Carbon-Based Fuels Dominate Global Energy Use
Crude oil, coal, and natural gas supply about 85% of the energy used in the world. Fossil fuels are valuable as sources of energy because they contain hydrocarbons and other carbon-based materials.
A racecar can be slowed with a constant acceleration of -14 m/s2. If the car is going 75 m/s, how many meters will take to stop?
Answer:
Explanation:
v² = u² + 2as
s = (v² - u²)/2a
s = (0² - 75²) / (2(-14))
s = 200.8928
s = 200 m
Answer:
Identification
Explanation:
:-;..........
Given that T= [L] * [m] Y [Lt-2]Z
Find
1) Y
2)X
3)Z
Answer:
2
Explanation:
please mark me as brainlist i need it so sorry
which statement is true about heat and light energy
Answer:
statements?
Explanation:
Name two environmental factors, one natural and one human-made, that could account for the trend or pattern in bird (quail, wren) and rodent (mice, rabbits) populations before 1997. Help!
Factors such as more crops, fewer predators, invasive species, and changes in the environment explain a decrease or increase in the birds and rodent populations.
Before 1997, both rodents and birds populations increased due to different factors. Moreover, these factors can be classified as natural or human-made factors. Here are the factors that mainly affected these animals:
Bird population:
Increase:
Decrease in predators that increase birds chances to survive (natural factor)
Increase in crops that improve food access for birds (human factor)
Decrease:
Invasive species or increase of predators (natural factor)
Destruction of natural habitats due to an increase in industry and extraction (human factor)
Rodent population:
Increase:
More adaptability that increased rodents chances to survive (natural factor)Increase in crops that improve food access for rodents (human factor)Decrease:
Invasive species or increase of predators (natural factor)Increase in rabbit consumption and use of rodent control products (human factor)Learn more in: https://brainly.com/question/18123593
In a thermodynamic process, the internal energy of a system in a container with adiabatic walls decreases by 800 J. Which statement is correct
The correct statement is, the system performed 800 J of work on its surroundings.
The given parameters:
change in internal energy, ΔU = 800 JApply first law of thermodynamics;
ΔU = Q + W
where;
Q is the heat gainedW is the work done on the system by the surroundingIn adiabatic process no heat is gained or lost by the system.
Q = 0
ΔU = W (work is positive when it is done on the system by its surrounding)
If work is done by the system to the surrounding, the new equation becomes;
ΔU = - W
W = - ΔU
W = -800 J
This implies that the system performed 800 J of work on its surroundings.
"Your question is not complete, it seems to be missing the following information";
In a thermodynamic process, the internal energy of a system in a container with adiabatic walls decreases by 800 J. Which statement is correct?
a. The system lost 800 J by heat transfer to its surroundings.
b. The system gained 800 J by heat transfer from its surroundings.
c. The system performed 800 J of work on its surroundings.
d. The surroundings performed 800 J of work on the system.
e. The 800 J of work done by the system was equal to the 800 J of heat transferred to the system from its surroundings.
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Another box of samples is hoisted up by the same rope. If the rope is shaken with the same frequency as before, and the wavelength is found to be 7.9 m , what is the mass of this box of samples
The motion of the rope which is perpendicular to the direction of the
propagation of the wave is a transverse wave motion.
