Answer:
4653
4565445655687568677667876
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
Why does the sky appear red and yellow during sunsets.
1. The reds and yellows are caused by the Sun's light having to pass through dust particles near closer to the horizon than when the Sun is higher in the sky.
2. At sunset the prismatic effect of the atmosphere at low angles produces longer red and yellow wavelengths.
3. At sunset blue light is scattered by air molecules at higher altitudes. The longer red and yellow wavelengths must travel further through the atmosphere when the Sun is near the horizon. At some point the reds and yellows finally scatter in the lower atmosphere causing the reds and yellows.
4. The red and yellows are appear at sunset because the light must pass through more atmosphere which slows down the wavelength frequency turning it to slower red and yellow frequencies.
Answer:
The answer should be 3-At sunset blue light is scattered by air molecules at higher altitudes. The longer red and yellow wavelengths must travel further through the atmosphere when the Sun is near the horizon. At some point the reds and yellows finally scatter in the lower atmosphere causing the reds and yellows.
Answer:
3. At sunset blue light is scattered by air molecules at higher altitudes. The longer red and yellow wavelengths must travel further through the atmosphere when the Sun is near the horizon. At some point the reds and yellows finally scatter in the lower atmosphere causing the reds and yellows.
A 5 kg bowling ball travelling at 2 m/s hits a motionless 10 kg bowling ball. If the smaller ball bounces back at a speed of -1 m/s, what will be the speed of larger ball after the collision? Hint: Use the conservation of momentum equation to solve this problem.
Answer:
1.5 m/s
Explanation:
Conservation of momentum means the momentum of the system before the collision is the same as after.
The before, after momentum of each ball is ...
5 kg ball: (5 kg)(2 m/s), (5 kg)(-1 m/s)
10 kg ball: (10 kg)(0 m/s), (10 kg)(v)
The sum of the "before" products is the same as the sum of the "after" products:
(5 kg)(2 m/s) +0 = (5 kg)(-1 m/s) +(10 kg)v
(10 +5) kg·m/s = (10 kg)·v . . . . . add (5 kg)(1 m/s) to both sides
v = (15 kg·m/s)/(10 kg) = 1.5 m/s
The speed of the larger ball will be 1.5 m/s. Its direction of motion will be the opposite of that of the 5 kg ball after the collision.
HELP ASAP!!!!! Choose all the answers that apply. Technology A)influences science
B)helps scientists observe fast phenomena
C)is the same as science
D) influences history
E)helps scientists observe slow phenomena
If work stays the same and the distance is increased, then less force is needed to do the work.
True
False
Answer:
False
Explanation:
This is because the work is the same so the force won't change
what were your preparetion before going the different physical fitness test?
Answer:
Avoid heavy strenuous exercise for the 24 hours prior to testing. Do not exercise at all on the day of testing to ensure you are well rested. Wear appropriate clothing for the conditions (e.g. shorts/track pants and t-shirt/singlet/sports top) and non-slip athletic footwear with laces securely fastened
I will mark brainlist
If a wave’s amplitude is 2cm, then its height is equal to:
5 cm.
0 cm
4 cm.
2 cm
Answer:
4 cm
Explanation:
Amplitude is the measure from the MIDLINE to the peak of the wave....so the wave HEIGHT is twice the ampltude
2 x 2 cm = 4 cm
the push up is dynamic or static
Answer:
Dynamic exercises
Explanation:
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.
We have seen in earlier readings how to determine the speed of a wave on a string. What will happen to the wavelength of a sinusoidal wave on a string if the tension in the string is increased (assuming we keep the frequency of the wave the same)
This question involves the concepts of tension in a string and the wavelength of the wave in a string.
The wavelength of a sinusoidal wave will "increase by square power" on a string if the tension in the string is increased when the frequency is kept constant.
The speed of a wave on a string is given by the following formula:
[tex]v=\sqrt{\frac{T}{\mu}}[/tex]
where,
v = speed of wave = fλ
f = frequency of the wave
λ = wavelength of the wave
T = tension force
μ = linear mass density of the string
Therefore,
[tex]f\lambda=\sqrt{\frac{T}{\mu}}\\\\T = f^2\lambda^2\mu[/tex]
It is given that the frequency is kept constant. The linear mass density is also constant for a string. Therefore,
[tex]T=(constant)\lambda^2\\T\ \alpha\ \lambda^2[/tex]
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Which light is most sensitive to the eyes?
Answer:
Our eyes are most sensitive to the wavelengths corresponding to the yellow and green colors of the spectrum. Flashy signs and some fire engines are painted in a yellowish-green color to attract our attention.
