Answer:
Explanation:
When the path difference is equal to wave length or its integral multiple, constructive interference occurs . If it is odd multiple of half wave length , then destructive interference occurs.
For constructive interference , path diff = n λ
For destructive interference path diff = ( 2n+ 1 ) λ /2
where λ is wave length of wave , n is an integer.
a )
path diff = 10 cm which is half the wavelength , so maximum destructive interference will occur.
b )
path diff = 15 cm which is neither half the wavelength nor full wavelength , so in between is the right option.
c )
path diff = 20 cm which is equal to the wavelength , so maximum constructive interference will occur.
d)
path diff = 30 cm which is 3 times half the wavelength , so maximum destructive interference will occur.
e)
path diff = 35 cm which is neither integral multiple of half the wavelength , nor integral multiple of wavelength so in between is th eright answer.
f )
path diff = 40 cm which is 2 times the wavelength , so maximum constructive interference will occur
If the kinetic energy of a particle has increased to 25 times its initial value, then the percentage of the change in the wavelength which is associated with the particle's motion is...
A) 80%
B) 60%
C) 40%
D) 20%
Develop a hypothesis regarding one factor you think might affect the period of a pendulum or an oscillating mass on a spring. Potential factors include the mass, the spring constant, and the length of the pendulum's string. Write down your hypothesis. 2. Design a controlled experiment to test your hypothesis. Take extreme care to keep all factors constant except the variable you are testing.
Answer:
A hypothesis for the period of a pendulum is:
"The period of the pendulum varies with its length"
Explanation:
A hypothesis for the period of a pendulum is:
"The period of the pendulum varies with its length"
To test this hypothesis we can carry out a measurement of a simple pendulum keeping the angle fixed, in general the angle used is about 5º since when placing this value in radiand and the sine of this angle they differ little <5%. therefore measured the time of some oscillations, for example about 10 oscillations, changing the length of the pendulum to test the hypothesis.
If the hypothesis and the model used is correct, the relationship to be tested is
T² =(4π² /g) L
by making a graph of the period squared against the length if obtaining, os a line, the hypothesis is tested.
What is a downside of nonprofit fitness centers?
They offer fewer luxury services than for-profit health centersThey offer fewer luxury services than for-profit health centers , ,
Very few people live near a nonprofit fitness center.Very few people live near a nonprofit fitness center. , ,
Members tend to be wealthy and lack diversity. Members tend to be wealthy and lack diversity. , ,
Membership costs more than it does in for-profit centers.Membership costs more than it does in for-profit centers. , ,
they offer fewer luxury services because they lack the money to provide them.
The primary purpose of a switch in a circuit is to ___________.
A)either open or close a conductive path
B)change a circuit from parallel to series
C)change a circuit from series to parallel
D)store a charge for later use
Answer:
store a charge for later use
A 0.160 kg glider is moving to the right on a frictionless, horizontal air track with a speed of 0.710 m/s. It has a head-on collision with a 0.296 kg glider that is moving to the left with a speed of 2.23 m/s. Suppose the collision is elastic.
Required:
a. Find the magnitude of the final velocity of the 0.157kg glider.
b. Find the magnitude of the final velocity of the 0.306kg glider.
The masses of the gliders provided in the question differ from the masses mentioned in the "Required" section. I'll use the first masses throughout.
Momentum is conserved, so the total momentum of the system is the same before and after the collision:
m₁ v₁ + m₂ v₂ = m₁ v₁' + m₂ v₂'
==>
(0.160 kg) (0.710 m/s) + (0.296 kg) (-2.23 m/s) = (0.160 kg) v₁' + (0.296 kg) v₂'
==>
-0.546 kg•m/s ≈ (0.160 kg) v₁' + (0.296 kg) v₂'
where v₁' and v₂' are the gliders' respective final velocities. Notice that we take rightward to be positive and leftward to be negative.
