Answer:
a. 24 m
Explanation:
Destructive interference occurs when two waves arrive at a point, out of phase. In a completely destructive interference, the two waves cancel out, but in a partially destructive interference, they produce a wave with a time varying amplitude, but maintain a wavelength the wavelength of one of the original waves. Since the two waves does not undergo complete destructive interference, then the possible value of the new wave formed can only be 24 m, from the options given.
"Neon signs need 12,000 V to operate. If a transformer operates off a 240 V source and has 1000 turns in its primary coil, how may turns must the secondary coil have
Answer:
50000 turns
Explanation:
Vp / Vs = Np / Ns
240 / 12000 = 1000 / Ns
Ns = 50000 turns
change in entropy of universe during 900g of ice at 0 degree celcus to water at 10 degree celcius at room temp=30 degree celcius
Answer:
4519.60 J/KExplanation:
Change in entropy is expressed as ΔS = ΔQ/T where;
ΔQ is the total heat change during conversion of ice to water.
T is the room temperature
First we need to calculate the total change in heat using the conversion formulae;
ΔQ = mL + mcΔθ (total heat energy absorbed during phase change)
m is the mass of ice/water = 900g = 0.9kg
L is the latent heat of fusion of ice = 3.33 x 10⁵J/kg
c is the specific heat capacity of water = 4200J/kgK
Δθ is the change in temperature of water = 10°C - 0C = 10°C = 283K
Substituting the given values into ΔQ;
ΔQ = 0.9(333000)+0.9(4200)(283)
ΔQ = 299700 + 1069740
ΔQ = 1,369,440 Joules
Since Change in entropy ΔS = ΔQ/T
ΔS = 1,369,440/30+273
ΔS = 1,369,440/303
ΔS = 4519.60 J/K
Hence, the change in entropy of the universe is 4519.60 J/K
What is the requirement for the photoelectric effect? Select one: a. The incident light must have enough intensity b. The incident light must have a wavelength shorter than visible light c. The incident light must have at least as much energy as the electron work function d. Both b and c
Answer:
c. The incident light must have at least as much energy as the electron work function
Explanation:
In photoelectric effect, electrons are emitted from a metal surface when a light ray or photon strikes it. An electron either absorbs one whole photon or it absorbs none. After absorbing a photon, an electron either leaves the surface of metal or dissipate its energy within the metal in such a short time interval that it has almost no chance to absorb a second photon. An increase in intensity of light source simply increase the number of photons and thus, the number of electrons, but the energy of electron remains same. However, increase in frequency of light increases the energy of photons and hence, the
energy of electrons too.
Therefore, the energy of photon decides whether the electron shall be emitted or not. The minimum energy required to eject an electron from the metal surface, i.e. to overcome the binding force of the nucleus is called ‘Work Function’
Hence, the correct option is:
c. The incident light must have at least as much energy as the electron work function
In the circuit shown, the galvanometer shows zero current. The value of resistance R is :
A) 1 W
B) 2 W
C) 4 W
D) 9 W
Answer:
its supposed to be (a) 1W
A wooden artifact from a Chinese temple has a 14C activity of 41.0 counts per minute as compared with an activity of 58.2 counts per minute for a standard of zero age. You may want to reference (Pages 913 - 916) Section 21.4 while completing this problem. Part A From the half-life for 14C decay, 5715 yr, determine the age of the artifact. Express your answer using two significant figures. t
Answer:
Explanation:
The relation between activity and number of radioactive atom in the sample is as follows
dN / dt = λ N where λ is disintegration constant and N is number of radioactive atoms
For the beginning period
dN₀ / dt = λ N₀
58.2 = λ N₀
similarly
41 = λ N
dividing
58.2 / 41 = N₀ / N
N = N₀ x .70446
formula of radioactive decay
[tex]N=N_0e^{-\lambda t }[/tex]
[tex].70446 =e^{-\lambda t }[/tex]
- λ t = ln .70446 = - .35
t = .35 / λ
λ = .693 / half life
= .693 / 5715
= .00012126
t = .35 / .00012126
= 2886.36
= 2900 years ( rounding it in two significant figures )
Two positive charges are located at x = 0, y = 0.3m and x = 0, y = -.3m respectively. Third point charge q3 = 4.0 μC is located at x = 0.4 m, y = 0.