Ryder is testing the change in motion for an object that weighs 25 kg and an object that weighs 30 kg. He will push each object with the same force. Ryder predicts the object that weighs 25 kg will have a greater change in motion. Is this correct?
Yes, the object with the smaller mass will have a greater change in motion.
No, the change in motion for both objects will be the same.
Yes, the mass is greater, and the object will have a greater change of motion.
No, neither of the objects will have a change in motion.
Ryder is testing the change in motion for an object that weighs 25 kg and an object that weighs 30 kg. Then, the greater change in motion is due to the greater mass of the object, and the object will have a greater change of motion. Thus, the correct option is C.
Motion is the phenomenon in which an object changes its position with respect to the time taken. Motion of an object is mathematically described in terms of the displacement covered by the object, the distance, velocity, acceleration, speed and the frame of reference to an observer and measurement of the change in the position of the body of an object relative to that particular frame with the change in time taken.
The motion of an object is proportional to the mass of the object. The mass is greater in this case, and therefore the objects will have a greater change of motion.
Therefore, the correct option is C.
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A pumpkin is launched in the air and travels at a horizontal velocity of 25 meters per second for 5 seconds. How far does it travel horizontally?
Answer:
30.3 meters, 172 degrees
Explanation:
To insure the most accurate solution, this problem is best solved using a calculator and trigonometric principles. The first step is to determine the sum of all the horizontal (east-west) displacements and the sum of all the vertical (north-south) displacements.
Horizontal: 2.0 meters, West + 31.0 meters, West + 3.0 meters, East = 30.0 meters, West
Vertical: 12.0 meters, North + 8.0 meters, South = 4.0 meters, North
The series of five displacements is equivalent to two displacements of 30 meters, West and 4 meters, North. The resultant of these two displacements can be found using the Pythagorean theorem (for the magnitude) and the tangent function (for the direction). A non-scaled sketch is useful for visualizing the situation.
Applying the Pythagorean theorem leads to the magnitude of the resultant (R).
R2 = (30.0 m)2 + (4.0 m)2 = 916 m2
R = Sqrt(916 m2)
R = 30.3 meters
The angle theta in the diagram above can be found using the tangent function.
tangent(theta) = opposite/adjacent = (4.0 m) / (30.0 m)
tangent(theta) = 0.1333
theta = invtan(0.1333)
theta = 7.59 degrees
This angle theta is the angle between west and the resultant. Directions of vectors are expressed as the counterclockwise angle of rotation relative to east. So the direction is 7.59 degrees short of 180 degrees. That is, the direction is ~172 degrees.
I will mark brainlist
A wave is disturbance that transfers energy and matter.
true
false
Answer:
False
Explanation:
A wave is a disturbance that transfers energy from one place to another without transferring matter.
Answer:
I'm pretty sure it true sorry if I'm wrong
Example 4.16
An object of mass 3 kg rests on a plane. The coefficient of static friction and that of kinein
friction are given by Hs = 0.3 and pk = 0.2.
The plane is inclined at angle o to the horizontal.
(i) Find the maximum value of 0 for which the object remains at rest on the plane.
(ii) Find the acceleration of the object if it started sliding from rest down the plane at
angle Omax to the horizontal.
(ii) How long does it take the object to move, from rest, a distance of Imetre under the
conditions of (ii).
Answer:
Explanation:
(i) μs = F/N = mgsinθ/mgcosθ = tanθ
tanθ = 0.3
θ = 16.7°
(ii) a = F/m
a = (mgsinθ - (μk)mgcosθ) / m
a = g(sinθ - (μk)cosθ)
a = 9.8(sin16.7 - (0.2)cos16.7)
a = 0.94 m/s²
(iii) s = ½at²
t = √(2s/a)
t = √(2(1)/0.94)
t = 1.5 s
The octopus’s tentacle keeps _ right after it is bitten off ? a. Moving b. Breathing c. Growing
Answer:
The answer is A
Explanation:
The octopus’s tentacle keeps moving right after it is bitten off
What is friction ???
Answer:
Frictional force is produced when two bodies are rubbed against each other. It is the force that oppose the motion and therefore it stops or slow down a moving body.It depends upon the roughness or smoothness of the surface of the body in contact.Rough surface have more friction that the smooth surface. Similarly, the heavier body produces more friction than a lighter body. Frictional force acts in the opposite direction of the motion of the body.
2.(01.01 LC)
Which of the following is true for gravitational force? (3 points)
Decreases with increase in mass
Increases with increase in mass
Increases with increase in distance
Decreases with decrease in distance
Answer:
Increases with increase in mass
Explanation:
gravity is proportional to mass and inversely proportional to the square of the distance between them
F = GMm/d²
Be able to list the three compounds that are formed as products of highly exothermic
reactions such as detonating nitrogen-based explosives?
