Answer:
X = 10.8387 cm²/s
Explanation:
In this exercise, you're required to convert a value from one unit to another.
Converting 42 ft²/hr to cm²/s;
Conversion:
1 ft² = 929.03 cm²
42 ft² = X cm²
Cross-multiplying, we have;
X = 42 * 929.03
X = 39019.26 cm²
Next, we would divide by time in seconds.
1 hour = 3600 seconds
X = 39019.26/3600
X = 10.8387 cm²/s
herical piece of candy is suspended in flowing water. The candy has a density of 1950 kg/m3 and has a 1.0 cm diameter. The water velocity is 1.0 m/s, the water density is assumed to be 1000.0 kg/m3, and the water viscosity is 1.010-3 kg/m/s. The diffusion coefficient of the candy solute in water is 2.010-9 m2/s, and the solubility of the candy solute in water is 2.0 kg/m3. Calculate the mass tran
Answer: Below is the complete question
A spherical piece of candy is suspended in flowing water. The candy has a density of 1950 kg/m3 and has a 1.0 cm diameter. The water velocity is 1.0 m/s, the water density is assumed to be 1000.0 kg/m3, and the water viscosity is 1.0x10-3 kg/m/s. The diffusion coefficient of the candy solute in water is 2.0x10-9 m2/s, and the solubility of the candy solute in water is 2.0 kg/m3. Calculate the mass transfer coefficient (m/s)
answer:
mass transfer coefficient = 9.56 * 10^-5 m/s
Explanation:
Candy density = 1950 kg/m^3
Candy diameter = 1 cm
Velocity of water = 1 m/s
water density = 1000 kg/m^3
Viscosity of water = 1 * 10^-3 kg/m/s
diffusion coefficient of candy in water = 2 * 10^-9 m^2/s
solubility of candy = 2 kg/m^3
Determine the mass transfer coefficient ( m/s )
( Sh) mass transfer coefficient ( flow across sphere ) = 2 + 0.6Re^1/2 * SC^1/3
where : Re = vdp / μ , Sh = KLd / Deff
attached below is the remaining solution .
mass transfer coefficient = 9.56 * 10^-5 m/s
The position of a particle is given by ~r(t) = (3.0 t2 ˆi + 5.0 ˆj j 6.0 t kˆ) m
Answer:
[tex]v=(6ti+6k)\ m/s[/tex]
Explanation:
Given that,
The position of a particle is given by :
[tex]r(t) = (3.0 t^2 i + 5.0j+ 6.0 tk) m[/tex]
Let us assume we need to find its velocity.
We know that,
[tex]v=\dfrac{dr}{dt}\\\\=\dfrac{d}{dt}(3.0 t^2 i + 5.0j+ 6.0 tk) \\\\=(6ti+6k)\ m/s[/tex]
So, the velocity of the particle is [tex](6ti+6k)\ m/s[/tex].
what is the dimensional formula of young modulas
Answer:
The dimensional formula of Young's modulus is [ML^-1T^-2]
Answer:
G.oogle : The dimensional formula for Young’s modulus is:
A. [ML−1T−2]A. [ML−1T−2]
B. [M0LT−2]B. [M0LT−2]
C. [MLT−2]C. [MLT−2]
D. [ML2T−2]
From the given picture What's the force? And where did it happen? (at least 2 forces)
Answer:
the force happens on the wall and couch
Explanation:
she is using her arm strength to lift and hold
why do atom absorb photon since it makes it unstable??
[tex]\textsf{When an electron is hit by a }[/tex] [tex]\textsf{photon of light, it absorbs the quanta}[/tex] [tex]\textsf{of energy the photon was carrying}[/tex] [tex]\textsf{and moves to a higher energya}[/tex] [tex]\textsf{ state. Electrons therefore have to }[/tex] [tex]\textsf{jump around within the atom as }[/tex] [tex]\textsf{they either gain or lose energy. }[/tex]
When an electron is hit by a photon of light, it absorbs the quanta of energy the photon was carrying and moves to a higher energy state. Electrons therefore have to jump around within the atom as they either gain or lose energy.