Kinetic energy is also conserved, so that
1/2 m₁ v₁² + 1/2 m₂ v₂² = 1/2 m₁ (v₁' )² + 1/2 m₂ (v₂' )²
or
m₁ v₁² + m₂ v₂² = m₁ (v₁' )² + m₂ (v₂' )²
==>
(0.160 kg) (0.710 m/s)² + (0.296 kg) (-2.23 m/s)² = (0.160 kg) (v₁' )² + (0.296 kg) (v₂' )²
==>
1.55 kg•m²/s² ≈ (0.160 kg) (v₁' )² + (0.296 kg) (v₂' )²
Solve for v₁' and v₂'. Using a calculator, you would find two solutions, one of which we throw out because it corresponds exactly to the initial velocities. The desired solution is
v₁' ≈ -3.11 m/s
v₂' ≈ -0.167 m/s
and take the absolute values to get the magnitudes.
If you want to instead use the masses from the "Required" section, you would end up with
v₁' ≈ -3.18 m/s
v₂' ≈ -0.236 m/s
A man standing in an elevator holds a spring scale with a load of 5 kg suspended from it. What would be the reading of the scale, if the elevator is accelerating downward with an acceleration 3.8 m/s?.
Answer:
3.1 kg
Explanation:
Applying,
R = m(g-a)..................... Equation 1
Where R = weight of the scale when the elevator is coming down, a = acceleration of the elevator, g = acceleration due to gravith.
From the question,
Given: m = 5 kg, a = 3.8 m/s²
Constant: g = 9.8 m/s²
Substitute these values into equation 1
R = 5(9.8-3.8)
R = 5(6)
R = 30 N
Hence the spring scale is
m' = R/g
m' = 30/9.8
m' = 3.1 kg
The New England Merchants Bank Building in Boston is 152 m high. On windy days it sways with a frequency of 0.18 Hz , and the acceleration of the top of the building can reach 1.9 % of the free-fall acceleration, enough to cause discomfort for occupants.
Required:
What is the total distance, side to side, that the top of the building moves during such an oscillation?
Answer:
The total distance, side to side, that the top of the building moves during such an oscillation is approximately 0.291 meters.
Explanation:
Let suppose that the building is experimenting a Simple Harmonic Motion due to the action of wind. First, we determine the angular frequency of the system ([tex]\omega[/tex]), in radians per second:
[tex]\omega = 2\pi\cdot f[/tex] (1)
Where [tex]f[/tex] is the frequency, in hertz.
If we know that [tex]f = 0.18\,hz[/tex], then the angular frequency of the system is:
[tex]\omega = 2\pi\cdot (0.18\,hz)[/tex]
[tex]\omega \approx 1.131\,\frac{rad}{s}[/tex]
The maximum acceleration experimented by the system is represented by the following formula, of which we estimate amplitude of the oscillation:
[tex]r\cdot g = \omega^{2}\cdot A[/tex] (2)
Where:
[tex]r[/tex] - Ratio of real acceleration to free-fall acceleration, no unit.
[tex]g[/tex] - Free-fall acceleration, in meters per square second.
[tex]A[/tex] - Amplitude, in meters.
If we know that [tex]\omega \approx 1.131\,\frac{rad}{s}[/tex], [tex]r = 0.019[/tex] and [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], then the amplitude of the oscillation is:
[tex]A = \frac{r\cdot g}{\omega^{2}}[/tex]
[tex]A = \frac{(0.019)\cdot \left(9.807\,\frac{m}{s^{2}} \right)}{\left(1.131\,\frac{rad}{s} \right)^{2}}[/tex]
[tex]A \approx 0.146\,m[/tex]
The total distance, side to side, is twice the amplitude, that is to say, a value of approximately 0.291 meters.
A 0.2 oz. bullet leaves the muzzle of a rifle with a speed of 1420 ft/s. If the length of the barrel is 24 inches, what is the magnitude of the force acting on the bullet while it travels down the barrel
Answer:
196 lbf
Explanation:
v² = u² + 2as
1420² = 0² + 2a(24/12)
a = 1420²/4 = 504,100 ft/s²
F = ma = 0.2oz(1lb/16 oz)(1slug/32.2 lb)(504,100) = 195.6909...