A) Make a careful sketch of decent size that illustrates all force vectors with directions and magnitudes.
B) What is the resulting vector of the total force on charge q1 exerted by the other two charges using vector algebra?
Answer:
0.46N
Explanation:
See attached file
A charged particle is moving with speed v perpendicular to a uniform magnetic field. A second identical charged particle is moving with speed 2v perpendicular to the same magnetic field. If the frequency of revolution of the first particle is f, the frequency of revolution of the second particle is
Answer:
the frequency of revolution of the second particle is f
Explanation:
centripetal force is balanced by the magnetic force for object under magnetic field is given as
Mv²/r= qvB
But v= omega x r
Omega= 2pi x f
f= qB/2pi x M
So since frequency does not depend on the velocity.therefore the frequency of revolution of the second particle remains the same and its equal to f
if a speed sound in air at o°c is 331m/s. what will be its value at 35 °c
Answer:
please brainliest!!!
Explanation:
V1/√T1 =V2/√T2
V1 = 331m/s
T1 = 0°C = 273k
V2 = ?
T2 = 35°c = 308k
331/√273 = V2/√308331/16.5 = V2/17.520.06 = V2/17.5V2 = 20.06 x 17.5 V2 = 351.05m/sA force acting on an object moving along the x axis is given by Fx = (14x - 3.0x2) N where x is in m. How much work is done by this force as the object moves from x = -1 m to x = +2 m?
Answer:
72J
Explanation:
distance moved is equal to 3m.then just substitute x with 3m.
Fx = (14(3) - 3.0(3)2)) N
Fx =(42-18)N
Fx =24N
W=Fx *S
W=24N*3m
W=72J
The answer is 72J.
Distance moved is equal to 3m.
Then just substitute x with 3m.
Fx = (14(3) - 3.0(3)2)) N
Fx =(42-18)N
Fx =24N
W=Fx *S
W=24N*3m
W=72J
Is there any definition of force?A force is a push or pulls upon an object resulting from the object's interaction with another object. Whenever there is an interaction between two objects, there is a force upon each of the objects.
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An appliance with a 20.0-2 resistor has a power rating of 15.0 W. Find the maximum current which can flow safely through the appliance g
Q: An appliance with a 20 Ω resistor has a power rating of 15.0 W. Find the maximum current which can flow safely through the appliance g
Answer:
0.866 A
Explanation:
From the question,
P = I²R............................. Equation 1
Where P = power, I = maximum current, R = Resistance.
Make I the subject of the equation
I = √(P/R).................... Equation 2
Given: P = 15 W, R = 20 Ω
Substitute these values into equation 2
I = √(15/20)
I = √(0.75)
I = 0.866 A
Hence the maximum current that can flow safely through the appliance = 0.866 A
A communications satellite orbiting the earth has solar panels that completely absorb all sunlight incident upon them. The total area A of the panels is 10m2.
1) The intensity of the sun's radiation incident upon the earth is about I=1.4kW/m2. Suppose this is the value for the intensity of sunlight incident upon the satellite's solar panels. What is the total solar power P absorbed by the panels?
Express your answer numerically in kilowatts to two significant figures.
2) What is the total force F on the panels exerted by radiation pressure from the sunlight?
Express the total force numerically, to two significant figures, in units of newtons.
Answer:
1) 14 kW
2) 4.67 x 10^-5 N
Explanation:
Area of solar panel = 10 m^2
Intensity of sun's radiation incident on earth = 1.4 kW/m^2
Solar power absorbed = ?
