Answer:1
Explanation:
When we boil the water, initially small bubbles started forming at the bottom of the vessel. These bubbles are air which enables the aquatic organisms to survive inside the water.
The solubility of gas started decreases as we increase the temperature, therefore, more and more bubbles formed. Around 373 K water started boiling and vapor started forming inside the water. At this Point water and its vapor is in Equilibrium and every molecule has the same tendency to form a bubble. So there is more and more tendency to form bubbles.
Final answer:
1) Initial bubbles formed in boiling water are due to dissolved gases escaping as the water heats up; the solubility of these gases decreases with temperature. When the water reaches its boiling point, water vapor within the bubbles matches atmospheric pressure, leading to boiling.
Explanation:
When you set a pot of tap water on the stove to boil, you will notice bubbles forming before the water reaches its boiling point. 1) This occurs because water contains dissolved air and other gases that start to come out of solution as the water heats up. The solubility of gases in water decreases with increasing temperature, leading to the formation of bubbles.
At the onset, these bubbles are composed mainly of air and water vapor. As the water temperature rises, more water vapor enters the bubbles, increasing the pressure inside to match the atmospheric pressure. Once the water reaches 100°C and the vapor pressure of water equals the atmospheric pressure, boiling occurs. The established bubbles grow in size and rise to the surface as part of the boiling process.
In summary, the initial bubbles that form when water is heated are due to gases coming out of solution due to decreased solubility with rising temperature. Eventually, as the temperature of the water continues to rise, these bubbles contribute to the process by which water transforms from a liquid to a gaseous state, resulting in boiling.
Assume that the equator of the Earth is 24,000 miles long (that’s its circumference). Now, pretend that you are standing somewhere on the equator, such as in the country of Ecuador. Now, if the Earth turns once, completely, in 24 hours, then, how fast would you be going (in miles per hour), even if you just stood still?
Answer:
1000 mph
Explanation:
P = Perimeter = 24000 mi
r = Radius of the equator
t = Time taken to complete one rotation = 24 h
Perimeter of a circle is given by
[tex]P=2\pi r\\\Rightarrow r=\dfrac{P}{2\pi}\\\Rightarrow r=\dfrac{24000}{2\pi}\ mi[/tex]
Angular speed is given by
[tex]\omega=\dfrac{2\pi}{t}\\\Rightarrow \omega=\dfrac{2\pi}{24}[/tex]
Velocity if given by
[tex]v=r\omega\\\Rightarrow v=\dfrac{24000}{2\pi}\times \dfrac{2\pi}{24}\\\Rightarrow v=1000\ mph[/tex]
The person would be going at a speed of 1000 mph
A baseball player throws a baseball with a velocity of 13 m/s North it is caught by a second player seven seconds later how far is the second player from the first player
The second player is 91 meters far from the first player.
Why?
First, let be the +y the North, so, to solve the problem we can use the following formula:
[tex]y=yo+v_o*t+\frac{1}{2}*a*t[/tex]
Now, subsituting the given information, we have(assuming that the speed is constant):
\\\\y-yo=13\frac{m}{s}*7s+\frac{1}{2}*0*7s\\\\y-yo=91m\\\\distance=91m[/tex]
Hence, we have that the second player is 91m far from the first player.
Have a nice day!
A satellite with an orbital period of exactly 24.0 h is always positioned over the same spot on Earth.This is known as a geosynchronous orbit. Television, communication, and weather satellites use geosynchronous orbits. At what distance would a satellite have to orbit Earth in order to have a geosynchronous orbit?
Answer:
35870474.30504 m
Explanation:
r = Distance from the surface
T = Time period = 24 h
G = Gravitational constant = 6.67 × 10⁻¹¹ m³/kgs²
m = Mass of the Earth = 5.98 × 10²⁴ kg
Radius of Earth = [tex]6.38\times 10^6\ m[/tex]
From Kepler's law we have relation
[tex]T^2=\dfrac{4\pi^2r^3}{GM}\\\Rightarrow r^3=\dfrac{T^2GM}{4\pi^2}\\\Rightarrow r=\left(\dfrac{(24\times 3600)^2\times 6.67\times 10^{-11}\times 5.98\times 10^{24}}{4\pi^2}\right)^{\dfrac{1}{3}}\\\Rightarrow r=42250474.30504\ m[/tex]
Distance from the center of the Earth would be
[tex]42250474.30504-6.38\times 10^6=\mathbf{35870474.30504\ m}[/tex]
35870474.30504 m
A satellite in geosynchronous orbit remains over the same point on Earth, requiring an altitude of approximately 35,793 kilometers. This is calculated using the orbital mechanics involving the gravitational constant and Earth's mass and radius.