Answer:
Ammonium perchlorate NH4ClO4
Ammonium Nitrate
Calcium Cyanamide
When detonated, the reaction products are all gases, such as water vapor, nitrogen gas, and oxides of nitrogen.
Hopefully this helps :)
What is First Aid.
I mark u brainliest answer
Answer:
First aid refers to the emergency or immediate care you should provide when a person is injured or ill until full medical treatment is available.
Explanation:
A wooden barrel full of water has a flat circular top of radius 25.0 cm with a small hole in it. A tube of height 8.00 m and inner radius 0.582 cm is suspended above the barrel with its lower end inserted snugly in the hole. Water is poured into the upper end of the tube until it is full. The density of water is 1.00 × 103 kg/m3.
What is the force with which the water in the barrel pushes up on the top of the barrel?
Answer:
that is all i know
Explanation:
radius= 25.0cm
height= 8m
inner radius= 0.582cm
density= 1.00 × 103kgm³f= m× a A wooden box with a mass of 10.0 kg rest on a ramp that is incline at an angle of 25° to the horizontal. A rope attached to the box runs parallel to the ramp and then passes over a frictionless bully. A bucket with a mass of M hangs at the end of the rope. The coefficient of static friction between the ramp in the box is 0.50. The coefficient of Connecticut friction between the ramp in the box is 0.35.
Suppose the box remains at rest relative to the ramp. What is the maximum magnitude of the friction force exerted on the box by the ramp?
The maximum magnitude of the friction force exerted on the box by the ramp is 44.41 N.
The given parameters;
Mass of the box, m = 10 kgInclination of the ramp, θ = 25⁰Coefficient of static friction, μ = 0.5 Coefficient of kinetic friction, μk = 0.35The normal force on the wooden box is calculated as follows;
[tex]F_n = mg \times cos(\theta)\\\\F_n = 10 \times 9.8 \times cos(25)\\\\F_n = 88.8 2 \ N[/tex]
The maximum magnitude of the friction force exerted on the box by the ramp is calculated as follows;
[tex]F_f = \mu \times F_n\\\\F_f = 0.5 \times 88.82 \\\\F_f = 44.41 \ N[/tex]
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A 500 kg cart is rolling to the right at 1.3 m/s. a 60 kg man is standing on the right end of the cart. what is the speed of the cart if tha man suddenly starts running to the left with a speed 10.0 m/s relative to the cart
Answer:
P1 = 1.3 (500 + 60) = 728 kg-m total momentum to right at start
P2 = (v2 - 10) 60 + 500 v2
total momentum after running at -10 with respect to cart = 728 where v2 is the new speed of the cart
728 = 560 v2 - 600
v2 = 1328 / 560 = 2.37 m/s new speed of cart
Check:
After: p2 for cart = 500 * 2.37 = 1186
p1 for man = (2.37 - 10) * 60 = -458
P2 = p1 + p2 = 728 total momentum unchanged
An electron is held up against the force of gravity by the attraction of a fixed proton some distance above it. How far above the electron is the proton
5.08 m
Explanation:
The weight of the electron is being counteracted by the attractive electrostatic force exerted by the proton above it. We can write the force equation as follows:
[tex]m_eg = k_e\dfrac{Q_pQ_e}{r^2}[/tex]
where the Q's are the charges of the proton and electron, r is the distance between the particles, g is the acceleration due to gravity, [tex]m_e[/tex] is the mass of the electrons and [tex]k_e[/tex] is the Coulomb constant. So solving for r, we get
[tex]r^2 = k_e\dfrac{Q_pQ_e}{m_eg}[/tex]
Taking the square root of r^2, we then get the distance as
[tex]r = \sqrt{k_e\dfrac{Q_pQ_e}{m_eg}}[/tex]
The values are given as follows:
[tex]m_e = 9.11×10^{-31}\:\text{kg}[/tex]
[tex]g = 9.8\:\text{m/s}^2[/tex]
[tex]Q_p = Q_e = 1.60×10^{-19}\:\text{C}[/tex]
[tex]k_e = 8.99×10^9\:\text{N-m}^2\text{/C}^2[/tex]
Putting in all of these values in our equation for r,
[tex]r = \sqrt{\dfrac{(8.99×10^9\:\text{N-m}^2\text{/C}^2)(1.60×10^{-19}\:\text{C})^2}{(9.11×10^{-31}\:\text{kg})(9.8\:\text{m/s}^2)}}[/tex]
[tex]\:\:\:\:\:= 5.08\:\text{m}[/tex]
of the following which is the largest body?
a. the moon
b. Pluto
c. Mercury
d. Ganymede
Answer:
Ganymede is the largest body
Explanation:
it is the satellite of jupiter
PLZ HELPPPP!!
this question is about popping microwave popcorn:
If you turn the microwave on for two minutes, is the rate of popping always the same, or does it change? Explain.