We know that the intensity of radiation on a given area is
[tex]I[/tex] = [tex]\frac{P}{A}[/tex]
where I is the intensity of the radiation
P is the power absorbed due to this intensity on a given area
A is the area on which this radiation is incident
From the equation, we have
P = IA
P = 1.4 kW/m^2 x 10 m^2 = 14 kW
b) For a perfect absorbing surface, the radiation pressure is given as
p = I/c
where p is the radiation pressure
I is the incident light intensity = 1.4 kW/m^2 = 1.4 x 10^3 kW/m^2
c is the speed of light = 3 x 10^8 m/s
substituting values, we have
p = (1.4 x 10^3)/(3 x 10^8) = 4.67 x 10^-6 Pa
we know that Force = pressure x area
therefore force on the solar panels is
F = 4.67 x 10^-6 x 10 = 4.67 x 10^-5 N
A wave travels at a consent speed. how does the frequency change if the wavelength is reduced by a factor of 4?
Answer:
The frequency increases by 4 because it is inversely proportional to the wavelength.
what is the mass of an oil drop having two extra electrons that is suspended motionless by the field between the plates
Answer:
m = 3,265 10⁻²⁰ E
Explanation:
For this exercise we can use Newton's second law applied to our system, which consists of a capacitor that creates the uniform electric field and the drop of oil with two extra electrons.
∑ F = 0
[tex]F_{e}[/tex] - W = 0
the electric force is
F_{e} = q E
as they indicate that the charge is two electrons
F_{e} = 2e E
The weight is given by the relationship
W = mg
we substitute in the first equation
2e E = m g
m = 2e E / g
let's put the value of the constants
m = (2 1.6 10⁻¹⁹ / 9.80) E
m = 3,265 10⁻²⁰ E
The value of the electric field if it is a theoretical problem must be given and if it is an experiment it can be calculated with measures of the spacing between plates and the applied voltage, so that the system is in equilibrium
Suppose a proton moves to the right and enters a uniform magnetic field into the page. It follows trajectory B with radius rp. An alpha particle (twice the charge and 4 times the mass) enters the same magnetic field in the same way and with the same velocity as the proton. Which path best represents the alpha particle’s trajectory?
Answer:
R = r_protón / 2
Explanation:
The alpha particle when entering the magnetic field experiences a force and with Newton's second law we can describe its movement
F = m a
Since the magnetic force is perpendicular, the acceleration is centripetal.
a = v² / R
the magnetic force is
F = q v x B = q v B sin θ
the field and the speed are perpendicular so the sin 90 = 1
we substitute
qv B = m v² / R
R = q v B / m v²
in the exercise they indicate
the charge q = 2 e
the mass m = 4 m_protón
R = 2e v B / 4m_protón v²
we refer the result to the movement of the proton
R = (e v B / m_proton) 1/2
the data in parentheses correspond to the radius of the proton's orbit
R = r_protón / 2
what is transmission of heat?
Answer:
Heat transfer is the transmission of heat energy from a body at higher temperature to lower temperature. The three mechanisms of heat transfer are
Conduction ConvectionRadiation.Example of Conduction:
Heating a metal
Example of Convection:
Sea Breeze
Example of Radiation:
Sun
Hope this helps ;) ❤❤❤
Answer:
Transmission of heat is the movement of thermal energy from one thing to another thing of different temperature.
There are three(3) different ways heat can transfer and they are:
a) Conduction (through direct contact).
b) Convection (through fluid movement).
c) Radiation (through electromagnetic waves).
Examples: 1.Heating a saucepan of water using a coalpot.(conduction&convection).
2. Baking a pie in an oven(radiation).
Hope it helps!!Please mark me as the brainliest!!!Thanks!!!!❤❤❤
The velocity of an object is given by the following function defined on a specified interval. Approximate the displacement of the object on this interval by sub-dividing the interval into the indicated number of sub-intervals. Use the left endpoint of each sub-interval to compute the height of the rectangles.
v= 4t + 5(m/s) for 3 < t < 7; n = 4
The approximate displacement of the object is______m.
