This question involves the concepts of density, volume, and mass.
(a) The volume occupied by this amount of water is "90 m³".
(b) The length of an edge of each small cube is "83 μm".
(c) The length of the edge of each small cube is "equal" to the diameter of a molecule.
(a)
The volume can be found using the following formula:
[tex]V = \frac{nM}{\rho}[/tex]
where,
V = volume occoupied = ?
n = no. of moles = 5 mol
M = molar mass = 18 g/mol
[tex]\rho[/tex] = density of water = 1 g/m³
Therefore,
[tex]V=\frac{(5\ mol)(18\ g/mol)}{1\ g/m^3}\\\\[/tex]
V = 90 m³
(b)
First, we will find the volume of an individual molecule:
[tex]V_i =\frac{V}{nN_A}[/tex]
where,
[tex]N_A[/tex] = Avogadro's number = 6.02 x 10²³ molecules/mol
Therefore,
[tex]V_i=\frac{90\ m^3}{5\ mol(6.02\ x\ 10^{23}\ molecules/mol)}[/tex]
Vi = 3 x 10⁻²³ m³
This volume can be given as a volume of the sphere:
[tex]V_i=\frac{4}{3}\pi r^3\\\\r=\sqrt[3]{\frac{3(3\ x\ 10^{-23}\ m^3)}{4\pi}}[/tex]
r = 4.15 x 10⁻⁷ m
Diameter = d = 2r = 2(4.15 x 10⁻⁷ m)
d = 8.3 x 10⁻⁷ m = 83 μm
since the cubes are adjacent to each other. Therefore, the diameter will be equal to the edge length.
Edge Length = L = d
L = 8.3 x 10⁻⁷ m = 83 μm
(c)
The edge length is equal to the diameter of the molecule.
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The Air Force is F 22 raptor fighter jets mass is 21,000 KG the F 22 is flying at a height of 26,000 miles what is its gravitational potential energy
Answer:
5460000000 J OR 5460000 KJ
Explanation:
GPE = mgh
21000*10*26000
=5460000000J OR 5460000KJ
What is the effect of erosion?
A. New land forms at the mouth of a river.
B. New land forms at the top of a mountain.
C. A mountain forms.
D. A fossil is created.
Comparing energy resources
Answer:
Sorry, This Photo is not clear.
If you were told an atom was an ion, you would know the atom must have a?
A neutral charge
B charge
C negative charge
D positive charge
Answer:
its b it must be charge
Explanation:
A block slides down a smooth ramp, starting from rest at a height h. When it reaches the bottom it's moving at speed v. It then continues to slide up a second smooth ramp. At what height is its speed equal to v/2
Answer:
3h/4
Explanation:
At speed v/2 height will be 3/4 h
What is equation of motion in kinematics?Equation that describes the motion of point , bodies , and system of bodies without considering the force that cause them to move is called equation of motion in kinematics
When block is at top of first ramp
u=0 ( block was at rest )
a = g ( acceleration due to gravity
using equation of motion
2as = v^2 - u^2
2gh = v^2
Then the block continued and reached a speed of v1 = v/2 on second ramp
now , final velocity = v= v1 =[tex]\sqrt{2gh}[/tex] / 2
u= [tex]\sqrt{2gh\\}[/tex]
s= h1
using equation of motion , we get
2as = v^2 - u^2
2(-g)h1 =( [tex]\sqrt{2gh}[/tex]/2)^2 - [tex]\sqrt{2gh}[/tex]
2(-g)h1 = (g h - 4 g h) / 2
h1 = 3/4 h
At speed v/2 height will be 3/4 h
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Easiest way to Find fahrenhiet to celsius please i need necessary 20 for the fastest correct answer
Answer:
thx for the points
Explanation:
no need brainliest
What is the volume of 150g of a substance that has a density of 150g of a substance that has a density of 11.3g/cm3
Answer:
25.0 cm3
Explanation:
The volume is 25.0 cm3 .
A weather forecaster uses a computational model on a Monday to predict the weather on Friday. Why might that forecast change? (1 point)
A. The forecaster may have done the calculations a second time because he made a mistake in the calculation.
B. Someone may have reported the weather incorrectly before the first computation.
C. The forecaster may have had the computer model do the calculations a second time, and he found that the prediction changed.
D. An area of low pressure might move more quickly on Tuesday and Wednesday than expected.
Answer:
D. An area of low pressure might move more quickly on Tuesday and Wednesday than expected.
Explanation:
I took the test and got it wrong :/
An object with mass 1.2 kg is moving at a constant speed in a circle with radius 2.5 m. If the
object makes exactly 12 revolutions in a 1 minute, what is the acceleration of the object?
