Answer:
The mass of the mercury remaining in the bottle is 497.57 grams.
Explanation:
The mass of the mercury remaining in the bottle is found by subtracting the mass expeled due to heating from initial mass inside the bottle. That is:
[tex]m_{f} = m_{o}-\Delta m[/tex] (1)
Where:
[tex]m_{o}[/tex] - Initial mass, in grams.
[tex]\Delta m[/tex] - Mass expelled due to heating, in grams.
[tex]m_{f}[/tex] - Final mass, in grams.
If we know that [tex]m_{o} = 500\,g[/tex] and [tex]\Delta m = 2.43\,g[/tex], then the mass of the mercury remaining in the bottle is:
[tex]m_{f} = m_{o}-\Delta m[/tex]
[tex]m_{f} = 497.57\,g[/tex]
The mass of the mercury remaining in the bottle is 497.57 grams.
Credit-Card Magnetic Strips Experiments carried out on the television show Mythbusters determined that a magnetic field of 1000 gauss is needed to corrupt the information on a credit card's magnetic strip. (They also busted the myth that a credit card can be demagnetized by an electric eel or an eelskin wallet.) Suppose a long, straight wire carries a current of 7.0A . How close can a credit card be held to this wire without damaging its magnetic strip?
Answer:
14 μm
Explanation:
The magnetic field due to a long straight wire is B = μ₀i/2πr where μ₀ = permeability of free space = 4π × 10⁻⁷ H/m, i = current = 7.0 A and r = distance of credit card from magnetic field.
So r = μ₀i/2πB since B = 1000 gauss = 1000 G × 1 T/10000 G = 0.1 T
r = 4π × 10⁻⁷ H/m × 7.0 A/(2π × 0.1 T)
r = 2 × 10⁻⁷ H/m × 7.0 A/0.1 T
r = 14 × 10⁻⁷ H/m × A/0.1 T
r = 140 × 10⁻⁷ m
r = 1.4 × 10⁻⁵ m
r = 14 × 10⁻⁶ m
r = 14 μm
A car driver spends 3hrs driving at an average speed of 80km/hr, stops for 30 minutes to
have some rest, and then drives at an average speed of 90km/hr for 2 hours.
Calculate the average speed during the whole journey
Answer:
The average speed throughout the journey was 76.36 kilometers per hour.
Explanation:
Given that a car driver spends 3hrs driving at an average speed of 80km / hr, stops for 30 minutes to have some rest, and then drives at an average speed of 90km / hr for 2 hours, to determine the average speed during the whole journey the following calculation must be performed:
80 km / h x 3 = 240 km
90 km / h x 2 = 180 km
240 + 180 = 420 km
3 + 2 + 0.5 = 5.5 hours
420 / 5.5 = 76.36
Thus, the average speed throughout the journey was 76.36 kilometers per hour.
You are testing a new car using crash test dummies. Consider two ways to slow the car from 90 km/h (56 mi/h) to a complete stop:
i) You let the car slam into a wall, bringing it to a sudden stop.
ii) You let the car plow into a giant tub of gelatin so that it comes to a gradual halt.
In which case is there a greater impulse of the net force on the car?
(Prove and select below.)
a) in case (i).
b) in case (ii).
c) The impulse is the same in both cases.
d) The impulse of the first case > the impulse of the second case.
e) The impulse of the first case < the impulse of the second case.
Answer:
c) The impulse is the same in both cases.
Explanation:
What drives the car is the amount of movement of the car during its journey. In this case, we can consider that in the car, presented in the question above, the movement is generated through speed and friction on the track, the impact that makes the car stop, is not part of the movement and therefore, not part of the impulse. In the two situations presented, the speed of the car is the same, the same car and the same track were also used, which shows us that the impulse, in both situations, is the same.