The mass of the box is approximately 9.93 kgReasons:
The given function for the wave speed is presented as follows;
[tex]\displaystyle v = \sqrt{\frac{T}{\mu} } \\[/tex]
Where;
[tex]\displaystyle \mu = \frac{Mass \ of \ rope }{Length \ of \ rope}[/tex]
Taking the mass of the rope as, m = 2.00 kg
The length of the rope, L = 80.0 m
The mass hanging on the rope, M = 20.0 kg
We have;
T = 20.0 kg × 9.81 m/s² = 196.2 N
[tex]\displaystyle \mu = \frac{2.0 }{80.0} = \frac{2.0 }{80.0} = 0.025[/tex]
Therefore;
Taking the wavelength as, λ = 7.9 m, and the frequency as 20 Hz, we have;
v = f × λ
Therefore;
v = 7.9 Hz × 7.9 m = 62.41 m/s
Which gives;
[tex]\displaystyle 62.41 = \sqrt{\frac{T}{0.025} }[/tex]
T = 62.41² × 0.025 = 97.3752025
[tex]\displaystyle Mass, \ m = \mathbf{\frac{T}{g}}[/tex]
Where;
g = The acceleration due to gravity which is approximately 9.81 m/s²
[tex]\displaystyle Mass, \ m = \frac{97.3752025}{9.81} \approx 9.93[/tex]
Therefore;
The mass of the box, m ≈ 9.93 kg
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The parameters obtained from a similar question online are;
[tex]\displaystyle The \ equation \ applied, \, v = \sqrt{\frac{T}{\mu} } \\[/tex]
Length of the rope, L = 80.0 m
Mass of the rope, m = 2.0 kg
Frequency of a point on the rope, f = 20 Hz
Find the work done by the force field F in moving an object from A to B. F(x, y) = 6y3/2i + 9x y j A(1, 1), B(3, 4)
Answer:
138
Explanation:
(since there's a blank next to "9x y j", I'm assuming the y of F(x, y) is 9x[tex]\sqrt{y}[/tex] j)
1) find the partial derivative of each:
[tex](6y^{\frac{3}{2} })i + (9x\sqrt{y} )j[/tex]
[tex]f_{x} =(6y^{\frac{3}{2} })_{x} = \int\limits{6y^{\frac{3}{2} }} \, dx = 6xy^{\frac{3}{2} } +c \\\\f_{y} = (9x\sqrt{y} )_{y} = \int\limits{9xy^{\frac{1}{2} } } \, dy = 9x(\frac{2}{3} )y^{\frac{3}{2} } +c = 6xy^{\frac{3}{2} } +c[/tex]
2) use partial integrals to make gradient of f:
take whatever you got from partial integral and add them together (if they repeat, just use it once)
[tex]F =[/tex] Vf (V = gradient of)
[tex]F(x, y) = 6xy^{\frac{3}{2} }[/tex]
3) Evaluate the integrals with given points:
Integral of F dotted with dr = F(point B) - F(point A) = F(3, 4) - F(1, 1)
[tex]F(point B) = 6(3)(4)^{\frac{3}{2} }\\F(Point A) = 6(1)(1)^{\frac{3}{2}}\\F(point B) - F(pointA) = 6(3)(4)^{\frac{3}{2} }-(6(1)(1)^{\frac{3}{2}})[/tex]
= 144 - 6 = 138 units of work
Work done by the force field F in moving an object from A to B = 138 J
Given data :
Force field F(x,y) = [tex]6y^{\frac{3}{2} }i + (9x\sqrt{y} ) j[/tex]
Step 1 : determine the partial derivatives of the vector quantity
Fx = ∫ [tex]6y^{\frac{3}{2} }i = 6xy^{\frac{3}{2} } + c[/tex]
Fy = ∫ [tex](9x \sqrt{y})_{y} = 9x(\frac{2}{3})y^{\frac{3}{2} } + c[/tex]
Equating the partial derivatives :
[tex]9x(\frac{2}{3})y^{\frac{3}{2} } + c[/tex] = [tex]6xy^{\frac{3}{2} } + c[/tex]
therefore the gradient of F i.e. F = vF = F( x,y ) = [tex]6xy^{\frac{3}{2} }[/tex]
Next step : Determine the work done
Work done ( F.dr ) = [ F(point b ) = F( 3,4 ) ] - [ F(point A) = F( 1,1 ) ]
F(3,4 ) = 6(3)(4)[tex]^{\frac{3}{2} }[/tex] = 144
F( 1,1 ) = 6(1)(1)[tex]^{\frac{3}{2} }[/tex] = 6
Therefore the work done by the force field = 144 - 6 = 138 J
Hence we can conclude that the work done by the force field F is = 138 J
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WILL MARK BRAINLIEST
1. What is the difference between distance and displacement?
A. Distance is measured in feet. Displacement is measured in meters
B. Distance is the difference between how quickly you move between two points. Displacement is the average of that motion
C. Distance is a measure of length of travel. Displacement is the measure of the time it takes to move that length
D. Distance is the length of travel. Displacement is the straight line distance between the starting point and the ending point
2. When forces are in opposite directions, they subtract from one another
A. True
B. False
3. When two forces are in opposite directions, and they are the exact same magnitude, the forces will
A. Subtract from each other
B. Cancel Out
C. Go on Infinitely
D. eventually reach equilibrium
4. Acceleration is a change in motion over time
True or False
Answer:
1. C
2. True
3. D
4. True
Explanation:
While distance refers to “how much land an object has traversed throughout its journey,” displacement measures “how far an object is out of place.” In this post, let's look at the difference between distance and displacement. Thus, option C is correct.
What are the distance and displacement?Displacement and distance are two distinct ideas. The total distance travelled is larger than the displacement between those two points if an object changes direction while travelling.
