Complete Question
Partial tides __________.
Question 7 options:
a. represent various components of local tides that are resolved mathematically
b. are added together to predict the height and timing of astronomical tides
c. consist of 4 components due to the influence of celestial bodies
d. consist of up to 60 components due to astronomical and non-astronomical factors
e. All of the above except c are correct.
Answer:
Option E
Explanation:
Generally
Partial tides represent various components of local tides that are resolved mathematically
Partial tides are added together to predict the height and timing of astronomical tides
Partial tides consist of up to 60 components due to astronomical and non-astronomical factors
But Partial tides do not consist of 4 components due to the influence of celestial bodies
Therefore
All of the above except c are correct.
Option E
Tay quay OB quay đều quanh trục cố định đi qua O với vận tốc góc không đổi ω. Con lăn A chuyển động trong rãnh thẳng đứng. Tại vị trí trên hình vẽ thì thanh OB thẳng đứng, OA có phương nằm ngang. Hãy xác định vận tốc góc thanh AB, vận tốc của con lăn A; gia tốc góc của thanh AB, gia tốc của con lăn A. Cho ω = 1,5 rad/s, r = 1 m.
In Young's double slit experiment, 402 nm light gives a fourth-order bright fringe at a certain location on a flat screen. What is the longest wavelength of visible light that would produce a dark fringe at the same location? Assume that the range of visible wavelengths extends from 380 to 750 nm.
Answer:
λ₂ = 357.3 nm
Explanation:
The expression for double-slit interference is
d sin θ = m λ constructive interference
d sin θ = (m + ½) λ destructive interference.
The initial data corresponds to a constructive interference, they indicate that we are in the fourth order (m = 4), let's look for the separation of the slits
d sin θ = m λ₁
now ask for destructive interference for m = 4
d sin θ = (m + ½) λ₂
we match these two expressions
m λ₁ = (m + ½) λ₂
λ₂ = ( m / m + ½) λλ₁
let's calculate
λ₂ =[tex]\frac{4}{(4.000 +0.5) \ 401}[/tex]
λ₂ = 357.3 nm
When you are standing on Earth, orbiting the Sun, and looking at a broken cell phone on the ground, there are gravitational pulls on the cell phone from you, the Earth, and the Sun. Rank the gravitational forces on the phone from largest to smallest. Assume the Sun is roughly 109 times further away from the phone than you are, and 1028 times more massive than you. Rank the following choices in order from largest gravitational pull on the phone to smallest. To rank items as equivalent, overlap them.
a. Pull phone from you
b. Pull on phone from earth
c. Pull on phone from sun
Answer:
The answer is "Option b, c, and a".
Explanation:
Here that the earth pulls on the phone, as it will accelerate towards Earth when we drop it.
We now understand the effects of gravity:
[tex]F \propto M\\\\F\propto \frac{1}{r^2}\\\\or\\\\F \propto \frac{M}{r^2}\\\\Sun (\frac{M}{r^2}) = \frac{10^{28}}{(10^9)^2} = 10^{10}[/tex]
The force of the sun is, therefore, [tex]10^{10}[/tex] times greater and the proper sequence, therefore, option steps are:
b. Pull-on phone from earth
c. Pull-on phone from sun
a. Pull phone from you
A bus starts from rest and accelerates at 1.5m/s squared until it reaches a velocity of 9m/s .the bus continues at this velocity and then deccelerate at -2m/s squared until it comes to stop 400m from it's starting point. how much time did the bus takes to cover the 400m?
Answer:
23s
Explanation:
s=ut+1/2at^2
the distance (s) is 400, initial velocity (u) is 0, acceleration (a) is 1.5 therefore
400=0t+1/2(1.5)t^2
400/0.75=0.75t^2/0.75
t^2=√533.33
t=23s
I hope this helps and sorry if it's wrong
Một loa phát ra với cường độ âm là 40 (W/m2
). Mức cường độ âm của loa thuộc phạm vi?
Answer:ew
Explanation:
qeeqw
A uniform disk turns at 3.6 rev/s around a frictionless spindle. A non rotating rod, of the same mass as the disk and length equal to the disk's diameter, is dropped onto the freely spinning disk . They then both turn around the spindle with their centers superposed.
What is the angular frequency in rev/s of the combination?
please express answer in proper significant figures and rounding.