A geosynchronous orbit means that the satellite has an orbital period of [tex]24[/tex] hours, remaining over the same point on Earth as it rotates. To find the distance at which a satellite must orbit to achieve this, we use the universal law of gravitation and centripetal force.
We know:
Orbital period [tex]T = 86400 \text{ seconds}[/tex]
Gravitational constant [tex]G = 6.67430 \times 10^{-11} \, \text{N} \left( \text{m/kg} \right)^2[/tex]
Mass of Earth [tex]M = 5.972 \times 10^{24} \, \text{kg}[/tex]
Radius of Earth [tex]R = 6.371 \times 10^{6} \, \text{meters}[/tex]
The formula for the geostationary orbit (orbital radius r) is given by:
[tex]r^3 = \frac{GMT^2}{4\pi^2}[/tex]
Substituting the values, we get:
[tex]GM T^2 = 6.67430 \times 10^{-11} \times 5.972 \times 10^{24} \times (86400)^2\\\\GM T^2 = 6.67430 \times 10^{-11} \times 5.972 \times 10^{24} \times 7464960000\\\\GM T^2 = 2.963 \times 10^{14}\\\\r^3 \approx \frac{2.963 \times 10^{14}}{4 \times 9.8696}\\\\r^3 \approx 7.507 \times 10^{12} \, \text{m}^3\\\\r \approx \sqrt[3]{7.507 \times 10^{12}}\\\\r \approx 42164 \, \text{km}\\\\[/tex]
To find the altitude (h) above Earth’s surface:
[tex]h = r - \text{Earth's radius} = 42164 \, \text{km} - 6371 \, \text{km} \approx 35793 \, \text{km}[/tex]
Thus, a satellite needs to orbit at an altitude of approximately [tex]35,793 km[/tex] to maintain a geosynchronous orbit.
In one case, a sports car, its engine running, is driven up a hill at a constant speed. In another case, a truck approaches a hill, and its driver turns off the engine at the bottom of the hill. The truck then coasts up the hill. Which vehicle is obeying the principle of conservation of mechanical energy? Ignore friction and air resistance.
The truck coasting up the hill with its engine turned off is obeying the principle of conservation of mechanical energy because the kinetic energy is converted into gravitational potential energy without external work, while the sports car with the running engine is not conserving mechanical energy since it is adding energy to the system.
Explanation:The principle of conservation of mechanical energy states that the total mechanical energy in a system remains constant if only conservative forces are doing work. When friction and air resistance are ignored, as requested, the truck that coasts up the hill with the engine turned off obeys this principle because its initial kinetic energy is converted into gravitational potential energy without any external work being done on or by the system. On the other hand, the sports car with its engine running is continually adding energy to the system to maintain a constant speed up the hill, which means it is not conserving mechanical energy as in the first case.
A 0.210-kg block along a horizontal track has a speed of 1.70 m/s immediately before colliding with a light spring of force constant 4.50 N/m located at the end of the track.
(a) What is the spring's maximum compression if the track is frictionless
(b) If the track is not frictionless, would the spring's maximum compression be greater than, less than, or equal to the value obtained in part (a)?greater lessequal
Answer
given,
mass of block = 0.21 Kg
speed = 1.70 m/s
spring constant = k = 4.50 N/m
using conservation of energy
a) K.E = P.E
[tex]\dfrac{1}{2}mv^2 = \dfrac{1}{2}kx^2[/tex]
[tex]\dfrac{1}{2}\times 0.21 \times 1.7^2 = \dfrac{1}{2}\times 4.5 \times x^2[/tex]
[tex]0.1348= x^2[/tex]
x = 0.367 m
b) if the track is not friction less the maximum compression will be same as the compression in the part a.
