Answer:
It takes 2.04s to get to the maximum height
Explanation:
This is a vertical throw problem so it can be treated as a uniform accelerated rectilinear motion. For computing time we are going to use the formula:
[tex]v_{f}=v_{o}+g*t[/tex]
where[tex]v_{f}[/tex] is the final velocity, [tex]v_{o}[/tex] is the initial velocity, [tex]t[/tex] is the time and, [tex]g[/tex] is the gravity.
For solving this kind of problems we need at least three values. The values we have are:
[tex]v_{o} = 20\dfrac{m}{s}[/tex][tex]g = -9.8\dfrac{m}{s^{2}} [/tex] (negative because gravity's direction is oposite from the object's moving direction)[tex]v_{f}=0[/tex] (final velocity equals zero because at maximun height the object stops moving)Now:
[tex]v_{f}=v_{o}+g*t[/tex]
[tex]v_{f}-v_{o}=g*t[/tex]
[tex]\dfrac{v_{f}-v_{o}}{g}=t[/tex]
[tex]\dfrac{0-20}{-9.8}=t[/tex]
[tex]t=2.04s[/tex]
At what point in the swing of the pendulum is the potential energy completely converted into Kinetic energy
Answer:
The correct answer is B
Explanation:
The motion of a pendulum is a classic example of mechanical energy conservation. So you can assume, ignoring air resistance, that total mechanical energy is:
[tex]E = K + U[/tex]
Where K is kinetic energy, U is potential energy and E is the resulting mechanical energy.
You know that:
[tex]U = mgh\\K = \frac{1}{2}mV^{2}[/tex]
From this, you can deduce that for potential energy to be transformed entirely to kinetic energy the height of the pendulum mass must be zero and traveling at its maximum velocity, wich is depicted as B in the attached picture.
During the rock cycle, sedimentary rocks are returned to the Earth's interior and become metamorphic rocks through metamorphism. Metamorphic rocks can then move even deeper into the Earth's interior and melt into magma.
What phenomenon is responsible for moving sedimentary and metamorphic rock into the Earth's interior?
A. plate tectonics
B. infiltration
C. condensation
D. the carbon cycle
give two ways in which the water vapour changes as it passes down the glass tube in the condenser
Water vapor in a condenser tube changes by losing heat energy and condensing from the gaseous to the liquid phase, while releasing latent heat of vaporization.
Explanation:As water vapor passes down the condenser tube, there are two primary changes that occur due to the process of condensation. First, the water vapor loses heat energy to the surrounding cooler surfaces of the condenser tube. This loss of energy decreases the kinetic energy of the water vapor molecules, causing them to slow down and become less spread out. Second, as the molecules lose energy, they begin to collect together due to the intermolecular forces between them, resulting in the transition from the gaseous phase to the liquid phase. Throughout this process, the latent heat of vaporization is released into the environment, which can be calculated using the formula Q = mLy, wherein 'Q' is the energy involved in the phase change, 'm' is the mass of the substance, and 'Ly' is the latent heat of vaporization.
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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?
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
Which force causes send dunes? A. Rivers B. Ice C. Wind D. Gravity
Answer: C. Wind
Explanation:
An erosion is a phenomena in which a physical agent takes away along with it soil and deposit it to some other place.
Sand dune is a kind of wind erosion of the loose soil called as sand. It is typically seen in desert regions. The high speed wind currents blow away huge amount of soil.
If the speed of an object increases, its kinetic energy _____.
increases
stays the same
decreases
If the speed of an object increases, its kinetic energy increases.
If the speed of an object increases, its kinetic energy increases. Kinetic energy is directly proportional to the square of the object's velocity. As the speed of an object increases, the kinetic energy increases at a faster rate. This relationship can be described by the equation:
Kinetic energy = (1/2) × mass × velocity²
Since the velocity is squared in the equation, any increase in velocity will have a greater impact on the kinetic energy.
Therefore, as the speed of an object increases, its kinetic energy also increases.
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The earth's radius is about 4000 miles. kampala, the capital of uganda, and singapore are both nearly on the equator. the distance between them is 5000 miles. the flight from kampala to singapore takes 9.0 hours. what is the plane's angular velocity with respect to the earth's surface? give your answer in °/h.
Jenny is studying a compound that has two oxygen atoms and one nitrogen atom. Which statement describes the subscripts she will use to write the chemical formula?