Hope this answer helps you..!!!
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Select it as the BRAINLIEST..!!
A standard bathroom scale is placed on an elevator. A 28 kg boy enters the elevator on the first floor and steps on the scale. What will the scale read (in newtons) when the elevator begins to accelerate upward at 0.5 m/s2
Answer:
Explanation:
Newton's Second Law is pretty much the standard for all motion that involves a force. It applies to gravitational force and torque and friction and weight on an elevator. The main formula for force is
F = ma. We have to adjust that to take into account that when the elevator is moving up, that "surge" of acceleration weighs down a bit on the scale, causing it to read higher than the actual weight until the acceleration evens out and there is no acceleration at all (no acceleration simply means that the velocity is constant; acceleration by definition is a change in velocity, and if there is no change in velocity, there is 0 acceleration). The force equation then becomes
[tex]F_n-w=ma[/tex] where [tex]F_n[/tex] is normal force. This is what the scale will read, which is what we are looking for in this problem (our unknown). Since we are looking for [tex]F_n[/tex], that is what we will solve this literal equation for:
[tex]F_n=ma+w[/tex] . m is the mass of the boy, a is the acceleration of the elevator (which is going up so we will call that acceleration positive), and w is weight. We have everything but the unknown and the weight of the boy. We find the weight:
w = mg so
w = 28(9.8) and
w = 274.4 N BUT rounding to the correct number of significance we have that the weight is actually
w = 270 N.
Filling in the elevator equation:
[tex]F_n=28(.50)+270[/tex] and according to the rules of significant digits, we have to multiply the 28(.50) {notice that I did add a 0 there for greater significance; if not that added 0 we are only looking at 1 significant digit which is pretty much useless}, round that to 2 sig fig's, and then add to 170:
[tex]F_n=14+270[/tex] and adding, by the rules, requires that we round to the tens place to get, finally:
[tex]F_n=280N[/tex] So you see that the surge in acceleration did in fact add a tiny bit to the weight read by the scale; conversely, if he were to have moved down at that same rate, the scale would have read a bit less than his actual weight). Isn't physics like the coolest thing ever!?
A student of mass 50kg takes 15seconds to run up a flight of 50 steps. If each step is 20cm, calculate the potential energy of the student at the maximum height
Answer:
the answer is 49000 joules at the maximum height
Explanation:
we know the mass (50kg)
we know the acceleration due to gravity(9.8m/s²)
we know the height too(maximum height meaning the 50th step so we multiply 50 with 20cm as each step is 20 cm and we get 1000 cm, convert to m it is 100 m
the formula is potential energy=mgh
m for mass
g for acceleration due to gravity
h for height
multiply them
50x9.8x100
we get 49000
the unit of potential energy is joules so the answer is
49000 joules
Answer:
49000 joules
Explanation:
hope it helpss
In these formulas, it is useful to understand which variables are parameters that specify the nature of the wave. The variables E0E0E_0 and B0B0B_0 are the __________ of the electric and magnetic fields. Choose the best answer to fill in the blank.
given A=4i-10j and B= 7i+5j find b such that A+bB is a vector pointing along the x-axis (i.e has no y component)
Answer:
-4/7
Explanation:
Given the following
A=4i-10j and B= 7i+5j
A+ bB = 4i-10j + (7i+5j)b
A+ bB = 4i-10j + 7ib+5jb
A+ bB =
The vector along the x-axis is expressed as i + 0j
If the vector A+ bB is pointing in the direction of the x-axis then;
[tex]A+ bB * \frac{i+0j}{|i+0j|} = 0 \\ (4+7b)i-(10-5b)j* \frac{i+0j}{\sqrt{1^2+0^2} } = 0\\(4+7b)i-(10-5b)j *(i+0j) = 0\\4+7b-0 =0\\7b=-4\\b = -4/7[/tex]
Hence the value of b is -4/7
The value of [tex]\beta[/tex] such that [tex]\vec C = \vec A + \beta \cdot \vec B = c\,\hat{i}[/tex] is 2.