You are at a train yard observing trains (because why not). You see a train car (let's call it car 1) moving to the right ( x direction) towards a stationary train car (let's call this one car 2). Car 1 has an initial velocity of 15.0 m/s. A helpful train employee tells you that Car 1 also has a mass of 1,825 kg and Car 2 has a mass of 2,645 kg. Car 1 gently collides with Car 2, allowing them to connect. After the collision the two train cars stay connected. You can assume that there is no friction in the system. If you have never see train cars connect, you can watch the first 25ish seconds of this video to see two train cars couple. However, these cars have friction, so they stop - unlike our problem. What is the Final Velocity of the system consisting of Car 1 and Car 2
Answer:
6.12 m/s
Explanation:
Using the law of conservation of momentum
momentum before collision = momentum after collision
m₁v₁ + m₂v₂ = (m₁ + m₂)V (since the train cars become attached to each other) where m₁ = mass of car 1 = 1,825 kg, m₂ = mass of car 2 = 2,645 kg, v₁ = initial velocity of car 1 = + 15.0 m/s (positive since it is moving in the positive x direction), v₂ = initial velocity of car 2 = 0 m/s (since it is initially stationary) and V = velocity of both cars after collision,
So, m₁v₁ + m₂v₂ = (m₁ + m₂)V
m₁v₁ + m₂(0 m/s) = (m₁ + m₂)V
m₁v₁ + 0 = (m₁ + m₂)V
V = m₁v₁/(m₁ + m₂)
substituting the values of the other variables into the equation, we have
V = 1,825 kg × 15.0 m/s/(1,825 kg + 2,645 kg)
V = 27375 kgm/s/ 4470kg
V = 6.124 m/s
V ≅ 6.12 m/s
Which two chemical equations show double-replacement reactions?
A. C+02 - CO2
B. 2Li + CaCl2 - 2LiCl + Ca
I C. Ca(OH)2 + H2S04 - CaSO4 + 2H20
D. Na2CO3 + H2S - H2CO3 + Na2S
The two chemical equations show double-replacement reactions are Ca(OH)2 + H2S04 - CaSO4 + 2H20 and Na2CO3 + H2S - H2CO3 + Na2S.
What is double replacement reaction?A double replacement reaction have two ionic compounds that are exchanging anions or cations.
From the given options, we can choose the following based on their exchange of anions or cations.
Ca(OH)2 + H2S04 - CaSO4 + 2H20Na2CO3 + H2S - H2CO3 + Na2SThus, the two chemical equations show double-replacement reactions are Ca(OH)2 + H2S04 - CaSO4 + 2H20 and Na2CO3 + H2S - H2CO3 + Na2S.
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An electron is travelling in the positive x direction. A uniform electric field is in the negative y direction. If a uniform magnetic field with the appropriate magnitude and direction also exists in the region, the total force on the electron will be zero. The appropriate direction for the magnetic field is:Group of answer choicesthe negative y directioninto the pageout of the pagethe negative x directionthe positive y direction
Answer:
into the page
Explanation:
Since the uniform electric field is in the negative y direction so its is -E and the electron is travelling in the positive x direction, it experiences an electric force F = -e × -E = + eE, so the electric force is in the positive y direction. Now since the net force on the electron is zero in the region of the magnetic field, it follows that the direction of the magnetic force is opposite to that of the electric force. Since the electric force is in the positive y direction, the magnetic force is in the negative y direction.
By the right hand rule, since the magnetic force is in the negative y direction and the electron moves in the positive x direction, it follows that the magnetic field is in the positive z direction, into the page.
The frequency and wavelength of EM waves can vary over a wide range of values. Scientists refer to the full range of frequencies that EM radiation can have as the electromagnetic spectrum. Electromagnetic waves are used extensively in modern wireless technology. Many devices are built to emit and/or receive EM waves at a very specific frequency, or within a narrow band of frequencies. Here are some examples followed by their frequencies of operation:__________.
