Answer:
v = [tex]n \frac{\hbar }{m r}[/tex]
the sppedof the electron is also quantized
Explanation:
The angular momentum of a rotating body is
L = m v r
in Bohr's atomic model the quantization postulate is that the angular momentum is equal to
L = n [tex]\hbar[/tex]
we substitute
n [tex]\hbar[/tex] = m v r
v = [tex]n \frac{\hbar }{m r}[/tex]
where n is an integer.
Therefore, the sppedof the electron is also quantized, that is, sol has some discrete values.
An open vertical tube has water in it. a tuning fork vibrates over its mouth. as the water level is lowered in the tube, a resonance is heard when the water level is 180 cm below the top of the tube, and again after the water level is 220 cm below the top of the tube a resonance is heard. what is the frequency of the tuning fork? the speed of sound in air is 343 m/s. answer in units
Answer:
[tex]428.75\ \text{Hz}[/tex]
Explanation:
[tex]\Delta y[/tex] = Change in water level = [tex]220-180=40\ \text{cm}[/tex]
[tex]\lambda[/tex] = Wavelength
[tex]v[/tex] = Speed of sound = 343 m/s
Between the points of resonance there exists [tex]\dfrac{1}{2}\lambda[/tex]
[tex]\dfrac{1}{2}\lambda=\Delta y\\\Rightarrow \lambda=2\Delta y\\\Rightarrow \lambda=2\times 40\\\Rightarrow \lambda=80\ \text{cm}[/tex]
Wavelength is given by
[tex]f=\dfrac{v}{\lambda}\\\Rightarrow f=\dfrac{343}{0.8}\\\Rightarrow f=428.75\ \text{Hz}[/tex]
The frequency of the tuning fork is [tex]428.75\ \text{Hz}[/tex].
A self-driving car traveling along a straight section of road starts from rest, accelerating at 2.00 m/s2 until it reaches a speed of 25.0 m/s. Then the vehicle travels for 39.0 s at constant speed until the brakes are applied, stopping the vehicle in a uniform manner in an additional 5.00 s.
(a) How long is the self-driving car in motion (in s)?
(b) What is the average velocity of the self-driving car for the motion described? (Enter the magnitude in m/s.) m/s
Answer:
[tex]56.5\ \text{s}[/tex]
[tex]21.13\ \text{m/s}[/tex]
Explanation:
v = Final velocity
u = Initial velocity
a = Acceleration
t = Time
s = Displacement
Here the kinematic equations of motion are used
[tex]v=u+at\\\Rightarrow t=\dfrac{v-u}{a}\\\Rightarrow t=\dfrac{25-0}{2}\\\Rightarrow t=12.5\ \text{s}[/tex]
Time the car is at constant velocity is 39 s
Time the car is decelerating is 5 s
Total time the car is in motion is [tex]12.5+39+5=56.5\ \text{s}[/tex]
Distance traveled
[tex]v^2-u^2=2as\\\Rightarrow s=\dfrac{v^2-u^2}{2a}\\\Rightarrow s=\dfrac{25^2-0}{2\times 2}\\\Rightarrow s=156.25\ \text{m}[/tex]
[tex]s=vt\\\Rightarrow s=25\times 39\\\Rightarrow s=975\ \text{m}[/tex]
[tex]v=u+at\\\Rightarrow a=\dfrac{v-u}{t}\\\Rightarrow a=\dfrac{0-25}{5}\\\Rightarrow a=-5\ \text{m/s}^2[/tex]
[tex]s=\dfrac{v^2-u^2}{2a}\\\Rightarrow s=\dfrac{0-25^2}{2\times -5}\\\Rightarrow s=62.5\ \text{m}[/tex]
The total displacement of the car is [tex]156.25+975+62.5=1193.75\ \text{m}[/tex]
Average velocity is given by
[tex]\dfrac{\text{Total displacement}}{\text{Total time}}=\dfrac{1193.75}{56.5}=21.13\ \text{m/s}[/tex]
The average velocity of the car is [tex]21.13\ \text{m/s}[/tex].
Which diagram shows magnets that will attract each other? 2 bar magnets side by side with their long axes vertical, both red S on top and blue N on bottom. 2 bar magnets top to bottom with their long axes vertical, the top one with red S on top and blue N on bottom and the bottom magnet with blue N on top and red S on bottom. 2 bar magnets top to bottom with their long axes vertical, the top one with blue N on top and red S on bottom and the bottom magnet with red S on top and blue N on bottom. 2 bar magnets top to bottom with their long axes vertical, the top one with red S on top and blue N on bottom and the bottom magnet with red S on top and blue N on bottom.
