Explanation:
The terms heavy and light are commonly used in two different ways. We refer to weight when we say that an adult is heavier than a child. On the other hand, something else is alluded to when we say that rock is heavier than soil. A small rock would obviously weigh less than a roomful of soil, but rock is heavier in the sense that a rock of given size weighs more than the same-size sample of soil. What we are actually comparing is the mass per unit volume, that is, the density. In order to determine these densities, we might weigh a cubic centimeter of each sample. If the rock sample weighed 2.71 g and the soil 1.20 g, we could describe the density of the rock as 2.71 g cm–3 and that of the soil as 1.20 g cm–3. Even though sand is made of rock fragments, its density is less because the porosity of sand lowers its bulk density (as shown below). (Note that the negative exponent in the units cubic centimeters indicates a reciprocal. Thus 1 cm–3 = 1/cm3 and the units for our densities could be written as g/cm3, or g cm–3. In each case the units are read as grams per cubic centimeter, the per indicating division.) We often abbreviate "cm3" as "cc", and 1 cm3 = 1 mL exactly, by definition.
Table 1 : Density of Soils and Rocks
sand
1.52
sandy loam
1.44
loam
1.36
silt loam
1.28
clay loam
1.28
clay
1.20
amphibolite
2.79–3.14
dolomite
2.72–2.84
gneiss
2.59–2.84
limestone
1.55–2.75
marble
2.67–2.75
schist
2.73–3.19
shale
2.06–2.67
slate
2.72–2.84
pyrite
5.0
gold
I need help making a chart (preferably a line graph) with this data. I'm not sure how to show that it will be a linear line in a graph. If you don't have excel can you explain to me how to do it.
note: these datasets are separate and arent related
Spring Constant (N/m) Displacement (m)
100 0.5m
300 0.167m
500 0.1m
700 0.071m
Applied force (N) Displacement (m)
20 0.2m
30 0.3m
50 0.5m
70 0.7m
Answer:
the graph plotted based on your data, you did not mention what are the variables on the x-axis and y- axis . and the graphs can be three small graphs of different material of one quality i.e. thickness or length
hope it helps
Explanation:
What is the correct definition of rarefaction
Answer:
Explanation:
A decrease in the density of something is rarefaction. ... Most of the time, rarefaction refers to air or other gases becoming less dense. When rarefaction occurs, the particles in a gas become more spread out. You may come across this word in the context of sound waves.
a force that is at rest or moves in at a constant speed and in a constant direction is called a what force
The question is fishing for "balanced force".
But the description in the question is terrible.
Anna used a rock to drive a peg into the ground to put up her tent. If the rock applied a 9. 5 N force in 0. 50 s, what is the impulse on the peg? 4. 8 N • s 9. 5 N • s 10 N • s 19 N • s.
Answer:
4.8 N*s
Explanation:
P = F*t
P = Impulse/ momentum
F = force (Newtons)
t = time (seconds)
P = 9.5N*0.5s
P = 4.75 N*s
Hi there!
[tex]\large\boxed{I = 4.8 N}[/tex]
Impulse = Force (N) × Time (s)
We can also express this as:
I = Δp = Ft
Plug in the given values:
I = 9.5(.50) = 4.75 ≈ 4.8 Ns
If a train travels 25 miles in 2 hours 15 minutes, what is its average speed in miles per hour? 11. 1 mph 12. 5 mph 10 mph 9. 1 mph.
Answer:
11.1
Explanation:
2 hours 15 mins = 2 1/4 hours
2 1/4 hours = 25 miles
1 hour = 25 ÷ 2 1/4 = 25 ÷ 9/4 = 25 x 4/9 = 11.1 miles
We are able to walk toward because…
a) we push on the air
b) we push on the ground
c) the air pushes on us
d) the ground pushes on us
We are able to walk toward because b) we push on the ground.
_
_
_
[tex] 😘 \: 😉 \: 😈 \; 😫 \: 😤 \: 😡 \: 😘[/tex]
A teacher has two radioactive sources, A and B.
Source A has a longer half-life than source B.
What can be deduced about the nuclei in source A compared with the nuclei
in source B?
Do not refer to isotopes in your answer.