Answer:
3.9m/s² is the acceleration
The acceleration of the object is 3.9m/s².
How do you find the acceleration of an object?Acceleration (a) is the change in velocity (Δv) over the change in time (Δt), represented by the equation a = Δv/Δt. This allows you to measure how fast velocity changes in meters per second squared (m/s^2).
What is an example of an object that has acceleration?When you fall off a bridge. The car turning at the corner is an example of acceleration because the direction is changing. The quicker the turns, the greater the acceleration.
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define moment of a force about a point and give the si unit of moment?
Answer:
Moment of a force is the product of the force and the perpendicular distance of force from axis of rotation.
The SI unit of force is newton (N).
Explanation:
Hope it helps you again!
What was different about the molecules you needed to make protein 3 compared to the molecules you used to make protein 2?
Answer:
the different about the molecules we needed to make protein 3 compared to the molecules we used to make protein 2 is that if we used 2 molecules than it will be smaller than using protein 3.
A power plant running at 31 % efficiency generates 270 MW of electric power. Part A At what rate (in MW) is heat energy exhausted to the river that cools the plant
The rate of heat energy exhausted to the river is 600.96 MW.
What is efficiency?The ratio of usable output to total input can be used to objectively measure efficiency. The efficiency of the device is defined as the ratio of energy converted to a useable form to the original amount of energy supplied.
Given parameters:
Efficiency of the power plant; η = 31 %
Output electric power; O = 270 MW.
We know that, Efficiency of the power plant;
η = (Output electric power/ input power)× 100%
⇒ input power = (Output electric power × 100)/η
⇒ input power = (270 × 100)/31 MW
= 870.96 MW.
So, the rate of heat energy exhausted to the river that cools the plant = Input power- output power
= (870.96 - 270) MW
= 600.96 MW.
Hence, heat energy exhausted to the river that cools the plant is 600.96 MW.
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help i am having a mental breakdown help! need this done...
*graph is below* help
1. What is Peter’s total distance traveled? What is Peter's displacement?
2. Is there a time when Peter is not moving? If so, when?
a certain hydraulic jack is used for lifting load. a force of 250N is applied on a small piston with a diameter of 80cm while the diameter of the larger piston is 2.0m.the applied force moves through a distance of 25cm.calculate the maximum mass of a load that can be lifted by the jack and the distance through which the load is lifted.(take g=9.81ms^-2)
The maximum mass of a load that can be lifted by the jack and the distance covered are:
m = 160.2 Kg
h = 25 cm
Given that a certain hydraulic jack is used for lifting load. a force of 250N is applied on a small piston with a diameter of 80cm while the diameter of the larger piston is 2.0m.
The parameters given are
[tex]F_{1}[/tex] = 250
[tex]A_{1}[/tex] = Area of the small piston = π[tex]r^{2}[/tex]
[tex]A_{1}[/tex] = 22/7 x [tex]0.4^{2}[/tex]
[tex]A_{1}[/tex] = 0.5 [tex]m^{2}[/tex]
[tex]F_{2}[/tex] = ?
[tex]A_{2}[/tex] = Area of the large piston = π[tex]r^{2}[/tex]
[tex]A_{2}[/tex] = π x 1
[tex]A_{2}[/tex] = 3.14 [tex]m^{2}[/tex]
To calculate the force on the large piston, we will use the below formula
[tex]F_{1}[/tex]/ [tex]A_{1}[/tex] = [tex]F_{2}[/tex] / [tex]A_{2}[/tex]
Substitute all the parameters into the equation
250/0.5 = [tex]F_{2}[/tex]/3.14
[tex]F_{2}[/tex] = 1570 N
To calculate the maximum mass of a load that can be lifted by the jack, let us apply Newton second law
F = mg
1570 = 9.8m
m = 1570/9.8
m = 160.2 Kg
.(take g=9.81ms^-2)
If the applied force moves through a distance of 25cm, the distance through which the load is lifted will be
[tex]F_{1}[/tex]/ 0.25[tex]A_{1}[/tex] = [tex]F_{2}[/tex] / [tex]A_{2}[/tex]h
250/0.125 = 1570/3.14h
make h the subject of the formula
6280h = 1570
h = 1570/6280
h = 0.25 m
Therefore, the distance through which the load is lifted is 25 cm
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In which of the following situations is the Doppler Effect absent?
The source and the observer are moving towards each other.
The observer is moving toward the source.