2 bar magnets top to bottom with their long axes vertical, the top one with red S on top and blue N on bottom and the bottom magnet with red S on top and blue N on bottom. this diagram shows magnets that will attract each other. Hence option D is correct.
What is Magnet ?A permanent magnet is an item constructed of magnetised material that generates its own persistent magnetic field. A refrigerator magnet, for example, is commonly used to hold notes on a refrigerator door. Ferromagnetic (or ferrimagnetic) materials are those that can be magnetised and are strongly attracted to a magnet. These include the elements iron, nickel, and cobalt, as well as their alloys, some rare-earth metal alloys, and naturally occurring minerals such as lodestone. Although ferromagnetic (and ferrimagnetic) materials are the only ones that are strongly attracted to a magnet and are widely thought to be magnetic, all other substances respond weakly to a magnetic field via one of many different forms of magnetism.
Hence option D is correct.
To know more about magnet :
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While diving in cancun Mexico where the seawater has a density of 1,015 kg/m3 Nana observed that her pressure meter device reading was 3.75 atm. The reading at sea level is standard 1.0 atm. At what depth is she diving when the meter read 3.75 atm g
Answer:
The depth of the diver is 28.01 m
Explanation:
Given;
density of the seawater, ρ = 1,015 kg/m³
standard sea level pressure, P₀ = 1.0 atm = 101,325 Pa
the final reading of her pressure, P₁ = 3.75 atm = 379968.75 Pa
acceleration due to gravity, g = 9.8 m/s²
Let the depth she was diving at the final pressure = h
This depth is calculated as;
P₁ = P₀ + ρgh
P₁ - P₀ = ρgh
[tex]h = \frac{ P_1 \ - \ P_o}{\rho g} = \frac{379968.75 \ - \ 101325}{1015 \ \times \ 9.8} = 28.01 \ m[/tex]
Therefore, the depth of the diver is 28.01 m
A 0.454-kg block is attached to a horizontal spring that is at its equilibrium length, and whose force constant is 25.0 N/m. The block rests on a frictionless surface. A 5.70×10−2-kg wad of putty is thrown horizontally at the block, hitting it with a speed of 8.99 m/s and bounced with the same speed of 8.99 m/s in opposite direction. How far does the block compresses the spring?
The total momentum of the block and putty prior to their collision is
(0.454 kg) (0 m/s) + (5.70 × 10⁻² kg) (8.99 m/s) ≈ 0.512 kg•m/s
and the total momentum after the collision is
(0.454 kg) v + (5.70 × 10⁻² kg) (-8.99 m/s)
where v is the velocity of the block. Momentum is conserved, so
(0.454 kg) v + (5.70 × 10⁻² kg) (-8.99 m/s) = 0.512 kg•m/s
==> v ≈ 2.26 m/s
The total work done on the block by the spring as it gets compressed by a distance x is equal to the change in the block's kinetic energy:
1/2 (25.0 N/m) x ² = 1/2 (0.454 kg) (2.26 m/s) - 0
==> x ≈ 0.202 m ≈ 20.2 cm
Plz help w answer 1:/ confused ash
Answer:
I would say d I had the same question yesterday and I got it correct so hope that helps
A measure of the disorder of a system is called
Answer:
Entropy is typically defined as either the level of randomness (or disorder) of a system or a measure of the energy dispersal of the molecules in the system.
Answer:
entropy
Explanation:
edge2021
How does competition affect population size? Use the terms carrying capacity and limiting factor with your example.
Limiting factors are resources or other factors in the environment that can lower the population growth rate. Competition for resources like food and space cause the growth rate to stop increasing, so the population levels off. This flat upper line on a growth curve is the carrying capacity.
Answer:
they can eat all of the food and kill off the population
Explanation:
if the competition eats all of the food there's no food for the population and they will die off.
Which of the following was used to provide the oldest measurement of Earth's age?
A) Isotopic dating of fossils from the Cambrian period
B) Isotopic dating of meteor fragments
C) Relative dating of fossils in stratigraphic layers
D) Carbon dating of fossils from the Cambrian period
Answer:
Isotopic dating of meteor fragments
Explanation:
Earth is so old that its age can only be determined using isotopic dating. The oldest measurement of Earth's age was determined using isotopic dating of meteor fragments.