Cho lực F ⃗=6x^3 i ⃗-4yj ⃗ tác dụng lên vật làm vật chuyển động từ A(-2,5) đến B(4,7). Vậy công của lực là:
The work done by [tex]\vec F[/tex] along the given path C from A to B is given by the line integral,
[tex]\displaystyle \int_C \mathbf F\cdot\mathrm d\mathbf r[/tex]
I assume the path itself is a line segment, which can be parameterized by
[tex]\vec r(t) = (1-t)(-2\,\vec\imath + 5\,\vec\jmath) + t(4\,\vec\imath+7\,\vec\jmath) \\\\ \vec r(t) = (6t-2)\,\vec\imath+(2t+5)\,\vec\jmath \\\\ \vec r(t) = x(t)\,\vec\imath + y(t)\,\vec\jmath[/tex]
with 0 ≤ t ≤ 1. Then the work performed by F along C is
[tex]\displaystyle \int_0^1 \left(6x(t)^3\,\vec\imath-4y(t)\,\vec\jmath\right)\cdot\frac{\mathrm d}{\mathrm dt}\left[x(t)\,\vec\imath + y(t)\,\vec\jmath\right]\,\mathrm dt \\\\ = \int_0^1 (288(3t-1)^3-8(2t+5)) \,\mathrm dt = \boxed{312}[/tex]
Light of the same wavelength passes through two diffraction gratings. One grating has 4000 lines/cm, and the other one has 6000 lines/cm. Which grating will spread the light through a larger angle in the first-order pattern
Answer:
6000 lines/cm
Explanation:
From the question we are told that:
Grating 1=4000 lines/cm
Grating 2=6000 lines/cm
Generally The Spread of fringes is Larger when the Grating are closer to each other
Therefore
Grating 2 will spread the the light through a larger angle in the first-order pattern because its the closest with 6000 lines/cm
If a jet travels 350 m/s, how far will it travel each second?
Answer:
It will travel 350 meters each second.
Explanation:
The unit rate, 350 m/s, tells us that the jet will travel 350 meters per every second elapsed.
Answer:
5.83 seconds
Explanation:
60 seconds in 1 minute
350 meters per second
350/60
=5.83
The armature of an AC generator has 200 turns, which are rectangular loops measuring 5 cm by 10 cm. The generator has a sinusoidal voltage output with an amplitude of 18 V. If the magnetic field of the generator is 300 mT, with what frequency does the armature turn
Answer:
[tex]f=9.55Hz[/tex]
Explanation:
From the question we are told that:
Number of Turns [tex]N=200[/tex]
Length [tex]l=5cm to 10cm[/tex]
Voltage [tex]V=18V[/tex]
Magnetic field [tex]B=300mT[/tex]
Generally, the equation for Frequncy of an amarture is mathematically given by
[tex]f =\frac{ V}{(N B A * 2 pi )}[/tex]
[tex]f =\frac{ 18}{(200 300*10^{-3} (10*10^-2)(5*10^{-2}) * 2 *3.142 )}[/tex]
[tex]f=9.55Hz[/tex]
A person pulls on a 9 kg crate against a 22 Newton frictional force, using a rope attached to the center of the crate. If the The crate began with a speed of 1.5 m/s and speeded up to 2.7 m/s while being pulled a horizontal distance of 2.0 meters. What is the work in J done by the force applied by the rope on the crate
Answer:
uninstell this apps nobody will give u ans its happening to me alsoi was in exam hall i thought this app will give answer but no
on a horizontal axis whose unit is the meter, a linear load ranging from 0 to 1 ma a linear load distribution = 2 nC / m.
determine the modulus of the electric field created by the previous loaded bar at the point A of abscissa 2m (we have to find the relation between l, which is the distance between the elementary bar and the point A and x which sweeps the segment [0: 1]
Answer:
The correct answer is - 8.99N/C
Explanation:
[tex]dE=k\dfrac{dq}{x^2}\\ dq=\lambda{dx}\\ \lambda=2nC/m\\ dq=2dxnC\\ dE=k\dfrac{2dx}{x^2}\\ E=2k\int_1^2\dfrac{dx}{x^2}\\ E=2k(\frac{-1}{x})_1^2=k\times10^{-9}N/C\\ E=8.99\times10^9\times10^{-9}N/C\\ E=8.99N/C\\dE=k[/tex]
What particles in an atom can increase and decrease in number without changing the identity of the elements
Answer:
The number of neutrons or electrons in an atom can change without changing the identity of the element.