According to the statement, we have following system of vectorial equations:
[tex]\vec A = 4\,\hat {i} - 10\,\hat{j}[/tex] (1)
[tex]\vec {B} = 7\,\hat{i} + 5\,\hat{j}[/tex] (2)
[tex]\vec C = \vec A + \beta \cdot \vec B = c\,\hat{i}[/tex] (3)
By applying (1) and (2) in (3):
[tex](4\,\hat{i}-10\,\hat{j}) + \beta\cdot (7\,\hat{i}+5\,\hat{j}) = c\,\hat{i}[/tex]
[tex](4+7\cdot \beta)\,\hat{i} +(-10+5\cdot \beta)\,\hat{j} = c\,\hat{i}[/tex]
And we get two scalar equations after analyzing each component:
[tex]4+7\cdot \beta = c[/tex] (4)
[tex]-10+5\cdot \beta = 0[/tex] (5)
We solve for [tex]\beta[/tex] in (5):
[tex]\beta = 2[/tex]
And for [tex]c[/tex] in (4):
[tex]c = 4+7\cdot (2)[/tex]
[tex]c = 18[/tex]
The value of [tex]\beta[/tex] such that [tex]\vec C = \vec A + \beta \cdot \vec B = c\,\hat{i}[/tex] is 2.
Please see this question related to Sum of Vectors for further details: https://brainly.com/question/11881720
Why are hydraulic brakes used?
Answer:
Hydraulic brake systems are used as the main braking system on almost all passenger vehicles and light trucks. Hydraulic brakes use brake fluid to transmit force when the brakes are applied.
Explanation:
What improvements were made in measuring system with the introduction of standard units?
Answer:
Standard units are commonly used units of measurement, which help us measure length, height, weight, temperature, mass and more. These units are standardised, which means that everyone gets the same understanding of the size, weight and other properties of objects and things.
Explanation:
What Are the type's of Tidal turbines?
Answer:
Types of tidal turbines
Axial turbines.
Crossflow turbines.
Flow augmented turbines.
Oscillating devices.
Venturi effect.
Tidal kite turbines.
Turbine power.
Resource assessment.
Answer:
Axial turbines
Crossflow turbines
flow augmented turbines
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]
a vector starts at the point (0.0) and ends at (2,-7) what is the magnitude of the displacement
Answer:
|x| = √53
Explanation:
We are told that the vector starts at the point (0.0) and ends at (2,-7) .
Thus, magnitude of displacement is;
|x| = √(((-7) - 0)² + (2 - 0)²)
|x| = √(49 + 4)
|x| = √53
Suppose you exert a force of 314 N tangential to a grindstone (a solid disk) with a radius of 0.281 m and a mass of 84.2 kg What is the resulting angular acceleration of the grindstone assuming negligible opposing friction
Answer:
The angular acceleration is 26.6 rad/s^2.
Explanation:
Force, F = 314 N
radius, r = 0.281 m
mass, m = 84.2 kg
The grindstone is a disc.
The torque is given by
torque = force x radius
Torque = 314 x 0.281 = 88.234 Nm
The torque is given by
Torque = Moment of inertia x angular acceleration
[tex]88.234 = 0.5 mr^2 \alpha \\\\88.234 = 0.5\times 84.2\times 0.281\times 0.281\times \alpha \\\\\alpha = 26.6 rad/s^2[/tex]
A motorcyclist start from rest to reaches 6m/s with uniform acceleration for 3s what his acceleration?
