Answer:
The frequency of the second wave is half of the frequency of first one.
Explanation:
The wavelength of the second wave is double is the first wave.
As we know that the frequency is inversely proportional to the wavelength of the velocity is same.
velocity = frequency x wavelength
So, the ratio of frequency of second wave to the first wave is
[tex]\frac{f_2}{f_1} =\frac{\lambda _1}{\lambda _2}\\\\\frac{f_2}{f_1} =\frac{\lambda _1}{2\lambda _1}\\\\\frac{f_2}{f_1} =\frac{1}{2}\\\\[/tex]
The frequency of the second wave is half of the frequency of first one.
A rope, under a tension of 221 N and fixed at both ends, oscillates in a second-harmonic standing wave pattern. The displacement of the rope is given by y = (0.10 m)(sin πx/2) sin 12πt, where x = 0 at one end of the rope, x is in meters, and t is in seconds.
What are:
a. the length of the rope.
b. the speed of the waves on the rope
c. the mass of the rope
d. If the rope oscillates in a third-harmonic standing wave pattern, what will be the period of oscillation.
Answer:
sup qwertyasdfghjk
Explanation:
Olympus Mons on Mars is the largest volcano in the solar system, at a height of 25 km and with a radius of 309 km. If you are standing on the summit, with what initial velocity would you have to fire a projectile from a cannon horizontally to clear the volcano and land on the surface of Mars
Answer:
The velocity is 2661.5 m/s.
Explanation:
Radius, horizontal distance, d = 309 km
height, h = 25 km
acceleration due to gravity on moon, g =3.71 m/s^2
Let the time taken is t and the horizontal velocity is u.
horizontal distance = horizontal velocity x time
309 x 1000 = u t .... (1)
Use second equation of motion in vertical direction.
[tex]h = u_yt +0.5 gt^2\\\\25000 = 0 + 0.5\times 3.71\times t^2\\\\t =116.1 s[/tex]
So, put in (1)
309 x 1000 = u x 116.1
u = 2661.5 m/s
10 A turning pork creates sound cares
with
Frequency of 170Hz: To the
speed of sound in is in 340mls
calculate the wave
wave length
of
in air is
the sound wales.
Answer:
2m
Explanation:
wavelength=speed/frequency
=340/170
=2m
1. Compare and contrast the SI and the English systems of measurement.
Answer:The SI system is based on the number 10 as well as multiples and products of 10. This makes it much easier to use, and so it has been the accepted system in scientific and technical applications. The English system is more complicated as relationships between units of the same quantity aren't uniform.
Explanation:
Answer:
The metric system is an internationally agreed decimal system of measurement while The International System of Units (SI) is the official system of measurement in almost every country in the world
2. How do the phytochemicals present in various foods help us?
Phytochemicals are compounds that are produced by plants ("phyto" means "plant"). They are found in fruits, vegetables, grains, beans, and other plants. Some of these phytochemicals are believed to protect cells from damage that could lead to cancer.
You are holding one end of a horizontal stretched string. Flicking your wrist will send a pulse down the string. Which actions will make the pulse travel faster
Answer:
Use a lighter string of the same length, under the same tension.
Stretch the string tighter to increase the tension
Explanation:
The wave speed depends on propertices of the medium, not on how you generate the wave. For a string
Increasing the tension or decreasing the linear density (lighter string) will increase the wave speed.
A pulse will be sent down a horizontal extended thread if a person flicks the wrist. To raise the tension, pull the string tighter.
Define pulse.Pulse is the same thing as monitoring heartbeat. The pulse rate could also be measured via auscultation, which includes hearing to the heart rhythm using a stethoscope then counting it for one minute.
A pulse will be sent down a horizontal extended thread if a person holds one end of the rope and flicks their wrist.
To raise the tension, pull the string tighter.
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Two substances, M and N, have specific heats c and 2c. if heats Q and 4Q are supɔlied to Mand N, respectively, their changes in temperature become equal. If substance M has mass m, find the mass of substance N in terms of m
Answer:
If the mass of B is m and the temperature change is the same, the mass of B will be 2m.
Explanation:
Q = mcT
T = mc/Q
M = 4Q/2cT........... (1)
T = Q/mc
Plug this in equation 1.