Answer:
she will use one 2 as a subscript
A hot-air balloon is accelerating upward under the influence of two forces, its weight and the buoyant force. for simplicity, consider the weight to be only that of the hot air within the balloon, thus ignoring the balloon fabric and the basket. the hot air inside the balloon has a density of ρhot air = 0.93 kg/m3, and the density of the cool air outside is ρcool air = 1.29 kg/m3. what is the acceleration of the rising balloon?
The radius of the earth's very nearly circular orbit around the sun is 1.5×1011m. find the magnitude of the earth's velocity. assume a year of 365 days.
The magnitude of Earth's velocity in its orbit around the sun is approximately 29,450 m/s.
Explanation:Given that the radius of Earth's orbit around the sun is 1.5 x 10^11 m, we can calculate the magnitude of Earth's velocity using the formula for the circumference of a circle: v = 2 × Pi × r / T, where v is the velocity, r is the radius of the orbit, and T is the period of the orbit. Since one year is equivalent to 365 days, we have T = 365 days. Plugging in the values, we get: v = 2 × 3.14159 × 1.5 × 10^11 m / (365 × 24 × 60 × 60 s)
Simplifying the expression gives us:v ≈ 29,450 m/s
How is the surface area to volume ratio maximized in lungs?
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.
Water is boiled at 1 atm pressure in a 25-cm-internal- diameter stainless steel pan on an electric range. if it is observed that the water level in the pan drops by 10 cm in 45 min, determine the rate of heat transfer to the pan.
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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a car is driving on the turnpike accelerates uniformly in a straight line from 88ft/s to 110 ft/s in 3.50s A) what is the cars acceleration? B) how far does the car travle while it accelerates
For steady laminar flow through a long tube, the axial velocity distribution is given by u = c(r2-r2), where r is the tube outer radius and c is a constant. find the total volume flow q through the tube
Which electron configuration represents the atom with the largest atomic radius?
a.1s1
b.1s22s2
c.1s22s1
d.1s22s22p1?
The electron configuration that represents the atom with the largest atomic radius among the given options is 1s22s22p1. This is due to the addition of an electron in a higher energy level, which increases the atomic radius.
Explanation:The electron configuration that represents the atom with the largest atomic radius would be 1s22s22p1. The atomic radius typically increases as you move down a group in the periodic table because more energy levels (shells) are being added. In the given options, 1s22s22p1 represents an atom with the largest number of energy levels. It also suggests that the atom has one more electron than 1s22s2 and that electron goes into the next higher energy level, making the atomic radius larger.The electron configuration that represents the atom with the largest atomic radius is option b. 1s22s2.The atomic radius increases as you move down a group and as you move from right to left across a period on the periodic table. So, the electron configuration with the highest energy level and the least number of electrons in the valence shell represents the atom with the largest atomic radius.In option b, 1s22s2, the atom has electrons in both the 1s and the 2s orbitals, and the 2s orbital is in a higher energy level compared to option c (which has fewer electrons) and option d (which has more electrons).
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A car accelerates uniformly from rest to a speed of 56.8 km/h (15.8 m/s) in 10 s. find the distance the car travels during this time.
A ring, seen from above, is pulled on by three forces. the ring is not moving. how big is the force f
The magnitude of the force F depends on the other two forces acting on the ring. It should be such that the vector sum of all three forces equals zero, due to the object's static state, which suggests a condition of equilibrium.
Explanation:In the question, you have a ring being pulled on by three forces and it remains stationary. This situation is described by the physics concept of equilibrium, where the sum of all forces acting on an object equals zero because the object isn't moving. The forces here are vectors, meaning they have both magnitude (how big they are) and direction (which way they're pulling). Each force acts in a specific direction, which we'll assume are different for each one.
Now, the size of the force F would depend on the other two forces. Without exact information on the magnitude and direction of the other two forces, we cannot precisely compute F. However, we can say that F will be such that the vector sum of the all forces (including F) will be zero. This happens because no movement implies no net force according to Newton's second law. This law states that an object at rest, like the one in your assignment, stays at rest unless acted upon by a non-zero net force.
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Which group of animals would be served best by the following adaptations?
Large ears to dissipate heat.
Kidneys adapted to check water loss.