Answer:
The approximate displacement of the object is 23 m.
Explanation:
Given that:
v = 4t + 5 (m/s) for 3< t< 7; n= 4
The approximate displacement of the object can be calculated as follows:
The velocities at the intervals of t are :
3
4
5
6
the velocity at the intervals of t = 7 will be left out due the fact that we are calculating the left endpoint Reimann sum
n = 4 since there are 4 values for t, Then there is no need to divide the velocity values
v(3) = 4(3)+5
v(3) = 12+5
v(3) = 17
v(4)= 4(4)+5
v(4) = 16 + 5
v(4) = 21
v(5)= 4(5)+5
v(5) = 20 + 5
v(5) = 25
v(6) = 4(6)+5
v(6) = 24 + 5
v(6) = 29
Using Left end point;
[tex]= \dfrac{1}{4}(17+21+25+29)[/tex]
= 23 m
Unpolarized light passes through a vertical polarizing filter, emerging with an intensity I0. The light then passes through a horizontal filter, which blocks all of the light; the intensity transmitted through the pair of filters is zero. Suppose a third polarizer with axis 45 ? from vertical is inserted between the first two.
What is the transmitted intensity now?
Express your answer in terms of I0. I got I0/8. But this is not right. I guess they want a number?
Answer:
I₂ = 0.25 I₀
Explanation:
To know the light transmitted by a filter we must use the law of Malus
I = I₀ cos² θ
In this case, the intensity of the light that passes through the first polarizer is I₀, it reaches the second polarized, which is at 45⁰, therefore the intensity I1 comes out of it.
I₁ = I₀ cos² 45
I₁ = I₀ 0.5
this is the light that reaches the third polarizer, which is at 45⁰ with respect to the second, from this comes the intensity I₂
I₂ = I₁ cos² 45
I₂ = (I₀ 0.5) 0.5
I₂ = 0.25 I₀
this is the intensity of the light transmitted by the set of polarizers
A jet transport with a landing speed of 200 km/h reduces its speed to 60 km/h with a negative thrust R from its jet thrust reversers in a distance of 425 m along the runway with constant deceleration. The total mass of the aircraft is 140 Mg with mass center at G. Compute the reaction N under the nose wheel B toward the end of the braking interval and prior to the application of mechanical braking. At lower speed, aerodynamic forces on the aircraft are small and may be neglected.
Answer:
257 kN.
Explanation:
So, we are given the following data or parameters or information in the following questions;
=> "A jet transport with a landing speed
= 200 km/h reduces its speed to = 60 km/h with a negative thrust R from its jet thrust reversers"
= > The distance = 425 m along the runway with constant deceleration."
=> "The total mass of the aircraft is 140 Mg with mass center at G. "
We are also give that the "aerodynamic forces on the aircraft are small and may be neglected at lower speed"
Step one: determine the acceleration;
=> Acceleration = 1/ (2 × distance along runway with constant deceleration) × { (landing speed A)^2 - (landing speed B)^2 × 1/(3.6)^2.
=> Acceleration = 1/ (2 × 425) × (200^2 - 60^2) × 1/(3.6)^2 = 3.3 m/s^2.
Thus, "the reaction N under the nose wheel B toward the end of the braking interval and prior to the application of mechanical braking" = The total mass of the aircraft × acceleration × 1.2 = 15N - (9.8 × 2.4 × 140).
= 140 × 3.3× 1.2 = 15N - (9.8 × 2.4 × 140).
= 257 kN.