M = 4Q/(2c × Q/mc) = 4Q ÷ 2Q/m = 4Q × m/2Q = 2m
A wheel rotates about a fixed axis with a constant angular acceleration of 3.3 rad/s2. The diameter of the wheel is 21 cm. What is the linear speed (in m/s) of a point on the rim of this wheel at an instant when that point has a total linear acceleration with a magnitude of 1.7 m/s2
Answer:
The the linear speed (in m/s) of a point on the rim of this wheel at an instant=0.418 m/s
Explanation:
We are given that
Angular acceleration, [tex]\alpha=3.3 rad/s^2[/tex]
Diameter of the wheel, d=21 cm
Radius of wheel, [tex]r=\frac{d}{2}=\frac{21}{2}[/tex] cm
Radius of wheel, [tex]r=\frac{21\times 10^{-2}}{2} m[/tex]
1m=100 cm
Magnitude of total linear acceleration, a=[tex]1.7 m/s^2[/tex]
We have to find the linear speed of a at an instant when that point has a total linear acceleration with a magnitude of 1.7 m/s2.
Tangential acceleration,[tex]a_t=\alpha r[/tex]
[tex]a_t=3.3\times \frac{21\times 10^{-2}}{2}[/tex]
[tex]a_t=34.65\times 10^{-2}m/s^2[/tex]
Radial acceleration,[tex]a_r=\frac{v^2}{r}[/tex]
We know that
[tex]a=\sqrt{a^2_t+a^2_r}[/tex]
Using the formula
[tex]1.7=\sqrt{(34.65\times 10^{-2})^2+(\frac{v^2}{r})^2}[/tex]
Squaring on both sides
we get
[tex]2.89=1200.6225\times 10^{-4}+\frac{v^4}{r^2}[/tex]
[tex]\frac{v^4}{r^2}=2.89-1200.6225\times 10^{-4}[/tex]
[tex]v^4=r^2\times 2.7699[/tex]
[tex]v^4=(10.5\times 10^{-2})^2\times 2.7699[/tex]
[tex]v=((10.5\times 10^{-2})^2\times 2.7699)^{\frac{1}{4}}[/tex]
[tex]v=0.418 m/s[/tex]
Hence, the the linear speed (in m/s) of a point on the rim of this wheel at an instant=0.418 m/s
The drawings show (in cross section) two solid spheres and two spherical shells. Each object is made from copper and has a net charge, as the plus and minus signs indicate. Which drawing correctly shows where the charges reside when they are in equilibrium?
a) shows a lot of negative signs in the interior of circle
b) shows a lot of positive signs in the interior of circle
c) shows a hollowed out "hole" in the interior of the circle, with negative signs surrounding the opening.
d) shows a hollowed out "hole" in the interior of the circle, with positive signs surrounding the exterior edge
Answer:
d
Explanation:
The minimum energy configuration in electrostatics states that Charges always reside on the surface of a conductor. If anyhow they were inside, an electric field would exist inside and would act to move them to the surface,
Therefore, the drawing that shows where the charges reside when they are in equilibrium is a hollowed-out "hole" in the interior of the circle, with positive signs surrounding the exterior edge. This means that the d part is the correct answer.
Investigators measure the size of fog droplets using the diffraction of light. A camera records the diffraction pattern on a screen as the droplets pass in front of a laser, and a measurement of the size of the central maximum gives the droplet size. In one test, a 690 nm laser creates a pattern on a screen 30 cm from the droplets. If the central maximum of the pattern is 0.24 cm in diameter, how large is the droplet?
Answer:
the diameter of the droplet is 0.021045 cm or 2.1 × 10⁻² cm
Explanation:
Given the data in the question;
Diameter of bright central maxima;
⇒ 2 × ( 1.22 × (λD/d) ) ⇒ 2.44( λD/d )
where D is the distance from the the droplet to the screen ( 30 cm )
d is the diameter of the droplet
λ is the wavelength of light ( 690 nm = 690 × 10⁻⁷ cm )
since the central maximum of the pattern is 0.24 cm in diameter,
we substitute
0.24 cm = 2.44( ( 690 × 10⁻⁷ cm × 30 cm ) / d )
solve for d
d = 2.44( ( 690 × 10⁻⁷ cm × 30 cm ) / 0.24 cm
d = 0.0050508 cm² / 0.24 cm
d = 0.021045 cm or 2.1 × 10⁻² cm
Therefore, the diameter of the droplet is 0.021045 cm or 2.1 × 10⁻² cm
During typical urination, a man releases about 400 mL of urine in about 30 seconds through the urethra, which we can model as a tube 4 mm in diameter and 20 cm long. Assume that urine has the same density as water, and that viscosity can be ignored for this flow.a. What is the flow speed in the urethra?b. If we assume that the fluid is released at the same height as the bladder and that the fluid is at rest in the bladder (a reasonable approximation), what bladder pressure would be necessary to produce this flow? (In fact, there are additional factors that require additional pressure; the actual pressure is higher than this.)
