?
By ideal gas approximation, the final to initial density ratio is 1/2.
We need to know about the ideal gas theory to solve this problem. The ideal gas is assumed that there is no interaction between particles in a gas. It can be determined by the equation
P . V = n . R . T
where P is pressure, V is volume, n is the number of moles gas, R is the ideal gas constant (8.31 J/mol.K) and T is temperature.
From the question above, we know that
P2 = 0.5 P1
Pressure is inversely proportional to pressure, hence it can be written as
P ~ (1/V)
Assuming the temperature does not change, the ratio of pressure is
P1 / P2 = (1/V1) / (1/V2)
P1 / P2 = V2 / V1
P1 / 0.5P1 = V2 / V1
V2 = 2 V1
Find the density ratio
ρ2 / ρ1 = (m/V2) / (m/V1)
ρ2 / ρ1 = V1 / V2
ρ2 / ρ1 = V1 / 2V1
ρ2 / ρ1 = 1 / 2
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The gravitational acceleration is 9.81 m/s2 here on earth at sea level. what is the gravitational acceleration on the top of mount everest?
The gravitational acceleration on the top of Mount Everest is 9.773 m/s^2.
What is gravitational acceleration?
Gravitational acceleration (symbolized g) is an expression used in physics to indicate the intensity of a gravitational field. It is expressed in meters per second squared. At the surface of the earth, 1 g is about 9.8 meter per squared second.
So, as mentioned in question g at surface of the earth is 9.81.
and the height of the Mount Everest is 8848 m.
Let, g' be the acceleration due to gravity on the top of Mount Everest.
By the equation,
[tex]g' = g(1-\frac{2h}{h})[/tex]
[tex]g' = 9.8(1-\frac{17696}{6500000})[/tex]
[tex]g' = 9.8(1-0.00276)[/tex]
[tex]g' = 9.8(0.99724)[/tex]
[tex]g' = 9.773[/tex]
Therefore, the gravitational acceleration at the top of the Mount Everest is [tex]9.773 \frac{m}{s^2}[/tex] (9.773 meter per squared second).
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Atoms, in which the number of electrons does not equal the number of protons, are called.
Atoms, in which the number of electrons does not equal the number of protons, are referred to what we call ions.
What is an Ion?This refers to an atom or molecule which has a net electric charge as a result of the gain or loss of electrons which is the part of the subatomic particle actively involved in a chemical reaction.
Due to the electrons involvement in a reaction, it doesn't equal the number of protons and results in the formation of ions in which the positive ions are the ones which donate electrons and negative ions accept electrons.
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QUESTION 17
IF a particular galaxy has a total mass of 10 to the 47 grams and the average star in that galaxy has a mass of 10 to the 33
grams, approximately how many stars are there in that galaxy? Give your answer to the nearest whole power of 10 by
entering the exponent only.
The number of stars present in the galaxy is 10 raised to the power of 14. So the exponent is equal to 14.
Given in the question
Total mass of the particular galaxy = 10 to the power of 47 grams
The average mass of the star in the galaxy = 10 to the power of 33 grams
We know that the total mass of the galaxy which is a collection of several thousand stars is equal to the sum of all the stars present in it. Also, the total mass of the galaxy can be calculated by multiplying the average mass of a star in the galaxy by the number of stars that are there in the galaxy.
So, Total mass = Average mass × Number of stars in the galaxy
Number of stars in the galaxy = Total mass of the galaxy/ Average mass of the star
Put in the values, we get
Number of stars in the galaxy = 10^47/10^33
Number of stars in the galaxy = 10^(47-33)
Number of stars in the galaxy = 10^14
So, the number of stars present in the galaxy is 10 raised to the power of 14. So the exponent is equal to 14.
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a skier travels 84m in 6s, what the average speed?
Answer:88.3 m/s
Explanation:
Uranus, one of the most distant planets in our solar system, is 2870 million kilometers from the sun. what is its distance from the sun in astronomical units?
The distance of Uranus from the Sun in astronomical units is 9.13 AU.
What is astronomical unit?
In our solar system, distances are really great. Therefore, astronomers frequently avoid using miles or kilometers when describing the distances to planets, asteroids, comets, or spacecraft. They substitute astronomical units, or AU, which represent the typical separation of the Earth from the sun. That translates to around 93 million miles, 150 million kilometers, or 8 light-minutes.
By above definition, we can say that
1 astronomical unit (AU) = 150 million kilometers (km)
In order to find the distance in astronomical units multiply the given distance expressed in kilometers of Uranus, which is 2870 million kilometers, by the unit ratio, 1 AU per 150 million kilometers, as shown on the expression:
[tex](2870 million Km)(\frac{1Au}{150millionKm})[/tex]
Therefore, the distance of Uranus from the Sun is 19.13 AU.
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a ball rolls off the top of a 1.75 m high ramp inclined at angle 30.0◦moving at 5.87 m/s. (note: the ball goes off the top of the ramp similarly to a bicycle going off a jump.)