The reaction N under the nose wheel B towards the end of the braking interval = 257 kN
Given data :
Landing speed of Jet = 200 km/h
Distance = 425 m
Total mass of aircraft = 140 Mg with mass center at G
Determine the reaction N under the nose of wheel B First step : calculate the value of the Jet accelerationJet acceleration = 1 / (2 *425) * (200² - 60² ) * 1 / (3.6)²
= 3.3 m/s²
Next step : determine the reaction N under the nose of WheelReaction N = Total mass of aircraft * jet acceleration* 1.2 = 15N - (9.8*2.4* 140). ----- ( 1 )
∴ Reaction N = 140 * 3.3 * 1.2 = 15 N - ( 9.8*2.4* 140 )
Hence Reaction N = 257 KN
We can conclude that the The reaction N under the nose wheel B towards the end of the braking interval = 257 kN
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What is the minimum magnitude of an electric field that balances the weight of a plasticsphere of mass 5.4 g that has been charged to -3.0 nC
Answer:
E = 17.64 x 10⁶ N/C = 17.64 MN/C
Explanation:
The electric field is given by the following formula:
E = F/q
E= W/q
E = mg/q
where,
E = magnitude of electric field = ?
m = mass of plastic sphere = 5.4 g = 5.4 x 10⁻³ kg
g = acceleration due to gravity = 9.8 m/s²
= charge = 3 nC = 3 x 10⁻⁹ C
Therefore,
E = (5.4 x 10⁻³ kg)(9.8 m/s²)/(3 x 10⁻⁹ C)
E = 17.64 x 10⁶ N/C = 17.64 MN/C
21. What is the most likely outcome of decreasing the frequency of incident light on a diffraction grating?
A. lines become narrower
B. distance between lines increases
C. lines become thicker
D. distance between lines decreases
Answer:
B.distance between lines increases
Answer:
A. Lines become narrower
Explanation:
I got it right on my quiz!
I hope this helps!! :))
You are fixing a transformer for a toy truck that uses an 8.0-V emf to run it. The primary coil of the transformer is broken; the secondary coil has 40 turns. The primary coil is connected to a 120-V wall outlet.
(a) How many turns should you have in the primary coil?
(b) If you then connect this primary coil to a 240-V source, what emf would be across the secondary coil?
Comments: The relevant equation is N1/N2 = V1/V2 where N is the number of turns and V is the voltage. I'm just not sure how to get the voltage of the secondary coil using emf.
Answer:
a. The primary turns is 60 turns
b. The secondary voltage will be 360 volts.
Explanation:
Given data
secondary turns N2= 40 turns
primary turns N1= ?
primary voltage V1= 120 volts
secondary voltage V2= 8 volts
Applying the transformer formula which is
[tex]\frac{N1}{N2} =\frac{V1}{V2}[/tex]
we can solve for N1 by substituting into the equation above
[tex]\frac{N1}{40} =\frac{120}{8} \\\ N1= \frac{40*120}{8} \\\ N1= \frac{4800}{8} \\\ N1= 60[/tex]
the primary turns is 60 turns
If the primary voltage is V1 240 volts hence the secondary voltage V2 will be (to get the voltage of the secondary coil using emf substitute the values of the previously gotten N1 and N2 using V1 as 240 volts)
[tex]\frac{40}{60} =\frac{240}{V2}\\\\V2= \frac{60*240}{40} \\\\V2=\frac{ 14400}{40} \\\\V2= 360[/tex]
the secondary voltage will be 360 volts.
(a) In the primary coil, you have "60 turns".
(b) The emf across the secondary coil would be "360 volts".
Transformer and VoltageAccording to the question,
Primary voltage, V₁ = 120 volts
Secondary voltage, V₂ = 8 volts
Secondary turns, N₂ = 40 turns
(a) By applying transformer formula,
→ [tex]\frac{N_1}{N_2} = \frac{V_1}{V_2}[/tex]
or,
N₁ = [tex]\frac{N_2\times V_1}{V_2}[/tex]
By substituting the values,
= [tex]\frac{40\times 120}{8}[/tex]
= [tex]\frac{4800}{8}[/tex]
= 60
(2) Again by using the above formula,
→ V₂ = [tex]\frac{60\times 240}{40}[/tex]
= [tex]\frac{14400}{40}[/tex]
= 360 volts.