Answer:
[tex]\boxed {\boxed {\sf 2 \ m/s^2}}[/tex]
Explanation:
Acceleration is the rate of change in velocity with respect to time. It is calculated by dividing the change in velocity by the change in time. The formula is:
[tex]a= \frac{ \Delta v}{\Delta t}[/tex] or [tex]a= \frac{v_f-v_i}{\Delta t}[/tex]
The change in velocity is the difference between the initial velocity and the final velocity. The motorcycle starts at rest, or 0 meters per second and reaches 6 meters per second. The change in time is 3 seconds.
[tex]\bullet \ v_f= 6 \ m/s\\\bullet \ v_i= 0 \ m/s \\\bullet \ \Delta t = 3 \ s[/tex]
Substitute the values into the formula
[tex]a= \frac { 6 \ m/s - 0 m/s}{3 \ s}[/tex]
Solve the numerator.
[tex]a= \frac{6 \ m/s}{3 \ s}[/tex]
[tex]a= 2 \ m/s^2[/tex]
The motorcyclist's acceleration is 2 meters per second squared.
The area around a charged object that can exert a force on other charged objects is an electric ___
A simple pendulum consists of a ball of mass 3 kg hanging from a uniform string of mass 0.05 kg and length L. If the period of oscillation of the pendulum is 2 s, determine the speed of a transverse wave in the string when the pendulum hangs vertically.
Answer:
v = 3.12 m/s
Explanation:
First, we will find the length of the string by using the formula of the time period:
[tex]T = 2\pi \sqrt{\frac{l}{g}}\\\\l = \frac{T^2g}{4\pi^2}\\\\[/tex]
where,
l = length of string = ?
T = time period = 2 s
g = acceleration due to gravity = 9.81 m/s²
Therefore,
[tex]l = \frac{(2\ s)^2(9.81\ m/s^2)}{4\pi^2}\\\\l = 0.99\ m[/tex]
Now, we will find tension in the string in the vertical position through the weight of the ball:
T = W = mg = (3 kg)(9.81 m/s²)
T = 29.43 N
Now, the speed of the transverse wave is given as follows:
[tex]v=\sqrt{\frac{Tl}{m}}\\\\v=\sqrt{\frac{(29.43\ N)(0.99\ m)}{3\ kg}}\\\\[/tex]
v = 3.12 m/s
After enjoying a tasty meal of the first moth, the bat goes after another moth. Flying with the same speed and emitting the same frequency, this time the bat detects a reflected frequency of 55.5 kHz. How fast is the second moth moving
This question is incomplete, the complete question is;
A bat flies towards a moth at 7.1 m/s while the moth is flying towards the bat at 4.4 m/s. The bat emits a sound wave of 51.7 kHz.
After enjoying a tasty meal of the first moth, the bat goes after another moth. Flying with the same speed and emitting the same frequency, this time the bat detects a reflected frequency of 55.5 kHz. How fast is the second moth moving
Answer:
the second moth is moving at 5.062 m/s
Explanation:
Given the data in the question;
Using doppler's effect
[tex]f_{moth[/tex] = f₀( [tex]v_{s[/tex] ± [tex]v_{observer[/tex] / [tex]v_{s[/tex] ± [tex]v_{source[/tex] )
f₁ = f₀( ([tex]v_{s[/tex] + v₂) / ( [tex]v_{s[/tex] - v₁ ) )
frequency reflected from the moth,
Now, moth is the source and the bat is the receiver
f₂ = f₁( ([tex]v_{s[/tex] + v₁ ) / ( [tex]v_{s[/tex] - v₂ ) )
hence, f = f₀[ ( ( [tex]v_{s[/tex] + v₁ ) / ( [tex]v_{s[/tex] - v₂ ) ) ( ( [tex]v_{s[/tex] + u₂ ) / ( [tex]v_{s[/tex] - u₁ ) )
we know that, the velocity of sound [tex]v_{s[/tex] = 343 m/s.
given that v₁ and v₂ { velocity of bat } = 7.1 m/s, f₀ = 51.7 kHz and f = 55.5 kHz.
