Answer:
the length of the pipe is 0.85 m or 85 cm
Explanation:
Given the data in the question;
The successive harmonics are; 700 Hz , 900 Hz , and 1100 H
Now, for a closed pipe,
length of pipe (L) = λ/4
Harmonics; 1x, 3x, 5x, 7x, 9x, 11x
1100Hz - 900Hz = 200Hz
⇒ 2x = 200Hz
x = 100Hz ( fundamental frequency )
λ = V/f = 340 /100 = 3.4 m
Now
Length L = λ / 4
L = 3.4 / 4
L = 0.85 m or 85 cm
Therefore, the length of the pipe is 0.85 m or 85 cm
how to get infinite thanks could you show me please ☺️
Answer: divide by zero, or square root of a negative
Explanation: If your question is how to get infinity as an answer to a problem, that generally means that the answer is undefined or doesn't exist.
A couple of ways to get that...
You try to divide by zero. In other words, a problem that asks you to perform something like this: 5/0= or 23/(4-4)= Such a problem will give you an error on a calculator because the answer is infinity or doesn't exist. Another way is to try to get the square root of a negative number. That answer doesn't exist as a real number, so [tex]\sqrt{-4\\}[/tex] will also give you an error on a calculator.
The tub of a washer goes into its spin-dry cycle, starting from rest and reaching an angular speed of 3.0 rev/s in 13.0 s. At this point, the person doing the laundry opens the lid, and a safety switch turns off the washer. The tub slows to rest in 12.0 s. Through how many revolutions does the tub turn during this 25 s interval
Answer:
The tub turns 37.520 revolutions during the 25-second interval.
Explanation:
The total number of revolutions done by the tub of the washer ([tex]\Delta n[/tex]), in revolutions, is the sum of the number of revolutions done in the acceleration ([tex]\Delta n_{1}[/tex]), in revolutions, and deceleration stages ([tex]\Delta n_{2}[/tex]), in revolutions:
[tex]\Delta n = \Delta n_{1} + \Delta n_{2}[/tex] (1)
Then, we expand the previous expression by kinematic equations for uniform accelerated motion:
[tex]\Delta n = \frac{1}{2}\cdot ( \ddot n_{1}\cdot t_{1}^{2} - \ddot n_{2} \cdot t_{2}^{2})[/tex] (1b)
Where:
[tex]\ddot n_{1}, \ddot n_{2}[/tex] - Angular accelerations for acceleration and deceleration stages, in revolutions per square second.
[tex]t_{1}, t_{2}[/tex] - Acceleration and deceleration times, in seconds.
And each acceleration is determined by the following formulas:
Acceleration
[tex]\ddot n_{1} = \frac{\dot n}{t_{1}}[/tex] (2)
Deceleration
[tex]\ddot n_{2} = -\frac{\dot n}{t_{2} }[/tex] (3)
Where [tex]\dot n[/tex] is the maximum angular velocity of the tub of the washer, in revolutions per second.
If we know that [tex]\dot n = 3\,\frac{rev}{s}[/tex], [tex]t_{1} = 13\,s[/tex] and [tex]t_{2} = 12\,s[/tex], then the quantity of revolutions done by the tub is:
[tex]\ddot n_{1} = \frac{3\,\frac{rev}{s} }{13\,s}[/tex]
[tex]\ddot n_{1} = 0.231\,\frac{rev}{s^{2}}[/tex]
[tex]\ddot n_{2} = -\frac{3\,\frac{rev}{s} }{12\,s}[/tex]
[tex]\ddot n_{2} = -0.25\,\frac{rev}{s^{2}}[/tex]
[tex]\Delta n = \frac{1}{2}\cdot ( \ddot n_{1}\cdot t_{1}^{2} + \ddot n_{2} \cdot t_{2}^{2})[/tex]
[tex]\Delta n = \frac{1}{2}\cdot \left[\left(0.231\,\frac{rev}{s^{2}} \right)\cdot (13\,s)^{2}-\left(-0.25\,\frac{rev}{s^{2}} \right)\cdot (12\,s)^{2}\right][/tex]
[tex]\Delta n = 37.520\,rev[/tex]
The tub turns 37.520 revolutions during the 25-second interval.