Thus the above approach is correct.
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You indicate that a symbol
is a vector by drawing
A. through the symbol.
B. over the symbol.
c. under the symbol.
D. before the symbol.
Answer:
B. over the symbol.
Explanation:
vectors are represented with a symbol carrying an arrow head with also indicates direction
An alternating current is supplied to an electronic component with a warning that the voltage across it should never exceed 12 V. What is the highest rms voltage that can be supplied to this component while staying below the voltage limit in the warning?
Answer:
The highest rms voltage will be 8.485 V
Explanation:
For alternating electric current, rms (root means square) is equal to the value of the direct current that would produce the same average power dissipation in a resistive load
If the peak or maximum voltage should not exceed 12 V, then from the relationship
[tex]V_{rms} = \frac{V_{p} }{\sqrt{2} }[/tex]
where [tex]V_{rms}[/tex] is the rms voltage
[tex]V_{p}[/tex] is the peak or maximum voltage
substituting values into the equation, we'll have
[tex]V_{rms} = \frac{12}{\sqrt{2} }[/tex] = 8.485 V
The copper wire to the motor is 6.0 mm in diameter and 1.1 m long. How far doesan individual electron travel along the wire while the starter motor is on for asingle start of the internal combustion engine
Answer:
0.306mm
Explanation:
The radius of the conductor is 3mm, or 0.003m
The area of the conductor is:
A = π*r^2 = π*(.003)^2 = 2.8*10^-5 m^2
The current density is:
J = 130/2.8*10^-5 = 4.64*10^6 A/m
According to the listed reference:
Vd = J/(n*e) = 4.64*10^6 / ( 8.46*10^28 * 1.6*10^-19 ) = 0.34*10^-6 m/s = 0.34mm/s
The distance traveled is:
x = v*t = 0.34 * .90 = 0.306 mm
A converging lens of focal length 7.40 cm is 18.0 cm to the left of a diverging lens of focal length -7.00 cm . A coin is placed 12.0 cm to the left of the converging lens.
A) Find the location of the coin's final image relative to the diverging lens.
B) Find the magnification of the coin's final image.
Answer:
Explanation:
The set up is a compound microscope. The converging lens is the objective lens while the diverging lens is the eyepiece lens.
In compound microscopes, the distance between the two lenses is expressed as L = v0+ue
v0 is the image distance of the objective lens and ue is the object distance of the eye piece lens.
Befre we can get the location of the coin's final image relative to the diverging lens (ve), we need to get ue first.
Given L = 18.0cm
Using the lens formula to get v0 where u0 = 12.0cm and f0 = 7.40cm
1/f0 = 1/u0+1/v0
f0 and u0 are the focal length and object distance of the converging lens (objective lens)respectively.
1/v0 = 1/7.4-1/12
1/v0 = 0.1351-0.0833
1/v0 = 0.0518
v0 = 1/0.2184
v0 = 19.31cm
Note that v0 = ue = 19.31cm
To get ve, we will use the lens formula 1/fe = 1/ue+1/ve
1/ve = 1/fe-1/ue
Given ue = 19.31cm and fe = -7.00cm
1/ve = -1/7.0-1/19.31
1/ve = -0.1429-0.0518
1/ve = -0.1947
ve = 1/-0.1947
ve = -5.14cm
Hence, the location of the coin's final image relative to the diverging lens is 5.14cm to the lens
b) Magnification of the final image M = ve/ue
M = 5.14/19.31
M = 0.27
Magnification of the final image is 0.27
23.15. Can an object carry a charge of 2.0 10-19 C?
Answer:
Ok, the minimal quantity of charge that we can find is on the electron or in the proton (the magnitude is the same, but the sign is different)
Where the charge of a single proton is:
C = 1.6x10^-19 C
Now, you need to remember that when we are working with charges, we are working with discrete math:
What means that?