we substitute
55.5 = 51.7[ ( ( 343 + 7.1 ) / ( 343 - 7.1 ) ) ( ( 343 + u ) / ( 343 - u ) ) ]
55.5 = 51.7[ ( 350.1 / 335.9 ) ( ( 343 + u ) / ( 343 - u ) ) ]
55.5 = 51.7[ 1.04227 ( ( 343 + u ) / ( 343 - u ) ) ]
55.5 = 53.885359 ( ( 343 + u ) / ( 343 - u ) ) ]
55.5 / 53.885359 = ( 343 + u ) / ( 343 - u )
1.02996 = ( 343 + u₂ ) / ( 343 - u )
( 343 + u₂ ) = 1.02996( 343 - u )
343 + u = 353.27628 - 1.02996u
u + 1.02996u = 353.27628 - 343
2.02996u = 10.27628
u = 10.27628 / 2.02996
u = 5.062 m/s
Therefore, the second moth is moving at 5.062 m/s
Rachel has good distant vision but has a touch of presbyopia. Her near point is 0.60 m. Part A When she wears 2.0 D reading glasses, what is her near point
Answer:
The right answer is "0.273 m".
Explanation:
Given:
Power (P),
[tex]\frac{1}{f} = 2D[/tex]
Near point,
u = 0.6 m
As we know,
⇒ [tex]\frac{1}{v} -\frac{1}{u}=\frac{1}{f} = 2[/tex]
By substituting the values, we get
⇒ [tex]\frac{1}{v} -\frac{1}{0.6} =2[/tex]
[tex]\frac{1}{v}=2+\frac{1}{0.6}[/tex]
[tex]\frac{1}{v} =\frac{1.2+1}{0.6}[/tex]
[tex]\frac{1}{v}=\frac{2.2}{0.6}[/tex]
By applying cross-multiplication, we get
[tex]0.6=2.2 \ v[/tex]
[tex]v = \frac{0.6}{2.2}[/tex]
[tex]S_{near} = 0.273 \ m[/tex]
There are 5640 lines per centimeter in a grating that is used with light whose wavelegth is 455 nm. A flat observation screen is located 0.661 m from the grating. What is the minimum width that the screen must have so the centers of all the principal maxima formed on either side of the central maximum fall on the screen
The minimum width of the screen is 34 cm.
For a diffraction grating, dsinθ = mλ where d = grating spacing = 1/5640 lines per cm = 1/5640 cm per line = 1/5640 × 10⁻² m per line, θ = angle between principal maximum and the center axis of the grating, m = order of maxima = 1 (since we require the position of the principal maximum) and λ = wavelength = 455 nm = 455 × 10⁻⁹ m
So, sinθ = mλ/d
Also tanθ = L/D where θ = angle between principal maximum and the center axis of the grating, L = distance between central maximum and principal maximum and D = distance between grating and screen = 0.661 m.
For small angles sinθ ≈ tanθ
So, mλ/d = L/D
making L subject of the formula, we have
L = mλD/d
L = 1 × 455 × 10⁻⁹ m × 0.661 m ÷ 1/5640 × 10⁻² m per line
L = 1 × 455 × 10⁻⁹ m × 0.661 m × 5640 × 10² line per m
L = 1696258.2 × 10⁻⁷ m
L = 0.16963 m
L ≅ 0.17 m
So, for centers of all the principal maxima formed on either side of the central maximum fall on the screen, the minimum width of the screen is w = 2L.
So, w = 2 × 0.17 m
w = 0.34 m
w = 34 cm
So for the centers of all the principal maxima formed on either side of the central maximum fall on the screen, the minimum width of the screen is 34 cm.