Answer:
The nuclei of source have greater stability than those of source B.
Balanced forces are forces that have equal amount of ________. (word choices: force, repel, attract, motion)
Answer:
a
Explanation:
if forces on an object are balanced a moving object continues to move at the same speed and in the same direction
5. SEP Construct Explanations If the line on
a distance-versus-time graph and the line on
a speed-versus-time graph are both straight
lines going through the origin, can the two
graphs be displaying the motion of the same
object? Explain.
Answer:
If and only if the object has 0 speed. Any other case is a no.
Explanation:
In increasing math complexity:
If the graph of a distance/time is a straight line with a non-zero slope it means that the velocity-time graph is an horizontal line where every point has the same second coordinate, equal to the value of the slope of the first line (ie, if the first line has equation [tex]y=2t[/tex] the second line will have equation [tex]y=2[/tex] - I'm using y to denote either the distance or the speed, read it as "the non-time variable")
If the graph of a speed/time is a straight line, the graph of the distance/time is the area of the region of space beneath the straight line in the first graph, ie a triangle. And as you increase the sides of that triangle keeping the angles fixed, you get a parabolic curve.
Bonus case: if [tex]v(t)= 0; d(t) = 0[/tex], they are both straight lines, they both pass through the origin, and they are the exact same line.
With calculus (or: math becomes fun after you add the alphabet to it)
You are asking if the answer of the IVP
[tex]\left \{ {{x'=x} \atop {x(0)=0}} \right.[/tex]
is a linear function. The answer is: besides the obvious [tex]x(t)'=x(t)=0 \ \forall t[/tex] there are no other solutions, linear or otherwise.
Find the velocity of a body of mass 100 g having kinetic energy of 20 J.
Answer:
Explanation:
Let v be the velocity of object ,
Given that mass of object , m=100 g=
1000
100
kg=0.1 kg
& kinetic energy of object =20 J
K.E =
2
1
mv
2
20=
2
1
×0.1v
2
⟹v
2
=400
⟹v=20 m/s
solve for the BMI weight 58kg Height 1.61
Answer:
Explanation:
BMI= weight/(height × height) ; weight in kilogram and height in metter
= 58kg / (1.61m × 1.61m )
= (58/ 2.5921) kg/[tex]m^{2}[/tex]
= 22.375 kg/[tex]m^{2}[/tex]
≈ 22.4 kg/[tex]m^{2}[/tex]
A balloon with a volume of 2. 0 L at 25°C is placed in a hot room at 35°C. The pressure on the balloon is constant at 1. 0 atm. How does the volume of the balloon change after moving it to the hot room?.
Answer:
25A°C=2L
35A°C=? More
=35/25×2L
=7/5×2L
=14L/5
=2.8L
a
(2) A 800 g block is pushed up an inclined plane (angled at 18°) with a velocity of 11.8 m/s. The first block slides up the
incline a distance of 2.2 m and strikes a second block with a mass of 300 g also moving at 3.4 m/s up the incline.
The two blocks hit and stick together. Determine the following:
(i) The maximum vertical height of the two blocks when they stop.
(ii) The time needed for the blocks to reach the bottom of the incline after the moment of impact.
(u = 0.19)
this is my attachment answer hope it's helpful to you
Calculate the force applied if 100 N/m² pressure is exerted over the area of 0.2m²
Answer: 100 pascals times 0.2 m^2= 20 N force is applied.
Explanation:
Consider the schematic of the molecule shown, with two hydrogen atoms, H, bonded to an oxygen atom, O. The angle between the two bonds is 106°. If the bond length r = 0.103 nm long, locate the center of mass of the molecule. The mass mH of the hydrogen atom is 1.008 u, and the mass mO of the oxygen atom is 15.9999 u. (Use a coordinate system centered in the oxygen atom, with the x-axis to the right and the y-axis upward. Give the coordinates of the center of mass in nm.)