Distance is the length of any path connecting any two places. Displacement is the straight-line distance between any two points when calculated along the shortest route.
Therefore, The direction is ignored when calculating distance. The direction is accounted for in the displacement calculation.
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A merry-go-round rotates from rest with an angular acceleration of 1.08 rad/s2. How long does it take to rotate through (a) the first 3.74 rev and (b) the next 3.74 rev
Answer:
Let ω1 be the initial angular speed and ω2 the final angular speed:
α = (ω2- α1) / t
corresponding to a = (v2 - v1) / t
S (distance corresponds to theta)
1 rev = 2 pi and 3.74 rev = 7.48 pi = 23.5 radians
S = 1/2 a t^2 linear or S = 1/2 α t^2 angular acceleration
23.5 = 1/2 * 1.08 t^2 and t = 6.60 sec for first 3.84 rev
b) ω1 = 1.08 * 6.6 = 7.13 rad/sec initial speed for second 3.74
23.5 = 7.13 t + .54 t^2 compare to S = v1 t + 1/2 a t^2
.54 t^2 + 7.13 t - 23.5 = 0
t^2 + 13.2 t - 43.5 = 0
t = 2.7 sec for next 3.74
Check:
7.13 * 2.7 + .54 * 2.7^2 = 23.2 rad = 3.7 rad
Which processes result in the release of carbon? Select three options.
Animals break down food molecules to obtain energy.
The remains of producers are broken down by decomposers.
The remains of consumers are broken down by soil decomposers.
Producers take in carbon dioxide.
Producers make sugars and starches.
Answer:
The remains of producers are broken down by decomposers.
The remains of consumers are broken down by soil decomposers.
Animals break down food molecules to obtain energy.
Explanation:
Answer:
It is A. B. and C.
Explanation:
Hope this helps
For a certain transverse standing wave on a long string, an antinode is at x = 0 and an adjacent node is at x = 0.20 m. The displacement y(t) of the string particle at x = 0 is shown in the figure, where the scale of the y axis is set by ys = 4.3 cm. When t = 0.90 s, what is the displacement of the string particle at (a) x = 0.30 m and (b) x = 0.40 m ? What is the transverse velocity of the string particle at x = 0.30 m at (c) t = 0.90 s and (d) t = 1.3 s?
The expressions for the traveling and standing wave to find the results for the questions about the displacement and speed of the particle are:
a) For time zero, the displacement at position x = 0.30 m is y = 3.04 cm
b) For time zero, the displacement at position x = 0.40 m is: y = 0
c) For the point x = 0.30 and time t = 0.9s, the velocity of the particle is:
v = 9.11 cm / s
d) For the point x = 0.30 and time t = 1.3s, the velocity of the particle is:
v = 9.65 cm / s
The traveling wave is a disturbance in the medium that moves at constant speed, in the case of a transverse wave the expression for the perpendicular oscillation is:
y = A sin (kx - wt)
Where y is the oscillation perpendicular to the direction of the displacement, A the amplitude, k in wave number and w the angular velocity.
Standing waves are formed when a traveling wave collides with an obstacle and is reflected, in this case the sum of the two waves gives a wave that does not shift in time and fulfills the relationship
[tex]\frac{\lambda}{2} = \frac{L}{n}[/tex]
Where λ is the wavelength, L the distance between the reflection points and n the number of nodes.
Indicates that for the standing wave the distance between an antinode and the node is x = 0.20 m, therefore
[tex]\frac{\lambda}{4} = \frac{L}{1}[/tex]
λ = 4L
λ = 4 0.20
λ = 0.80 m
The wave number.
k = [tex]\frac{2\pi }{\lambda }[/tex]
k = [tex]\frac{2 \pi }{0.80 }[/tex]
k = 2.5π i m⁻¹
In the associated traveling wave, from the graph we can see that the period of the wave is:
T = 2.8 s
the angular velocity is related to the period.
[tex]w=\frac{2\pi}{T} \\w = \frac{2\pi }{2.8}[/tex]
w = 0.714π rad/s
indicate the maximum displacement that is the amplitude of the wave.