Answer: Isotopic dating of meteor fragments
Explanation:
Determine how would the frequency of the pendulum change if it was taken to the moon by finding the ratio of its frequency on the moon fM to its frequency on the earth fE. Suppose that gE is the free-fall acceleration on the earth and gM is the free-fall acceleration on the moon.
Express your answer in terms of some or all of the variables l, m, gE, gM.
fM/fE = ?
For the pendulum taken to the moon, The frequency change that would occur is mathematically given as
[tex]\frac{Fmoon}{Fearth}=0.408[/tex]
What frequency change would occur to the pendulum if it was taken to the moon?Generally, the equation for the Time period is mathematically given as
[tex]T=2\pi\sqrt{L/g}[/tex]
Therefore
[tex]\frac{Fmoon}{Fearth}=\frac{\sqrt{g/6L}}{\sqrt{g/6L}}\\\\\frac{Fmoon}{Fearth}=\sqrt{1/6}[/tex]
[tex]\frac{Fmoon}{Fearth}=0.408[/tex]
In conclusion, The frequency change
[tex]\frac{Fmoon}{Fearth}=0.408[/tex]
Read more about frequency
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Answer:
.408
Explanation:
A ball is thrown downward from the top of a building with an initial speed of 25 m/s.
It strikes the ground after 2.0 s. How high is the building?
20 m
30 m
50 m
70 m
Answer:
h = 69.6 m
Explanation:
Data:
Vo = 25 m/st = 2.0 sg = 9.8 m/s²h = ?Formula:
[tex]\boxed{\bold{h=V_{0}*t+\frac{g*(t)^{2}}{2}}}[/tex]Replace and solve:
[tex]\boxed{\bold{h=25\frac{m}{s}*2.0\ s+\frac{9.8\frac{m}{s^{2}}*(2.0\ s)^{2}}{2}}}[/tex][tex]\boxed{\bold{h=50\frac{m}{s^{2}}+\frac{9.8\frac{m}{s^{2}}*4\ s^{2}}{2}}}[/tex][tex]\boxed{\bold{h=50\ m+\frac{39.2\ m}{2}}}[/tex][tex]\boxed{\bold{h=50\ m+19.6\ m}}[/tex][tex]\boxed{\boxed{\bold{h=69.6\ m}}}[/tex]The building has a height of 69.6 meters.
Greetings.
A student wants to determine the speed of sound at an elevation of one mile. To do this the student performs an experiment to determine the resonance frequencies of a tube that is closed at one end. The student takes measurements every day for a week and gets different results on different days. Which of the following experiments would help the student determine the reason for the different results?
a. Repeating the experiment on several 10 degree C days and several 20 degree C days
b. Repeating the experiment using a wider range of frequencies of sound
c. Repeating the original experiment for an additional week
d. Repeating the experiment using a longer tube
Answer:
The correct answer is a
Explanation:
The speed of a sound wave depends on the square root of the modulus of compressibility and the density of the medium.
For the same medium, the speed of sound depends on the temperature of the fora
v = [tex]v_o \ \sqrt{1 + \frac{T}{273} }[/tex]
Therefore, the different results that are obtained are due to changes in temperature. The correct answer is a
since this way it has the values of the speed of sound for each temperature, for which it can compare with the results obtained from the trip.
A spring is hung from the ceiling. When a coffee mug is attached to its end, it stretches 2.5 cm before reaching its new equilibrium length. The block is then pulled down slightly and released. What is the frequency of oscillation
Answer:
Explanation:
In equilibrium , weight of mug is equal to restoring force .
mg = kx where m is mass of mug , k is spring constant and x is extension .
k / m = g / x = 9.8 ms⁻² / .025 m
= 392
frequency of oscillation n = [tex]\frac{1}{2\pi}\sqrt{\frac{k}{m} }[/tex]
[tex]n=\frac{1}{2\pi}\sqrt{392 }[/tex]
= 4.46 per second.
The figure below shows regions of the electromagnetic
spectrum
Gamma
Radio Microwaves Infrared Visible Ultraviolet X-rays Rays
Which of the following waves has the highest frequency?
{GAMMA RAYS}
Answer:
The electromagnetic spectrum in order of increasing frequency is - radio waves, microwaves, infrared radiation, visible light, ultraviolet radiation, X-rays and gamma rays.
The frequency of the gamma rays is >3×10
19
m.
Hence, the gamma rays has the maximum frequency in the electromagnetic spectrum.
During a football game player A (mass 120kg velocity 5.5m/s [L]) tackles and grabs player B (mass 105kg velocity 7.9m/s [R]). What is the final speed of the two players immediately after the collision?