PLEASE HELP ME WITH THIS ONE QUESTION
What is the rest energy of a proton? (c = 2.9979 x 10^9 m/s, mp = 1.6726 x 10^-27)
A) 8.18 x 10^-14 J
B) 2.73 x 10^-22 J
C) 1.5053 x 10^-10 J
D) 1.5032 x 10^-10 J
Answer:
D) 1.5032 x 10^-10 J
Explanation:
The rest energy of a proton, E₀, follows the equation:
E₀ = mp*rate²
Where mass of proton, mp = 1.6726 x 10^-27kg
rate = 2.9979 x 10^8 m/s (2.9979 x 10^9 m/s is not the speed light)
E₀ = 1.6726 x 10^-27kg * (2.9979 x 10^8 m/s)²
E₀ =1.5032 x 10^-10 J
Right answer is:
D) 1.5032 x 10^-10 Jcaculate the component of a force of 200 ns
at a direction of 60° to the force
Answer:
[tex]F_x = 100N[/tex]
[tex]F_y = 100\sqrt 3 \ N[/tex]
Explanation:
Given
[tex]F = 200N[/tex]
[tex]\theta = 60^o[/tex]
Required
The component of the force in F direction
To do this, we simply calculate the force in the vertical and horizontal direction.
This is calculated as:
[tex]F_x = F * \cos(\theta)[/tex] --- Horizontal
[tex]F_y = F * \cos(\theta)[/tex] ---- Vertical
So, we have:
[tex]F_x = F * \cos(\theta)[/tex] --- Horizontal
[tex]F_x = 200N * \cos(60^o)[/tex]
[tex]F_x = 200N * 0.5[/tex]
[tex]F_x = 100N[/tex]
[tex]F_y = F * \cos(\theta)[/tex] ---- Vertical
[tex]F_y = 200N * \sin(60^o)[/tex]
[tex]F_y = 200N * \frac{\sqrt 3}{2}[/tex]
[tex]F_y = 100\sqrt 3 \ N[/tex]
50 POINTS‼️‼️‼️‼️‼️
A 4.88 x 10-6 C charge moves 265 m/s
perpendicular (at 90°) to a magnetic
field of 0.0579 T. What is the magnetic
force on the charge?
Answer: 0
Explanation: Trust
The magnetic force on the charge is approximately 6.47 x 10^(-4) Newtons.
The magnetic force on a charged particle moving through a magnetic field can be calculated using the formula:
F = q * v * B * sin(θ)
Where:
F is the magnetic force,
q is the charge of the particle (in this case, 4.88 x 10^(-6) C),
v is the velocity of the particle (in this case, 265 m/s),
B is the magnetic field strength (in this case, 0.0579 T),
θ is the angle between the velocity vector and the magnetic field vector (in this case, 90 degrees).
Plugging in the values:
F = (4.88 x 10^(-6) C) * (265 m/s) * (0.0579 T) * sin(90°)
Since sin(90°) is equal to 1, the equation simplifies to:
F = (4.88 x 10^(-6) C) * (265 m/s) * (0.0579 T) * 1
Calculating the value:
F = 6.47 x 10^(-4) N
Therefore, the magnetic force on the charge is approximately 6.47 x 10^(-4) Newtons.
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4. You currently have TWO 9 volt batteries connected to a 82 resistor. How
much current is flowing through the circuit? *
1.125 A
5 A
72 A
2.25 A
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Explanation:
Total voltage = 9 × 2 = 18v
Resistance = 82 Ω
Ohm's law::
V = IR
18v = 82 Ω × I
18v /82 /Ω = I
18/82 Ampere is the current
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?
A) 1.67 x 10^-4 s^-1
B) 5.43 x 10^-4 s^-1
C) 1.40 x 10^-4 s^-1
D) 2.22 x 10^-4 s^-1
OPTION C is the correct answer.
What is the most commonly used semiconducting material? Nickel Aluminum b. Cobalt d. Silicon
Answer:
i believe the answer is D
Explanation:
Hope it works
Answer:
D. SiliconExplanation:
Silicon is the most widely used type of semiconductor material. Its major advantage is that it is easy to fabricate and provides good general electrical and mechanical properties.does net force stay the same when a massless pulley is replaced by a pulley with mass
During hockey practice, two pucks are sliding across the ice in the same direction. At one instant, a 0.18-kg puck is moving at 16 m/s while the other puck has a mass of 0.14 kg and a speed of 3.8 m/s. What is the velocity of the center of mass of the two pucks?