Answer:
Explanation:
Given:
volume of urine discharged, [tex]V=400~mL=0.4~L=4\times 10^{-4}~m^3[/tex]
time taken for the discharge, [tex]t=30~s[/tex]
diameter of cylindrical urethra, [tex]d=4\times10^{-3}~m[/tex]
length of cylindrical urethra, [tex]l=0.2~m[/tex]
density of urine, [tex]\rho=1000~kg/m^3[/tex]
a)
we have volume flow rate Q:
[tex]Q=A.v[/tex] & [tex]Q=\frac{V}{t}[/tex]
where:
[tex]A=[/tex] cross-sectional area of urethra
[tex]v=[/tex] velocity of flow
[tex]A.v=\frac{V}{t}[/tex]
[tex]\frac{\pi d^2}{4}\times v=\frac{4\times 10^{-4}}{30}[/tex]
[tex]v=\frac{4\times4\times 10^{-4}}{30\times \pi (4\times 10^{-3})^2}[/tex]
[tex]v=1.06~m/s[/tex]
b)
The pressure required when the fluid is released at the same height as the bladder and that the fluid is at rest in the bladder:
[tex]P=\rho.g.l[/tex]
[tex]P=1000\times 9.8\times 0.2[/tex]
[tex]P=1960~Pa[/tex]
An object with mass m is located halfway between an object of mass M and an object of mass 3M that are separated by a distance d. What is the magnitude of the force on the object with mass m?A) 8GMm/d^2B) GMm/(4d^2)C) 4GMm/d^2D) GMm/(2d^2)E) 3GMm/2d^2
Answer:
A) 8GMm/d^2
Explanation:
We are given that
[tex]m_1=M[/tex]
[tex]m_2=3M[/tex]
[tex]m_3=m[/tex]
Distance between m1 and m2=d
Distance of object of mass m from m1 and m2=d/2
Gravitational force formula
[tex]F=\frac{Gm_1m_2}{r^2}[/tex]
Using the formula
Force acting between m and M is given by
[tex]F_1=\frac{GmM}{d^2/4}[/tex]
Force acting between m and 3M is given by
[tex]F_2=\frac{Gm(3M)}{d^2/4}[/tex]
Now, net force acting on object of mass is given by
[tex]F=F_2-F_1[/tex]
[tex]F=\frac{Gm(3M)}{d^2/4}-\frac{GmM}{d^2/4}[/tex]
[tex]F=\frac{12GmM}{d^2}-\frac{4GmM}{d^2}[/tex]
[tex]F=\frac{12GmM-4GmM}{d^2}[/tex]
[tex]F=\frac{8GmM}{d^2}[/tex]
Hence, the magnitude of the force on the object with mass m=[tex]\frac{8GmM}{d^2}[/tex]
Option A is correct.
a concrete has a height of 5m and has unit area 3m² supports a mass of 30000kg.
Determine the stress, strain and change in height
Answer:
stress = 98000 N/m^2
strain = 3.92 x 10^-6
change in height = 0.0196 mm
Explanation:
Height, h = 5 m
Area, A = 3 m²
mass, m = 30000 kg
Stress is defined as the force per unit area.
[tex]stress = \frac{mg}{A}\\\\stress = \frac{30000\times 9.8}{3}\\\\stress = 98000 N/m^2[/tex]
Young's modulus of concrete is Y = 2.5 x 10^10 N/m^2
Young's modulus is defined as the ratio of stress to the strain.
[tex]Y = \frac{stress}{strain}\\\\2.5\times 10^{10}= \frac{98000}{strain}\\\\strain = 3.92\times 10^{-6}[/tex]
let the change in height is h'.
Strain is defined as the ratio of change in height to the original height.
[tex]3.92\times 10^{-6} = \frac{h'}{5}\\\\h' = 1.96\times 10^{-5}m = 0.0196 mm[/tex]
Why are objects measured?
In order to find out how long/wide/heavy/high/dense/deep/ massive/voluminous/reflective/opaque/ tansparent/warm/cold/hard/soft/ malleable/flexible/rigid/radioactive/old/ valuable/symmetrical/flat/regular/ irregular they are.