In a frictionless track, the maximum compression of the spring is 0.494 m, calculated based on the energy conservation principle. If there's friction on the track, the maximum compression will be less, because part of the block's kinetic energy is used to overcome friction, leaving less to compress the spring.
Explanation:This physics problem is rooted in concepts of energy conservation, specifically kinetic energy and potential energy. In part (a), when the block collides with the spring, the kinetic energy of the block is converted into potential energy stored in the compressed spring. The formula we need here is the conservative energy formula which states that: kinetic energy + potential energy = constant. Hence the kinetic energy before collision equals the potential energy at maximum compression.
So, 1/2 * mass * velocity^2 = 1/2 * k * compression^2. Substituting given values: 1/2 * 0.210 kg * (1.70 m/s)^2 = 1/2 * 4.50 N/m * x^2. Solving this equation yields x = 0.494 m, which is the maximum compression of the spring.
In part (b), if there is friction on the track, then some of the block's kinetic energy is used to overcome friction, which means less energy is available to compress the spring. Therefore, the friction on the track would result in a less than value for the maximum compression of the spring compared to the frictionless situation in part (a).
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The Bernoulli effect can have important consequences for the design of buildings. For example, wind can blow around a skyscraper at remarkably high speed, creating low pres- sure. The higher atmospheric pressure in the still air inside the buildings can cause windows to pop out. As originally constructed, the John Hancock Building in Boston popped windowpanes that fell many stories to the sidewalk below. (a) Suppose a horizontal wind blows with a speed of 11.2 m/s outside a large pane of plate glass with dimensions 4.00 m 3 1.50 m. Assume the density of the air to be constant at 1.20 kg/m3. The air inside the building is at atmospheric pressure. What is the total force exerted by air on the win- dowpane?
Answer:
9483.26399 N
Explanation:
[tex]\rho[/tex] = Density of air = 1.2 kg/m³
v = Velocity of wind = 11.2 m/s
A = Area = [tex]4\times 31.5\ m^2[/tex]
The force on the pane is
[tex]F=\dfrac{1}{2}\rho v^2A\\\Rightarrow F=\dfrac{1}{2}\times 1.2\times 11.2^2\times 4\times 31.5\\\Rightarrow F=9483.26399\ N[/tex]
The force on the pane of glass is 9483.26399 N
As part of your daily workout, you lie on your back and push with your feet against a platform attached to two stiff springs arranged side by side so that they are parallel to each other. When you push the platform, you compress the springs. You do an amount of work of 85.0 J when you compress the springs a distance of 0.250 m from their uncompressed length. What magnitude of force must you apply to hold the platform in this position? How much additional work must you do to move the platform a distance 0.220 m farther? What maximum force must you apply to move the platform a distance 0.220 m farther?
Answer:
a) F = 680 N, b) W = 215 .4 J , c) F = 1278.4 N
Explanation:
a) Hooke's law is
F = k x
To find the displacement (x) let's use the elastic energy equation
[tex]K_{e}[/tex] = ½ k x²
k = 2 [tex]K_{e}[/tex] / x²
k = 2 85.0 / 0.250²
k = 2720 N / m
We replace and look for elastic force
F = 2720 0.250
F = 680 N
b) The definition of work is
W = ΔEm
W = [tex]K_{ef}[/tex] - [tex]K_{eo}[/tex]
W = ½ k ( [tex]x_{f}[/tex]² - x₀²)
The final distance
[tex]x_{f}[/tex] = 0.250 +0.220
[tex]x_{f}[/tex] = 0.4750 m
We calculate the work
W = ½ 2720 (0.47² - 0.25²)
W = 215 .4 J
We calculate the strength
F = k [tex]x_{f}[/tex]
F = 2720 0.470
F = 1278.4 N
A bullet with mass 4.53 g is fired horizontally into a 1.953-kg block attached to a horizontal spring. The spring has a constant 5.73 102 N/m and reaches a maximum compression of 6.20 cm. Find the initial speed of the bullet-block system.
Answer:
The initial speed of the bullet-block system is 113 m/s
Explanation:
Hi there!