Complete question is;
The frequency and wavelength of EM waves can vary over a wide range of values. Scientists refer to the full range of frequencies that EM radiation can have as the electromagnetic spectrum. Electromagnetic waves are used extensively in modern wireless technology. Many devices are built to emit and/or receive EM waves at a very specific frequency, or within a narrow band of frequencies. Here are some examples followed by their frequencies of operation:
garage door openers: 40.0 MHz
standard cordless phones: 40.0 to 50.0 MHz
baby monitors: 49.0 MHz
FM radio stations: 88.0 to 108 MHz
cell phones: 800 to 900 MHz
Global Positioning System: 1227 to 1575 MHz
microwave ovens: 2450 MHz
wireless internet technology: 2.4 to 2.6 GHz
Which of the following statements correctly describe the various applications listed above? Check all that apply.
a.) All these technologies use radio waves, including low-frequency microwaves.
b.) All these technologies use radio waves, including high-frequency microwaves.
c.) All these technologies use a combination of infrared waves and high-frequency microwaves.
d.) Microwave ovens emit in the same frequency band as some wireless Internet devices.
e.) The radiation emitted by wireless Internet devices has the shortest wavelength of all the technologies listed above.
f.) All these technologies emit waves with a wavelength in the range of 0.10 to 10.0 m.
g.) All the technologies emit waves with a wavelength in the range of 0.01 to 10.0 km.
Answer:
B, D, E, F are the correct statements.
Explanation:
Looking at the options;
A) This is true because radio waves are electromagnetic radiation being used today in television, mobile phones, radios and other areas of communication technologies. And the examples given to us fall in the category of technologies that use radio waves.
B) microwaves usually have long wavelengths and low frequencies. However, sometimes they could have high frequencies usually more than radio waves. Thus, this option is correct.
C) This option is wrong because it's not all the listed technologies that use combination of infrared waves and high-frequency microwaves.
D) we are given the frequency of microwave ovens as 2450 MHz.
Converting to GHz gives; 2.45 GHz.
We are told that wireless internet technology has frequency between 2.4 to 2.6 GHz. Thus, microwave frequency falls in the same range as wireless internet technology and thus the statement is true.
E) we know that wavelength is inversely proportional to frequency. This means that the higher the frequency, the shorter the wavelength.
In the frequencies given to us, wireless internet technology have the highest frequency which means they have the shortest wavelength. The statement is true.
F) from the frequencies given to us, the smallest is garage door openers = 40.0 MHz = 40 × 10^(6) Hz while the biggest is 2.6 GHz = 2.6 × 10^(9) Hz
Formula for wavelength is;
Wavelength = speed of light/frequency
Speed of light = 3 × 10^(8) m/s
Thus;
Wavelength = (3 × 10^(8))/(40 × 10^(6))
Or wavelength = (3 × 10^(8))/(2.6 × 10^(9))
So,wavelength = 7.5 m or 0.12 m
This falls into the given range of 0.10 to 10.0 m.
Thus, the statement is true.
a Ferris wheel with a diameter of 35 m starts from rest and achieves its maximum operational tangential speed of 2.3 m/s in a time of 15 s. what is the magnitude of the wheels angular acceleration?
b. what is the magnitude of the tangential acceleration after the maximum operational speed is reached?
(a) The magnitude of the wheels angular acceleration is 0.0088 rad/s².
(b) The magnitude of the tangential acceleration after the maximum operational speed is reached is 0.153 m/s².
Angular acceleration of the wheel
The angular acceleration of the wheel is calculated as follows;
α = ω/t
ω = v/r
α = v/(rt)
α = (2.3)/(17.5 x 15)
α = 0.0088 rad/s²
Tangential acceleration of the wheela = v/t
a = (2.3)/15
a = 0.153 m/s²
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PLEASE HELP
the graph shows a plot of an objects velocity versus time for 15 seconds. is the acceleration of the object constant or changing? how do you know? what does this tell you about the net force on the object?
Answer:
It cannot be constant because if it does not change and each time it increases its strength and speed.
Explanation:
A vertical straight conductor X of length 0.5m is held along the positive X-axis and situated in a uniform horizontal magnetic field of 0.1T which is pointing towards the positive Y-axis. (i) Calculate the magnitude and direction of force on X, when a current of 4A is passed through it. (ii) Through what angle must X be turned in a vertical plane so that the force on X is halved
Answer:
i. 0.2 N ii. 30°
Explanation:
(i) Calculate the magnitude and direction of force on X, when a current of 4A is passed through it.