The source is moving away from the observer.
The source and the observer are both moving with the same velocity.
Answer:
Correct option is
C
The listener is moving towards the source of sound
The Doppler effect (or the Doppler shift) is the change in frequency or wavelength of a wave (or another periodic event) for an observer moving relative to its source.
So if both listener and source of sound are stationary or are moving with the same velocity that is relative velocity is zero, there won't be any change in frequency.
Thus, in this case, the effect will arise when a listener is moving towards the source of the sound.
When measuring the critical angle, in which medium do we need the refracted ray to be (air or glass)
Answer:
N1 sin theta1 = N2 sin theta2 Snell's Law
For the refracted ray to be reflected then sin theta2 = 1
N1 sin theta 1 = N2
Also N2 must be less than N1 for complete reflection
sin theta1 = N2/N1
If you are considering air and glass then N2 = 1 (for air)
sin theta1 = 1 / N2 where N2 must be for glass in this case
What is the maximum height above ground a projectile of mass 0.79 kg, launched from ground level, can achieve if you are able to give it an initial speed of 80.3 m/s?
From kinematic, the maximum height above ground a projectile of mass 0.79 kg, launched from ground level can achieve is 329 meters approximately.
The parameters given from the question are
mass m = 0.79kg
initial speed U = 80.3 m/s
Maximum height H = ?
The object will be going up under the influence of gravity. Acceleration due to gravity g = -9.8m/[tex]s^{2}[/tex]
At maximum height, final velocity V = 0
From kinematic formula, the best formula to use is
[tex]V^{2} = U^{2} - 2gH[/tex]
Substitute all the parameters into the equation
0 = [tex]80.3^{2}[/tex] - 2 x 9.8 x H
19.6H = 6448.09
H = 6448.09 / 19.6
H = 328.98 m
Therefore, the maximum height above ground a projectile of mass 0.79 kg, launched from ground level can achieve is 329 meters approximately.
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What is the acceleration of a box weighing 596 N if a force of 777 N is applied to it?
Answer:
[tex]12.79 m/s^2[/tex]
Explanation:
1. Apply Newton's 2nd law:
[tex]F=ma[/tex]
2. Use the definition of weight to find the mass of the box:
[tex]W=mg=m=\frac{W}{g}[/tex]
3. Solve for acceleration in Newton's 2nd law and substitute for mass:
[tex]a=\frac{F}{\frac{W}{g} }[/tex]
4. Plug in the given values from the question and assume [tex]g=9.81m/s^2[/tex]:
[tex]a=\frac{777N}{\frac{596N}{9.81m/s^2} } } =12.79m/s^2[/tex]
Suppose that Hubble's constant were H0 = 51 km/s/Mly (which is not its actual value). What would the approximate age of the universe be in that case?
Given the Hubble's constant, the approximate age of the universe is 5.88 × 10⁹ Years.
Given the data in the question;
Hubble's constant; [tex]H_0 = 51km/s/Mly[/tex]
Age of the universe; [tex]t = \ ?[/tex]
We know that, the reciprocal of the Hubble's constant ( [tex]H_0[/tex] ) gives an estimate of the age of the universe ( [tex]t[/tex] ). It is expressed as:
[tex]Age\ of\ Universe; t = \frac{1}{H_0}[/tex]
Now,
Hubble's constant; [tex]H_0 = 51km/s/Mly[/tex]
We know that;
[tex]1\ light\ years = 9.46*10^{15}m[/tex]
so
[tex]1\ Million\ light\ years = [9.46 * 10^{15}m] * 10^6 = 9.46 * 10^{21}m[/tex]
Therefore;
[tex]H_0 = 51\frac{km}{\frac{s}{Mly} } = 51000\frac{m}{s\ *\ Mly} \\\\H_0 = 51000\frac{m}{s\ *\ (9.46*10^{21}m)} \\\\H_0 = 5.39 *10^{-18}s^{-1}\\[/tex]
Now, we input this Hubble's constant value into our equation;
[tex]Age\ of\ Universe; t = \frac{1}{H_0}\\\\t = \frac{1}{ 5.39 *10^{-18}s^{-1}} \\\\t = 1.855 * 10^{17}s\\\\We\ convert\ to\ years\\\\t = \frac{ 1.855 * 10^{17}}{60*60*24*365}yrs \\\\t = \frac{ 1.855 * 10^{17}}{31536000}yrs\\\\t = 5.88 *10^9 years[/tex]
Therefore, given the Hubble's constant, the approximate age of the universe is 5.88 × 10⁹ Years.