Let´s apply the theorem of conservation of energy. If we neglect friction, all the kinetic energy of the bullet-block system is used to compress the spring. In other words, the kinetic energy is converted into elastic potential energy. Then, the initial kinetic (KE) energy of the bullet-block system will be equal to the final elastic potential energy (EPE) of the spring:
KE = EPE
1/2 · m · v² = 1/2 · k · x²
Where:
m = mass of the bullet-block system.
v = initial speed of the bullet-block system.
k = spring constant.
x = compression of the spring
Then, solving for v:
v² = k · x² / m
v² = 5.73 × 10² N/m · (0.620 m)² / 1.957 kg
v = 113 m/s
A green object will absorb ____________________ light and reflect ____________________ light.
Answer:Magenta , Green
Explanation:
A green object will absorb its complementary color i.e. Magenta (mixture of red and blue) and thus reflect the remaining green color.
This can be understood by the fact that a particular light filter absorb its complementary color and reflects its color for example
Blue filter will absorb yellow light (mixture of red and green)and reflects blue color.
Final answer:
A green object absorbs light in the red portion of the visible spectrum and reflects green light, as the observed color is complementary to the color that is absorbed.
Explanation:
The question is asking which colors of light a green object will absorb and which it will reflect. A green object will absorb light in the red portion of the visible spectrum. This is due to the fact that the color observed from an object comes from the light that is transmitted or reflected, not the light that is absorbed. The reflected or transmitted light is always complementary in color to the light that is absorbed. Thus, for a green object, the absorbed light is predominantly red while it reflects green light.
An example that helps explain this concept is how different complexes absorb light depending on their ligand field strength and the structure of the complex, which influences the observable color. For instance, the complex [Cr(NH3)6]3+ has strong-field ligands which cause it to absorb high-energy photons corresponding to blue-violet light, making the complex appear yellow, which is the complementary color to blue-violet.
From the list below, identify the disease and its classification that is mismatched.
A) polio; meningitis
B) African sleeping sickness; encephalitis
C) La Crosse Virus infection; encephalitis
D) Naegleria fowleri infection; meningoencephalitis
E) Valley Fever; meningitis
Answer:E)Valley Fever;meningitis
Explanation:Valley fever is a a fungal lung infection while the meningitis is a viral or bacterial disease that affect the meninges.
Which of the following statements is true?
1. The 2s orbital in the hydrogen atom is larger than the 3s orbital also in the hydrogen atom.
2. The hydrogen atom has quantized energy levels.
3. The Bohr model of the hydrogen atom has been found to be incorrect.
4. An orbital is the same as a Bohr orbit.
5. The third energy level has three sublevels - the s, p, and d sublevels.
Answer:
2. The hydrogen atom has quantized energy levels.
Explanation:
The Bohr model of the atom states that the structure of the atom is quantized, that is, that electrons can only orbit the nucleus in specific orbits with a fixed radius. Therefore, the electron cannot be in energy levels that do not correspond to these quantized levels.
The Royal Gorge bridge over the Arkansas River is 393 m above the river. A bungee jumper of mass 150 kg has an elastic cord of length 78 m attached to her feet. Assume the cord acts like a spring of force constant k. The jumper leaps, barely touches the water, and after numerous ups and downs comes to rest at a height h above the water. The acceleration of gravity is 9.81 m/s². Find h. Answer in units of m.