A = [tex]y_s[/tex]
A = 4.3 cm
Let's write the equation of the traveling wave.
y = 4.3 sin [π (2.5 x - 0.714 t)]
with this expression we can answer the questions.
a) the displacement of the particle for x = 0.30 m
y = 4.3 sin (π (2.5 0.30 - 0.714 t))
y = 4.3 sin π( 0.75 - 0.714 t(
Remember that the angles must be in radians. For time t = 0 the displacement is
y = 4.3 0.707
y = 3.04 cm
b) The displacement for x = 0.4m
y = 4.3 sin (π 2.5 0.4)
y = 0 cm
c) the transverse velocity of the wave at x = 0.30 m for the time of t = 0.90s
the speed of the wave is
[tex]v= \frac{dy}{dt} \\v= A w cos ( kx - wt)[/tex]
v = 4.3 0.714π cos π(2.5 0.3 - 0.714 t)
v = 9.65 cos π(0.75 - 0.714 t)
For time t = 0.90 s the velocity is:
v = 9.65 cos π(0.75 - 0.714 0.9)
v = 9.65 0.9436
v = 9.11 cm / s
d) The velocity for time t = 1.3 s
v = 9.65 cos π(0.75 - 0.714 1.3)
v = 9.65 0.9999
v = 9.65 cm / s
In conclusion, using the expressions for the traveling and standing wave, we can find the results for the questions about the displacement and speed of the particle are:
a) For time zero, the displacement at position x = 0.30 m is y = 3.04 cm
b) For time zero, the displacement at position x = 0.40 m is: y = 0
c) For the point x = 0.30 and time t = 0.9s, the velocity of the particle is:
v = 9.11 cm / s
d) For the point x = 0.30 and time t = 1.3s, the velocity of the particle is:
v = 9.65 cm / s
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What are the two factors that affect the frictional force between objects
A spring has a length of 0.200 m when a 0.300-kg mass hangs from it, and a length of 0.750 m when a 1.95-kg mass hangs from it. (a) What is the force constant of the spring
Answer:
Explanation:
The change in force = 9.81(1.95 - 0.300) = 16.2 N
The change in length is 0.750 - 0.200 = 0.550 m
K = ΔF/Δx = 16.2/0.550 = 29.4 N/m
A load of 20N is lifted through a height of 8m by a machine.If the effort of 80N moves a distance of 6m.Calculate the efficiency percentage of the machine
A load of 20N is lifted through a height of 8m by a machine.If the effort of 80N moves a distance of 6m.Calculate the efficiency percentage of the machine
Explanation:
A load of 20N is lifted through a height of 8m by a machine.If the effort of 80N moves a distance of 6m.Calculate the efficiency percentage of the machine
Answer:

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Join / LoginQuestionIn a lifting machine an effort of 500 N is to be moved by a distance of 20 m to raise a load of 10000 N by a distance of 0.8 m Determine the velocity ratio and mechanical advantage
Determine the velocity ratio and mechanical advantageA
Determine the velocity ratio and mechanical advantageA10 and 35
Determine the velocity ratio and mechanical advantageA10 and 35B
Determine the velocity ratio and mechanical advantageA10 and 35B20 and 35
Determine the velocity ratio and mechanical advantageA10 and 35B20 and 35C
Determine the velocity ratio and mechanical advantageA10 and 35B20 and 35C10 and 25
Determine the velocity ratio and mechanical advantageA10 and 35B20 and 35C10 and 25D
Determine the velocity ratio and mechanical advantageA10 and 35B20 and 35C10 and 25D25 and 20
Determine the velocity ratio and mechanical advantageA10 and 35B20 and 35C10 and 25D25 and 20Easy
Determine the velocity ratio and mechanical advantageA10 and 35B20 and 35C10 and 25D25 and 20EasyOpen in App
Determine the velocity ratio and mechanical advantageA10 and 35B20 and 35C10 and 25D25 and 20EasyOpen in AppSolution
Determine the velocity ratio and mechanical advantageA10 and 35B20 and 35C10 and 25D25 and 20EasyOpen in AppSolution
Determine the velocity ratio and mechanical advantageA10 and 35B20 and 35C10 and 25D25 and 20EasyOpen in AppSolutionVerified by Toppr
Determine the velocity ratio and mechanical advantageA10 and 35B20 and 35C10 and 25D25 and 20EasyOpen in AppSolutionVerified by TopprCorrect option is
Determine the velocity ratio and mechanical advantageA10 and 35B20 and 35C10 and 25D25 and 20EasyOpen in AppSolutionVerified by TopprCorrect option isD
Determine the velocity ratio and mechanical advantageA10 and 35B20 and 35C10 and 25D25 and 20EasyOpen in AppSolutionVerified by TopprCorrect option isD25 and 20