The magnetic force F = BILsinФ where B = magnetic field strength = 0.1 T, I = current = 4 A and L= length of conductor = 0.5 m. Since the conductor X of length 0.5m is held along the positive X-axis and situated in a uniform horizontal magnetic field of 0.1T which is pointing towards the positive Y-axis, both B and L are perpendicular to each other. So, Ф = 90°
So, F = BILsinФ
F = 0.1 T × 4 A ×0.5 m × sin90°
F = 0.1 T × 4 A ×0.5 m × 1
F = 0.2 N
(ii) Through what angle must X be turned in a vertical plane so that the force on X is halved
If F' = BILsinФ' where Ф'=the new angle, and BIL = F
F'/F = sinФ'
Since F'/F = 1/2
sinФ' = 1/2
Ф' = sin⁻¹(1/2) = 30°
An electron is confined in the 1s state of Hydrogen atom. Find ∆. Given that
∆ = ( in pic )
Help me pls!
The electron's distance from the nucleus in the 1s state of a hydrogen atom is A = [(r * (a^3 / 4√π) * ∫ e^(-2r/a) dr) - (a^3 / 4√π) * (∫ e^(-2r/a) dr)^2] from 0 to ∞ = [(r * (a^3 / 4√π) * (a/2)) - (a^3 / 4√π) * (a/2)^2].
To find the expectation value of the electron's distance from the nucleus in the 1s state of a hydrogen atom, we need to calculate the quantity Ar, where A is given by:
A = ⟨r⟩ = ∫ r * ψ(1s) * r^2 dr from 0 to ∞
First, let's express ψ(1s) in terms of the radial wavefunction R(r):
ψ(1s) = R(r) * Y(0,0) [Since the 1s state has azimuthal quantum number l = 0 and magnetic quantum number m = 0]
The radial wavefunction R(r) for the 1s state is given by:
R(r) = (1 / √π * a^(3/2)) * e^(-r/a)
Now, let's calculate A:
A = ∫ r * ψ(1s) * r^2 dr from 0 to ∞
= ∫ r * (1 / √π * a^(3/2)) * e^(-r/a) * r^2 dr from 0 to ∞
We can use integration by parts to simplify this integral. Let's define u = r and dv = (1 / √π * a^(3/2)) * e^(-r/a) * r^2 dr.
Taking the differentials, we have du = dr and v = ∫ (1 / √π * a^(3/2)) * e^(-r/a) * r^2 dr.
Integrating v, we can use the given integrals:
v = ∫ (1 / √π * a^(3/2)) * e^(-r/a) * r^2 dr
= (a^3 / 4√π) * ∫ e^(-2r/a) dr [Using the given integral]
Now, applying integration by parts:
∫ u dv = uv - ∫ v du
∫ r * (1 / √π * a^(3/2)) * e^(-r/a) * r^2 dr
= (r * v) - ∫ v du
= (r * (a^3 / 4√π) * ∫ e^(-2r/a) dr) - ∫ [(a^3 / 4√π) * ∫ e^(-2r/a) dr] dr
Simplifying the expression:
A = [(r * (a^3 / 4√π) * ∫ e^(-2r/a) dr) - (a^3 / 4√π) * ∫ ∫ e^(-2r/a) dr dr] from 0 to ∞
The integral ∫ ∫ e^(-2r/a) dr dr can be expressed in terms of the given integrals:
∫ ∫ e^(-2r/a) dr dr = (∫ e^(-2r/a) dr)^2
= (a/2)^2 [Using the given integral]
Now, substituting the values and evaluating the limits:
A = [(r * (a^3 / 4√π) * ∫ e^(-2r/a) dr) - (a^3 / 4√π) * (∫ e^(-2r/a) dr)^2] from 0 to ∞
= [(r * (a^3 / 4√π) * (a/2)) - (a^3 / 4√π) * (a/2)^2]
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Irrigation channels that require regular flow monitoring are often equipped with electromagnetic flowmeters in which the magnetic field is produced by horizontal coils embedded in the bottom of the channel. A particular coil has 100 turns and a diameter of 6.0 m. When it's time for a measurement, a 4.5 A current is turned on. The large diameter of the coil means that the field in the water flowing directly above the center of the coil is approximately equal to the field in the center of the coil. The field is directed downward and the water is flowing east. The water is flowing above the center of the coil at 1.5 m/s
Required:
a. What is the magnitude of the field at the center of the coil?
b. What is the direction of the force on a positive ion in the water above the center of the coil?