Answer:
188.7 m
Explanation:
height of bridge above water (h) = 393 m
mass of bungee jumper (m) = 150 kg
length of cord (L) = 78 m
acceleration due to gravity (g) = 9.8 m/s
initial energy = mgh = 150 x 9.8 x 393 = 577,710 J
since the jumper barely touches the water, the maximum extension of the cord (x) = 393 - 78 = 315 m
from the conservation of energy mgh = [tex](\frac{1}{2})kx^{2}[/tex]
therefore
577,710 = [tex](\frac{1}{2})kx315^{2}[/tex]
k = 11.64 N/m
from Hooke's law, force (f) = kx' ⇒ mg = kx'
where x' is the extension of the cord when it comes to rest
150 x 9.8 = 11.64 × x'
x' = 126.3 m
the final height at which the cord comes to a rest = height of the bridge - length of the cord - extension of the cord when it comes to rest
the final height at which the cord comes to a rest = 393 - 78 - 126.3 = 188.7 m
What is the distance from the moon to the point between Earth and the Moon where the gravitational pulls of Earth and Moon are equal? The mass of Earth is 5.97 x 10^24 kg, the mass of the Moon is 7.35 x 10^22 kg, the distance between Earth and the Moon is 3.84 x 10^8 m, and >G= 6.67x10^-11N x m^2/kg^2
A)3.83 x 10^6 mB)3.83 x 10^7 mC)4.69 x 10^6 mD)4.69 x 10^7 mE)3.45 x 10^8 m
Answer:
the point between Earth and the Moon where the gravitational pulls of Earth and Moon are equal is E)3.45 × 10⁸ m
Explanation:
The force that the Earth exerts on a mass m is
F_e = (G M_e m) / R_e²
where
G is the universal gravitational constantM_e is the mass of EarthR_e is the radius of EarthThe force that the Moon exerts on a mass m is
F_m = (G M_m m) / R_m²
where
G is the universal gravitational constantM_m is the mass of the MoonR_m is the radius of the MoonTherefore, the point where the gravitational pulls of Earth and Moon are equal is:
F_e = F_m
R_e + R_m = R = 3.84×10⁸ m
Thus,
(G M_e m) / R_e² = (G M_m m) / R_m²
M_e / R_e² = M_m / (R - R_e²)
(R - R_e²) / R_e² = M_m / M_e
(R - R_e) / R_e = (M_m / M_e)^1/2
R_e(R/R_e -1) / R_e = (M_m / M_e)^1/2
R/ R_e = (M_m / M_e)^1/2 + 1
R_e = R / [(M_m / M_e)^1/2 + 1]
R_e = (3.84×10⁸ m) / [(7.35 x 10²² kg / 5.97 x 10²⁴ kg )^1/2 + 1]
R_e = 3.45 × 10⁸ m
Therefore, the point between Earth and the Moon where the gravitational pulls of Earth and Moon are equal is 3.45 × 10⁸ m.
How did the theory of plate tectonics come to be a theory?
Explanation:
Plate tectonics is the idea that the earth's outer solid crust (the lithosphere) is fragmented into a few dozen "plates" that pass relative to each other across the earth's surface, like ice slabs on a lake.
A scientist named Alfred Wegener in 1915 proposed that the continents rammed through ocean basin crust, which would clarify why the shapes of many coastlines (such as South America and Africa) seem to match together like a puzzle.
When you throw a ball up in the air, it travels up and then stops instantaneously before falling back down. At the point where it stops and changes directions to fall back down is
Answer:
The ball stops instantaneously at the topmost point of the motion.
Explanation:
Assume we have thrown a ball up in the air. For that we have given a force on the ball and it acquires an initial velocity in the upward direction.
The forces that resist the motion of the ball in the upward direction are the force of gravity and air resistance. The ball will instantaneously come to rest when the velocity of the ball reduces to zero.
The two forces acting in the downward direction reduces its speed continuously and it becomes zero at the topmost point.
Final answer:
The highest point of a ball's upward trajectory is when its velocity is zero, and it momentarily stops before falling back to Earth under the influence of gravity. This is the point where the ball has its maximum potential energy and is the peak of the trajectory.
Explanation:
When you throw a ball up in the air, it travels upwards and slows down under the influence of gravity. At the highest point of its trajectory, the ball's velocity momentarily becomes zero before it reverses direction and falls back to the ground. This moment is known as the peak or the apex of the ball's trajectory. The ball's vertical velocity increases in the downward direction as it descends due to the acceleration caused by gravity. At the peak, not only does the velocity become zero, but this is also the point where the ball has its maximum potential energy.
When calculating the time it takes for the ball to reach its highest point or how high it goes, you can use the initial velocity and the acceleration due to gravity. For example, if a ball is thrown upwards with a velocity of 10 m/s, you can calculate the maximum height using the formula for the displacement under constant acceleration, considering that the final velocity at the highest point is zero and the acceleration due to gravity is -9.8 m/s² (negative as it is in the opposite direction to the ball's initial motion).