The definition of the center of mass allows to find the result for the position of the mass center of more than the H₂O molecule is;
[tex]x_{cm} = 0 \ and \ y_{cm} = 6.9 10^{-3 } nm[/tex]
the concept of center of mass of a system is the point where external forces are applied, it is given by the expression
[tex]\frac{x}{y} =\frac{1}{M_{total}} \sum m_i r_i[/tex]
Where M is the total mass of the systemr_i and m_i sums the position and masses of the element i of the system
In the attachment we have a diagram of the system where the axis and coordinates of the molecules are shown, in this case it is indicated that the origin is in the oxygen atom, so its distance is zero.
[tex]r_{cm} = \frac{1}{2 m + M} \ (2 m r )[/tex] )
They indicate the mass of the hydrogen atom m = 1.008 amu, the bond length r = 0.103 nm and there is an angle 106º between the two hydrogens, therefore the angle from the vertical is:
θ = 106/2 = 53º
Let's find the position of the center of mass for each axis.
x-axis
[tex]x_{cm} = \frac{1}{2m+ M} \ ( m x_1 - m x_2)[/tex]
y-axis
[tex]y_{cm} = \frac{1}{2m + M} \ ( m y_1 + m_2)[/tex]
Let's use trigonometry to find the components of the bond length.
cos θ = [tex]\frac{y}{L}[/tex]
sin θ = [tex]\frac{x}{L}[/tex]
y = L cos θ
x = L sin θ
We substitute.
[tex]x_{cm} = \frac{1}{2m+M} \ (mL (sin 53 + sin (-53)) \\y_{cm} = \frac{1}{2m + M} \ ( mL cos 53 + mL sin 53)[/tex]
we use.
sin θ = - sin -θ
cos θ = cos -θ
Let's calculate.
[tex]x_{cm} = 0 \\y_{cm} = \frac{1}{2 \ 1.008 + 15.9999} \ ( 2 \ 1.008 \ 0.103 cos 53)[/tex]
We see that the center of mass is on the x axis and at a distance from the y-axis of 6.9 10-3 nm
In conclusion using the definition of the center of mass we can find the result for the position of the center of mass of the H₂O molecule is;
[tex]x_{cm}=0[/tex] and [tex]y_{cm}[/tex]cm = 6.9 10⁻³ nm
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Suppose measure the mass of an orange. Then you peel the orange and separate out the slices. If you measure the orange peel and all the slices, will that mass equal the mass of the whole orange?
Answer:
Yes
Explanation:
It's still the same orange with the same amount of stuff.
A 3.00 kg toy falls from a height of 10.0 m. Just before hitting the ground, what will its kinetic energy be? (Disregard air resistance; g = 9.81 m/s2.)
The kinetic energy of the toy before hitting the ground is 294.3 J
From the question given above, the following data were obtained:
Mass (m) = 3 Kg
Height (h) = 10 m
Acceleration due to gravity (g) = 9.81 m/s²
Kinetic energy (KE) =?Before the toy will hit the ground, the potential and kinetic energy of the toy will be the same.
KE = PE = mgh
KE = 3 × 9.81 × 10
KE = 294.3 JTherefore, the kinetic energy of the toy is 294.3 J
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what is the chemical and physical changes if making a starchs
Answer:
Chemical modification of starch is based on the chemical reactivity of the constituent glucose monomers which are polyhydroxyl and can undergo several reactions. Starch can undergo reactions such as hydrolysis, esterification, etherification and oxidation.
Explanation:
Hope this helps!
Part B
Based on the information in the article, how could plants be used in the future to help counteract some human
emissions?
Answer:
Having more plants performing photosynthesis could help reduce the amount of carbon dioxide in the atmosphere.
Explanation:
sample answer
The plants can be used in the future to help counteract some human
emissions they absorb carbon dioxide from the air and release oxygen.
What is photosynthesis?It is a process carried out by photosynthetic autotrophic organisms, such as plants, algae and some prokaryotes. These organisms capture sunlight, convert it into chemical energy, which will be used to produce organic compounds, based on water and carbon dioxide.
Trees, like other plants, in the process called photosynthesis, absorb carbon dioxide from the air and release oxygen. They also make the air wetter and cooler. In this way, plants contribute to the good quality of the air we breathe.