Answer:
A) B = 9.425 × 10^(-5) T
B) North direction
Explanation:
A) We are given;
Current in coil; I = 4.5 A
Number of turns; N = 100 turns
Radius;R = diameter/2 = 6/2 = 3 m
Formula for the magnetic field at the center of the coil is given by;
B = (μ_o•N•I)/2R
Where μ_o is a constant = 4π × 10^(-7) H/m
Thus;
B = (4π × 10^(-7) × 100 × 4.5)/(2 × 3)
B = 9.425 × 10^(-5) T
B) The direction of the force on a positive ion in water can be gotten by the application of flemmings right hand rule.
From flemmings right hand rule, we know that;
- The thumb indicates the direction of the motion of the force which is in the north direction.
- The Index finger indicates the direction of the magnetic field which is in the east direction
- The middle finger indicates the direction of magnetic field which is downwards in the west direction.
Therefore, the direction of the force as seen from flemmings right hand rule is in the north direction
Two protons move with uniform circular motion in the presence of uniform magnetic fields. Proton one moves twice as fast as proton two. The magnitude of the magnetic field in which proton one is immersed is twice the magnitude of the magnetic field in which proton two is immersed. The radius of the circle around which proton one moves is r. What is the radius of the circle around which proton two moves
Answer:
r₂ = 4 r
Explanation:
For this exercise let's use Newton's second law with the magnetic force
F = q v x B
bold letters indicate vectors, the magnitude of this expression is
F = q v B sin θ
in this case we assume that the angle is 90º between the speed and the magnetic field.
If we use the rule of the right hand with the positive charge, the thumb in the direction of the speed, the fingers extended in the direction of the magnetic field, the palm points in the direction of the force, which is towards the center of the circle, therefore the force is radial and the acceleration is centripetal
a = v² / r
let's use Newton's second law
F = ma
q v B = m v² / r
r = [tex]\frac{qB}{mv}[/tex]
Let's apply this expression to our case.
Proton 1
r = \frac{qB_1}{mv_1}
Proton 2
r₂ = [tex]\frac{q \ B_2}{m \ v_2}[/tex]
in the exercise indicate some relationships between the two protons
* v₁ = 2 v₂
v₂ = v₁ / 2
* B₂ = 2B₁
we substitute
r₂ = [tex]\frac{q \ 2B_1}{m \ \frac{v_1}{2} }[/tex]
r₂ = 4 [tex]\frac{qB_1}{mv_1}[/tex]
r₂ = 4 r
When a car makes a sharp left turn, what causes the passengers to move toward the right side of the car? *
A centrifugal force
B inertia
C centripetal acceleration
D centripetal force
B, the body at rest becomes reluctant to start moving or a body in motion becomes
reluctant and stop moving once in motion in a straight line
A go-cart is traveling at a rate of 25 m/sec for 20 seconds. How far will the go cart travel?
Answer:
Distance travel by go-cart = 500 meter
Explanation:
Given:
Speed of go cart = 25 m/s
Time travel = 20 seconds
Find:
Distance travel by go-cart
Computation:
Distance = Speed x time
Distance travel by go-cart = Speed of go cart x Time travel
Distance travel by go-cart = 25 x 20
Distance travel by go-cart = 500 meter
should scientist be allowed to do anything that they can?
No, because some scientist are bad and they must have a scientific attitude like honesty, curiosity, open-mindedness, optimistic and more. And they should follow others sometimes because for their own good.
Answer:
No
Explanation:
Science operates in the context of national and international laws, agreements and conventions. It requires scientists to conduct and communicate scientific work for the benefit of society, with excellence, integrity, respect, fairness, trustworthiness, clarity, and transparency.