In a Young’s double-slit experiment, a thin sheet of mica is placed over one of the two slits. As a result, the center of the fringe pattern (on the screen) shifts by an amount corresponding to 30 dark bands. The wavelength of the light in this experiment is 480 nm and the index of the mica is 1.60. The mica thickness is:
Answer:0.024 mm
Explanation:
Given
Fringe shifts by an amount to 30 dark bands i.e. [tex]m=30[/tex]
Wavelength [tex]\lambda =480 nm[/tex]
refractive index of mica [tex]n_2=1.6[/tex]
refractive index of air [tex]n_1=1[/tex]
Phase difference is given by
[tex]m=\frac{t\left [ n_2-n_1\right ]}{\lambda }[/tex]
[tex]30=\frac{t\left [ 1.6-1\right ]}{480\times 10^{-9}}[/tex]
[tex]t=\frac{30\times 480\times 10^{-9}}{1.6-1}[/tex]
[tex]t=0.024\ mm[/tex]
To determine the thickness of the mica in a Young's double-slit experiment, we can use the formula d = mλ / sinθ, where d is the thickness of the mica, λ is the wavelength of the light, θ is the angle of the shift in the fringe pattern, and m is the number of dark bands shifted. By substituting the given values into the formula and solving for d, we can calculate the thickness of the mica.
Explanation:In a Young’s double-slit experiment, the shift in the center of the fringe pattern on the screen is caused by the introduction of a thin sheet of mica over one of the slits. This shift corresponds to 30 dark bands. To determine the thickness of the mica, we can use the formula for double-slit interference:
d sinθ = mλ,
where d is the distance between the slits, θ is the angle of the shift in the fringe pattern, m is the number of dark bands shifted, and λ is the wavelength of the light. Rearranging the formula, we have:
d = mλ / sinθ.
Given that the wavelength of the light is 480 nm and the index of refraction for the mica is 1.60, we can calculate the thickness of the mica by substituting the values into the formula and solving for d.
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Energy degradation takes place in the energy transformations which occur in the generation of electrical power. Explain what is meant in this context by energy degradation.
Answer:
The degradation of energy is the loss of useful energy: energy is conserved in changes, but tends to be transformed into thermal energy, which is a less usable form of energy.
Explanation:
Some forms of energy can be transformed into others. In these transformations, energy degrades, loses quality. In any transformation, part of the energy is converted into heat or heat energy.
Any type of energy can be completely transformed into heat; but, this cannot be completely transformed into another type of energy. It is said, then, that heat is a degraded form of energy. Are examples:
- The electrical energy, when going through a resistance.
- Chemical energy, in the combustion of some substances.
- Mechanical energy, by shock or friction.
Therefore, the Yield is defined as the ratio (in% percent) between the useful energy obtained and the energy contributed in a transformation.
R = ( useful energy / total energy) * 100
where R is the yield
Final answer:
Energy degradation during the generation of electrical power reflects the decrease in usability of energy as it is transformed, with much of it becoming waste heat. This affects the efficiency of power systems, indicating that more energy must be input to produce desired effects.
Explanation:
In the context of generating electrical power, energy degradation refers to the decrease in the quality or usefulness of energy as it undergoes various transformations. Although energy is conserved according to the first law of thermodynamics, during transformations, some energy is inevitably converted to forms that are less useful for doing work, primarily heat energy.
The phenomenon of energy degradation is a fundamental concept in thermodynamics and has significant implications for the efficiency of power generation systems. It points to the reality that while energy is not destroyed, its capacity to do work diminishes, requiring more energy inputs to achieve the desired outputs in practical applications.
A mxiture of n2 and H2 has mole fraction of 0.4 and 0.6 respectively. Determine the density of the mixture at one bar and 0 c.
Answer:
The density of the mixture is 0.55kg/m^3
Explanation:
P = 1bar = 100kN/m^2, T = 0°C = 273K, n = 0.4+0.6 = 1mole
PV = nRT
V = nRT/P = 1×8.314×273/100 = 22.70m^3
Mass of N2 = 0.4×28 = 11.2kg
Mass of H2 = 0.6×2 = 1.2kg
Mass of mixture = 11.2 + 1.2 = 12.4kg
Density of mixture = mass/volume = 12.4/22.7 = 0.55kg/m^3
A person is straining to lift a large crate, without success because it is too heavy. We denote the forces on the crate as follows: P is the upward force the person exerts on the crate, C is the vertical contact force exerted on the crate by the floor, and W is the weight of the crate. How are the magnitudes of these forces related while the person is trying unsuccessfully to lift the crate?