Answer:
10.66 m/s
Explanation:
Applying,
The law of conservation of momentum,
Total momentum before collision = Total momentum after collision
mu+m'u' = V(m+m').............. Equation 1
Where m = mass of the first puck, u = initial velocity of the first puck, m' = mass of the second puck, u' = initial velocity of the second puck, V = Velocity of the center of the two pucks
make V the subject of the equation
V = (mu+m'u')/(m+m').............. Equation 2
Given: m = 0.18 kg, u = 16 m/s, m' = 0.14 kg, u' = 3.8 m/s
Substitute these values into equation 2
V = [(0.18×16)+(0.14×3.8)]/(0.18+0.14)
V = (2.88+0.532)/(0.32)
V = 3.412/0.32
V = 10.66 m/s
The large scale structure of the universe has been carefully mapped using redshift surveys of a very large number of galaxies. Answer the following: Suppose that a large telescope with modern equipment can measure the redshift to a galaxy in just 10 minutes. And suppose we want to spend no more than a year mapping the distribution of galaxies. How many redshifts can be surveyed
Answer:
26280
Explanation:
In current time, good telescope can measure redshift to a galaxy in 10 minutes.
Thus, in one year that has on an average 365 days, the total time taken to measure redshifts is = ( 365 *12 *60) minute
= 262800 minutes .
Hence, the number of redshifts observed in a year = (262800/10) = 26280
A bug crawls 3.0 mm east, 4.0mm north, and then 5.0 mm at 45 north of east. Draw a diagram showing its displacements and determine its resultant displacement vector by use of the diagram.
Answer:
Explanation is given
Mark Me as Branliest
The magnetic field at the center of a 1.0-cm-diameter loop is 2.5 mT.
a. What is the current in the loop?
b. A long straight wire carries the same current from part a. At what distance from the wire is the magnetic field 2.5 mT?
Answer:
(a) The current in the wire is 19.89 A
(b) The distance from the wire is 0.159 cm
Explanation:
Given;
magnetic field, B = 2.5 mT
diameter of the wire, d = 1 cm
radius of the wire, r = 0.5 cm = 0.005 m
(a) The current in the wire is calculated as;
[tex]I = \frac{2Br}{\mu_0} \\\\I = \frac{2\times 2.5 \times 10^{-3} \times 0.005 }{4\pi \times 10^{-7} } \\\\I = 19.89 \ A[/tex]
(b) The distance from the wire where the magnetic field is 2.5 mT is calculated as;
[tex]B = \frac{\mu_0 I}{2\pi d} \\\\where;\\\\d \ is \ the \ distance \ from \ the \ wire\\\\d = \frac{\mu_0 I}{2\pi B} = \frac{4 \pi \times 10^{-7} \times 19.89}{2\pi \times 2.5 \times 10^{-3}} = 0.00159 \ m = 0.159 \ cm[/tex]
A 280-m-wide river flows due east at a uniform speed of 4.7m/s. A boat with a speed of 7.1m/s relative to the water leaves the south bank pointed in a direction 26o west of north. What is the (a) magnitude and (b) direction of the boat's velocity relative to the ground
Answer:
(a) The speed is 7.96 m/s
(b) The direction is 76 degree from positive X axis in counter clockwise direction.
Explanation:
Width of river = 280 m
speed of river, vR = 4.7 m/s towards east
speed of boat with respect to water, v(B,R) = 7.1 m/s at 26 degree west of north
[tex]vR = 4.7 i \\\\v(B,R) = 7.1 (- sin 26 i + cos 26 j) = - 3.1 i + 6.4 j[/tex]
(a) The velocity of boat with respect to ground is
[tex]\overrightarrow{v}_{(B,R)}=\overrightarrow{v}_{(B,G)}-\overrightarrow{v}_{(R,G)}\\\\- 3.1 \widehat{i} +6.4 \widehat{j}=\overrightarrow{v}_{(B,G)} - 4.7 \widehat{i}\\\\\overrightarrow{v}_{(B,G)} = 1.6 \widehat{i} + 6.4 \widehat{j}\\\\{v}_{(B,G)} = \sqrt{1.6^2 + 6.4^2}=6.96 m/s[/tex]
(b) The direction is given by
[tex]tan\theta = \frac{6.4}{1.6} =4\\\\\theta = 76^o[/tex]
find the resistance of wire of
0.65m Radius 0.25
and
resistivity 3x10-6 OHM
Complete Question:
Find the resistance of a wire of length 0.65 m, radius 0.25 mm and resistivity 3 * 10^{-6} ohm-metre.