If the minimum positive is the charge of one proton, then the consecutive charge will be the charge of two protons (there is no somethin in between)
So the consecutive charge will be:
C = 2*1.6x10^-19 C = 3.2x10^-19 C.
So, because we are working in discrete math, we can not have any object that has charge between 1.6x10^-19 C and 3.2x10^-19 C.
Particularly, 2.0x10^-19 C is in that range, so we can conclude that:
No, an object can not carry a charge of 2.0x10^-19 C.
A ball with a mass of 275 g is dropped from rest, hits the floor and rebounds upward. If the ball hits the floor with a speed of 2.10 m/s and rebounds with a speed of 1.90 m/s, determine the following.
a. magnitude of the change in the ball's momentum (Let up be in the positive direction.)
________ kg - m/s
b. change in the magnitude of the ball's momentum (Let negative values indicate a decrease in magnitude.)
_______ kg - m/s
c. Which of the two quantities calculated in parts (a) and (b) is more directly related to the net force acting on the ball during its collision with the floor?
A. Neither are related to the net force acting on the ball.
B. They both are equally related to the net force acting on the ball.
C. The change in the magnitude of the ball's momentum
D. The magnitude of the change in the ball's momentum
Answer:
a) The magnitude of the change in the ball's momentum is 1.1 kilogram-meters per second, b) The change in the magnitude of the ball's momentum is -0.055 kilogram-meters per second, c) D. The magnitude of the change in the ball's momentum.
Explanation:
a) This phenomenon can be modelled by means of the Principle of Momentum Conservation and the Impact Theorem, whose vectorial form is:
[tex]\vec p_{o} + Imp = \vec p_{f}[/tex]
Where:
[tex]\vec p_{o}[/tex], [tex]\vec p_{f}[/tex] - Initial and final momentums, measured in kilogram-meters per second.
[tex]Imp[/tex] - Impact due to collision, measured in kilogram-meters per second.
The impact experimented by the ball due to collision is:
[tex]Imp = \vec p_{f} - \vec p_{o}[/tex]
By using the definition of momentum, the expression is therefore expanded:
[tex]Imp = m \cdot (\vec v_{f}-\vec v_{o})[/tex]
Where:
[tex]m[/tex] - Mass of the ball, measured in kilograms.
[tex]\vec v_{o}[/tex], [tex]\vec v_{f}[/tex] - Initial and final velocities, measured in meters per second.
If [tex]m = 0.275\,kg[/tex], [tex]\vec v_{o} = -2.10\,j\,\left [\frac{m}{s} \right][/tex] and [tex]\vec v_{f} = 1.90\,j\,\left [\frac{m}{s} \right][/tex], the vectorial change of the linear momentum is:
[tex]Imp = (0.275\,kg)\cdot \left[1.90\,j+2.10\,j\right]\,\left[\frac{m}{s} \right][/tex]
[tex]Imp = 1.1\,j\,\left[\frac{kg\cdot m}{s} \right][/tex]
The magnitude of the change in the ball's momentum is 1.1 kilogram-meters per second.
b) The magnitudes of initial and final momentums of the ball are, respectively:
[tex]p_{o} = (0.275\,kg)\cdot \left(2.10\,\frac{m}{s} \right)[/tex]
[tex]p_{o} = 0.578\,\frac{kg\cdot m}{s}[/tex]
[tex]p_{f} = (0.275\,kg)\cdot \left(1.90\,\frac{m}{s} \right)[/tex]
[tex]p_{o} = 0.523\,\frac{kg\cdot m}{s}[/tex]
The change in the magnitude of the ball's momentum is:
[tex]\Delta p = p_{f}-p_{o}[/tex]
[tex]\Delta p = 0.523\,\frac{kg\cdot m}{s} - 0.578\,\frac{kg\cdot m}{s}[/tex]
[tex]\Delta p = -0.055\,\frac{kg\cdot m}{s}[/tex]
The change in the magnitude of the ball's momentum is -0.055 kilogram-meters per second.
c) The quantity calculated in part a) is more related to the net force acting on the ball during its collision with the floor, since impact is the product of net force, a vector, and time, a scalar, and net force is the product of the ball's mass and net acceleration, which creates a change on velocity.