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The ______________ is all the surroundings of a living organism.
genetics
environment
temperature
heredity
1. A rocket is launched from a 300 cm rail. The upper Launch Lug is placed 1 point 150 cm from the bottom of the rocket. What is the effective distance of the Launch Rail? 300 cm 150 cm 3 m 300 m O 1m O 1.5 m
Answer:
i think its going to be 150 because its half of 300
Explanation:
60W light bulb is left on for 5mins. How much energy does it use in KWH
Answer:
Energy = Power × Time
time must be hoursPower must be in kilowatts[tex]{ \tt{E = ( \frac{60}{1000} ) \times ( \frac{5}{60}) }} \\ \\ { \tt{E = 0.005 \: kwh {}^{ } }}[/tex]
1 point
A roller coaster is 20 m high and traveling 11 m/s. If the roller coaster has a
mass of 150 kg, what is its total energy? (only put the number, no units or
commas)
Answer:
ETotal=38505J
Explanation:
Known values:
m=150kg
h=20m
v=11m/s
KE=mv^2/2
PE=mgh
ETotal=PE+KE
KE=(150(11^2))/2
KE=9075
PE=150(9.81)(20)
PE=29430
ETotal=9075+29430
ETotal=38505J
Which statement indicates that motion has occurred?
A. The reference point has changed.
B. The position of the object has changed.
C.The object has not changed.
D. The object being described has changed
A long-distance runner runs at a constant speed of 4.8 m/s. How long does it take them to run 1.5 km?
Convert km to meters:
1km = 1000 m
1.5 km x 1000 = 1500 m
1500m / 4.8 m/s = 312.5 seconds
312.5 seconds / 60 seconds per minute =5.2 minutes = 5 minutes 12.5 seconds
How is "speed" defined in terms of physics?
Explanation:
In everyday use and in kinematics, the speed of an object is the magnitude of the rate of change of its position with time or the magnitude of the change of its position per unit of time; it is thus a scalar quantity.
3. If John leaves his house to walk his dog and he walks around the block, what is
his distance?
180 Meters
180 Meters
180 Meters
180 Meters
Answer:
180 meters
Explanation:
If a 5.00 kg box slides down a ramp inclined at 60.0° above the horizontal, what is the
acceleration of the box
Answer:
A 70 kg box is slid along the floor by a 400 n force. The coefficient of friction between the box and the floor is 0. 50 when the box is sliding
A wrecking ball has a mass of 650 kg. It swings from a cable that is 21 m long. The centripetal force on the ball is 56 N. What is the tangential speed of the wrecking ball? 1. 3 m/s 1. 8 m/s 46 m/s 870 m/s.
Hi there!
For an object moving in a circle:
Centripetal force = ∑F = mv²/r
Thus:
56 = mv²/r
Rearrange to solve for velocity:
√(56r/m) = v
√(56 × 21)/650 = 1.345 m/s
Answer:
AExplanation:
have a good day
Part IV Objects on an incline w/ Tension + Friction
1. A small 63 kg sleigh is pulled by a rein attached to horse up a 15'angle hill to the
horizontal. The tension of the rein is 510 N. The coefficient of kinetic friction is
0.25
a. What is the normal force that the sleigh exerts on the hill?
b. What are the magnitude and direction of the sleigh's acceleration?
(a) The normal force on the sleigh is 596.36 N.
(b) The magnitude and direction of acceleration of the sleigh is 3.2 m/s² upwards.
The given parameters;
mass of the sleigh, m = 63 kginclination of the hill, θ = 15⁰tension on the rein, F = 510 Ncoefficient of friction, μ = 0.25The normal force on the sleigh is calculated as follows;
[tex]F_n = mg \times cos(\theta)\\\\F_n = 63 \times 9.8 \times cos(15)\\\\F_n = 596.36 \ N[/tex]
The magnitude and direction of acceleration of the sleigh is calculated as follows;
[tex]\Sigma F= ma\\\\F - mgsin(\theta) - F_f = ma\\\\F - mgsin(\theta) - \mu F_n = ma\\\\510\ - \ 63 \times 9.8 \times sin15 \ -\ 0.25\times 596.36 = 63a\\\\201 .11 = 63a\\\\a = \frac{201.11}{63} \\\\a = 3.2 \ m/s^2 \ upwards[/tex]
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