Estimate the monthly cost of using a 700-W refrigerator that runs for 10 h a day if the cost per kWh is $0.20.
Answer:
Energy in KWH = Power in KW * Time in hours
Cost is # of KWH * the cost per KWH
350 kwh
Explanation:
The monthly cost of using a 700-W refrigerator that runs for 10 h a day if the cost per kWh is $0.20 is $42.
What is cost price per watt?The cost price per watt is the amount of money required to use one watt or kilo watt electricity,
To find the total cost of electricity used, multiply the cost price of electricity per watt to total watts of electricity used.
Given information-
The power used by the refrigerator is 700- W per hour.
Total run time of the refrigerator is 10 hours per day.
The cost of the electricity is $0.20 kWh.
As, the power used by the refrigerator is 700- W per hour and total run time of the refrigerator is 10 hours per day. The power used by the refrigerator per day is,
[tex]P=700\times10\\P=7000\rm Wh\\P=7\rm kWh[/tex]
There is total 30 days in a month. Thus the total power used by the refrigerator per month is,
[tex]P=7\times30\\P=210\rm kWh[/tex]
As the cost of the electricity is $0.20 kWh. Thus the cost of 210 kWh electricity is,
[tex]C=0.2\times210\\C=42[/tex]
Thus the monthly cost of using a 700-W refrigerator that runs for 10 h a day if the cost per kWh is $0.20 is $42.
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What best describes the difference between the second sound wave and the first
Second sound is a quantum mechanical phenomenon in which heat transfer occurs by wave-like motion, rather than by the more usual mechanism of diffusion. This leads to a very high thermal conductivity. It is known as "second sound" because the wave motion of heat is similar to the propagation of pressure waves in air (sound).
Normal sound waves are fluctuations in the density of molecules in a substance; second sound waves are fluctuations in the density of particle-like thermal excitations (rotons and phonons[1]). Second sound can be observed in any system in which most phonon-phonon collisions conserve momentum. This occurs in superfluids,[2] and also in some dielectric crystals[3][4][5] when Umklapp scattering is small. (Umklapp phonon-phonon scattering exchanges momentum with the crystal lattice, so phonon momentum is not conserved.)
wikipedia sourced to
Answer:
Its a
Explanation:
a ball was kicked horizontally off a clif at 15m/s, how high was the cliff if the ball landed 83 m from base of the cliff
The ball's horizontal position [tex]x[/tex] at time [tex]t[/tex] is
[tex]x=v_{0x}t[/tex]
The ball has initial velocity in the horizontal direction only, so [tex]v_{0x}=15\,\frac{\mathrm m}{\mathrm s}[/tex]. Then the time it takes for the ball to travel 83 m horizontally [tex]t[/tex] is given by
[tex]83\,\mathrm m=\left(15\,\dfrac{\mathrm m}{\mathrm s}\right)t\implies t=5.5\,\mathrm s[/tex]
Meanwhile, the ball's vertical position [tex]y[/tex] at time [tex]t[/tex], starting at the height of the cliff [tex]h[/tex], is given by
[tex]y=h-\dfrac12gt^2[/tex]
where [tex]g=9.8\,\frac{\mathrm m}{\mathrm s^2}[/tex] is the acceleration due to gravity. After 5.5 seconds, the ball hits the ground, so that [tex]y=0[/tex] when [tex]t=5.5\,\mathrm s[/tex], and we use this to solve for [tex]h[/tex]:
[tex]0=h-\dfrac12g(5.5\,\mathrm s)^2\implies h\approx150\,\mathrm m[/tex]
Please answer this! Answer #14
I would say B because not wearing a seatbelt is an example of risky behavior, which they’re were trying to prevent
These three members of the Noble gas family all have one property in common because they are gases at room temperature. That is they A) are combustible. B) are compressible. C) are relatively dense. D) have no definite shape but a definite volume.
B) are compressible.
gasses are compressible
Answer: Option (B) is the correct answer.
Explanation:
Noble gases group 18 elements. Members of this group are He, Ne, Ar, Kr, Xe, and Rn.
All these elements have completely filled sub-shells and hence they are stable in nature. All the elements of group 18 are gases, this means that their particles are loosely held due to weak intermolecular interaction.
As a result, their particles can move freely from one place to another. These gases can be compressed in order to bring their particles together. It is also known that all gases are compressible in nature.
Thus, we can conclude that noble gases have one property in common is that they are compressible.
Select the correct words from the drop-down menus to complete the sentence. The work of a machine can never exceed the work because uses some of the work.
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This is the same answer.
Answer: The work OUTPUT of a simple machine can never exceed the work INPUT because FRICTION uses some of the work.
Explanation:
This concept is related to ideal machines and efficiency.
Only ideal machines can convert all the work input into useful work (work ouput).
The efficiency is a concept that tells you the proportion of input work that is transformed into output work.
Efficiency = (output work / input work) × 100
Ideal machines, those where friction does not exist, have a 100% efficiency, and so ideal machines convert all the input work into useful output work.
Real machines have a efficiency less than 100%. Friction lowers the efficiency and the work output is less than the work input, because friction uses work which is converted, mostly, into heat energy or sound energy.
I’m gym class you run 22m horizontal then climb a rope vertically for 4.8m. What is the direction angle of your total displacement, as measured from the horizontal?
The answer is: 12.30 degrees.
To solve this problem you must apply the proccedure shown below:
1. You run 22 meters horizontally and then you climb a rope vertically for 4.8 meters.
2. So. let's call the angle [tex]\alpha[/tex], therefore, you must find it using [tex]tan^{-1}[/tex], as following:
[tex]tan^{-1}(\alpha)=\frac{opposite}{adjacent} \\opposite=4.8\\adjacent=22[/tex]
3. Substitute values and solve for the angle:
[tex]tan^{-1}(\alpha)=\frac{4.8}{22}\\\alpha=12.30degrees[/tex]
What is the terminal velocity of a 6.00-kg mass object in falling with a drag force with a magnitude that depends on speed, v, as Fdrag = (30.0 N·s/m)v?
When object reached the terminal speed then its acceleration is zero
So as per Newton's II law we can say
[tex]F_{net} = 0[/tex]
now in that case we can say that net force is zero so here weight of the object is counter balanced by the drag force when it will reach at terminal speed
so we can write
[tex]mg - F_d = 0[/tex]
so here we are given that
[tex]F_d = 30[/tex]
[tex]6*9.8 - 30*v = 0[/tex]
[tex]58.8 - 30 *v = 0[/tex]
[tex]v = \frac{58.8}{30} [/tex]
[tex]v = 1.96 m/s[/tex]
so terminal speed will be nearly 2 m/s
What modifications did Kepler make to Copernicus model
Planets move in elliptical orbits and this theory was formulated by Johannes Kepler. Kepler's laws of planetary motion defined a set of rules followed by the planets while revolve around the sun and determines their orbits.
This modified the earlier theory of circular orbits from Nicholas Copernicus.
Kepler made two key modifications to Copernicus's model:
1. Kepler modified the circular orbits of planets proposed by Copernicus into elliptical orbits.
2. Kepler introduced the concept of varying planetary speeds along their orbits, rather than assuming constant speeds as Copernicus did.
To calculate the modifications Kepler made:
1. Elliptical Orbits:
- Kepler's first law states that planets move in elliptical orbits with the Sun at one of the foci. The formula for an ellipse is [tex]\( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \)[/tex], where [tex]\( a \)[/tex] is the semi-major axis and [tex]\( b \)[/tex] is the semi-minor axis.
- In contrast, Copernicus's model assumed perfect circular orbits with the Sun at the center.
2. Varying Planetary Speeds:
- Kepler's second law states that a line segment joining a planet and the Sun sweeps out equal areas during equal intervals of time. Mathematically, this is expressed as [tex]\( \frac{dA}{dt} = \text{constant} \),[/tex]where [tex]\( A \)[/tex] is the area swept and [tex]\( t \)[/tex] is time.
- Copernicus's model assumed uniform orbital speeds for planets.
Kepler's modifications were revolutionary as they better explained observations such as varying planetary brightness and retrograde motion. These changes laid the foundation for modern celestial mechanics and our understanding of planetary motion.
Complete Question:
What modifications did Kepler make to Copernicus model?
A bowling ball is whizzing down the bowling lane at 4 m/s. If the mass of the bowling ball is 7-kg, what is its kinetic energy? 52 J 48 J 56 J 66 J
[tex]Given:\\m=7kg\\v=4\frac{m}{s} \\\\Find:\\E_k=?\\\\Solution:\\\\E_k=\frac{mv^2}{2} \\\\E_k=\frac{7kg\cdot (4\frac{m}{s})^2}{2} =\frac{7kg\cdot 16\frac{m^2}{s^2} }{2} =7kg\cdot 8\frac{m^2}{s^2} =56J[/tex]
A 1-kg iron frying pan is placed on a stove. The pan increases from 20°C to 250°C. If the same amount of heat is added to a pan with a greater specific heat, what can you predict about the temperature of this second pan? A) The second pan would reach a lower temperature. B) The second pan would reach a higher temperature. C) The second pan would reach the same temperature. D) No conclusion can be made from this information.
Here mass of the iron pan is given as 1 kg
now let say its specific heat capacity is given as "s"
also its temperature rise is given from 20 degree C to 250 degree C
so heat required to change its temperature will be given as
[tex]Q = ms \Delta T[/tex]
[tex]Q = 1*s*(250 - 20)[/tex]
[tex]Q = 1*s*230[/tex]
now if we give same amount of heat to another pan of greater specific heat
so let say the specific heat of another pan is s'
now the increase in temperature of another pan will be given as
[tex]Q = ms'\Delta T[/tex]
[tex]1*s*230 = 1* s' * \Delta T[/tex]
now we have
[tex]\Delta T = (\frac{s}{s'})*230[/tex]
now as we know that s' is more than s so the ratio of s and s' will be less than 1
And hence here we can say that change in temperature of second pan will be less than 230 degree C which shows that final temperature of second pan will reach to lower temperature
So correct answer is
A) The second pan would reach a lower temperature.
As the specific heat increases, the temperature change decrease. The second pan would reach a lower temperature.
a the average distance from the Sun to Pluto is approximately 6.10 x 1 0 ^9. how long does it take light from the sun to reach Pluto?
6.10 x 10^9 / 3 x 10^5 = 2.033 x 10^4 seconds = 20.33 / 3.6 = nearly 6 hours .
Light travels from the sun to Pluto in approximately 5.5 hours. This is calculated using the speed of light in vacuum and the average distance from the sun to Pluto.
To calculate the time it takes for light from the sun to reach Pluto, we first need to understand the concept of light speed. Light speed is the fastest speed at which light can travel in vacuum, and it is broadly accepted as being about 299,792 kilometers per second.
Given that the average distance from the Sun to Pluto is approximately 6.1 x 10^9 kilometers, we can utilize the equation
speed = distance / time where time = distance / speed.
In this case, we divide the distance 6.1 x 10^9 kilometers by the speed of light 299,792 kilometers per second, resulting in approximately 5.5 hours.
Therefore, the time it takes for light from the Sun to reach Pluto is around 5.5 hours.
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The most reactive metals are located at the of the periodic table.
Answer:
the elements towards the bottom left corner
Explanation:
Metals are the substances which tend to lose their valence electrons in order to attain stability.
For example, metals of group 1 are lithium, sodium, potassium, rubidium and cesium.
And, each of these elements have only 1 valence electron present. So, it is much easier to lose just one electron rather than losing 2 or 3 valence electrons.
Hence, alkali metals are the most reactive substances of the periodic table and they are present at extreme left side of periodic table.
Therefore, we can conclude that the most reactive metals are located at the extreme left of the periodic table.
the earths magnetic field is associated with the
The Earth's magnetic field is believed to be generated by electric currents in the conductive material of its core, created by convection currents due to heat escaping from the core.
which planet formed near the sun where the solar system temperatures were very hugh
mercury is the closest planet from the sun
The terrestrial planets formed near the Sun due to the high temperatures that allowed only heavy materials like silicates and metals to condense into solid forms. Beyond the asteroid belt, colder temperatures in the outer solar system permitted gases and ices to condense, leading to the formation of gas giants like Jupiter and Saturn.
The planets that formed near the Sun where the solar system temperatures were very high are known as the terrestrial planets. Close to the Sun, temperatures were too great for light gases like hydrogen and helium to condense; only heavier materials such as silicates and metals could form solid clumps. This is why the inner planets, including Mercury, Venus, Earth, and Mars, are composed mostly of rock and metal.
In contrast, in the outer solar system, colder temperatures allowed gases to clump together and form the gas giants like Jupiter and Saturn. These giant planets attracted and retained hydrogen and helium, becoming much larger than their terrestrial counterparts. With greater volume and mass, the gas giants were able to grow very large, very quickly, and generate significant internal heat through contraction, even after their formation.
The difference in planet composition and formation is a result of the thermal gradients in the solar nebula, being hotter closer to the Sun and cooler further away. This led to terrestrial planets forming close to the Sun and gas giants beyond the asteroid belt, where it was cold enough for ice and gases to condense into large massive planets.
Describe what happens when you jump from a small boat onto a dock from the perspective of the 3rd Law.
according to newton's third law, every action has equal and opposite reaction. in this scenario of making a jump from the boat onto a dock, as i jump my feet in contact with the boat push the boat in backward direction. hence the action force is the push by my feet on the boat. the boat reacts by applying a reaction force on my feet pushing the feet in forward direction. hence reaction force here is the force by the boat on the feet. due to the reaction force of the boat on feet, i am pushed in forward direction to reach the the dock.
When jumping from a boat to a dock, Newton's Third Law of Motion explains that the force you use to push off the boat also pushes the boat in the opposite direction.
When you jump from a small boat onto a dock, Newton's Third Law of Motion is in full effect. According to this law, for every action, there is an equal and opposite reaction. When you push yourself off the boat to jump onto the dock, you exert a force on the boat. The boat, in turn, reacts to your force by moving in the opposite direction.
The same principle is evidenced if a man fixed on the shore pulls a boat by a rope; the man pulls on the rope, exerting a force on the boat. At the same time, the boat exerts an equal and opposite force back on the man through the rope. Neither of these interactions violate the third law of motion.
Applying this concept further, let's consider the swimmer pushing off the side of a pool. The swimmer exerts a force against the pool wall, and the wall exerts an equal and opposite force that propels the swimmer forward. Essentially, this thrust is similar in nature to how rockets and other vehicles generate propulsion.
What makes some chemical bonds more stable than others
it occurs when the energy of the combination has lower energy than the separated atoms
The use of force to move an object is "_____".
The use of force to move an object is defined as 'work' in physics. Work is done when a force moves an object. The amount of work is measured as force multiplied by distance.
Explanation:The use of force to move an object is called work. This is a fundamental concept in Physics, specifically in the field of mechanics. According to this principle, work is done when a force that is applied to an object moves that object. For instance, when you push a box and it moves, you're doing work on the box. The amount of work done is calculated by multiplying the force by the distance over which it is applied (Work = Force x Distance). If there is no movement despite the force applied, no work is done according to physics.
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Please select the word from the list that best fits the definition Used to compare two specific variables
A. line graph B. circle graph C. bar graph
please help.
i would say line graph
Answer: a. line graph
Explanation: A line graph is formed by joining the points given by the data with straight lines.
A line graph is usually used to show the change of information over a period of time using two variables . This means that the horizontal axis is usually a time scale, for example minutes, days, months or years; and the vertical axis shows the value of the information that varies in time.
An 80.0-kg object is falling and experiences a drag force due to air resistance. The magnitude of this drag force depends on its speed, v, and obeys the equation Fdrag=(12.0N⋅s/m)v+(4.00N⋅s2/m2)v2. What is the terminal speed of this object?
Answer:
Terminal velocity of object = 12.58 m/s
Explanation:
We know that the terminal velocity is attained when drag force and gravitational force are of the same magnitude.
Gravitational force = mg = 80 * 9.8 = 784 N
Drag force = [tex]12.0v+4.00v^2[/tex]
Equating both, we have
[tex]784=12.0v+4.00v^2\\ \\ v^2+3v-196=0\\ \\ (v-12.58)(v+15.58)=0[/tex]
So v = 12.58 m/s or v = -15.58 m/s ( not possible)
So terminal velocity of object = 12.58 m/s
The terminal speed of the object is approximately [tex]\( 12.59 \, \text{m/s} \).[/tex]
To find the terminal speed of the object, we need to set the drag force equal to the gravitational force acting on the object. At terminal speed, the net force on the object is zero, meaning the drag force and the gravitational force are balanced.
The gravitational force [tex]\( F_g \)[/tex] acting on the object is given by the mass of the object m times the acceleration due to gravity g , which is approximately[tex]\( 9.81 \, \text{m/s}^2 \):[/tex]
[tex]\[ F_g = m \cdot g = 80.0 \, \text{kg} \times 9.81 \, \text{m/s}^2 \][/tex]
[tex]\[ F_g = 784.8 \, \text{N} \][/tex]
The drag force [tex]\( F_{\text{drag}} \)[/tex] is given by the equation:
[tex]\[ F_{\text{drag}} = (12.0 \, \text{N} \cdot \text{s/m})v + (4.00 \, \text{N} \cdot \text{s}^2/\text{m}^2)v^2 \][/tex]
At terminal speed,[tex]\( F_{\text{drag}} = F_g \)[/tex], so we have:
[tex]\[ (12.0 \, \text{N} \cdot \text{s/m})v_{\text{terminal}} + (4.00 \, \text{N} \cdot \text{s}^2/\text{m}^2)v_{\text{terminal}}^2 = 784.8 \, \text{N} \][/tex]
Now, we can use the quadratic formula to solve for [tex]\( v_{\text{terminal}} \):[/tex]
[tex]\[ v_{\text{terminal}} = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
where [tex]\( a = 4.00 \, \text{N} \cdot \text{s}^2/\text{m}^2 \), \( b = 12.0 \, \text{N} \cdot \text{s/m} \)[/tex], and [tex]\( c = -784.8 \, \text{N} \).[/tex]
Plugging in the values:
[tex]\[ v_{\text{terminal}} = \frac{-12.0 \pm \sqrt{(12.0)^2 - 4 \cdot 4.00 \cdot (-784.8)}}{2 \cdot 4.00} \][/tex]
[tex]\[ v_{\text{terminal}} = \frac{-12.0 \pm \sqrt{144 + 12556.8}}{8} \][/tex]
[tex]\[ v_{\text{terminal}} = \frac{-12.0 \pm \sqrt{12700.8}}{8} \][/tex]
[tex]\[ v_{\text{terminal}} = \frac{-12.0 \pm 112.7}{8} \][/tex]
Since speed cannot be negative, we take the positive root:
[tex]\[ v_{\text{terminal}} = \frac{-12.0 + 112.7}{8} \][/tex]
[tex]\[ v_{\text{terminal}} = \frac{100.7}{8} \][/tex]
[tex]\[ v_{\text{terminal}} = 12.5875 \, \text{m/s} \][/tex]
___Fe + ___O2 →___Fe2O3
Your old bike left out in the rain starts to rust. What coefficients fill in the blanks to balance the equation?
A) 1,2,3
B) 2,3,1
C) 2,3,2
D) 4,3,2
The answer is D) 4,3,2
Just had this question
for typical rubber-on-concrete friction, what is the shortest time in which a car could accelerate from 0 to 80 mph? Suppose that μs=1.00 and μk=0.80.
Answer: 3.63 s
The car would accelerate due to force of friction. Initially the car would have to overcome static force of friction.
Static force of friction:
[tex]F_s=\mu_smg[/tex]
Where[tex]\mu_s[/tex] is the coefficient of static friction, mg is the weight.
Let m be the mass of the car and a be the acceleration, then
[tex]ma=\mu_smg \Rightarrow a=\mu_sg[/tex]
It is given that, [tex]\mu_s=1.00[/tex]
[tex]a=1.00\times 9.8 m/s^2= 9.8 m/s^2[/tex]
Now, using the equation of motion:
[tex] v-u=at[/tex] we can find the shortest time in which the car would accelerate from[tex]u= 0 \hspace{1mm} to \hspace{1mm} v=80 mph=80\frac{miles}{h} \times 1.6 \frac{km}{mile} \times \frac{5 m/s}{18km/h}=35.56 m/s[/tex]
[tex] \Rightarrow t=\frac{v-u}{a}=\frac{35.56 m/s-0}{9.8m/s^2}=3.63 s[/tex]
For a typical rubber-on-concrete friction, the shortest time in which a car could accelerate from 0 to 80 mph is 3.65 seconds.
Given the data in the question;
Since the car accelerated from 0 to 80.
Initial velocity; [tex]u = 0m/s[/tex]
Final velocity; [tex]v = 80mph = 35.7632 m/s[/tex]
To determine the shortest time
We use the expression from the Force of Static Friction:
[tex]f_s = u_sN[/tex]
[tex]f_s = u_smg[/tex]
Also, From Newton's Second Law of Motion:
[tex]F = ma[/tex]
Now, from the diagram ( free body )
[tex]F = f_s[/tex]
Hence,
[tex]ma = u_smg\\a = u_sg[/tex]
We know that acceleration due to gravity;[tex]g = 9.8m/s^2[/tex] and in the question, [tex]u_s = 1.00[/tex]
So we substitute into the equation
[tex]a = 1.00 \ *\ 9.8m/s^2\\a = 9.8m/s^2[/tex]
Next, from the First Equation of Motion;
[tex]v = u + at[/tex]
Where v is the final velocity, u is the initial velocity, a is the acceleration.
We make time "t", the subject of the formula
[tex]t = \frac{v-u}{a}[/tex]
We substitute in our value
[tex]t = \frac{35.7632m/s- 0m/s}{9.8m/s^2} \\\\t = \frac{35.7632m/s}{9.8m/s^2} \\\\t = 3.649s\\\\t = 3.65s[/tex]
Therefore, for a typical rubber-on-concrete friction, the shortest time in which a car could accelerate from 0 to 80 mph is 3.65 seconds.
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An object is moving at constant speed in a straight line. Which of the following would cause it to move differently? A. It is acted on by balanced forces. B. It is acted on by unbalanced forces. C. It is pushed by two new, exactly opposing forces. D. It ceases to be subject to any force.
An object moving at constant speed in a straight line would move differently if acted on by unbalanced forces, as indicated by Newton's first and second laws. Unbalanced forces would cause changes in the object's velocity, either in speed, direction, or both.
Explanation:An object moving at constant speed in a straight line would move differently if it was acted on by unbalanced forces. According to Newton's first law, an object will continue in its state of motion at constant speed in a straight line unless it is compelled to change by an external force. This principle refers to the inertia of the object.
For instance, the answer option 'B. It is acted on by unbalanced forces.' would make the object move differently. Unbalanced forces will cause the object to either speed up, slow down, or change its direction of motion. If there are exactly opposing forces (option 'C'), assuming they are equal in magnitude, they would balance each other out and the object will continue to move at the same speed in the same direction.
Newton's second law further explains that the rate of change of momentum of an object is directly proportional to the unbalanced force acting on it and takes place in the direction in which the force acts. As such, any presence of unbalanced forces will result in a change in the object's velocity - either speed, direction, or both - meaning the object's motion would be different.
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What is the magnitude of force required to accelerate a car of mass
1.7 × 103 kilograms by 4.75 meters/second2?
Solution is attached.
Answer:
C. 8.1 × 103 newtons
Explanation:
an archer shoots an arrow horizontally that hits a target 100m away. A bystander times the flight of the arrow at 2.2s. a) what was the initial velocity of the arrow? b) How far did the arrow drop in the vertical direction?
The archer is fired horizontally.
It hits the target which is at 100 m distance.
The horizontal velocity of the archer is always constant for a projectile in this situation.
The total time of flight is given as 2.2 s
Let the initial velocity is u .
Hence the target distance = ut
⇒ 100 m = u×2.2 s
⇒u = 100÷2.2 m/s
=45.45 m/s
We are asked to calculate the vertical distance travelled.
The vertical distance travelled is calculated as-
[tex]s= ut +\frac{1}{2} at^2[/tex] [ s is the distance and a is the acceleration]
[tex]s = 0*2.2 -\frac{1}{2} g[2.2]^2[/tex] [g is te acceleration due to gravity]
[tex]s=\frac{1}{2} 9.8*[2.2]^2[/tex] [here we have taken only magnitude]
=23.716 m [ans]
Bethany had a beaker with a small amount of baking soda. She added a few drops of pickle juice to the baking soda and observed fizzing and bubbling. Based on her observation, which of these can Bethany determine about the new substance formed by mixing the baking soda and pickle juice?
A chemical reaction produced a gas.
A chemical reaction produced a solid.
No chemical reaction took place.
Only a physical change happened.
A chemical reaction produced a gas. This was the observation done for the experiment performed by Bethany.
Answer: A
Explanation
Baking soda which is chemically known as sodium bicarbonate reacts with acidic agents like lemon juice or vinegar forming a bubbling like thing.
The sodium bicarbonate undergoes a chemical reaction with the acidic solution and the product of this reaction is the formation of carbon di oxide gas as bubbles and fizzing.
In this case, the pickle solution contains acidic substance like vinegar and lemon juice extract so when it reacts with sodium bicarbonate or baking soda, the carbon-di-oxide present in the sodium bicarbonate (NaHCo3) will be released as gas in the form of bubbles and fizzes on the surface of the solution.
Help Please!! An object is dropped from a height h meters. From the time the object is h/2 meters above the ground, it takes 1 second until the object strikes the ground. From what height was the object dropped?
Here object is dropped from height "h"
so we can say its initial speed is zero
and it will accelerate downwards due to gravity
now we can say it will take time T to hit the ground
now we can use
[tex]h = v_i*t + \frac{1}{2}gt^2[/tex]
[tex]h = 0 + \frac{1}{2}*gT^2[/tex]
now it is given that it will take 1 second to drop h/2 height to strike the ground
so here we have can say that in "T - 1" s it will cover the h/2 distance from start
[tex]h/2 = 0 + \frac{1}{2}g(T-1)^2[/tex]
now we can say
[tex]h = g(T-1)^2[/tex]
from above two equations we have
[tex]\frac{1}{2}gT^2 = g(T-1)^2[/tex]
[tex]T^2 = 2(T^2 + 1 -2T)[/tex]
[tex]T^2 - 4T + 2 = 0[/tex]
[tex]T = 3.41 s[/tex]
now we can find total height of the drop by first equation
[tex]h = \frac{1}{2}gT^2[/tex]
[tex]h = \frac{1}{2}*9.81*3.41^2[/tex]
[tex]h = 57.2 m[/tex]
Scientists suspect that modern horses have different dietary habits than horses that lived long ago. What is the most reliable source that scientists can use to prove that horses have changed over time? A. interviews with veterinarians B. fossil evidence C. cave drawings D. fables
Answer:
I think it would be B
Explanation:
What is the difference between a data table and a graph?
At the same time, however, you get less detail or less precision in a chart or graph than you do in the table. Imagine the difference between a table of sales figures for a ten-year period and a line graph for that same data. You get a better sense of the overall trend in the graph but not the precise dollar amount.
A data tables presents information and data in a table, usually in rows. A graph presents statistics and data in graph, or coordinate grid or plane.
In the past, it was common for waste rocks from mines to be dumped in piles, known as tailings piles. These piles were sometimes located near bodies of moving water. how likely are tailings piles affect the water's natural processes?
A.
It increased erosion upstream.
B.
It increased erosion downstream.
C.
It decreased deposition upstream.
D.
It increased deposition downstream.
D.
It increased deposition downstream.
Answer:
D. It increased deposition downstream.
Explanation:
Tailing is the material leftover after separation of valuable material from ore. Tailing ponds are made near moving water are not good for water as they increase water pollution. In case of waste rocks from mine dumped near water body such as river, may reduce the depth of river bed and increase chances of flood. These deposits also increase deposition downstream and form big deltas.