Answer:
6.62m/s
Explanation:
Given data
M1= 120kg
U1= 5.5m/s
M2= 105kg
U2= 7.9m/s
The system experiences an inelastic collision, the expression for inelastic collision is
M1U1+ M2U1= (M1+M2)V
Subsitute
120*5.5+ 105*7.9= (120+105)*V
660+ 829.5= (225)*V
1489.5= (225)*V
Divide both sides by 225
V= 1489.5/225
V= 6.62m/s
Hence the common velocity is 6.62m/s
The disk weights 40 lb and has a radius of gyration is 0.6 ft. A 15 lb/ft moment is applied and the spring has a spring constant of 10 lb/ft. The system was initially at rest and the disk is rolling without slipping. The spring is initially unstretched. Find the angular velocity of the wheel when disk moves to the right 0.5 ft.
Answer:
angular velocity = 2.6543 rad/s
Explanation:
To find the angular velocity of the wheel when the disk moves to the right 0.5 ft, we need to be aware that the spring will stretch twice the value of gyration with any slight change in the position or movement of gyration since the top of the wheel is holding the spring.
The work done here:
= ((distance moved by the wheel) X spring constant X (Final displacement^2 - Initial displacement^2)) + Mass (q2 – q1)
Where q2 = 0.5ft
q1 = 0.8 lb
Note that linear velocity = radius X angular velocity
= -0.5(10)(1^2 – 0) + 15(0.5/0.8) = 4.375 ft·lb
Then, the kinetic energy :
Since the spring is initially unstretched, the initial tension in the spring = 0
So the final tension = ((distance moved by the wheel) X (linear velocity)^2 X (angular velocity) ^2 + (distance moved by the wheel) X (linear velocity) X ( radius of gyration) ^2 X (angular velocity) ^2
= 0.5(40/32.2)(0.8w) ^2 + 0.5(40/32.2)(0.6)^2 X w2
final tension = 0.621 w2
So the angular velocity of the wheel when disk moves to the right 0.5 ft = The Initial workdone + Initial kinetic energy will be equal to the final workdone + the final Kinetic energy
0 + 4.375 ft·lb = 0.621 w2
angular velocity = 2.6543 rad/s
g 2. In a laboratory experiment on standing waves a string 3.0 ft long is attached to the prong of an electrically driven tuning fork which vibrates perpendicular to the length of the string at a frequency of 60 Hz. The weight (not mass) of the string is 0.096 lb. a) [5 pts] What tension must the string be under (weights are attached to the other end) if it is to vibrate in four loops
Answer:
The tension in string will be "3.62 N".
Explanation:
The given values are:
Length of string:
l = 3 ft
or,
= 0.9144 m
frequency,
f = 60 Hz
Weight,
= 0.096 lb
or,
= 0.0435 kgm/s²
Now,
The mass will be:
= [tex]\frac{0.0435}{9.8}[/tex]
= [tex]0.0044 \ kg[/tex]
As we know,
⇒ [tex]\lambda=\frac{2L}{n}[/tex]
On substituting the values, we get
⇒ [tex]=\frac{2\times 0.9144}{4}[/tex]
⇒ [tex]=0.4572 \ m[/tex]
or,
⇒ [tex]v=f \lambda[/tex]
⇒ [tex]=0.4572\times 60[/tex]
⇒ [tex]=27.432 \ m/s[/tex]
Now,
⇒ [tex]v=\sqrt{\frac{T}{\mu} }[/tex]
or,
⇒ [tex]T=\frac{m}{l}\times v^2[/tex]
On putting the above given values, we get
⇒ [tex]=\frac{0.0044}{0.9144}\times (27.432)^2[/tex]
⇒ [tex]=\frac{752.51\times 0.0044}{0.9144}[/tex]
⇒ [tex]=3.62 \ N[/tex]
A satellite of mass m is in a circular orbit of radius R2 around a spherical planet of radius R1 made of a material with density ρ. ( R2 is measured from the center of the planet, not its surface.) Use G for the universal gravitational constant.
A) Find the kinetic energy of this satellite, K
Express the satellite's kinetic energy in terms of G, m, π, R1, R2, and ρ.
B) Find U, the gravitational potential energy of the satellite. Take the gravitational potential energy to be zero for an object infinitely far away from the planet.
Express the satellite's gravitational potential energy in terms of G, m, π, R1, R2, and ρ.
C) What is the ratio of the kinetic energy of this satellite to its potential energy?