A nocturnal lifestyle
A) animals that live in deserts
B) animals that live in the tundra
C) animals that live in grasslands
D) animals that live in the rain forest
Can an object be in mechanical equilibrium when only a single force acts on it? explain.
What is a Punnett square for a dragon
To maintain a constant speed, the force provided by a car’s engine must equal the drag force plus the force of friction of the road (the rolling resistance). what are the drag forces at 50 km/h and 80 km/h for a toyota camry? (drag area is 0.67 m2.)
Gold and silicon are mutually insoluble in the solid state and form a eutectic system with a eutectic temperature of 636 k and a eutectic composition of xsi=0.186. calculate the gibbs free energy of the eutectic melt relative to (a) unmixed liquid au and liquid si, and (b) unmixed solid au and solid si.
To calculate the Gibbs free energy of the eutectic melt relative to unmixed liquid gold and liquid silicon, you need to use the equation for Gibbs free energy. For part (a), assume that the Gibbs free energy change is zero for the unmixed liquids, and for part (b), assume it is zero for the unmixed solids. Substitute the given values into the equation to find the answer.
Explanation:In a eutectic system, gold and silicon are mutually insoluble in the solid state. The eutectic system has a eutectic temperature of 636 K and a eutectic composition of xsi=0.186.
To calculate the Gibbs free energy of the eutectic melt relative to (a) unmixed liquid gold (Au) and liquid silicon (Si), and (b) unmixed solid gold (Au) and solid silicon (Si), we need to use the equation for Gibbs free energy (ΔG = ΔH - TΔS).
For (a), we can assume that the Gibbs free energy change (ΔG) is equal to zero for the unmixed liquids. The equation becomes ΔG(melt) = ΔH(melt) - TΔS(melt). Since the eutectic composition is given as xsi=0.186, we can assume that 0.186 moles of gold and 0.814 moles of silicon are present in the eutectic melt.
The change in enthalpy (ΔH) can be calculated by taking the eutectic temperature (636 K) and multiplying it by the change in entropy of mixing (ΔS(mix)). The ΔS(mix) can be calculated using the equation: ΔS(mix) = xAu · R · ln(xAu) + xSi · R ·
ln(xSi), where xAu is the mole fraction of gold and xSi is the mole fraction of silicon. Substituting the values, we can calculate ΔH(melt) = 1717.28 J/mol and ΔS(mix) = -2.25 J/(mol·K). Finally, we can substitute all the values into the Gibbs free energy equation to find the answer.
For (b), we can assume that the Gibbs free energy change (ΔG) is equal to zero for the unmixed solids. The equation becomes ΔG(melt) = ΔH(melt) - TΔS(melt).
The change in enthalpy (ΔH) for the unmixed solids can be calculated using the heat of fusion (ΔH(fus)) and the mass of gold and silicon in the eutectic melt. The ΔS(mix) for the unmixed solids can be calculated using the equation:
ΔS(mix) = R · ln(1 - xsi) + R · ln(1 - xAu), where xsi is the eutectic composition and xAu is the mole fraction of gold. Substituting the values, we can calculate ΔH(melt) = 27.39 J/mol and ΔS(mix) = 6.72 J/(mol·K). Finally, we can substitute all the values into the Gibbs free energy equation to find the answer.
When measuring an object with a ruler, the end of the object should be lined up with the _______of the ruler.
end
0 mark
1 mark
middle
Answer:
0 Mark
Explanation:
For all the measurements that we make we always need some reference to find the value of the reading
So all the scales have its reference marked as 0 mark
When we put one end of our measurement at that reference mark then the other end of the measurement will give the reading of our measurements.
so here the correct answer should be
0 MARK
For each of the following systems and time intervals, select the appropriate version of
ΔK + ΔU + ΔEint = W + Q + TMW + TET + TER,
the conservation of energy equation.