19 point please please answer right need help
Block on an incline
A block of mass m1 = 3.9 kg on a smooth inclined plane of angle 38is connected by a cord over a small frictionless
pulley to a second block of mass m2 = 2.6 kg hanging vertically. Take the positive direction up the incline and use 9.81
m/s2 for g.
What is the tension in the cord to the nearest whole number?
Explanation:
We can write Newton's 2nd law as applied to the sliding mass [tex]m_1[/tex] as
[tex]T - m_1g\sin38 = m_1a\:\:\:\:\:\:\:(1)[/tex]
For the hanging mass [tex]m_2,[/tex] we can write NSL as
[tex]T - m_2g = -m_2a\:\:\:\:\:\:\:(2)[/tex]
We need to solve for a first before we can solve the tension T. So combining Eqns(1) & (2), we get
[tex](m_1 + m_2)a = m_2g - m_1g\sin38[/tex]
or
[tex]a = \left(\dfrac{m_2 - m_1\sin38}{m_1 + m_2}\right)g[/tex]
[tex]\:\:\:\:= 0.30\:\text{m/s}^2[/tex]
Using this value for the acceleration on Eqn(2), we find that the tension T is
[tex]T = m_2(g - a) = (2.6\:\text{kg})(9.51\:\text{m/s}^2)[/tex]
[tex]\:\:\:\:=24.7\:\text{N}[/tex]
What are the systems of units? Explain each of them.
THERE ARE COMMONLY THREE SYSTEMS OF UNIT. THEY ARE:-
• CGS System- (Centimeter-Gram-Second system) A metric system of measurement that uses the centimeter, gram and second for length, mass and time.
• FPS System- (Foot–Pound–Second system).
The system of units in which length is measured in foot , mass in pound and time in second is called FPS system. It is also known as British system of units.
• MKS System- (Meter-Kilogram-Second system) A metric system of measurement that uses the meter, kilogram, gram and second for length, mass and time. The units of force and energy are the "newton" and "joule."
A capacitor consists of a set of two parallel plates of area A separated by a distance d. This capacitor is connected to a battery and charged until its plates carry charges If the separation between the plates is doubled, the electrical energy stored in the capacitor will
Answer:
The electrical energy stored in the capacitor will be cut in half.
Explanation:
The energy in a capacitator is given by E=C[tex]V^{2}[/tex]/2 and the formula for the Capacitance in a capacitator is C= [tex]\frac{Q}{V}[/tex] = ε[tex]\frac{A}{d}[/tex] .
So if we replace C = ε[tex]\frac{A}{d}[/tex] in the first equation we have:
E = ε[tex]\frac{AV^{2} }{2d}[/tex]
A spring whose stiffness is 3500 N/m is used to launch a 4 kg block straight up in the classroom. The spring is initially compressed 0.2 m, and the block is initially at rest when it is released. When the block is 1.3 m above its starting position, what is its speed
Answer:
the speed of the block at the given position is 21.33 m/s.
Explanation:
Given;
spring constant, k = 3500 N/m
mass of the block, m = 4 kg
extension of the spring, x = 0.2 m
initial velocity of the block, u = 0
displacement of the block, d =1.3 m
The force applied to the block by the spring is calculated as;
F = ma = kx
where;
a is the acceleration of the block
[tex]a = \frac{kx}{m} \\\\a = \frac{(3500) \times (0.2)}{4} \\\\a = 175 \ m/s^2[/tex]
The final velocity of the block at 1.3 m is calculated as;
v² = u² + 2ad
v² = 0 + 2ad
v² = 2ad
v = √2ad
v = √(2 x 175 x 1.3)
v = 21.33 m/s
Therefore, the speed of the block at the given position is 21.33 m/s.
The speed of the block at a height of 1.3 m above the starting position is 21.33 m/s
To solve this question, we'll begin by calculating the acceleration of the block.
How to determine the acceleration Spring constant (K) = 3500 N/m Mass (m) = 4 KgCompression (e) = 0.2 mAcceleration (a) =?F = Ke
Also,
F = ma
Thus,
ma = Ke
Divide both side by m
a = Ke / m
a = (3500 × 0.2) / 4
a = 175 m/s²
How to determine the speed Initial velocity (u) = 0 m/sAcceleration (a) = 175 m/s²Distance (s) = 1.3 mFinal velocity (v) =?v² = u² + 2as
v² = 0² + (2 × 175 × 1.3)
v² = 455
Take the square root of both side
v = √455
v = 21.33 m/s
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On her camping trip, Penelope was in charge of collecting firewood. The firewood she found had a mass of 120 g and a volume of 480 cm3. What is the density of the firewood? Explain the steps you took to solve this problem.