Determine the velocity ratio and mechanical advantageA10 and 35B20 and 35C10 and 25D25 and 20EasyOpen in AppSolutionVerified by TopprCorrect option isD25 and 20Distance moved by effort is 20m,that of load is 0.8m
Determine the velocity ratio and mechanical advantageA10 and 35B20 and 35C10 and 25D25 and 20EasyOpen in AppSolutionVerified by TopprCorrect option isD25 and 20Distance moved by effort is 20m,that of load is 0.8mVR=LdEd
Determine the velocity ratio and mechanical advantageA10 and 35B20 and 35C10 and 25D25 and 20EasyOpen in AppSolutionVerified by TopprCorrect option isD25 and 20Distance moved by effort is 20m,that of load is 0.8mVR=LdEdVR=0.820
Determine the velocity ratio and mechanical advantageA10 and 35B20 and 35C10 and 25D25 and 20EasyOpen in AppSolutionVerified by TopprCorrect option isD25 and 20Distance moved by effort is 20m,that of load is 0.8mVR=LdEdVR=0.820VR=25
Determine the velocity ratio and mechanical advantageA10 and 35B20 and 35C10 and 25D25 and 20EasyOpen in AppSolutionVerified by TopprCorrect option isD25 and 20Distance moved by effort is 20m,that of load is 0.8mVR=LdEdVR=0.820VR=25the load is 10,000N and effort=500N
MA=effortdloadd
MA=effortdloaddMA=50010000
MA=effortdloaddMA=50010000MA=20
What’s 1 + 1 many half a window
Answer:
1
+
1
is
2
Explanation:
because if you add one to a one it would end up being 2 <3
Answer: 2
Explanation: 1+1=2
This was THE hardest question I've ever decided to answer
A tow truck pulls a car 5.00 km along a horizontal roadway using a cable having a tension of 850 N. (a.) How much work does the cable do on the car if it pulls horizontally? If it pulls at 35.0° above the horizontal? (b.) How much work does the cable do on the tow truck in both cases of part (a)? (c.) How much work does gravity do on the car in part (a)?
I think 1980is the answer because you add???
Identify:
In each case the forces are constant and the displacement is along a straight line, so
[tex]$$W=F s \cos \phi \text {. }$$[/tex]
Set-Up:
In part (a), when the cable pulls horizontally [tex]$\phi=0^{\circ}$[/tex] and when it pulls [tex]$35.0^{\circ}$[/tex] above the horizontal [tex]$\phi=35.0^{\circ}$[/tex]. In part (b), if the cable pulls horizontally[tex]$\phi=180^{\circ}$[/tex]. If the cable pulls on the car [tex]$35.0^{\circ}$[/tex] above the horizontal it pulls on the truck at below the horizontal and [tex]$\phi=145.0^{\circ}$[/tex]. For the gravity force [tex]$\phi=90^{\circ}$[/tex], since the force is vertical and the displacement is horizontal.
Execute:
(a) When the cable is horizontal, [tex]$W=(850 \mathrm{~N})\left(5.00 \times 10^{3} \mathrm{~m}\right) \cos 0^{\circ}=4.25 \times 10^{6} \mathrm{~J}$[/tex].
When the cable is[tex]$35.0^{\circ}$[/tex] above the horizontal,[tex]$W=(850 \mathrm{~N})\left(5.00 \times 10^{3} \mathrm{~m}\right) \cos 35.0^{\circ}=3.48 \times 10^{6} \mathrm{~J}$[/tex].
(b)[tex]$\cos 180^{\circ}=-\cos 0^{\circ}$[/tex] and [tex]$\cos 145.0^{\circ}=-\cos 35.0^{\circ}$[/tex],
So the answers are [tex]$-4.26 \times 10^{6} \mathrm{~J}$[/tex] and [tex]$-3.48 \times 10^{6} \mathrm{~J}$[/tex].
(c) Since [tex]$\cos \phi=\cos 90^{\circ}=0, W=0$[/tex] in both cases.
Evaluate: If the car and truck are taken together as the system, the tension in the cable does no net wnetwork
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which of the following statements describes a perfectly inelastic collision
Answer: An ice hockey player picks up a trophy as he slides past it.
Two pool balls colliding precisely inelastically and rebounding off one another is an example. So, the correct option is A.
What is an Inelastic collision?Inelastic collisions are those in which the total kinetic energy is lower after the impact than it was before. The stick is travelling quickly in the direction of the ball before to the contact. The stick comes to a stop following the accident. It transfers some of its kinetic energy to the cue ball, which rolls forward.
The type of collision mentioned in the given example is known as an inelastic collision is one in which the system's momentum is preserved but its kinetic energy is not. In a ballistic pendulum, an example of an inelastic collision. Dropped ball of clay is another illustration of an inelastic collision.