Answer:
Resistance = 9.95 Ohms
Explanation:
Given the following data;
Length = 0.65 m
Radius = 0.25 mm to meters = 0.00025 m
Resistivity = 3 * 10^{-6} ohm-metre.
To find the resistance of the wire;
Mathematically, resistance is given by the formula;
[tex] Resistance = P \frac {L}{A} [/tex]
Where;
P is the resistivity of the material. L is the length of the material.A is the cross-sectional area of the material.First of all, we would find the cross-sectional area of the wire.
Area of circle = πr²
Substituting into the equation, we have;
Area = 3.142 * (0.00025)²
Area = 3.142 * 6.25 * 10^{-8}
Area = 1.96 * 10^{-7} m²
Now, to find the resistance of the wire;
[tex] Resistance = 3 * 10^{-6} * \frac {0.65}{1.96 * 10^{-7}} [/tex]
[tex] Resistance = 3 * 10^{-6} * 3316326.531 [/tex]
Resistance = 9.95 Ohms
Two argon atoms form the molecule Ar2 as a result of a van der Waals interaction with U0 = 1.68×10-21 J and R0= 3.82×10 the frequency of small oscillations of one Ar atom about its equilibrium position.
Answer:
[tex]\mathbf{f_o =1.87 \times 10^{11} \ Hz}[/tex]
Explanation:
From the given information:
The elastic potential energy can be calculated by using the formula:
[tex]U_o = \dfrac{1}{2}kR_o^2[/tex]
Making K the subject;
[tex]K = \dfrac{2 U_o}{R_o^2}[/tex]
[tex]k = \dfrac{2\times 1.68 \times 10^{-21}}{(3.82\times 10^{-10})^2}[/tex]
k = 2.3 × 10⁻² N/m
Now; the frequency of the small oscillation can be determined by using the formula:
[tex]f_o = \dfrac{1}{2 \pi}\sqrt{\dfrac{k}{m}}[/tex]
where;
m = mass of each atom = 1.66 × 10⁻²⁶ kg
[tex]f_o = \dfrac{1}{2 \pi}\sqrt{\dfrac{2.3 \times 10^{-2} N/m}{1.66 \times 10^{-26} \ kg}}[/tex]
[tex]\mathbf{f_o =1.87 \times 10^{11} \ Hz}[/tex]
Scientific theories are deductive in nature.?
Answer:
deductive reasoning usually follows steps .
That is, how we predict what the observations should be if the theory were correctWhich Circut Model is more efficient, why?
Answer:
right one is more efficient
Explanation:
Darwin believed that emotional expressions began as ________ that came to have evolutionary value because they ________. Select one: a. physiological reactions; increased the efficiency of bodily reactions b. communication devices; increased the efficiency of bodily reactions Incorrect c. physiological reactions; convey emotional states to other members of the species d. random mutations; convey expectations to other members of the species
Answer:
c. physiological reactions; convey emotional states to other members of the species
Explanation:
Darwin believed that emotional expressions began as physiological reactions that came to have evolutionary value because they convey emotional states to other members of the species. These reactions are used by many different types of species to convey various things.
A reservoir located in the mountain 250 m above sea level flows through a pipe to a hydroelectric plant in a town at sea level. Assuming the pressure in both locations are the same and the density of water is 1000 kg/m3. How fast will the water flow into the plant?
Answer:
v₂ = 70 m / s
Explanation:
For this exercise let's use Bernoulli's equation
where subscript 1 is for the top of the mountain and subscript 2 is for Tuesday's level
P₁ + ½ ρ v₁² + ρ g y₁ = P₂ +1/2 ρ v₂² + ρ g y₂
indicate that the pressure in the two points is the same, y₁ = 250 m, y₂ = 0 m, the water in the upper part, because it is a reservoir, is very large for which the velocity is very small, we will approximate it to 0 (v₁ = 0), we substitute
ρ g y₁ = ½ ρ v₂²
v₂ = [tex]\sqrt {2g \ y_1}[/tex]
let's calculate
v₂ = √( 2 9.8 250)
v₂ = 70 m / s
Electricity is the result of moving electrons, so it's classified as
A. Kinetic Energy
B. Gravitational Energy
C. Potential Energy
D. Elastic Energy
How can a small spark start a huge explosion
Answer:
The explosion is set off by an electrostatic spark. When the mixture ignites, the rapid increase in temperature brings about a huge increase in gas pressure. If the burning vapour were to be confined the resulting rise in pressure could destroy the chamber with a loud explosion.