Answer:
D = .25g/cm³
Explanation:
D = m/V
D = 120g/480cm³
D = .25g/cm³
a. The molecules of a magnet are independent...
Answer:
variable
Explanation:
An electron is moving at speed of 6.3 x 10^4 m/s in a circular path of radius of 1.7 cm inside a solenoid the magnetic field of the solenoid is perpendicular to the plane of the electron's path. Find its relevatn motion.
Answer:
Here, m=9×10
−31
kg,
q=1.6×10
−19
C,v=3×10
7
ms
−1
,
b=6×10
−4
T
r=
qB
mv
=
(1.6×10
−19
)(6×10
−4
)
(9×10
−31
)×(3×10
7
)
=0.28m
v=
2πr
v
=
2πm
Bq
=
2×(22/7)×9×10
−31
(6×10
−4
)×(1.6×10
−19
)
=1.7×10
7
Hz
Ek=
2
1
mv
2
=
2
1
×(9×10
−31
)×(3×10
7
)
2
J
=40.5×10
−17
J=
1.6×10
−16
40.5×10
−17
keV
=2.53keV
A spherical, concave shaving mirror has a radius of curvature of 0.983 m. What is the magnification of a person's face when it is 0.155 m from the vertex of the mirror (answer sign and magnitude)
Answer:
Magnification = 1
Explanation:
given data
radius of curvature r = - 0.983 m
image distance u = - 0.155
solution
we get here first focal length that is
Focal length, f = R/2 ...................1
f = -0.4915 m
we use here formula that is
[tex]\frac{1}{v} + \frac{1}{u} + \frac{1}{f}[/tex] .................2
put here value and we get
[tex]\frac{1}{v} = \frac{1}{0.155} - \frac{1}{4915}[/tex]
v = 0.155 mso
Magnification will be here as
m = [tex]- \frac{v}{u}[/tex]
m = [tex]\frac{0.155}{0.155}[/tex]
m = 1Answer:
The magnification is 1.5.
Explanation:
radius of curvature, R = - 0.983 m
distance of object, u = - 0.155 m
Let the distance of image is v.
focal length, f = R/2 = - 0.492 m
Use the mirror equation
[tex]\frac{1}{f}=\frac{1}{v}+\frac {1}{u}\\\\\frac{-1}{0.492}=\frac{1}{v}-\frac{1}{0.155}\\\\\frac{1}{v}=\frac{1}{0.155}-\frac{1}{0.492}\\\\\frac{1}{v}=\frac{0.492-0.155}{0.155\times 0.492}\\\\\frac{1}{v}=\frac{0.337}{0.07626}\\ \\v = 0.226 m[/tex]
The magnification is given by
m = - v/u
m = 0.226/0.155
m = 1.5
A woman pulls on a 6.00-kg crate, which in turn is connected to a 4.00-kg
crate by a light rope. The light rope remains taut. Compared to the 6.00-kg crate,
the lighter 4.00-kg crate
Please explain why any of these multiple choices is correct!
Answer:
B. is subject to a smaller net force but same acceleration.
Explanation:
F = m*a
So because our force applied is constant from the women pulling on the rope which means the acceleration is the same on both the 4kg create and the 6kg create. The only thing that changes here is the mass of the creates, so there is more tension force between the women and the 6kg create then there is between the 4kg create and the 6kg. It takes less force to move the 4kg create therefore the tension force is less between the two creates.
The net force on both crates is the same and the acceleration of both crates is the same.
The given parameters;
mass of the crate, m = 6 kgmass of the second crate, = 4 kgThe force on the 4kg crate is calculated as follows;
[tex]F_{4kg } = T + F[/tex]
The force on the 6kg crate is calculated as follows;
[tex]F_{6 kg} = -T + F[/tex]
The net force on both crates is calculated as follows;
[tex]\Sigma F= -T + F - (T + F)\\\\\Sigma F= -2T[/tex]
Thus, we can conclude that the net force on both crates is the same and the acceleration of both crates is the same.