Learn more about diffraction grating here:
https://brainly.com/question/15712101
In a double-slit experiment, the slit separation is 1.75 mm, and two coherent wavelengths of light, 425 nm and 510 nm, illuminate the slits. At what angle from the centerline on either side of the central maximum will a bright fringe from one pattern first coincide with a bright fringe from the other pattern
Answer:
the required angle is 0.0834879⁰
Explanation:
Given the data in the question;
slit separation; d = 1.75 mm = 1.75 × 10⁻³ m
wavelength λ₁ = 425 nm = 425 × 10⁻⁹ m
wavelength λ₂ 510 nm = 510 × 10⁻⁹ m
Now, we know that, the angle at which a particular bright fringe occurs on either side of the central bright fringe will be;
tanθ = [tex]y_m[/tex] / D = mλ/d
since they both coincides;
tanθ₁ = tanθ₂
m₁λ₁/d = m₂λ₂/d
multiply both sides by d
so,
m₁/m₂ = λ₂/λ₁
we substitute
m₁/m₂ = 510 nm / 425 nm
m₁/m₂ = 510 nm / 425 nm
divide through by 85
m₁/m₂ = 6 / 5
hence m₁ and m₂ are 6 and 5
so, from the previous formula
tanθ₂ = m₂λ₂/d
we substitute
tanθ₂ = [ 5 × ( 510 × 10⁻⁹ m ) ] / 1.75 × 10⁻³ m
tanθ₂ = 255 × 10⁻⁸ m / 1.75 × 10⁻³ m
tanθ₂ = 255 × 10⁻⁸ m / 1.75 × 10⁻³ m
tanθ₂ = 0.00145714
θ₂ = tan⁻¹( 0.00145714 )
θ₂ = 0.0834879⁰
Therefore, the required angle is 0.0834879⁰
Thorium-232 goes through multiple types of decay in order to reach a stable isotope. What isotope is created after the first two decays if it first goes through an alpha decay and then a beta decay?
A)uranium-236
B)protactinium-232
C)radon-224
D)Astinium-228
Answer:
The answer would be D), if the decay is beta negative.
Explanation:
Thorium-232 goes through alpha decay:
Thorium-232 --> Helium-4 + Radium-228
Radium-228 then can undergo beta positive or beta negative decay:
Beta positive = Radium-228 --> Electron + Francium-228
Beta negative = Radium-228 --> Positron + Actinium-228
Therefore, the isotope that is created is Actinium-228
A 100 kg man is one fourth of the way up a 4.0 m ladder that is resting against a smooth, frictionless wall. The ladder has mass 25 kg and makes an angle of 56 degrees with the ground. What is the magnitude of the force of the wall on the ladder at the point of contact, if this force acts perpendicular to the wall and points away from the wall
Answer:
[tex]N_f=248N[/tex]
Explanation:
From the question we are told that:
Mass [tex]m=100kg[/tex]
Ladder Length [tex]l=4.0m[/tex]
Mass of Ladder [tex]m_l=25kg[/tex]
Angle [tex]\theta=56 \textdegree[/tex]
Generally the equation for Co planar forces is mathematically given by
[tex]mgcos \theta *2+Mgcos\theta*1 -N_fsin \theta*4=0[/tex]
Therefore
[tex]25*9.81cos 56 *2+100*9.81cos56*1 -N_fsin 56*4=0[/tex]
[tex]N_f=248N[/tex]
How many types of physics?
Answer:
Two Main Branches of Physics
it is Classical Physics and Modern Physics.
Explanation:
Further sub Physics branches are Mechanics, Electromagnetism, Thermodynamics, Optics, etc. The rapid progress in science during recent years has become possible due to discoveries and inventions in the field of physics.
hope it helped
A 15kg mass suspended from a ceiling is pulled aside with a horizontal force, F. Calculate the value of the tension.
Answer:
147 Newtons
Explanation:
To find tension, you can use the formula Tension = (mass)(gravity)
*Gravity's acceleration = 9.8 m/s^2 because of Newton's law of universal gravitation*
T = (15kg)(9.8m/s^2)
= 147 Newtons
Hope this helps! Best of luck <3
A crude approximation is that the Earth travels in a circular orbit about the Sun at constant speed, at a distance of 150,000,000 km from the Sun. Which of the following is the closest for the acceleration of the Earth in this orbit?