A rifle can shoot a 4.00 g bullet at a speed of 998 m/s. Find the kinetic energy of the bullet. What work is done on the bullet if it starts from rest?
Answer:
1992.008J
Explanation:
Only one setup below will light the bulb. Which setup will light the bulb?
Answer:
A. it's got everything set.... correct connection
The setup that will light the bulb in the given different circuit diagram is circuit A, because it is the only circuit is that is closed.
What is a closed circuit?A complete circuit or closed circuit is the circuit that will allow the flow of electric current through it.
When a circuit is closed or complete, the connection of the different components of the circuit is complete.
From the image given, only option A has complete connection of different components of the circuit.
Thus, the setup that will light the bulb in the given different circuit diagram is circuit A, because it is the only circuit is that is closed.
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A 45 kg object is lifted vertically at a constant speed to a height of 9.0 m by a 7.5x10² w electric motor. If this motor is 25% efficient in converting electric energy to mechanical energy, how long does the motor take to lift the object?
Answer:
the time taken for the motor to lift the object is 21.17 s.
Explanation:
Given;
mass of the object, m = 45 kg
height through which the object was lifted, h = 9.0 m
electrical power used by the motor, P = 750 W
Efficiency of the electrical motor, n = 25% = 0.25
The electrical energy used by the motor in lifting the object is calculated as;
E = P x t
where;
t is the time taken for the motor to lift the object
E = 750 x t
E = 750t
The electrical energy converted by the motor to mechanical energy is calculated as;
P.E = 0.25(750t)
P.E = 187.5t
Recall, P.E = mgh
mgh = 187.5t
45 x 9.8 x 9 = 187.5t
3969 = 187.5t
t = 3969/187.5
t = 21.17 s
Therefore, the time taken for the motor to lift the object is 21.17 s.
[tex]\\ x^{2} \int\limits^a_b {x} \, dx \int\limits^a_b {x} \, dx \neq \pi \pi \left[\begin{array}{ccc}1&2&3\\4&5&6\\7&8&9\end{array}\right] \left[\begin{array}{ccc}1&2&3\\4&5&6\\7&8&9\end{array}\right] \left[\begin{array}{ccc}1&2&3\\4&5&6\\7&8&9\end{array}\right][/tex]
Answer:
I think you have a problem trying to write the question down, can you re write it because it seems glitched.
Explanation:
describe what will happen to our sun in the future
Answer: After fusing helium in its core to carbon, the Sun will begin to collapse again, evolving into a compact white dwarf star after ejecting its outer atmosphere as a planetary nebula. The predicted final mass is 54.1% of the present value, most likely consisting primarily of carbon and oxygen.
Answer: the Sun will begin to collapse
Explanation: "After fusing helium in its core to carbon, the Sun will begin to collapse again, evolving into a compact white dwarf star after ejecting its outer atmosphere as a planetary nebula. The predicted final mass is 54.1% of the present value, most likely consisting primarily of carbon and oxygen."
What is the term used to describe the
force pushing on an area or surface?
A. density
B. pressure
C. surface tension
D. volume
In an experiment, a disk is set into motion such that it rotates with a constant angular speed. As the disk spins, a small sphere of clay is dropped onto the disk, and the sphere sticks to the disk. All frictional forces are negligible. What would happened to the angular momentum and the total kinetic energy of the disk-sphere system immediately before and after the collision?
Answer:
L₀ = L_f , K_f < K₀
Explanation:
For this exercise we start as the angular momentum, with the friction force they are negligible and if we define the system as formed by the disk and the clay sphere, the forces during the collision are internal and therefore the angular momentum is conserved.
This means that the angular momentum before and after the collision changes.