The acceleration ax = 3.03 m/s^2 and the ay = 1.75 m/s^2.
CalculationIt is given that
h = 1.75m
[tex]\theta[/tex]= 30°
Vo = 5.87 m/s
Now Vox= Vo cos[tex]\theta[/tex]
= 5.87 x cos30° = 5.08 m/s
Voy = Vo sin[tex]\theta[/tex] = 2.94 m/s
Tan[tex]\theta[/tex] = h/x
⇒ x =h/tan[tex]\theta[/tex]
⇒ x = 3.03m
y = h = 1.75m
Now, v for the ball is given by
V = Vo + [tex]\sqrt{10gh/7}[/tex]
= Vo + [tex]\sqrt{10*9.81*1.75/7}[/tex]
V = 5.87 + 4.95 = 10.82m/s
⇒ Vx = Vcos[tex]\theta[/tex] = 9.37 m/s
and Vy = 5.41 m/s
a = 5/7 gsin[tex]\theta[/tex]
So, ax = 5/7 gsin[tex]\theta[/tex] x cos[tex]\theta[/tex]
ax = 5/14 gsin(2[tex]\theta[/tex])
ax = 5/14 x 9.81 x sin60°
ax= 3.03 m/s^2
and ay = 5/7gsin[tex]\theta[/tex] x sin[tex]\theta[/tex] = 5/7 gsin^2[tex]\theta[/tex]
= 5/7 x 9.81 x sin^2 30°
⇒ ay = 1.75 m/s^2
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What is the change in energy experienced by an electron in a hydrogen atom that undergoes an n = 2 to n = 1 transition?
The change in energy is -10.2 eV.
We need to know about transition energy to solve this problem. Electrons in a hydrogen atom can move to another level of energy. It is also called a transition. In this process, electrons will absorb or release their energy to move. The transition energy can be determined as
ΔE = E2 - E1
where ΔE is transition energy, E2 is the final state of energy and E1 is the initial state of energy.
The state energy of a hydrogen atom can be calculated by
E = -(13.6) / n² eV
where n is the number of energy states.
From the question above, we know that
n1 = 2
n2 = 1
By substituting the following parameter, we get
ΔE = E2 - E1
ΔE = -(13.6) / n2² - (-(13.6) / n1²)
ΔE = -(13.6) / 1² - (-(13.6) / 2²)
ΔE = 3.4 - 13.6
ΔE = -10.2 eV
The transition energy is minus because electrons emit the energy to move.
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a tire has a tread pattern with a crevice every 2.20 cm. each crevice makes a single vibration as the tire moves. what is the frequency of these vibrations if the car moves at 43.0 m/s?
The frequency of these vibrations if the car moves at 43.0 m/s is 1954.54 Hz.
Given that,
The speed of the car = 43m/s.
or the speed of the car = 4300cm/s
A tire has a tread pattern with a crevice = 2.20cm
Since the crevice hits every 2.20cm,
This happens,
= 4300cm / 2.20
= 1954.54.
Therefore,
The frequency is 1954.54Hz
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if a penny has mass 2.56 g, what charge (magnitude and sign) on the penny is necessary to levitate the penny? (2)
The charge on the penny is necessary to levitate the penny, the magnitude is 3*10^-4C and sign is negative.
What is charge?Charged matter experiences a force when it is exposed to an electromagnetic field due to the physical property of electric charge. Positive or negative charges can exist in an electric field (commonly carried by protons and electrons respectively). A neutral object is one that carries no net charge. The early understanding of how charged particles interact is today known as classical electrodynamics, and it is still true for issues that do not demand for taking into account quantum phenomena. Electric charge is a conserved property; the net charge of an isolated system, which is the sum of the positive and negative charges, cannot change.
According to Newton's second law -
∑fy= ma
fe-mg=ma
Fe= ma + mg
qE = 3*10^-3*0.19 + 3*10^3*9.81
qE= 30 * 10^-3
q= 30*10^3/100 = 30*10^-5
q= 3*10^-4C
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the greatest magnification on the compound light microscope can be achieved by using the high power objective lens.
The statement is false, that the greatest magnification on the compound light microscope can be achieved by the use of the high-power objective lens.
What is a Compound Light microscope?A microscope with numerous lenses and its own light source is referred to as a compound light microscope. In this type of microscope, the rotating nosepiece is positioned closer to the specimen and contains the objective lenses, while the binocular eyepieces house the ocular lenses.
The compound binocular microscope is now more often utilized, even though it can occasionally be found as monocular with only one ocular lens.
Zacharias Jansen's 1595 invention of the compound microscope with collapsible tubes and a 9X magnification produced the first light microscope.
Since then, microscopes have advanced significantly; the strongest compound microscopes available today have magnifications of 1,000 to 2,000X.
Hence, the statement is false that it can be achieved by using the high power objective lens.
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