(a) the heating coils in your toaster during the first five seconds after you turn the toaster on
ΔEint = Q + TET + TER
ΔK + ΔU + ΔEint = W + Q + TMW + TMT
ΔU = Q + TMT
0 = Q + TMT + TET + TER
ΔK = W + Q
ΔEint = W + Q + TET
ΔU = W + Q + TMW + TMT
(b) your automobile from just before you fill it with gas until you pull away from the gas station at speed v
ΔEint = Q + TET + TER
ΔK + ΔU + ΔEint = W + Q + TMW + TMT
ΔU = Q + TMT
0 = Q + TMT + TET + TER
ΔK = W + Q
ΔEint = W + Q + TET
ΔU = W + Q + TMW + TMT
(c) your body while you sit quietly and eat a peanut butter and jelly sandwich for lunch
ΔEint = Q + TET + TER
ΔK + ΔU + ΔEint = W + Q + TMW + TMT
ΔU = Q + TMT
0 = Q + TMT + TET + TER
ΔK = W + Q
ΔEint = W + Q + TET
ΔU = W + Q + TMW + TMT
(d) your home during five minutes of a sunny afternoon while the temperature in the home remains fixed
ΔEint = Q + TET + TER
ΔK + ΔU + ΔEint = W + Q + TMW + TMT
ΔU = Q + TMT
0 = Q + TMT + TET + TER
ΔK = W + Q
ΔEint = W + Q + TET
ΔU = W + Q + TMW + TMT
The correct forms of the conservation of energy equation are chosen for different scenarios based on the type of energy transformations occurring. For a toaster, ΔEint = Q is more appropriate. For a car moving from a gas station, ΔK = W + Q and ΔU = W + Q are relevant. When eating a sandwich, ΔEint = W + Q is appropriate. For a house with constant temperature, 0 = Q + TMT + TET + TER is relevant.
Explanation:The conservation of energy equation in physics considers various energy forms in the system, including kinetic energy (represented as ΔK), potential energy (represented as ΔU), and internal energy (represented as ΔEint). In each scenario, different energy forms come into play.
(a) For the toaster, the main transformation involves electrical energy is being converted into heat energy due to the heating coils' resistance. Therefore, the equation best suited for this scenario is ΔEint = Q , which signifies the increase in internal energy as equivalent to the heat gained.(b) For a car moving from a gas station, the potential and kinetic energies are of relevance. Therefore, ΔK = W + Q and ΔU = W + Q are more relevant as they represent the changes in kinetic and potential energies as equivalent to work done and heat transferred.(c) When you're eating a sandwich, the body is primarily converting the sandwich's chemical energy into internal energy, making ΔEint = W + Q the best equation.(d) When the temperature inside the house remains constant, and there is no kinetic or potential energy involved, the conservation of energy equation simplifies to 0 = Q + TMT + TET + TER as there is no energy being added or lost.Learn more about Conservation of Energy here:https://brainly.com/question/35373077
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Answer for brainliest and mad points pllllssssssss
Two balls undergo inelastic collision. The y-momentum after the collision is 98 kilogram meters/second, and the x-momentum after the collision is 100 kilogram meters/second. What is the magnitude of the resultant momentum after the collision?
To find the magnitude of the resultant momentum after an inelastic collision with given x and y components, one applies the Pythagorean theorem. The resultant momentum is approximately 140.01 kilogram meters/second.
The question involves calculating the magnitude of the resultant momentum after two balls undergo an inelastic collision. The given momentum components are 98 kilogram meters/second in the y-direction and 100 kilogram meters/second in the x-direction. To find the resultant momentum, we use the Pythagorean theorem since momentum is a vector quantity.
Using the values given:
(resultant momentum) = (100^2 + 98^2)
(resultant momentum) = (10000 + 9604)
(resultant momentum) = (19604)
The magnitude of the resultant momentum is approximately 140.01 kilogram meters/second.
A dry cell does 7.5 j of work through chemical energy transfer 5.00C between terminals of the cell . What is the electric potential between the two terminals?
Final answer:
The electric potential between the terminals of a dry cell that does 7.5 joules of work to transfer 5.00 coulombs of charge is 1.5 volts.
Explanation:
The student asked for the electric potential between the terminals of a dry cell that does 7.5 joules (J) of work to transfer 5.00 coulombs (C) of charge. To find the electric potential (also known as voltage) across the terminals, we use the relationship that the work done (W) by the cell through electrical energy is equal to the charge (Q) multiplied by the potential difference (V), which can be expressed as W = QV. Rearranging this equation for V gives us V = W/Q.
Substituting the given values into the equation, we have:
V = 7.5 J / 5.00 C = 1.5 volts (V)
Therefore, the electric potential between the two terminals of the dry cell is 1.5 V. This result is the voltage of the dry cell when it is not supplying current and is therefore at its electromotive force (emf).
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.