Thus, the correct option is A.
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Your question is incomplete, most probably the complete question is:
Which of the following statements describes a perfectly inelastic collision?
A. Two billiard balls bounce off of each other,
B. A car crashes into a tree and rebounds in the other direction,
C. A wad of chewing gum is thrown and sticks to a moving truck,
D. Two ice skaters hit each other and fall over in opposite directions,
Explain different layers of atmosphere and the pressure in each layer. Draw diagram
Answer:
Our atmosphere has five different layers. They are:
1. Troposphere: This is the most important layer of the atmosphere with an average height of 13 km from the earth. It is in this layer that we find the air that we breathe. Almost all the weather phenomena such as rainfall, fog and hailstorm occur here.
2. Stratosphere: This layer extends up to a height of 50 km. It presents the most ideal condition for flying airplanes. It contains a layer of ozone gas which protects us from the harmful effect of the sun rays.
3. Mesosphere: This layer extends up to a height of 80 km. Meteorites bum up in this layer on entering from the space.
4. Thermosphere: In this layer, the temperature rises very rapidly with increasing height. The ionosphere is a part of this layer. It extends between 80-400 km. This layer helps in radio transmission. Radio waves transmitted from the earth the reflected back to the earth by this layer.
5. Exosphere: It is the uppermost layer where there is very thin air. Light gases such as helium and hydrogen float into space from here.
A double-pane glass window is 60.0 cm x 90.0 cm and has 3.00-mm window panes. If the temperature difference between inside and outside is 24.0 K, how far apart should the panes be to have a heat loss of 4.09 W? Assume there is air in the gap.
The distance between the glass to have the given heat loss is 2.54 m.
The given parameters:
dimension of the window, = 60 cm by 90 cmtemperature, T = 24 Kheat lost, Q = 4.09 Wthermal conductivity of glass, k = 0.8 W/mKThe area of the glass window is calculated as follows;
[tex]A = 0.6 \times 0.9\\\\A = 0.54 \ m^2[/tex]
The distance between the glass is calculated as follows;
[tex]Q = \frac{KA \Delta T}{\Delta x} \\\\\Delta x = \frac{kA \Delta T}{Q} \\\\\Delta x = \frac{0.8 \times 0.54 \times 24 }{4.09} \\\\\Delta x = 2.54 \ m[/tex]
Thus, the distance between the glass to have the given heat loss is 2.54 m.
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What will be the speed of these waves (in terms of V) if we increase M by a factor of 18.0, which stretches the wire to double its original length
The speed of the chord wave allows finding a new speed when the mass is increased and the chord length is:
The velocity is: v = [tex]\frac{v_o}{3}[/tex]
A wave is a periodic movement of the particles that carries energy, but not matter, the speed of the waves is related to the properties of the medium by the relationship.
[tex]v = \sqrt{\frac{T}{\mu } }[/tex]
Where v is the speed of the wave, T the force and μ is the linear density.
Indicates that the applied mass increases by a factor of 18.0 and the length of the head is increased to twice the original.
The linear density of the cable is
[tex]\mu = \frac{m}{l}[/tex]
Where m is the mass of the cable and l is the length.
Let's use the subscript "o" for the initial conditions.
M = 18.0 m₀
l = 2 l₀
We look for the density.
[tex]\mu = \frac{18.0 m_o}{2.0 l_o}[/tex]mu = 18.0 mo / 2 lo
[tex]\mu = 9.0 \mu_o[/tex]
We substitute in the expression for the velocity assuming that the tension is kept constant.
[tex]v= \sqrt{\frac{T}{9.0 \mu} }[/tex]
[tex]v= \frac{v_o}{3}[/tex]
In conclusion, using the speed of the chord wave we can find a new speed when the mass is increased and the chord length is:
The velocity is: v = [tex]\frac{v_o}{3}[/tex]
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Check all true statements about nonmetals. Group of answer choices
All nonmetals are gaseous at room temperature.
Nonmetals have low melting and boiling points. ✅
Nonmetals are brittle and have a relatively low density. ✅
Nonmetals are good at thermal and electrical conduction.
Using the colored periodic table below, match the group or area to its color.
alkali metals red alkaline ✅
earth metals orange ✅
transition metals white ✅
halogens purple ✅
Nobel gasses teal✅
How many of each subatomic particle are in a neutral atom of Potassium-39? What is its mass number? Potassium is in group 1 and has the symbol "K".
protons 19 ✅
neutrons 20 ✅
electrons 19 ✅
mass number 39✅
Match the ion with its charge.