In a way that you can easily and conveniently describe to other people.
What is the work done if a Boulder of mass 100 kilogram is rolled 40 meter up slope an angle of 20 degrees assuming the force of friction is negligible
Answer:
The work done is 13680.8 J.
Explanation:
The work done can be calculated as follows:
[tex] W = F*d [/tex]
Where:
F: is the force
d: is the displacement = 40 m
The force acting on the boulder is given by:
[tex] F = mgsin(\theta) [/tex]
Where:
m: is the mass = 100 kg
g: is the acceleration due to gravity = 10 m/s²
θ: is the angle = 20°
Then, the work is:
[tex] W = mgsin(\theta)d = 100 kg*10 m/s^{2}*sin(20)*40 m = 13680.8 J [/tex]
Therefore, the work done is 13680.8 J.
I hope it helps you!
why clinical thermometer cannot be used to measure the boiling point of water
Answer:
: No, a clinical thermometer cannot be used to measure the temperature of boiling water because it has a small range and might break due to extreme heat. ... The temperature is around 100 degrees Celsius.
In a
DC source, which has more cuwent?
(i)R L Circuit
(ii)RC Circuit (series)
(iii)LC Cirenit (series)
(iv)RLC Circuit (series)
Answer:
Answer is LC Cirenit (seres)
MCQ
................
Answer:
I think it would be (-7 C )..
) Efficiency of a lever is always less than hundred percent.
Yes. Because it opposes the law of friction
I hope this helps.
Explanation:
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If the length of the rod is 2.65 m, and the mass of the bob and the rod are both 1.4 kg, what is the period of this pendulum
Answer:
T = 5.66 s
Explanation:
The system formed by the bar plus ball forms a physical pendulum
w = [tex]\sqrt{mgd/I}[/tex]
the moment of inertia of a rod held at one end is
I = [tex]\frac{1}{3}[/tex] m L²
we substitute
w = [tex]\sqrt{\frac{d \ d}{ 3 L^2 } }[/tex]
in this case the turning distance and the length of the rod are equal
d = L
w = [tex]\sqrt{\frac{g}{3L} }[/tex]
angular velocity and period are related
w = 2π / T
2π / T = [tex]\sqrt{\frac{g}{3L} }[/tex]
T = 2π [tex]\sqrt{3L/g}[/tex]
let's calculate
T = 2π [tex]\sqrt{3 \ 2.65 / 9.8}[/tex]
T = 5.66 s
Find the refractive index of a medium
having a velocity of 1.5 x 10^8*
Explanation:
refractive index ,is the ratio of velocity of light in vacuum to the velocity of light a medium
A 0.20 mass on a horizontal spring is pulled back a certain distance and released. The maximum speed of the mass is measured to be 0.20 . If, instead, a 0.40 mass were used in this same experiment, choose the correct value for the maximum speed.
a. 0.40 m/s.
b. 0.20 m/s.
c. 0.28 m/s.
d. 0.14 m/s.
e. 0.10 m/s.
Answer:
d
Explanation:
Ya gon find the Kenitic Energy first
K=½mv²===> K=½×0.2×(0.2)²===> 0.1(0.04)===> 0.004
and now the replacement:
0.004=½×0.4V²====> v²=0.02===> V=0.14m/s
1.Lõi thép máy biến áp được ghép từcác lá thép là để:
(a) Giảm tổn hao công suất do dòng điện xoáy
(b) Giảm tổn hao công suất do từ trễ
(c) Giảm tổn hao công suất do dòng điện chạy qua dây quấn
(d) Giảm tất cảcác loại tổn hao công suất.
Answer:
Option (c)
Explanation:
1.The transformer core is assembled from steel sheets to:
(a) Reduced power loss due to eddy current
(b) Reduced power loss due to hysteresis
(c) Reduced power loss due to current flowing through the winding
(d) Reduce all types of power loss.
A transformer is a device which converts the low voltage into high and vice versa.
There are two types of a transformer.
Step up: It is used to convert low voltage into high.
Step down It is used to convert high voltage into high.
It depends on the number of turns in primary and the secondary coil.
The core of the transformer is laminated and it is in the form of sheets.
By using such type of core, the power loss due to the windings is reduced.
option (c) .
If an object of a constant mass experiences a constant net force, it will have a constant what?
Explanation:
hope it helps !!!!!!!!!!!!!