A. exactly 0 m/s2.
B. 0.006 m/s2.
C. 0.6 m/s2.
D. 6 m/s2.
E. 10 m/s2.
Answer:
The answer is "Option B".
Explanation:
[tex]r=15\times 10^{7}\ km\ = 15\times 10^{10}\ m\\\\w=\frac{2\pi}{1\ year}\\\\=\frac{2\pi}{1\times 365.24 \times 24 \times 60 \times 60\ sec}\\\\a=w^2r\\\\=(\frac{2\pi}{1\times 365.24 \times 24 \times 60 \times 60\ sec})^2 \times 15 \times 10^{10}\ \frac{m}{s^2}\\\\[/tex]
[tex]=5.940 \times 10^{-3} \ \frac{m}{s^2}\\\\=6 \times 10^{-3} \ \frac{m}{s^2}\\\\=0.006\ \frac{m}{s^2}\\\\[/tex]
Air contained in a rigid, insulated tank fitted with a paddle wheel, initially at 300 K, 2 bar, and a volume of 2 m3, is stirred until its temperature is 500 K. Assuming the ideal gas model, for the air, and ignoring kinetic and potential energy, determine
Answer:
The final pressure in bar will be "[tex]\frac{10}{3} \ Bar[/tex]".
Explanation:
As we know,
PV = nRT
[tex]\frac{P_1}{T_1} =\frac{P_2}{T_2} =CONST[/tex]
then,
⇒ [tex]\frac{2 \ bar}{300 \ K} = \frac{P_2}{500 \ K}[/tex]
⇒ [tex]P_2=(\frac{2}{300}\times 500 )Bar[/tex]
[tex]=\frac{10}{3} \ Bar[/tex]
Thus the above is the correct answer.
What is the mass of the diver in (Figure 1) if she exerts a torque of 2200 N⋅m on the board, relative to the left (A) support post?
A-->B = 1.0m
B--> end of board = 3.0m
Answer:
56.1 kg
Explanation:
Given
[tex]T = 2200Nm[/tex] -- torque
[tex]d_1 = 1.0m[/tex]
[tex]d_2 = 3.0m[/tex]
Required
The mass of the diver
From the question, we understand that the diver is at the extreme of the board.
So, we make use of the following torque equation
[tex]T = F * (d_1 + d_2)[/tex]
Where:
[tex]F \to Force[/tex]
So, we have:
[tex]2200 = F * (1.0 + 3.0)[/tex]
[tex]2200 = F * 4.0[/tex]
Divide both sides by 4.0
[tex]550 = F[/tex]
[tex]F = 550 N[/tex] --- This is the force exerted by the diver (in other words, the weight).
To calculate the mass, we use:
[tex]F = mg[/tex]
Make m the subject
[tex]m = \frac{F}{g}[/tex]
This gives:
[tex]m = \frac{550}{9.8}[/tex]
[tex]m = 56.1kg[/tex]
Twin skaters approach each other with identical speeds. Then, the skaters lock hands and spin. Calculate their final angular velocity, given each had an initial speed of 2.50 m/s relative to the ice. Each has a mass of 70.0 kg, and each has a center of mass located 0.800 m from their locked hands. You may approximate
Answer:
[tex]\omega=3.135rad/s[/tex]
Explanation:
From the question we are told that:
initial Speed [tex]V_1=2.50[/tex]
Mass [tex]m=70.0kg[/tex]
Center of mass [tex]d=0.0.800m\[/tex]
Generally the equation for angular velocity is is mathematically given by
[tex]\omega=\frac{v}{r}\\\\\omega=\frac{2.50}{0.0800}[/tex]
[tex]\omega=3.135rad/s[/tex]