11 protons, 12 neutrons, 10 electrons = +1 ✅
Groups 17, period 2 = -1 ✅
31 protons, 39 neutrons, 28 electrons = +3✅
Group 2, period 5 = +2
Which best describes a metal such as Silver (Ag)?
Answer: Lustrous, malleable, and forms cations.
Determine if the "quoted" word(s) makes each statement True or False.
Elements that share the most characteristics are found on the periodic table in the same "horizontal period". False ✅
If matter has a higher temperature, that means there will be "more" molecular motion. True✅
If you put a balloon in a freezer, its volume would "increase". False ✅
If you put the balloon into a chamber where there is half the pressure, its volume will "double". True ✅
If you cool down a propane tank, there will be "less" pressure in it. True ✅
The average atomic mass of Aluminum (Al) is "16.00 amu".
False✅
Oxygen's atomic number is "8" therefore it has "8" protons in its nucleus. True✅
An "electron" has the same mass as a neutron. False✅
The nucleus contains "protons and neutrons", virtually all the mass of an atom.✅
Metals are found towards the left hand side of the periodic table while Nonmetals are found towards the right hand side of the periodic table.
Non metals are found towards the right hand side of the periodic table. They are not all gaseous at room temperature some nonmetals such as iodine are solid at room temperature.
Nonmetals usually have low melting and boiling points, are brittle, have relatively low density and are not good at thermal and electrical conduction.
Potassium - 39 contains 19 protons, 20 neutrons and 19 electrons. Recall that the number of protons and electrons are equal in a neutral atom.
The following is an accurate matching of the ions;
11 protons, 12 neutrons, 10 electrons = +1 - Na^+
31 protons, 39 neutrons, 28 electrons = +3 - Ga^2+
Group 2, period 5 = +2 - Sr^2+
Silver is a lustrous metal.
Elements that share the most characteristics are found on the periodic table in the same "horizontal period". FalseIf matter has a higher temperature, that means there will be "more" molecular motion. TrueIf you cool down a propane tank, there will be "less" pressure in it. TrueIf you put a balloon in a freezer, its volume would "increase". FalseIf you put the balloon into a chamber where there is half the pressure, its volume will "double". TrueThe average atomic mass of Aluminum (Al) is "16.00 amu". FalseOxygen's atomic number is "8" therefore it has "8" protons in its nucleus. TrueAn "electron" has the same mass as a neutron. FalseThe nucleus contains "protons and neutrons", virtually all the mass of an atom - True
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Answer:
If your answers are the green check marks all of them are correct
Explanation: I took the test
)Suppose you have two insulated buckets containing the same amount of water at room temperature. You also happen to have two blocks of metal of the same mass, both at the same temperature, warmer than the water in the buckets. One block is made of aluminum and one is made of copper. You put the aluminum block into one bucket of water, and the copper block into the other. After waiting a while you measure the temperature of the water in both buckets. Which is warmer
The water in the bucket containing the aluminium block is warmer than the bucket containing the copper block.
The specific heat is the amount of heat needed or required to elevate the temperature of 1 gram of a substance by 1° C.
At standard conditions;
the specific heat of aluminium = 0.215 cal/gm/° Cthe specific heat of copper = 0.0923 cal/gm/° CWe know that:
Heat loss by the metal block = Heat gained by the water bucketTherefore, since the specific heat of aluminium is higher than that of copper, the water bucket containing aluminium block will be warmer than the bucket containing the copper block.
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what is translator disadvantage?
Machine translators:
Level of accuracy can be very low.
Accuracy is also very inconsistent across different languages.
Machines can't translate context.
Mistakes are sometimes costly.
Sometimes translation simply doesn't work.
Human translators:
Turnaround time is longer.
Translators rarely work for free.
Unless you use a translation agency, with access to thousands of translators, you're limited to the languages any one translator can work with.
What is Rest in physics ?
Rest in physics generally refers to the state in which an object is stationary.
Rest in physics refers to a situation in which an object does not move from one point to another. Usually, an object is at rest when it is acted upon by equal and opposite forces. For example, a book lying on a table.In a nutshell, the state of rest in physics generally refers to the state in which an object is stationary and does not translate.Hence, in Physics, an object is in a state of rest when it does not move from one point to another.