Answer:
ω₁ = 2.2 rev/s
Explanation:
Conservation of angular momentum
moment of inertia uniform disk is ½mR²
moment of inertia uniform rod about an end mL²/3
We can think of our rod as two rods of mass m/2 and length R
L = ½mR²ω₀
L = (½mR² + 2(m/2)R²/3)ω₁
ω₁ = ω₀(½mR² / (½mR² + mR²/3))
ω₁ = ω₀(½ / (½ + 1/3))
ω₁ = 0.6ω₀
ω₁ = 2.16
Để sử dụng nguồn điện xoay chiều 220V/50Hz thắp sáng bóng đèn 12V/3W, ta chọn điện trở giảm áp có giá trị:
Explanation:
Hi Linda,
How's it going?
Sorry I haven't been in touch for such a long time but I've had exams so I've been studying every free minute. Anyway, I'd love to hear all your news and I'm hoping we can get together soon to catch up. We just moved to a bigger flat so maybe you can come and visit one weekend?
How's the new job?
Looking forward to hearing from you!
Helga
Two harmonic sound waves reach an overseveer simulatenouslt. the obsever hears the sound intensity rise and fall with a time of 0.2 between the maximmu intensity and the successing minimum intensity. What is the difference in frequency of the two sound waves?
Answer:
[tex]dF=2.5Hz[/tex]
Explanation:
From the question we are told that:
Time [tex]T=0.2sec[/tex]
Generally the Period is given as
[tex]T= 2 * 0.2 = 0.4[/tex]
Therefore difference in frequency dF
[tex]dF=\frac{1}{T}[/tex]
[tex]dF=\frac{1}{0.4}[/tex]
[tex]dF=2.5Hz[/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
A 5.85-mm-high firefly sits on the axis of, and 13.7 cm in front of, the thin lens A, whose focal length is 5.01 cm. Behind lens A there is another thin lens, lens B, with focal length 25.9 cm. The two lenses share a common axis and are 62.5 cm apart. 1. Is the image of the firefly that lens B forms real or virtual?
a. Real
b. Vrtual
2. How far from lens B is this image located (expressed as a positive number)?
3. What is the height of this image (as a positive number)?
4. Is this image upright or inverted with respect to the firefly?
a. Upright
b. Inverted
Answer:
1. The image is real
2. 5.85
3. h' = 3.05 mm
4. The image is upright
Explanation:
1. Start with the first lens and apply 1/f = 1/p + 1/q
1/5.01 = 1/13.7 + 1/q
q = 7.90 cm
Since that distance is behind the first lens, and the second lens is 62.5 cm behind the first lens, that distance is 62.5 - 7.90 = 54.6 cm in front of the second lens, and becomes the object for that lens, thus,
1/25.9 = 1/54.6 + 1/q
q = 49.3 cm behind the second lens
Using that information, since q is positive, the image is real
2. Also, using that information, you have the second answer, which is 49.3 cm
The height can be found from the two magnifications.
m = -q/p
m1 = -7.9/13.7 = -.577
m2 = -49.3/54.6 = -.903
Net m = (-.577)(-.903) = .521
Then, m = h'/h
.521 = h'/5.85
3. h' = 3.05 mm
4. For the fourth answer, since the overall magnification is positive, the final image is upright
A parallal capacitor consists of two Squere plates each of Side 25cm, 3. Omm apart. If a potential difference of 2000volts is applied, calculate the change in the plate with
1.air
2. paper of relative permittity 2.5, fully the space between them E=8.9×10^-12
Answer:poop
Explanation:
poop
In a mass spectrometer chlorine ions of mass 35u and charge +5e are emitted from a source and accelerated through a potential difference of 250 kV. They then enter a region with a magnetic field that is perpendicular to their original direction of motion. The chlorine ions exit the spectrometer after being bent along a path with radius of curvature 3.5 m. What is the speed of the chlorine ions as they enter the magnetic field region?
(u = 1.66 × 10^(–27) kg, e = 1.6 × 10^(–19) C)
2.6 × 106 m/s
1.2 × 106 m/s
1.5 × 107 m/s
Answer:
v=26.23*105 m/s
or 2.6 × 106 m/s
Explanation:
Force generated by magnetic field will only provide centripetal acceleration thus the entering speed will be same as the exit speed
so,
.5mv2=eV potential differnce*charge= kinetic energy
.5*35*1.66*10-27*v2= 1.6*10-19*5*250000
v2=68.84*1011
v=26.23*105 m/s
or 2.6 × 106 m/s
.. Solve: 91
Find the half angular width of the central bright maximum in the Fraunhofer diffraction pattern of
a slit of width 12x10^-5 cm when the slit illuminated by monochromatic light of wave length
6000 A
[KUET’10-11)
(a) 30°
(b) 60°
(c) 15°
(d) None of these
Solution
Explanation:
bro I have no idea fam......