If an object of a constant mass experiences a constant net force, it will have a constant acceleration.
What is force?The definition of force in physics is: The push or pull on a massed object changes its velocity. An external force is an agent that has the power to alter the resting or moving condition of a body. It has a direction and a magnitude.
The application of force is the location at which force is applied, and the direction in which the force is applied is known as the direction of the force. A spring balance can be used to calculate the Force. The Newton is the SI unit of force.
According to Newton's second law of motion:
Applied force = mass × acceleration.
Hence, if an object of a constant mass experiences a constant net force, it will have a constant acceleration.
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A farmhand pushes a 26-kg bale of hay 3.9 m across the floor of a barn. If she exerts a horizontal force of 88 N on the hay, how much work has she done
Answer:
W = 343.2 J
Explanation:
Given that,
Mass of bale of hay = 26 kg
Horizontal force exerted = 88 N
Distance moved, d = 3.9 m
Work done, W = Fd
Put all the values,
W = 88 N × 3.9 m
= 343.2 J
So, the work done is 343.2 J.
suppose you have a block resting on a horizontal smooth surface. th block with a mass m is attached to a horizontal spring which is fixed at one end. the spring can be compressed and stretched. the mass is pulled to one side then released what is the formula required
The time period of the spring is 2[tex]\pi[/tex][√(m/k)].
What is meant by spring constant ?The spring constant of a spring is defined as the measurement of ratio of the force that is exerted on the spring to the displacement caused by it.
Here,
The mass of the block = m
Let F be the applied force on the spring and k be the spring constant.
When the mass attached to the spring is pulled to one side then released, it executes SHM.
Therefore we can write that, the applied force,
F = kx
Restoring force = -kx
According to Newton's law, we know that,
F = ma
So,
ma = -kx
Therefore, the acceleration,
a = (-k/m) x
For an SHM, the acceleration is given as,
a = -ω²x
Therefore, we can write that,
-ω²x = (-k/m) x
ω² = k/m
So, the time period of the spring,
T = 2[tex]\pi[/tex]/ω
T = 2[tex]\pi[/tex][√(m/k)]
Hence,
The time period of the spring is 2[tex]\pi[/tex][√(m/k)].
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At what distance x from the center of the coil, on the axis of the coil, is the magnetic field half its value at the center?
Answer:
The value of x is 2.1 cm from the center of the coil.
Explanation:
Radius, R = 2.7 cm
Number of turns, N = 800
The magnetic field at the axis is half of the magnetic field at the center.
[tex]B_{axis}=\frac{B_{center}}{2}\\\\\frac{\mu o}{4\pi}\times \frac{2 \pi I N R^2}{\left (R^2 + x^2 \right )^{\frac{3}{2}}} = 0.5\frac{\mu o}{4\pi}\times\frac{2\pi N I}{R}\\\\\frac{R^2}{(R^2 + x^2)^\frac{3}{2}} = \frac{1}{2R}\\\\4R^6 = (R^2+x^2)^3\\\\1.6 R^2 = R^2 + x^2\\\\x^2 = 0.6 \times 2.7\times 2.7 \\\\x = 2.1 cm[/tex]
PLEASE HELP ME WITH THIS ONE QUESTION
The half-life of Barium-139 is 4.96 x 10^3 seconds. A sample contains 3.21 x 10^17 nuclei. What is the decay constant for this decay?
Explanation:
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A commuter backs her car out of her garage with an acceleration of . (a) How long does it take her to reach a speed of 2.00 m/s
Question: A commuter backs her car out of her garage with an acceleration of 1.4 m/s² (a) How long does it take her to reach a speed of 2.00 m/s
Answer:
1.43 s
Explanation:
Applying,
a = (v-u)/t........... Equation 1
Where a = acceleration, v = final velocity, u = initial velocity, t = time
make t the subject of the equation
t = (v-u)/a........... Equation 2
From the question,
Given: v = 2 m/s, u = 0m/s (from rest), a = 1.4 m/s²
Substitute into equation 2
t = (2-0)/1.4
t = 1.43 s
Characteristics or properties of matter or energy that can be measured
Answer:
Physical properties are properties that can be measured or observed without changing the chemical nature of the substance. Some examples of physical properties are:
color (intensive)
density (intensive)
volume (extensive)
mass (extensive)
boiling point (intensive): the temperature at which a substance boils
melting point (intensive): the temperature at which a substance melts
Explanation: