The mixture you separated was mixture of iron filings, sand, and salt. Based on your understanding of matter, is this mixture a homogeneous mixture or heterogeneous mixture? How do you know?
The answer is "Heterogeneous mixture because the filings and particles of sand and salt are distributed randomly in the mixture."
A mixture is a mix of at least two substances in any extent. This is not quite the same as a compound, which comprises of substances in settled extents.
A heterogeneous mixture is formed when the chemicals in the blend are not consistently conveyed all through the blend. For example, on the off chance that you add sand to salt, no measure of crushing can make sand and salt frame a homogeneous blend like salt and water. In this way, we can state that a blend of sand and salt will shape a heterogeneous mixture.
-Heterogeneous mixture because the filings and particles of sand and salt are distributed randomly in the mixture
Explanation;A mixture is a substance that is made up of two or more substances. Mixtures are not pure substances like the case of elements and compounds. A mixture may be either homogeneous or heterogeneous. A homogeneous mixture is a mixture in which the components are uniformly distributed. This means there is only one phase of matter, hence a homogeneous mixture can exist as a solid, gas or liquid. A heterogeneous mixture on the other hand is a type of mixture in which the components making the mixture are not uniform. A heterogeneous mixture contains more than one phases of matter. These types of mixtures may be either suspensions or colloids.A 0.24-kg stone is held 1.2 m above the top edge of a water well and then dropped into it. the well has a depth of 4.8 m. (a) relative to the configuration with the stone at the top edge of the well, what is the gravitational potential energy of the stoneâearth system before the stone is released? 14.112 incorrect: your answer is incorrect. your response differs from the correct answer by more than 100%. j (b) relative to the configuration with the stone at the top edge of the well, what is the gravitational potential energy of the stoneâearth system when it reaches the bottom of the well? 0 incorrect: your answer is incorrect. the correct answer is not zero. j (c) what is the change in gravitational potential energy of the system from release to reaching the bottom of the well?
When the cab is carrying its maximum capacity, at what average rate must the motor deliver energy to get the cab up to cruising speed?
Rank the tensions in the ropes, t1, t2, and t3, from smallest to largest, when the boxes are in motion and there is no friction between the boxes and the horizontal surface.
The tensions in the ropes, from smallest to largest, are T1, T3, and T2.
When the boxes are in motion and there is no friction between the boxes and the horizontal surface, the tensions in the ropes can be ranked from smallest to largest as follows:
1. Tension in rope T1:
This is the smallest tension because it only needs to support the weight of box 1.
As long as box 1 is not accelerating vertically, the tension in T1 is equal to the weight of box 1.
2. Tension in rope T3:
This tension is greater than the tension in T1 because it needs to support the weight of both box 1 and box 2.
Since the two boxes are connected by T3, the tension in T3 is equal to the sum of the weights of box 1 and box 2.
3. Tension in rope T2:
This is the largest tension because it needs to support the weight of box 3, as well as the combined weight of box 1 and box 2.
Since both box 1 and box 2 are connected to box 3 by T2, the tension in T2 is equal to the sum of the weights of box 1, box 2, and box 3.
Hence, the tensions in the ropes, from smallest to largest, are T1, T3, and T2.
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The tensions in the ropes (t1, t2, and t3) when the boxes are in motion and frictionless can be ranked as follows: t1 < t2 < t3. by the principles of Newton's second law of motion.
When the boxes are in motion on a frictionless surface, the net force acting on each box is equal to its mass multiplied by its acceleration
(F = ma). Since all boxes experience the same acceleration, the ranking of tensions can be determined by comparing the magnitudes of the forces.
t1 corresponds to the box with the smallest mass. Its tension is just enough to overcome the gravitational force pulling it downward.
t2 corresponds to the middle box. It experiences a tension slightly greater than t1, as it needs to overcome both its own weight and the weight of the box below it.
t3 corresponds to the largest box. It experiences the highest tension among the three ropes since it needs to overcome its own weight and the combined weight of the other two boxes above it.
Therefore, the ranking of tensions from smallest to largest is t1 < t2 < t3, reflecting the relationship between mass, acceleration, and force in accordance with Newton's laws of motion.
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The findings of Henri Becquerel and the Curies helped to discredit what notion held for a very long time? That matter can never be created or destroyed. Matter is made up of any smaller parts. Matter is radioactive and decays over time. That matter is invisible and indestructible.
The answer is: That matter is indivisible and indestructible
Through his experiment, Henri Becquerel found that matters can be stored within another object, (He took potassium uranyl sulfate and put it under the sun and then put the substance within photographic plates that he cover with black paper, ) . Curie found that over time matters would decay and eventually gone.
If x=450 mm, determine the mass of the counterweight s required to balance a 90-kg load, l.
To balance the 90-kg load, a counterweight with a mass of 90 kg is required.
Explanation:To balance the system, the torque exerted by the 90-kg load, l, must be equal to the torque exerted by the counterweight, s. The torque is given by the product of the force and the lever arm. Since the distance from the fulcrum to the load is x=450 mm, the torque exerted by the load is (90 kg)(9.8 m/s^2)(0.45 m). To balance this torque, the counterweight must exert an equal torque but in the opposite direction. Therefore, the mass of the counterweight can be calculated using the equation:
torque load = torque counterweight
(90 kg)(9.8 m/s^2)(0.45 m) = mass counterweight(9.8 m/s^2)(0.45 m)
Simplifying and solving for the mass of the counterweight, we have:
mass counterweight = (90 kg)(0.45 m) / 0.45 m = 90 kg
Therefore, the mass of the counterweight required to balance the 90-kg load is 90 kg.
G: assume that, at a certain angular speed ω2, the radius r becomes twice l. find ω2.
The angular speed at twice the initial radius becomes half of the initial angular speed.
Further Explanation:
Speed is the measure of a quantity of an object the tells how fast the object is moving in the other words we can define the speed that it is the distance covered by an body divided by the time taken to cover that distance. It is a quantity with only magnitude so it is a scalar quantity.
Given:
The certain angular speed is [tex]{\omega _2}[/tex].
The radius is [tex]r[/tex].
Concept:
The expression for the linear motion can be written as:
[tex]\fbox{\begin\\v=\dfrac{s}{t}\end{minispace}}[/tex] …… (1)
Here, [tex]v[/tex] is the linear speed, [tex]s[/tex] is the total distance covered and [tex]t[/tex] is the time taken to cover to distance.
The expression for the circular speed can be written as:
[tex]\fbox{\begin\\\omega=\dfrac{\theta }{t}\end{minispace}}[/tex]
Here, [tex]\omega[/tex] is the circular speed and [tex]\theta[/tex] is the angular displacement.
The expression for the total distance covered in term of angular displacement is:
[tex]s=\theta r[/tex]
Substitute [tex]\theta r[/tex] for [tex]s[/tex] in equation (1)
[tex]\begin{aligned}v&=\frac{{\theta r}}{t}\hfill\\v&=\frac{\theta }{t}\cdot r\hfill\\v&=\omega\cdot r\hfill\\\omega&=\frac{v}{r} \hfill\\ \end{aligned}[/tex]
From above expression the angular speed is inversely proportional to the radius.
[tex]\fbox{\begin\\\omega\propto\dfrac{1}{r}\end{minispace}}[/tex]
That is if the radius increases the angular speed decreases and if the radius decreases the angular speed increases.
Considered the linear speed remain content.
Case 1:
The angular speed is given by
[tex]{\omega _1}=\dfrac{v}{r}[/tex] …… (2)
Case 2:
The radius is double that is [tex]2r[/tex].
The angular speed is given by
[tex]{\omega _2}=\dfrac{v}{{2r}}[/tex] …… (3)
Divide equation (2) by equation (3).
[tex]\begin{aligned}\frac{{{\omega _2}}}{{{\omega _1}}}&=\frac{{\frac{v}{{2r}}}}{{\frac{v}{r}}}\hfill\\\frac{{{\omega _2}}}{{{\omega _1}}}&=\frac{1}{2}\hfill\\{\omega _2}&=\frac{{{\omega _1}}}{2}\hfill\\\end{aligned}[/tex]
Therefore, the angular speed at twice the initial radius becomes half of the initial angular speed.
Learn more:
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3. Angular speed https://brainly.com/question/4721004
Answer Details:
Grade: college
Subject: Physics
Chapter: Kinematics
Keywords:
Certain, angular speed, w2, radius, r, twice, w1/2, half, initial angular speed.
where do you bump into neon in everyday life
An object is suspended from the ceiling with two wires that make an angle of 40° with the ceiling. The weight of the body is 150 N. What is the tension in each wire?
41.6 newtons
82.3 newtons
150 newtons
117 newtons
Answer:
the Answer is D 117 newtons
Explanation:
A car speeds up as it rolls down a hill. which is this an example of? positive acceleration negative acceleration relative velocity
How are all paths that have a displacement of zero similar?
Can an object be in mechanical equilibrium when only a single force acts on it? explain.
A position vector with magnitude 10 m points to the right and up. its x-component is 6.0 m. part a what is the value of its y-component?
The position vector with magnitude [tex]10 \ m[/tex] points to the right and up. its x-component is [tex]6.0\ m[/tex] and its y-component is [tex]8\ m[/tex].
A vector is a physics term for a quantity with both magnitude (size) and direction. Physical quantities including each of these properties are represented by vectors, such as displacement, velocity, force, and acceleration.
The resultant vector is the vector that is created when two or more vectors are added together. It depicts the final result of mixing the various vectors. The vector addition of the various vectors yields the resultant vector.
Given:
Magnitude = [tex]10 \ m[/tex]
x- component = [tex]6\ m[/tex]
The resultant vector is given as:
[tex]r = sqrt(x^2+y^2)\\10 = sqrt(6^2+y^2)\\y^2 = 100-36\\y = 8\ m[/tex]
Hence, the position vector with magnitude [tex]10 \ m[/tex] points to the right and up. its x-component is [tex]6.0\ m[/tex] and its y-component is [tex]8\ m[/tex].
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A blue car pulls away from a red stop-light just after it has turned green with a constant acceleration of 0.3 m/s2. a green car arrives at the position of the stop-light 5 s after the light had turned green. what is the slowest constant speed which the green car can maintain and still catch up to the blue car?
Final answer:
The slowest constant speed that the green car can maintain and still catch up to the blue car is 3.75 meters per second.
Explanation:
To find the slowest constant speed which the green car can maintain and still catch up to the blue car, we need to determine the distance that the blue car travels during the 5-second interval before the green car arrives at the position of the stop-light.
Using the equation of motion for constant acceleration, we can calculate the distance traveled by the blue car with an initial velocity of 0 and an acceleration of 0.3 m/s²:
d = ut + 1/2at²
Where d is the distance, u is the initial velocity, t is the time, and a is the acceleration.
Substituting the given values into the equation, we have:
d = 0(5) + 1/2(0.3)(5)²
d = 0 + 1/2(0.3)(25)
d = 1/2(7.5)
d = 3.75 meters
Therefore, the slowest constant speed that the green car can maintain and still catch up to the blue car is 3.75 meters per second.
The pressure 75.0 meters underwater is 837 kpa. what is this pressure in atm
For the rotational part of the experiment, a student measures a force of 1.8 n when the radius is 12 cm and the angular velocity 10 rad/s.
a.what is the total mass of the object undergoing uniform circular motion?
b.if the same mass (at the same radius) is rotated at 5 rad/s, what will be the measured force?
What is the defining feature of a system? A. properties that don’t change B. collision C. interaction D. gravitational force
Why might it be necessary to ignore some of the data points just before and just after the collision?
In physics, some data points may need to be ignored just before and after a collision. This is done to remove errors and focus on the specific moment of collision.
Explanation:In physics, it may be necessary to ignore some of the data points just before and just after a collision for various reasons. One reason is to remove outliers or errors in the measurements that could skew the results. For example, if there is a sudden spike or dip in the data due to a measurement error, it would be necessary to exclude those points to ensure accurate analysis.
Another reason is to focus on the specific moment of collision. By excluding the data points surrounding the collision, it allows for a clearer observation and analysis of the collision itself. This is important for understanding the dynamics and effects of the collision.
Therefore, it is not necessary to ignore both sets of data equally, but rather selectively remove data points that may introduce errors or hinder the analysis of the collision.
A typical set of stairs is angled at 38 degrees. You climb a set of stairs at a speed of 3.1m/s. 1.) How much height will you gain in 4.4 seconds? 2.) How much horizontal distance will you cover in 4.4 second
An earth satellite moves in a circular orbit at a speed of 5000 m/s. part a what is its orbital period? express your answer using two significant figures
Its orbital period is about 5.6 hours
[tex]\texttt{ }[/tex]
Further explanationNewton's gravitational law states that the force of attraction between two objects can be formulated as follows:
[tex]\large {\boxed {F = G \frac{m_1 ~ m_2}{R^2}} }[/tex]
F = Gravitational Force ( Newton )
G = Gravitational Constant ( 6.67 × 10⁻¹¹ Nm² / kg² )
m = Object's Mass ( kg )
R = Distance Between Objects ( m )
Let us now tackle the problem !
[tex]\texttt{ }[/tex]
Given:
speed of satellite = v = 5000 m/s
mass of earth = M = 6 × 10²⁴ kg
Asked:
orbital period = T = ?
Solution:
[tex]\Sigma F = ma[/tex]
[tex]G \frac{ M m} { R^2 } = m \frac{v^2}{R}[/tex]
[tex]G \frac{ M } { R^2 } = \frac{v^2}{R}[/tex]
[tex]G \frac{ M } { R } = v^2 [/tex]
[tex]R = G \frac{ M } { v^2 } [/tex]
[tex]2 \pi R = 2 \pi G \frac{ M } { v^2 } [/tex]
[tex]\frac{ 2 \pi R }{ v } = (2 \pi G \frac{ M } { v^2 } ) \div v[/tex]
[tex]T = 2 \pi G \frac{ M } { v^3 }[/tex]
[tex]T = 2 \pi \times 6.67 \times 10^{-11} \times \frac{ 6 \times 10^{24} } { 5000^3 }[/tex]
[tex]T \approx 20116 \texttt{ s}[/tex]
[tex]T \approx 5.6 \texttt{ hours}[/tex]
[tex]\texttt{ }[/tex]
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Answer detailsGrade: High School
Subject: Physics
Chapter: Gravitational Fields
The orbital period of a satellite moving at 5000 m/s in a circular orbit around Earth is approximately 8500 seconds.
To find the orbital period (T) of a satellite moving in a circular orbit at a speed of 5000 m/s, we need to use the formula for orbital speed:
[tex]v = \frac{2\pi r}{T}[/tex]
Here, v is the orbital speed, and r is the radius of the orbit. We can rearrange this formula to solve for the period (T):
[tex]T = \frac{2\pi r}{v}[/tex]
First, we need to find the radius of the satellite's orbit. Assuming the satellite is close to Earth's surface, we'll use Earth's radius plus an additional small altitude:
r = 6371 km + 400km = 6.771 x 10⁶ m
Now, substitute the values into the formula:
[tex]T = \frac{2\pi (6.771 \times 10^6 \, \text{m})}{5000 \, \text{m/s}}[/tex]
Calculating this gives:
T ≈ 8515.68 seconds
Expressing the answer with two significant figures:
T ≈ 8500 seconds
The orbital period of a satellite moving at 5000 m/s in a circular orbit around Earth is approximately 8500 seconds. This calculation uses the Earth's radius and the orbital speed provided.
One mole of helium atoms has a mass of 4 grams. if a helium atom in a balloon has a kinetic energy of 1.224e-21 j, what is the speed of the helium atom? (the speed is much lower than the speed of light.) v = 6.12e-19 incorrect: your answer is incorrect. m/s
Part b suppose the magnitude of the gravitational force between two spherical objects is 2000 n when they are 100 km apart. what is the magnitude of the gravitational force fg between the objects if the distance between them is 150 km ? express your answer in newtons to three significant figures. hints fg = 889 n submitmy answersgive up correct significant figures feedback: your answer 890 n was either rounded differently or used a different number of significant figures than required for this part. part c what is the gravitational force fg between the two objects described in part b if the distance between them is only 50 km ?
The formula for gravitational force is:
F = G m1 m2 / r^2
where G m1 m2 are constants, therefore:
F r^2 = constant
Part b. Given F1 = 2000 N, r1 = 100 km
Find F2 = ?, r2 = 150 km
(2000 N) * (100 km)^2 = F2 * (150 km)^2
F2 = 888.89 N
Part c. Given F1 = 2000 N, r1 = 100 km
Find F2 = ?, r2 = 50 km
(2000 N) * (100 km)^2 = F2 * (50 km)^2
F2 = 8000 N
A 0.14-kg baseball is dropped from rest from a height of 2.0 m above the ground. what is the magnitude of its momentum just before it hits the ground if we neglect air resistance?
If a 0.14-kg baseball is dropped from rest from a height of 2.0 m above the ground, then the magnitude of its momentum just before it hits the ground would be 0.8764 kg - m / s.
What is momentum?It can be defined as the product of the mass and the speed of the particle, it represents the combined effect of mass and the speed of any particle, and the momentum of any particle is expressed in Kg m/s unit.
As given in the problem if a 0.14-kg baseball is dropped from rest from a height of 2.0 m above the ground.
The velocity just before hitting the ground = √ ( 2 × 9.8 × 2)
= 6.26 m / s
The magnitude of its momentum just before it hits the ground = 0.14 × 6.26
= 0.8764 kg - m / s
Thus, the magnitude of its momentum just before it hits the ground would be 0.8764 kg - m/s
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How can you make a solution saturated
A. Add more solvent
B. Add more solute
C. Add more solution
D. Decrease the concentration
Answer: The answer is B. Add more solute (took test)
Explanation:
Answer:
It's definitely B add more solute.
Explanation:
Suppose that data are stored on 8.54-gbyte single-sided, double-layer dvds that weigh 15 g each. suppose that an eurostar rail service train, london to paris via chunnel, carries 10 4 kg of these dvds. the great circle distance of the line is 640 km and the traveling time is 2 hours, 15 minutes. what is the data transmission rate in bits per second of this system?
The data transmission rate of the Eurostar rail service train, London to Paris via Chunnel is 11516003 bite/second.
What is transfer rate?Transfer rate is a common metric for gauging how quickly data or information moves from one place to another.
Amount of of data stored in single-side = 8.54-gbyte.
Mass of the one double-layer DVDs= 0.015 g.
Total mass of DVDs = 10⁴ kg.
Total number of DVDs = 10⁴ ÷ 0.015 = 666667
Total amount of data = 666667 × 2 × 8.54 × 1024 × 8 bite
Total time taken = 2 hour 15 min = 2.25 hour = (2.25 × 3600) s = 8100 second
The data transmission rate is = (666667 × 2 × 8.54 × 1024 × 8) ÷ 8100 bite/second
= 11516003 bite/second.
Therefore, the data transmission rate of the Eurostar rail service train, London to Paris via Chunnel is 11516003 bite/second.
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When there is no number in front of a chemical formula in a chemical equation, what number is understood?
Final answer:
In a chemical equation, an absent coefficient in front of a chemical formula indicates that the understood number is '1', signifying one mole of the substance in the reaction.
Explanation:
When there is no number in front of a chemical formula in a chemical equation, it is understood that the number is '1'. In a balanced chemical equation, coefficients are used to indicate the mole ratio in which the reactants combine to form products. If a chemical formula has no coefficient written before it, we assume the coefficient is 1. This means one mole of the substance is involved in the reaction. Coefficients are particularly important as they reflect the relative numbers of molecules or formula units in the reactants and products.
What is the magnitude of the frictional force exerted on the mug?
Which material are you most likely to find in the D horizon? A.Plant roots B.Bedrock C.Humus D.Iron oxide
Bedrock
Explanation;Soil is made up of several main horizons, namely; A, B, and C. Some soil may have O horizons which may be a thin or a thick layer and lack in some other soils. other layers are D, R and E horizons,When all the layers are put together they form a soil profile. D- horizon is a layer of soil that may occur below the B-horizon or C-horizon when it is present. Materials such as bedrock are found on this layer.Which of the following is the SI unit used in measuring the temperature of a hot cup of coffee?
A. Joule
B. centigrade
C. Celsius
D. kelvin
kelvin is the right answer
Mountain climbers have climbed to the top of mt. everest at an altitude of 8.85 km above sea level. the air pressure at this level is approximately _________ mb, and there is _______% of the atmosphere above the mountain climbers.
Final answer:
The air pressure at the top of Mt. Everest is around 253 mm Hg (33.7 kPa) due to the high altitude, resulting in significantly lower oxygen levels and causing extreme drying of breathing passages due to the cold, thin air.
Explanation:
The air pressure at the top of Mt. Everest (8.85 km above sea level) is significantly lower than at sea level due to the altitude. Specifically, the atmospheric pressure on the summit of Mt. Everest can be as low as 253 mm Hg, which corresponds to about 33.7 kPa (kiloPascals) or 0.308 atm (atmospheres).
This reduced pressure means that the oxygen content is much lower as well, posing significant challenges to climbers such as reduced oxygen availability for breathing and the extreme drying of breathing passages.
The partial pressure of oxygen at this altitude, considering that it comprises 20.9% of the atmospheric composition, would be considerably less than at sea level.
Climbers often need supplemental oxygen to compensate for the lower oxygen levels. The extreme drying experienced by climbers at high altitudes occurs because the cold, thin air contains very little moisture, leading to rapid evaporation of moisture from the breathing passages.