Initial instant. Before the crash
L₀ = I₀ w₀
Final moment. Right after the crash
L_f = (I₀ + mr²) w
we treat the clay sphere as a point particle
how the angular momentum is conserved
L₀ = L_f
I₀ w₀ = (I₀ + mr²) w
w = [tex]\frac{I_o}{I_o + m r^2}[/tex] w₀
having the angular velocities we can calculate the kinetic energy
starting point. Before the crash
K₀ = ½ I₀ w₀²
final point. After the crash
K_f = ½ (I₀ + mr²) w²
sustitute
K_f = ½ (I₀ + mr²) ( [tex]\frac{I_o}{I_o + m r^2}[/tex] w₀)²
Kf = ½ [tex]\frac{I_o^2}{ I_o + m r^2}[/tex] w₀²
we look for the relationship between the kinetic energy
[tex]\frac{K_f}{K_o}[/tex]= [tex]\frac{I_o}{I_o + m r^2}[/tex]
[tex]\frac{K_f}{K_o } < 1[/tex]
K_f < K₀
we see that the kinetic energy is not constant in the process, this implies that part of the energy is transformed into potential energy during the collision
What fuel does a main-sequence star use for nuclear fusion?
oxygen (0)
petroleum
helium (He)
hydrogen (H)
Answer:
A main sequence star is powered by fusion of hydrogen into helium in its core
Explanation:
Please help I will mark you brainliest
I believe the answer is a
How many excess electrons must be distributed uniformly within the volume of an isolated plastic sphere 30.0 cmcm in diameter to produce an electric field of 1440 N/CN/C just outside the surface of the sphere
Answer:
1.78×10×10^10 electron
Explanation:
Electric field outside the sphere can be calculated using below expression
E= kq/ r^2..........eqn(1)
Where k= 9 × 10^9 NM^2/C^2
q= charge
E= 1440 N/C
Diameter= 30.0 cm= 0.3 m
r= radius= 0.3/2= 0.15m
If we make q subject of formula from eqn(1) we have
q= Er^2/k............eqn(2)
q= 1440 × (0.15)^2 /(9 × 10^9 )
= 2.85×10^-9C
Total charge is an integer of electron charge , then we can calculate the number of the electron using the expression below
q= Ne
Where
N = number of electron
Making N subject of the formula we have
N= q/e
Where e= electron value= 1.6× 10^-19
N=2.85×10^-9 /1.6× 10^-19
= 1.78×10×10^10 electron
What happens to a light wave that is absorbed by matter
Answer:
In absorption, the frequency of the incoming light wave is at or near the energy levels of the electrons in the matter. The electrons will absorb the energy of the light wave and change their energy state.
Explanation:
Rasheeda made a model of a plant cell and labeled parts of the cell that are not found in animal cells. Which other part of the cell should Rasheeda label?
A) cytoplasm
B) large central vacuole
C) lysosomes
D) mitochondria
Answer:
Cytoplasm
Explanation:
A car braked with a constant deceleration of 36 ft/s2, producing skid marks measuring 50 ft before coming to a stop. How fast was the car traveling when the brakes were first applied
Answer:
Initial velocity u = 60 ft/s
Explanation:
Given:
Deceleration a = -36 ft/s²
Distance covered s =50ft
Final velocity v = 0 ft/s
Find:
Initial velocity u
Computation:
Using third equation of motion;
v² = u² + 2as
0² = u² + 2(-36)(50)
0 = u² - 3600
u² = 3600
u = 60 ft/s
Initial velocity u = 60 ft/s
A 20 ft ladder leans against a wall. The bottom of the ladder is 3 ft from the wall at time t=0 and slides away from the wall at a rate of 2ft/sec. Find the velocity of the top of the ladder at time t=1.
Answer: 0.516 ft/s
Explanation:
Given
Length of ladder L=20 ft
The speed at which the ladder moving away is v=2 ft/s
after 1 sec, the ladder is 5 ft away from the wall
So, the other end of the ladder is at
[tex]\Rightarrow y=\sqrt{20^2-5^2}=19.36\ ft[/tex]
Also, at any instant t
[tex]\Rightarrow l^2=x^2+y^2[/tex]
differentiate w.r.t.
[tex]\Rightarrow 0=2xv+2yv_y\\\\\Rightarrow v_y=-\dfrac{x}{y}\times v\\\\\Rightarrow v_y=-\dfrac{5}{19.36}\times 2=0.516\ ft/s[/tex]