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8.92 A 45.0 kg woman stands up in a 60.0 kg canoe 5.00 m long. She walks from a point 1.00 m from one end to a point I .00 m from the other end (Fig. P8.92). If you ignore resistance to motion Of the canoe in the water, how far does the canoe move during fis process
The distance that the canoe moves in this process is 1.29 meters.
We first have to find the center of mass
[tex]X = \frac{MsXs+McXc}{Mw+Mc} \\\\[/tex]
Where
Ms = Woman's mass = 45
Mc = Canoe's mass = 60kg
Xs = position from left= 1 cm
Xc = position from left end of canoe's mass = 2.5cm
When we put these values into the equation we have:
[tex]X=\frac{45*1+60*2.5}{45+60} \\\\= 1.857\\[/tex]
The center of gravity lies at the center of this boat. Therefore,
[tex]Xc = \frac{L}{2} \\\\L = 5 m long\\\\\frac{5}{2} =2.5[/tex]
5.00 - 1. 00 = 4 meters
[tex]\frac{45*4+60*2.5}{45+60} = 3.14meters\\\\[/tex]
To get the distance that is moved by this canoe
distance = 3.143-1.857
= 1.286
≈ 1.29 meters
The distance that the canoe moves in this process is 1.29 meters.
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Calculate the following
Earth's mass: Man's Mass
Earth's Radiusi man Radius
Earth's Density: mon Density
Farth's granty: man granty
How did the study of the atom contribute to our understanding of the periodic table of the elements? (1 point) Atoms are representative of elements, so scientists scaled up O atomic char- derstand elemental characteristics, allowing sc Highlight e elements in a periodic table. The determination of electron charge led to an understanding of O how atoms interact with one another, which facilitated the organization of the periodic table. Elements are made of atoms, so understanding atoms provided O information about elements, which led to the organization of the periodic table. Experiments that identified characteristics of atoms provided O scientists with atomic weights and atomic numbers, which were used to organize the periodic table.
The study of the atom contribute d to our understanding of the periodic table of the elements by virtue of the following;
Elements are made of atoms, so understanding atoms provided information about elements, which led to the organization of the periodic tableExperiments that identified characteristics of atoms provided scientists with atomic weights and atomic numbers, which were used to organize the periodic table.As we know; the periodic table is an array of elements in order of their atomic number.
In which case, the periodic table is made up of 7 rows otherwise called Periods and 8 columns otherwise called Groups.
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A less than youthful 82.6 kg physics professor decides to run the 26.2 mile (42.195 km) Los Angeles Marathon. During his months of training, he realizes that one important component in running a successful marathon is carbo-loading, the consumption of a sufficient quantity of carbohydrates prior to the race that the body can store as glycogen to burn during the race. The typical energy requirement for runners is 1 kcal/km per kilogram of body weight, and each mole of oxygen intake allows for the release of 120 kcal of energy by oxidizing (burning) glycogen.
(a) If the professor finishes the marathon in 4:45:00 h, what is the professor's oxygen intake rate, in liters per minute, during the race if he metabolizes all of the carbo-loaded glycogen during the race and the ambient temperature is 21.5°C? 2.28 Read the problem statement again carefully. Is the air at standard temperature and pressure during the marathon? How would this affect the volume of 1 mol of oxygen? L/min
(b) The human body has an efficiency of 25.0%. Only 25.0% of the energy released from oxidizing glycogen is used as macroscopic mechanical energy, and the remaining 75.0% is used for body processes such as pumping blood and respiration, and then leaves the body through the skin via radiation, evaporative cooling, and other processes. What is the average mechanical power (in W) generated by the professor during the run? 197.561 What is the total energy required by the professor during the run? How efficient is the human body, and how long did the race last? W
(c) What is the change in entropy (in J/K) of the professor's body if his core temperature has risen to 38.3°C during the run and his skin temperature is at 36.0°C during the marathon? J/K
(d) What is the change of the entropy (in J/K) of the air surrounding the professor during the race if the ambient temperature remains constant at 21.5°C? J/K
Find V. in the circuit of the following figure
Answer:
A5 A20 A32 5G A1 G A20 5G T mOBIL
3. pinapatupad ng inyong lungsod ang pagpapanatili ng mga tao sa bahay sa araw ng Linggo
dahil sa nararanasang pandemya. Alin ang dapat gawin ng mga mamamayan?
a. Maagang pumunta sa dagat at doon magpalipas ng maghapon.
b. Magpunta sa simbahan at doon manatili sa buong araw.
a Pumunta sa kamag anak sa kabilang lungsod.
d. Manatili sa loob ng bahay sa buong araw.
Answer:
d because u now
Explanation:
sana makatulong