Express K/U in terms of parameters given in the introduction.
Answer:
a)
get mass of planet:
ρ = M / V
V = 4/3 * R_1^3
M = ρ * V
M = ρ * 4/3 * R_1^3
equate force equations:
F = (GMm) / r^2 // r = R_2
F = ma
a = v^2/R_2
F = m * (v^2/R_2)
m * (v^2/R_2) = (GMm) / R_2^2
plug in and solve v^2:
m * (v^2/R_2) = (G * (ρ * 4/3 * R_1^3) *m) / R_2^2
v^2 = (G * ρ * (4/3) * π * R_1^3) / R_2
put into kinetic energy equation:
K = 1/2 * m * v^2
K = 1/2 * m * (G * ρ * (4/3) * π * R_1^3) / R_2
B)
givens:
U = -(GmM) / R_2
plug in mass of planet:
U = -(G * m * ρ * 4/3 * R_1^3) / R_2
C)
use previous equations and do some algebra:
K/U = (1/2 * m * (G * ρ * (4/3) * π * R_1^3) / R_2) * -(R_2 / (G * m * ρ * 4/3 * R_1^3))
K/U = -1/2
If a magnet can cut into a small piece , each piece will remain a magnet ? Explain this phenomenon
yes ,each piece will remain a magnet because the small pieces will cut into north and south
Learning Goal: To understand the concept of moment of inertia and how it depends on mass, radius, and mass distribution.
In rigid-body rotational dynamics, the role analogous to the mass of a body (when one is considering translational motion) is played by the body's moment of inertia. For this reason, conceptual understanding of the motion of a rigid body requires some understanding of moments of inertia. This problem should help you develop such an understanding.
The moment of inertia of a body about some specified axis is I = cmr^2, where c is a dimensionless constant, m is the mass of the body, and r is the perpendicular distance from the axis of rotation. Therefore, if you have two similarly shaped objects of the same size but with one twice as massive as the other, the more massive object should have a moment of inertia twice that of the less massive one. Furthermore, if you have two similarly shaped objects of the same mass, but one has twice the size of the other, the larger object should have a moment of inertia that is four times that of the smaller one.
Two spherical shells have their mass uniformly distrubuted over the spherical surface. One of the shells has a diameter of 2 meters and a mass of 1 kilogram. The other shell has a diameter of 1 meter. What must the mass m of the 1-meter shell be for both shells to have the same moment of inertia about their centers of mass?
Answer:
m₂ = 4 kg
Explanation:
The moment of inertia is defined by
I = ∫ r² dm
for bodies with high symmetry it is tabulated, for a spherical shell
I = 2/3 m r²
in this case the first sphere has a radius of r₁ = 2m and a mass of m₁ = 1 kg, the second sphere has a radius r₂ = 1m.
They ask what is the masses of the second spherical shell so that the moment of inertia of the two is the same.
I₁ = ⅔ m₁ r₁²
I₂ = ⅔ m₂ r₂²
They ask that the two moments have been equal
I₁ = I₂
⅔ m₁ r₁² = ⅔ m₂ r₂²
m₂ = (r₁ / r₂) ² m₁
let's calculate
m₂ = (2/1) ² 1
m₂ = 4 kg
Ice is placed in cool water. What happens to the temperature of the ice and the water?
Answer:
Explanation:
ice absorbs heat from the water. As the water molecules lose energy, they begin to slow down, and consequently to cool. So, it's kind of the opposite of what we might think: when we put ice in water, the ice doesn't give its cold to the water, it takes heat from the water.
Your ear is capable of differentiating sounds that arrive at each ear just 0.34 ms apart, which is useful in determining where low frequency sound is originating from. (a) Suppose a low-frequency sound source is placed to the right of a person, whose ears are approximately 18 cm apart, and the speed of sound generated is 340 m/s. How long is the interval between when the sound arrives at the right ear and the sound arrives at the left ear
Answer:
Δt = 5.29 x 10⁻⁴ s = 0.529 ms
Explanation:
The simple formula of the distance covered in uniform motion can be used to find the interval between when the sound arrives at the right ear and the sound arrives at the left ear.
[tex]\Delta s = v\Delta t\\\\\Delta t = \frac{\Delta s}{v}[/tex]
where,
Δt = required time interval = ?
Δs = distance between ears = 18 cm = 0.18 m
v = speed of sound = 340 m/s
Therefore,
[tex]\Delta t = \frac{0.18\ m}{340\ m/s}[/tex]
Δt = 5.29 x 10⁻⁴ s = 0.529 ms
If you lived on Saturn, which planets would exhibit retrograde motion like that observed for Mars from Earth? (Select all that apply.)