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A vector has a magnitude of 30m at an angle of 225 degrees with respect to the positive x-axis and has a magnitude of 13m.what are the Component of vectors B and it's direction
Answer:
I'm pretty sure this is not a complete question. My guess is that you are trying to add/subtract vectors. Vectors have both magnitude and direction, so vector A is pretty clear, but a magnitude of 13 (i'm guessing a resultant) without a direction is weird.
IF 13 is the magnitude of the resultant, vector B added to vector A could have any magnitude 17 ≤ B ≤ 43
It could have any direction of
θ = (225 - 180) ± arcsin(13/30)
θ = 45 ± 25.679...
70.679 ≤ θ ≤ 19.321
components of vector B would be
Bx = |B|cosθ
By = |B|sinθ
My guess is that you are trying to add/subtract vectors. Vectors have both magnitude and direction, so vector A is pretty clear, but a magnitude of 13 (i'm guessing a resultant) without a direction is weird.
What is magnitude of the resultant?IF 13 is the magnitude of the resultant, vector B added to vector A could have any magnitude 17 ≤ B ≤ 43
It could have any direction of
θ = (225 - 180) ± arcsin(13/30)
θ = 45 ± 25.679...
70.679 ≤ θ ≤ 19.321
components of vector B would be
Bx = |B|cosθ
By = |B|sinθ
My guess is that you are trying to add/subtract vectors. Vectors have both magnitude and direction, so vector A is pretty clear, but a magnitude of 13 (i'm guessing a resultant) without a direction is weird. A vector has a magnitude of 30m at an angle of 225 degrees with respect to the positive x-axis and has a magnitude of 13m.
Therefore, My guess is that you are trying to add/subtract vectors. Vectors have both magnitude and direction, so vector A is pretty clear, but a magnitude of 13 (i'm guessing a resultant) without a direction is weird.
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About the pulmonary surfactant it is true that: I. It reduces the surface tension of water. II. It is important for generating a pressure gradient between small and large alveoli.
C, I = False, Il = True
B, I = True, Il False
D, I = False, Il = False
A, I = True, II = True
Answer:
a
Explanation:
Look up the specs on a c6 rocket engine. How many C6 engines would it take to launch Mr. Blazey (90kg)
from the elevation of Boulder to the elevation of longs peak? (3 points) What would the max power be? (3
points) The max acceleration? (3 points) For all 12 points, you'll need to account for the mass of the added
rockets (you can assume their mass declines linearly over their burn and is O after they finish burning).
Answer:
The Estes C6-0 engine is a booster stage engine designed for model rocket flight and has to be used with a standard engine. This engine is for flights in rockets weighing less than 4 ounces, including the engines. Each package includes 3 engines, 4 starters and 4 plugs.
A solid sphere starts from rest and rolls down a slope that is 6.4 m long. If its speed at the bottom of the slope is 5.3 m/s, what is the angle of the slope
From the relationship between acceleration a and g on an inclined plane, the angle of the slope is 13 degrees
Given that a solid sphere starts from rest and rolls down a slope that is 6.4 m long. The speed at the bottom of the slope is 5.3 m/s, the distance travelled is 6.4 m. That is,
Initial velocity U = 0 ( since it starts from rest)
Final velocity V = 5.3 m/s
distance S = 6.4 m
Let us first calculate its acceleration by using third equation of motion.
[tex]V^{2}[/tex] = [tex]U^{2}[/tex] + 2aS
[tex]5.3^{2}[/tex] = 0 + 2 x 6.4a
28.09 = 12.8a
a = 28.09 / 12.8
a = 2.2 m / [tex]s^{2}[/tex]
To calculate the angle of the slope, let us use the relationship between acceleration a and g on an inclined plane.
acceleration a = gsin∅
substitute all the relevant parameters
2.2 = 9.8 sin∅
sin∅ = 2.2/9.8
sin∅ = 0.224
∅ = [tex]Sin^{-1}[/tex](0.224)
∅ = 12.97 degrees
∅ = 13 degrees (approximately)
Therefore, the angle of the slope is 13 degrees
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Which of the following correctly describes electromagnetic waves?
A. transverse waves
B. Longitudinal waves
C. Have a constant wavelength
D. Need a medium to transfer energy
Answer:
answer
transverse waves
organize the three states of matter from least kinetic energy to greatest kinetic energy
Particle arrangement and movement
⇒ In terms of relative energy, gas particles have the most energy, solid particles have the least energy and liquid particles are somewhere in between. (All compared at the same temperature.)