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state any 3 properties of an ideal gas as assumed by the kinetic theory.
Answer:
The simplest kinetic model is based on the assumptions that: (1) the gas is composed of a large number of identical molecules moving in random directions, separated by distances that are large compared with their size; (2) the molecules undergo perfectly elastic collisions (no energy loss) with each other and with the walls of the container, but otherwise do not interact; and (3) the transfer of kinetic energy between molecules is heat.
A 207-kg merry-go-round in the shape of a uniform, solid, horizontal disk of radius 1.50 m is set in motion by wrapping a rope about the rim of the disk and pulling on the rope. What constant force would have to be exerted on the rope to bring the merry-go-round from rest to an angular speed of 0.750 rev/s in 2.00 s
Answer:
366 N
Explanation:
τ = Iα
FR = ½mR²α
F = ½mR(Δω/t)
F = ½(207)(1.50)(0.75)(2π) /2.00
F = 365.79919...
If ∆H = + VE , THEN WHAT REACTION IT IS
1) exothermic
2) endothermic
Answer:
endothermic
Explanation:
An endothermic is any process with an increase in the enthalpy H (or internal energy U) of the system. In such a process, a closed system usually absorbs thermal energy from its surroundings, which is heat transfer into the system.
Convert the unit of 0.00023 kilograms into grams. (Answer in scientific notation)
Answer:
2.3 × [tex]10^{-1}[/tex]
Explanation:
1 kg = 1000 g.
0.00023 kg x 1000 g = 0.23 grams
Answer:
0.23×10⁴
Explanation:
kilogram to gram ÷ 1000
0.00023kg ÷ 1000
=0.23g
scientific notation=0.23×10⁴
Two train 75 km apart approach each other on parallel tracks, each moving at
15km/h. A bird flies back and forth between the trains at 20km/h until the trains pass
each other. How far does the bird fly?
Answer:
The correct solution is "37.5 km".
Explanation:
Given:
Distance between the trains,
d = 75 km
Speed of each train,
= 15 km/h
The relative speed will be:
= [tex]15 + (-15)[/tex]
= [tex]30 \ km/h[/tex]
The speed of the bird,
V = 15 km/h
Now,
The time taken to meet will be:
[tex]t=\frac{Distance}{Relative \ speed}[/tex]
[tex]=\frac{75}{30}[/tex]
[tex]=2.5 \ h[/tex]
hence,
The distance travelled by the bird in 2.5 h will be:
⇒ [tex]D = V t[/tex]
[tex]=15\times 2.5[/tex]
[tex]=37.5 \ km[/tex]
A point charge of +35 nC is above a point charge of –35 nC on a vertical line. The distance between the charges is 4.0 mm. What are the magnitude and direction of the dipole moment ?
Answer:
Magnitude = 140 x 10⁻¹² Cm
Direction = upwards
Explanation:
A pair of two equal and opposite point charges forms an electric dipole.
The magnitude of the moment of such dipole is the product of the magnitude of any of the charges (since the charges are the same in magnitude) and the distance of separation between them. i.e
p = q x d ----------(i)
Where;
p = dipole moment
q = magnitude of any of the charges
d = distance between the charges.
The direction of the dipole moment is from the negative charge to the positive charge.
(a) From the question, the charges are +35 nC and -35 nC, and the distance between them is 4.00mm.
This implies that;
q = 35 nC = 35 x 10⁻⁹C
d = 4.00mm = 4.0 x 10⁻³ m
Substitute the values of q and d into equation (i) to give;
p = 35 x 10⁻⁹C x 4.00 x 10⁻³ m
p = (35 x 4.0) x (10⁻⁹ x 10⁻³) C m
p = 140 x 10⁻¹² Cm
The magnitude of the dipole moment is 140 x 10⁻¹² Cm
(b) From the question, the +35nC charge is above the -35nC charge on a vertical line as shown below;
o +35nC
|
|
|
|
|
|
o -35nC
Since the direction should point from the negative charge to the positive charge, this means that the direction of the dipole moment of the two charges is upwards (due North).
o +35nC
↑
|
|
|
|
|
|
o -35nC