Mercury
Venus
Earth
Mars
Jupiter
Uranus
Neptune
Answer:
earth , mercury , and neptune
Explanation:
pls mark brainless
Consider the system consisting of the box and the spring, but not Earth. How does the energy of the system when the spring is fully compressed compare to the energy of the system at the moment immediately before the box hits the ground? Justify your answer.
Answer:
the energy when it reaches the ground is equal to the energy when the spring is compressed.
Explanation:
For this comparison let's use the conservation of energy theorem.
Starting point. Compressed spring
Em₀ = K_e = ½ k x²
Final point. When the box hits the ground
Em_f = K = ½ m v²
since friction is zero, energy is conserved
Em₀ = Em_f
1 / 2k x² = ½ m v²
v = [tex]\sqrt{ \frac{k}{m} }[/tex] x
Therefore, the energy when it reaches the ground is equal to the energy when the spring is compressed.
Based on the law of conservation of energy, the elastic potential energy of the system when the spring is fully compressed is equal to the kinetic energy of the system at the moment immediately before the box hits the ground.
What is the energy in a compressed spring?The energy in a compressed spring is elastic potential energy given by the formula:
Ek = 1/2 Kx^2where
K is spring constant x is displacement of the springWhat is the kinetic energy of a body?The kinetic energy of a body is the energy the body the has due to it's motion.
Kinetic energy, KE, is givenby the formula below:
KE = 1/2mv^2How does the energy of the system when the spring is fully compressed compare to the energy of the system at the moment immediately before the box hits the ground?From the law of conservation of energy, the total energy in a closed system is conserved.
Based on this law, all the energy in the compressed spring is converted to the kinetic energy of the box just before it reaches the ground.
Therefore, the elastic potential energy of the system when the spring is fully compressed is equal to the kinetic energy of the system at the moment immediately before the box hits the ground.
Learn more about conservation of energy at: https://brainly.com/question/381281
what is the best structure for a egg dropping project you will be name brainiest
Answer:
bubble wrap in stuff animal
Explanation:
did it
Answer:
i would say putting like pillows around it i had to do it once and i won like that so
Explanation:
why do blades come in different lengths in a jig saw sander?
Answer:
To determine the minimum blade length, add 1" to the workpiece thickness. One type of material, and some materials can be cut by more than one type of blade. No matter the material, there's likely a jigsaw blade designed specifically for. Armed with the right blade, follow these pointers to make your work go (and cut) .
Explanation:
Biodiversity decline poses a problem in an ecosystem because
Answer:
Biodiversity decline continues due to a rapidly expanding human population. Habitat is damaged in order to meet growing needs for agriculture, urban development, water and materials. Fish, wildlife and plants are overharvested, despite mounting evidence that many harvesting practices are unsustainable.
A ball is kicked at 10.4 m/s at an angle of 32 degrees to the horizontal
how long (time) is the ball in the air?
find the horizontal displacement (range) of the ball
Answer:
3M/S
Explanation:
Directions: Fill in the blank with the correct word. Choose from the list of possible answers in the word bank below:
atoms
compound
element
heterogeneous
homogeneous
mixture
molecule
particles
solution
suspension
1. An _______________ is made up of just one kind of atom.
2. Matter is made up of tiny _______________.
3. In a mixture called a ________________ the particles are larger and can settle out.
4. A _______________ of water is made up of two hydrogen atoms and one oxygen atom.
5. Two or more elements chemically combined form a _______________.
6. _______________ are the building blocks of matter.
7. Seawater is an example of a _______________ mixture because it is the same throughout.
8. Vegetable soup is an example of a _______________.
9. In a _______________ one substance is dissolved in another.
10. A mixture that is not the same throughout is called a _______________ mixture.
Answer:
1).atoms (3). mixture. (5). Element
2). particles (4). molecules (6). suspension
Explanation:
(7). Homogeneous (8). Heterogeneous
(9). compound (10). solutions
What is the answer to question 4
In order to make it all the way from the opening to the detector, a wave has to travel through air, glass, water, plastic, and vacuum.
-- The siren and the tuning fork are sounds.
-- Sound cannot travel through vacuum.
-- That knocks out choices B, C, and D.
The answer is A.
Note: Making a big exception here. We don't do test questions on Brainly. That would be cheating. Don't let me catch you doing it again.