No,the poping of microvave is not same there time is different
There are 270 students and teachers going on a field trip to a science center. If each school bus holds 54 people, how many buses are needed?
9
are things that you can achieve quickly.
O A.
Dreams
OB.
Long-term goals
O c.
Short-term goals
OD.
Plans
Answer:
d
Explanation:
A dreams
.......wkkwkwkwkwkwnwnsksk
Answer:
C. short term goal
Explanation:
took a midterm quiz on canvas, Hope this helps <3
Una turbina de vapor
recibe vapor con un flujo másico de 30 kg/s a 6205 kPa, 811 K, con una velocidad a la
entrara de 10 m/s. El vapor a la entrada tiene una energía interna específica de 3150.3
kJ/kg y un volumen específico de 0.05789 m3
/kg. El vapor sale de la turbina a 9.859 kPa,
318.8 K. El vapor sale a 200 m/s con una energía interna específica de 2211.8 kJ/kg y
un volumen específico de 13.36 m3
/kg. Encuentre la potencia producida por la turbina
si ésta pierde calor a una tasa de 30 kW.
Este problema está describiendo una turbina de vapor a la que entra vapor a 30 kg/s, 6.205 kPa y 811 K con una velocidad de 10 m/s y sale a 9.859 kPa, 318.8 K y con una velocidad de 200 m/s. Adicionalmente, tanto el volumen específico como la energía interna son dados para ambas corrientes.
Con lo anterior, resulta posible escribir un balance de energía para esta turbina, despreciando todo efecto por energía potencial ya que no hay diferencia significativa entre la altura de la entrada (1) y la salida (2), pues están practicamente al mismo nivel:
[tex]mh_1+\frac{1}{2} mv^2_1=mh_2+\frac{1}{2} mv^2_2+Q_2+W_2[/tex]
Aquí vemos que la incógnita es [tex]W_2[/tex] como la potencia que produce la turbina. Ahora, el primer cáculo a realizar es el de las entalpías de las corrientes de entrada y salida, dada la energía interna, presión y volumen específico:
[tex]h_1=3150.3\frac{kJ}{kg}+6205kPa*0.05789\frac{m^3}{kg} =3509.51\frac{kJ}{kg}\\\\h_2=2211.8\frac{kJ}{kg}+9.859kPa*13.36\frac{m^3}{kg} =2342.72\frac{kJ}{kg}[/tex]
Ahora, podemos reacomodar el balance de energía con el fin de resolver [tex]W_2[/tex]:
[tex]W_2=m(h_1-h_2)+\frac{1}{2} m(v^2_1-v^2_2)-Q_2[/tex]
Finalmente, reemplazamos los valores para obtener:
[tex]W_2=10\frac{kg}{s} (3509.51-2342.72)\frac{kJ}{kg} +\frac{1}{2} *10\frac{kg}{s} [(10\frac{m}{s}) ^2-(200\frac{m}{s} )^2]*\frac{1kJ}{1000J} -30\frac{kJ}{s}\\\\W_2=11438.4 kJ/s=11438.4kW[/tex]
Es de precisar que la energía cinética como 1/2 m*v² resulta en Joules, por lo que hay que convertir a kilojoules para tener unidades consistentes de kilowatts al final.
Aprende más:
https://brainly.com/question/21902769https://brainly.com/question/243223505) You pull a 10.0 kg wagon along a flat road. You exert a force of 80.0 N at an angle of 30.0 degrees above the horizontal while you move the wagon 10.0 m forward. The coefficient of friction between the wagon and road is 0.500. Calculate the work down by you and the work done by friction.
Consult the attached free body diagram. The only forces doing work on the wagon are the frictional force opposing the wagon's motion and the horizontal component of the applied force.
By Newton's second law, the net vertical force is
• ∑ F [v] = n + (80.0 N) sin(30.0°) - mg = 0
where a is the acceleration of the wagon.
Solve for n (the magnitude of the normal force) :
n = (10.0 kg) g - (80.0 N) sin(30.0°) = 58.0 N
Then
f = 0.500 (58.0 N) = 29.0 N
Meanwhile, the horizontal component of the applied force has magnitude
(80.0 N) cos(30.0°) ≈ 69.3 N
Now calculate the work done by either force.
• friction: -(29.0 N) (10.0 m) = -290. J
• pull: (69.3 N) (10.0 m) = 693 J