Explanation:
50 POINTS/BRAINLIEST TO CORRECT!
A 1.25 x 10-4 C charge is moving
5200 m/s at 37.0° to a magnetic field
of 8.49 x 10-4 T. What is the magnetic
force on the charge?
Answer: 3.32x10^-4
Explanation: Works for Acellus
Magnitude of magnetic force F= qvB Sin0®
q is the magnitude of charge moving with speed v in magnetic field B. Theta is the angle between velocity and magnetic field.
F=1.25×10⁻⁴C×5200m/s×8.49×10⁻⁴T(sin37deg).
F=3.32×10⁻⁴N.
What is charge?An electric charge is the property of matter where it has more or fewer electrons than protons in its atoms. Electrons carry a negative charge and protons carry a positive charge.
Matter is positively charged if it contains more protons than electrons, and negatively charged if it contains more electrons than protons.
Thus, the magnetic force on the charge is 3.32×10⁻⁴N.
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What effect does increased blood flow have on the body when performing exercises? A. delivers more sugar to organs B. delivers more energy to muscles C. delivers more oxygen to the body D. delivers more protein to muscles Please select the best answer from the choices provided. O A . OB ос OD Next Submit Save and Exit Mark this and return
A small object with mass 0.200 kg moves with constant speed in a vertical circle of radius 0.500 m. It takes the object 0.500 s to complete one revolution. (a) What is the translational speed of the object
Answer:
6.28 m/s.
Explanation:
Given that,
The mass of the object, m = 0.2 kg
The radius of the circle, r = 0.5 m
It takes the object 0.500 s to complete one revolution.
We need to find the translational speed of the object. Let it is v. We know that,
[tex]v=\dfrac{2\pi r}{t}\\\\v=\dfrac{2\pi \times 0.5}{0.5}\\\\v=6.28\ m/s[/tex]
So, the transalational speed of the object is 6.28 m/s.
What is the main way in which heat transfer occurs in liquids and gases?
Calculate the current flowing when the voltage across is 35V and the resistance is 7ohms.
Explanation:
V= IR
35=I×7
I=35/7
I=5amperes
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Un objeto tiene una velocidad de vi=3i-4j m/s, luego duplica su velocidad en 12 segundos, calcula la magnitud de la distancia que recorre en metros.
Answer:
Explanation:
This is an exercise in kinematics, the speed they give is in two dimensions, let's work on each component
X axis
initial velocity v₀ₓ = 3 m / s in a time of t = 12 s, the velocity is doubled, the final velocity is vₓ = 6 m / s
acceleration is
vₓ = v₀ₓ + aₓ t
aₓ = [tex]\frac{v_x - v_{ox}} {t}[/tex]
aₓ = 6 - 3/12
aₓ = 0.25 m / s²
the distance traveled is
vₓ² = v₀ₓ² + 2 aₓx x
x = vx² - vox² / 2a
x = 6² - 3² / 2 0.25
x = 54 m
Y axis
we look for acceleration
v_y = v_{oy} + a_y t
a_y = [tex]\frac{v_y - v_{oy} }{t}[/tex]
a_y = [tex]\frac{8 -4} {12}[/tex]
ay = 0.3333 m / s²
the distance is
v_y² = v_{oy}² + 2 a⁷y
y = vy² - voy² / 2 0.25
y = 8² - 4² / 2 0/3333
y = 72 m
the distance traveled is
r = (54 i + 72j) m / s
A 12 kg hanging sculpture is suspended by a 80-cm-long, 6.0 g steel wire. When the wind blows hard, the wire hums at its fundamental frequency. What is the frequency of the hum
Answer:
[tex]F=78.3hz[/tex]
Explanation:
From the question we are told that:
Mass [tex]m=12[/tex]
Length [tex]l=80cm=0.8m[/tex]
Linear density [tex]\mu= 6.0g[/tex]
Generally the equation for Frequency is mathematically given by
[tex]F=\frac{1}{2l}\sqrt{\frac{T}{K}}[/tex]
[tex]F=\frac{1}{2(0.8)}\sqrt{\frac{12*9.8*0.8}{6*10^{-3}}}[/tex]
[tex]F=78.3hz[/tex]
Which of the following best describes the upper respiratory tract?
O It takes air in from outside the body.
O It is where oxygen and carbon dioxide are exchanged.
O It is located inside the thorax.
O It is not directly involved in respiration.