Find the ratio of the diameter of aluminium to copper wire, if they have the same
resistance per unit length. Take the resistivity values of aluminium and copper to
be 2.65× 10−8 Ω m and 1.72 × 10−8 Ω m respectively
Answer:
1.24
Explanation:
The resistivity of copper[tex]\rho_1=2.65\times 10^{-8}\ \Omega-m[/tex]
The resistivity of Aluminum,[tex]\rho_2=1.72\times 10^{-8}\ \Omega-m[/tex]
The wires have same resistance per unit length.
The resistance of a wire is given by :
[tex]R=\rho \dfrac{l}{A}\\\\R=\rho \dfrac{l}{\pi (\dfrac{d}{2})^2}\\\\\dfrac{R}{l}=\rho \dfrac{1}{\pi (\dfrac{d}{2})^2}[/tex]
According to given condition,
[tex]\rho_1 \dfrac{1}{\pi (\dfrac{d_1}{2})^2}=\rho_2 \dfrac{1}{\pi (\dfrac{d_2}{2})^2}\\\\\rho_1 \dfrac{1}{{d_1}^2}=\rho_2 \dfrac{1}{{d_2}^2}\\\\(\dfrac{d_2}{d_1})^2=\dfrac{\rho_1}{\rho_2}\\\\\dfrac{d_2}{d_1}=\sqrt{\dfrac{\rho_1}{\rho_2}}\\\\\dfrac{d_2}{d_1}=\sqrt{\dfrac{2.65\times 10^{-8}}{1.72\times 10^{-8}}}\\\\=1.24[/tex]
So, the required ratio of the diameter of Aluminum to Copper wire is 1.24.
Which of the following behaviors would best describe someone who is listening and paying attention? a) Leaning toward the speaker O b) Interrupting the speaker to share their opinion c) Avoiding eye contact d) Asking questions to make sure they understand what's being said
Answer:
D
Explanation:
Anyone could be leaning forward toward the speaker but be distracted and I believe if you're paying attention to the speaker, you would ask questions to make sure you're understanding what they are speaking
Answer:
A
.............................
Help me in my hw,A train starts from rest.Its velocity becomes 90km/hr after 1 min,Calculate the acceleration of train and distance covered by the train.Answer it ASAP
Answer:
I am serious about that
Explanation:
.......
A magnetic field is passing through a loop of wire whose area is 0.015 m2. The direction of the magnetic field is parallel to the normal to the loop, and the magnitude of the field is increasing at the rate of 0.20 T/s. (a) Determine the magnitude of the emf induced in the loop. (b) Suppose the area of the loop can be enlarged or shrunk. If the magnetic field is increasing as in part (a), at what rate (in m^2/s) should the area be changed at the instant when B
This question is incomplete, the complete question is;
A magnetic field is passing through a loop of wire whose area is 0.015 m2. The direction of the magnetic field is parallel to the normal to the loop, and the magnitude of the field is increasing at the rate of 0.20 T/s.
(a) Determine the magnitude of the emf induced in the loop.
(b) Suppose the area of the loop can be enlarged or shrunk. If the magnetic field is increasing as in part (a), at what rate (in m^2/s) should the area be changed at the instant when B = 1.5 T, if the induced emf is to be zero? (Give the magnitude of the rate of change of the area.) (m2/s)
Answer:
a) the magnitude of the emf induced in the loop is 0.003 V
b) dA/dt = 0.002 m²/s
Explanation:
Area of the loop wire A = 0.015 m²
magnitude of the field is increasing dB/dt = 0.20 T/s
a)
Determine the magnitude of the emf induced in the loop.
V = A( dB/dt )
we substitute
V = 0.015 m² × 0.20 T/s
V = 0.003 V
Therefore, the magnitude of the emf induced in the loop is 0.003 V
b) the induced emf is;
V = B( dA/dt ) + A( dB/dt )
given that; induced emf is 0, B = 1.5
so we substitute
0 = [ 1.5T × ( dA/dt ) ] + [ 0.015 m² × 0.20 T/s ]
-[ 1.5T × ( dA/dt )] = 0.003 m²T/s
dA/dt = -[ 0.003 m²T/s / 1.5T ]
dA/dt = -0.002 m²/s
the negative shows that the area is decreasing
hence, dA/dt = 0.002 m²/s
A man throw a ball vertically up word with an intial speed 20m/s. What is the maximum height rich by the ball and how long does it take to return to the point it was trow
Answer:
u=20 m/s, T=4s
Explanation:
Given final velocity v= 0 m/s and displacement h= 20 m; acceleration due to gravity = 10 m/ s 2
From equation of motion
v2=u2+2gs−u2=−2(10).20u=20m/s
and time t can be determined by the formula
t=gv−u=−10−20=2s
total time = 2× time of ascend=2×2=4s
it is helpful for you
Particle A has less mass than particle B. Both are pushed forward across a frictionless surface by equal forces for 1 s. Both start from rest. Which is true? A. A has more momentum. B. B has more momentum. C. A and B have the same momentum D. Not enough information.
Answer:
Both will have the same momentum.
P = M v momentum
v = a t for uniform acceleration
P = M a t
But a = F / M
P = M (F / M) t = F t so both have the same momentum
What is the primary purpose of politics?
Answer:
a politician is a person active in party politic or person holding os seeking an elected seat in government.politicans purpose, support,and create laws that govern the land and by extension ,it's people.
difference between wavefront and wavelets
Answer:
A wavefront is the locus of all the particles which are in phase. A wavelet is an oscilation that starts from zero, then the amplitude increases and later decreases to zero
write down the unit of mass ,temperature ,power and density
Explanation:
mass=kilogram,temperature=Klevin,power=watt,density=kilogram per cubic metre
Explanation:
the unit of mass is kg , temperature is kelvin ,power is watt and density is kilogram per cubic meter.
A boy walks from point C to point D which is 50 m apart. Then, he walks back to point C. what is his displacement of his whole journey ?
A.25 m
B.75 m
C.50 m
D.0 m
Answer: D. 0 m
Explanation:
Concept:
Here, we need to know the concept of displacement.
Displacement is defined to be the change in position of an object.
The difference between displacement and distance is the total movement of an object without any regard to direction, while displacement is the pure change of position.
If you are still confused, please refer to the attachment below for a graphical explanation.
Solve:
STEP ONE: the boy walks from point C to point D (a distance of 50 m)
C ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ D
50 m
STEP TWO: the boy walks from point D to point C (a distance of 50 m)
D ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ C
50 m
STEP THREE: find the displacement
The boy started with point C
The boy ended with point C
He did not change his position throughout the journey.
Therefore, his displacement is 0 m.
Hope this helps!! :)
Please let me know if you have any questions
15 . A scientist who studies the whole environment as a working unit .
Botanist
Chemist
Ecologist
Entomologist
Answer:
Ecologist.
Your answer is Ecologist.
(Ecologist) is a scientist who studies the whole environment as a working unit.
A redox reaction is always a single-displacement reaction, but a single-
displacement reaction isn't always a redox reaction.
A. True
B. False
SUBMIT
The graph below shows a cycle of a heat engine. Add the following labels to the graph. Some labels are used more than once.
Labels: Isobaric process; W= 0J; Work done on the system; Work done by the system.
I will give brainliest!
P.S. AL2006 if you see this please help!
I'm not very good at this material. I'll try it, but if I were you, I wouldn't bet money on these answers.
"Isobaric" means constant pressure. So those are the horizontal lines, where every point on the line is at the same pressure. Those are the processes 1>2 and 3>4 .
I'm going around and around in my mind with the other labels, and I can't decide. So I'm afraid I can't answer any more of them ... they might be wrong.
Answer:
1 -> 2 & 3 -> 4: Isobaric process
4 -> 1: Work done BY the system
2 -> 3: Work done ON the system
W(total): W = 0J
Explanation:
The two horizontal lines (1 -> 2 & 3 -> 4) are "Isobaric" since isobaric processes take place at constant pressure. I believe 4 -> 1 is "Work done BY the system" since pressure increases when there is an increase of thermal energy, in other words, the system is absorbing heat. This is why the volume increases from 1 -> 2 after the system has absorbed heat in 4 -> 1. Following the directions of the arrows, 2 -> 3 would be "Work done ON the system" since pressure is DECREASING, meaning temperature is also exiting the system. That's why the next step (3 -> 4) shows a decrease in volume. This model depicts a process that has a W(total) of 0 J because this is a cycle.
I hope this helps :))
the lamp cord is 85cm long and comprises cupper wire. Calculate the wire‘s resistance?
radius of a wire is 1.8mmm,Use value of resistivity for Cu as 1.75 × 10-8Ωm.
Answer:
R = 0.0015Ω
Explanation:
The formula for calculating the resistivity of a material is expressed as;
ρ = RA/l
R is the resistance
ρ is the resistivity
A is the area of the wire
l is the length of the wire
Given
l = 85cm = 0.85m
A = πr²
A = 3.14*0.0018²
A = 0.0000101736m²
ρ = 1.75 × 10-8Ωm.
Substitute into the formula
1.75 × 10-8 = 0.0000101736R/0.85
1.4875× 10-8 = 0.0000101736R
R = 1.4875× 10-8/0.0000101736
R = 0.0015Ω
An object is made of glass and has the shape of a cube 0.13 m on a side, according to an observer at rest relative to it. However, an observer moving at high speed parallel to one of the object's edges and knowing that the object's mass is 3.3 kg determines its density to be 8100 kg/m3, which is much greater than the density of glass. What is the moving observer's speed (in units of c) relative to the cube
Answer:
[tex]v=0.9833\ c[/tex]
Explanation:
The density changes means that the length in the direction of the motion is changed.
Therefore,
[tex]$\text{Density} = \frac{m}{lwh}$[/tex]
Given :
Side, b = h = 0.13 m
Mass, m = 3.3 kg
Density = 8100 [tex]kg/m^3[/tex]
So,
[tex]$8100=\frac{3.3}{l \times 0.13 \times 0.13}$[/tex]
[tex]$l=\frac{3.3}{8100 \times 0.13 \times 0.13}$[/tex]
l = 0.024 m
Then for relativistic length contraction,
[tex]$l= l' \sqrt{1-\frac{v^2}{c^2}}$[/tex]
[tex]$0.024= 0.13 \sqrt{1-\frac{v^2}{c^2}}$[/tex]
[tex]$0.184= \sqrt{1-\frac{v^2}{c^2}}$[/tex]
[tex]$0.033= 1-\frac{v^2}{c^2}}$[/tex]
[tex]$\frac{v^2}{c^2}= 0.967$[/tex]
[tex]$\frac{v}{c}=0.9833$[/tex]
[tex]v=0.9833\ c[/tex]
Therefore, the speed of the observer relative to the cube is 0.9833 c (in the units of c).
b. Block on an incline
A block of mass mı = 3.9 kg on a smooth inclined plane of angle 38° is 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?
Block on the incline:
• net force parallel to the incline
∑ F (para.) = m₁ g sin(38°) - T = m₁ a
where T is the magnitude of tension in the cord.
Notice that we take down-the-incline to be the positive direction, so that if the 3.9-kg block pulls the 2.6-kg block upwards, then the acceleration of the system is positive.
Suspended block:
• net vertical force
∑ F (vert.) = T - m₂ g = m₂ a
Solve both equations for the acceleration a, set the results equal to one another, and solve for T :
a = g sin(38°) - T/m₁
a = T/m₂ - g
==> g sin(38°) - T/m₁ = T/m₂ - g
==> T (1/m₂ + 1/m₁) = g (sin(38°) + 1)
==> T = g (sin(38°) + 1) / (1/m₂ + 1/m₁)
==> T = (9.81 m/s²) (sin(38°) + 1) / (1/(2.6 kg) + 1/(3.9 kg)) ≈ 25 N
A car is driving towards an intersection when the light turns red. The brakes apply a constant force of 1,398 newtons to bring the car to a complete stop in 25 meters. If the weight of the car is 4,729 newtons, how fast was the car going initially
Answer:
the initial velocity of the car is 12.04 m/s
Explanation:
Given;
force applied by the break, f = 1,398 N
distance moved by the car before stopping, d = 25 m
weight of the car, W = 4,729 N
The mass of the car is calculated as;
W = mg
m = W/g
m = (4,729) / (9.81)
m = 482.06 kg
The deceleration of the car when the force was applied;
-F = ma
a = -F/m
a = -1,398 / 482.06
a = -2.9 m/s²
The initial velocity of the car is calculated as;
v² = u² + 2ad
where;
v is the final velocity of the car at the point it stops = 0
u is the initial velocity of the car before the break was applied
0 = u² + 2(-a)d
0 = u² - 2ad
u² = 2ad
u = √2ad
u = √(2 x 2.9 x 25)
u =√(145)
u = 12.04 m/s
Therefore, the initial velocity of the car is 12.04 m/s