In a nutshell, the right choice is option D.
Experts, ACE, Genius... can anybody calculate for the Reactions at supports A and B please? Will give brainliest! Given: fb = 300 kN/m, fc = 100 kN/m, Dy = 300 kN, spanAB = 6m, span BC = 6m, spanCD = 6m
Answer:
Support at Cy = 1.3 x 10³ k-N
Support at Ay = 200 k-N
Explanation:
given:
fb = 300 k-N/m
fc = 100 k-N/m
D = 300 k-N
L ab = 6 m
L bc = 6 m
L cd = 6 m
To get the reaction A or C.
take summation of moment either A or C.
Support Cy:
∑ M at Ay = 0
(( x1 * F ) + ( D * Lab ) + ( D * L bc + D * L cd )
Cy = -------------------------------------------------------------------
( L ab + L bc )
Cy = 1.3 x 10³ k-N
Support Ay:
Since ∑ F = 0, A + C - F - D = 0
A = F + D - C
Ay = 200 k-N
Answer:
i was going to but its to late
Explanation:
An electron and a proton each have a thermal kinetic energy of 3kBT/2. Calculate the de Broglie wavelength of each particle at a temperature of 1950 K. (kb is Boltzmann's constant, 1.38x10-23 J/K).
Answer:
The de Broglie wavelength of electron βe = 2.443422 × 10⁻⁹ m
The de Broglie wavelength of proton βp = 5.70 × 10⁻¹¹ m
Explanation:
Thermal kinetic energy of electron or proton = KE
∴ KE = 3kbT/2
given that; kb = 1.38 x 10⁻²³ J/K , T = 1950 K
so we substitute
KE = ( 3 × 1.38 x 10⁻²³ × 1950 ) / 2
kE = 4.0365 × 10⁻²⁰ ( is the kinetic energy for both electron and proton at temperature T )
Now we know that
mass of electron M'e = 9.109 × 10⁻³¹
mass of proton M'p = 1.6726 × 10⁻²⁷
We also know that
KE = p₂ / 2m
from the equation, p = √ (2mKE)
{ p is momentum, m is mass }
de Broglie wavelength = β
so β = h / p = h / √ (2mKE)
h = Planck's constant = 6.626 × 10⁻³⁴
∴ βe = h / √ (2m'e × KE)
βe = 6.626 × 10⁻³⁴ / √ (2 × 9.109 × 10⁻³¹ × 4.0365 × 10⁻²⁰ )
βe = 6.626 × 10⁻³⁴ / √ 7.3536957 × 10⁻⁵⁰
βe = 6.626 × 10⁻³⁴ / 2.71176984642871 × 10⁻²⁵
βe = 2.443422 × 10⁻⁹ m
βp = h / √ (2m'p ×KE)
βp = 6.626 × 10⁻³⁴ / √ (2 × 1.6726 × 10⁻²⁷ × 4.0365 × 10⁻²⁰ )
βp = 6.626 × 10⁻³⁴ / √ 1.35028998 × 10⁻⁴⁶
βp = 6.626 × 10⁻³⁴ / 1.16201978468527 × 10⁻²³
βp = 5.702140 × 10⁻¹¹ m
What is the change in internal energy of the system (∆U) in a process in which 10 kJ of heat energy is absorbed by the system and 70 kJ of work is done by the system?
Answer:
Explanation:
According to first law of thermodynamics:
∆U= q + w
= 10kj+(-70kJ)
-60kJ
, w = + 70 kJ
(work done on the system is positive)
q = -10kJ ( heat is given out, so negative)
∆U = -10 + (+70) = +60 kJ
Thus, the internal energy of the system decreases by 60 kJ.