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Kinesha and her friend were watching a solar eclipse. Kinesha explains to her friend that a solar eclipse means that Earth is located between the Sun and the Moon. Her friend tells Kinesha that her explanation is incorrect. Why?
this is where the sun and moon line up where you asleep only a tiny bit of the sun it's pretty cool to see
Explanation:
a solar eclipse means when the moon goes infront of the sun and the earth turns dark
Ignoring any effects of dc resistance, what is the total reactance of a 250 mH coil in series with a 4.7 microfarad capacitor at a signal frequency of 60 Hz
The total reactance of the inductor and the capacitor is 470.1 ohms.
The given parameters;
inductance of the coil, L = 250 mHcapacitance, C = 4.7 μfrequency of the circuit, f = 60 HzThe inductive reactance of the coil is calculated as follows;
[tex]X_l = \omega L\\\\X_l = 2\pi f L\\\\X_l = 2\pi \times 60 \times 250 \times 10^{-3}\\\\X_l = 94.26 \ ohms[/tex]
The capacitive reactance of the capacitor is calculated as follows;
[tex]X_c = \frac{1}{\omega C} \\\\X_c = \frac{1}{2\pi f C} \\\\X_c = \frac{1}{2\pi \times 60 \times 4.7 \times 10^{-6}} \\\\X_c = 564.31 \ ohms[/tex]
The impedance of the circuit is calculated as follows;
[tex]Z = \sqrt{(X_c - X_l)^2} \\\\Z = X_c - X_l\\\\Z = 564.31 - 94.26\\\\Z = 470.1 \ ohms[/tex]
Thus, the total reactance of the inductor and the capacitor is 470.1 ohms.
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A 45 kg figure skater is spinning on the toes of her skates at 1.1 rev/s. Her arms are outstretched as far as they will go. In this orientation, the skater can be modeled as a cylindrical torso (40 kg, 20 cm average diameter, 160 cm tall) plus two rod-like arms (2.5 kg each, 71 cm long) attached to the outside of the torso. The skater then raises her arms straight above her head. In this latter orientation, she can be modeled as a 45 kg, 20-cm-diameter, 200-cm-tall cylinder.
What is her new rotation frequency, in revolutions per second?
Answer: HOPE THIS HLEP YOU
Explanation:
Please help
A man stands on a freely rotating platform with his arms extended, his rotational frequency is 0.25rev/s. But when he draws them in, his frequency is 0.80revs/s. Find the ratio of his moment of inertia in the first case to that in the second.
Answer:
sorry for you
The ratio of the man's moment of inertia in the first case (arms extended) to that in the second case (arms drawn in) is 3.2.
The relationship between the rotational frequency [tex](\(\omega\))[/tex] and moment of inertia (I) is given by the equation:
[tex]\[I_1\omega_1 = I_2\omega_2\][/tex]
where [tex]\(I_1\)[/tex]and [tex]\(I_2\)[/tex] are the moments of inertia in the two cases, and [tex]\(\omega_1\) and \(\omega_2\)[/tex] are the corresponding rotational frequencies.
Let's denote the moment of inertia in the first case (arms extended) as [tex]\(I_1\)[/tex] and in the second case (arms drawn in) as [tex]\(I_2\)[/tex]. The given rotational frequencies are [tex]\(\omega_1 = 0.25 \, \text{rev/s}\) and \(\omega_2 = 0.80 \, \text{rev/s}\)[/tex].
Using the equation [tex]\(I_1\omega_1 = I_2\omega_2\)[/tex], we can rearrange it to solve for the ratio of moments of inertia:
[tex]\[\frac{I_1}{I_2} = \frac{\omega_2}{\omega_1}\][/tex]
Substituting the given values, we have:
[tex]\[\frac{I_1}{I_2} = \frac{0.80 \, \text{rev/s}}{0.25 \, \text{rev/s}}\][/tex]
Simplifying the expression, we get:
[tex]\[\frac{I_1}{I_2} = 3.2\][/tex]
Therefore, the ratio of the man's moment of inertia in the first case (arms extended) to that in the second case (arms drawn in) is 3.2.
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A child at the top of a slide de has a gravitational store of 1800J. What is the child's maximum kinetic store as he slides down? Explain why
Hi there!
We know that:
Initial Total Mechanical Energy = Final Total Mechanical Energy
(Ei = Ef)
In this instance:
Ei = Gravitational Potential Energy
Ef = Kinetic Energy
In the absence of friction, ALL of the initial potential energy will be changed into kinetic energy at the bottom of the slide. Thus, the maximum kinetic energy of the child will be 1800 J.