Answer:
P + C = W if in equilibrium?
P + C - W = 0; P + C = W
Explanation:
Action and reaction are equal and opposite according to newton's third law of motion. a force is that which tends to change a body's state of rest or uniform motion in a straight line, recall
P is the upward force the person exerts on the crate,
C is the vertical contact force exerted on the crate by the floor,
W is the weight of the crate
let the sum of upward forces be on the left hand side of the equation and the sum of downward forces on the right hand side.
P+C=W
P+C-W=0 if they are in equilibrium
the forces are related by the above.
When a person is unsuccessfully attempting to lift a crate, the forces acting on the crate are balanced. The sum of the upward force (P) applied by the person and the contact force (C) from the floor equals the weight (W) of the crate, represented by the equation P + C = W.
Explanation:In the situation described, the person is trying to lift a crate but is unable to do so. This means that the forces acting on the crate are in equilibrium, or balanced, as there is no resulting upward or downward motion of the crate. In this case, the force exerted by the person (P) and the vertical contact force from the floor (C) is opposing the weight of the crate (W).
In terms of magnitudes, the sum of P and C equals the weight W. This is based on the principle of equilibrium which states that an object at rest has equal and opposite forces acting on it. Therefore, P + C = W. This means the upward forces are equal to the downward force, hence the person can't lift the crate successfully.
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Marcie wants to work as a freelance editor. She purchases a desk, computer, widescreen monitor, and ergonomic keyboard. What kind of resources are these?
Answer:Capital Resources
Explanation:
Desk, computer, widescreen monitor, and ergonomic keyboard are an example of capital resources.
Capital resources are goods produced and used for the production of other goods and services. Basic items in capital goods are tools, machinery, and building. However, any good being utilized by a business to produce other good and services are under the category of capital goods.
Here Desk, computer and other items help Marcie to design her artwork as freelancer.
Two point sources, vibrating in phase, produce an interferencepattern in a ripple tank. If the frequency is increased by 20%, thenumber of nodal lines:
A) is increased by 20%
B) is increased by 40%
C) remains the same
D) is decreased by 20%
E) is decreased by 40%
Answer:
option A
Explanation:
given,
frequency is increased by 20%
we know,
[tex]\dfrac{x_n}{L}d = (n-\dfrac{1}{2})\times \lambda[/tex]...........(1)
where
x_n is the perpendicular distance between the point the interference pattern is obtained,
L is the distance between the center of the two point sources
and λ is the wavelength of light.
If the frequency is increased by 20%, then the number of nodal lines is increased by 20%.
From equation (1),we observe that the frequency is directly proportional to the number of nodal lines.
Hence, the correct answer is option A
dante is leading a parade across the main street in front of city hall. Starting at city hall, he marches the parade 4 blocks east, then 3 blocks south. From there, the parade marches 1 block west and 9 blocks north and finally stops. What is the vector displacement and direction of the the parade, starting from the city hall and the stopping point? (1 point)
Displacement: 6.71 m, Direction: 63.4 degrees north of east
Displacement: 8.01 m, Direction: 21.9 degrees north of east
Displacement: 2.56 m, Direction: 39.7 degrees north of east
Displacement: 4.31 m, Direction: 88.1 degrees north of east
Answer:
The correct option is A)
Displacement: 6.71 m, Direction: 63.4 degrees north of east
Explanation:
Given that Dante is leading a parade across the main street in front of city hall.
Let, Initial location of parade is 0i+0j
One block of city is one units on the XY- graph
Statement 1: Parade marches the parade 4 blocks east, then 3 blocks south
New location of parade is 4i-3j
Statement 2: The parade marches 1 block west and 9 blocks north and finally stops.
Final location of parade is (4i-3j)+(-1i+9j)=3i+6j
Displacement is given by
Displacement = (Final destination)-(Initial destination)
Displacement = (3i+6j)-(0i+0j)=3i+6j
Thus,
Magnitude of displacement = [tex]\sqrt{3^{2}+6^{2}}[/tex]
= 6.71 m
Direction of displacement = [tex]tan^{-1}(\frac{Y}{X} )[/tex]
= [tex]tan^{-1}(\frac{6}{3} )[/tex]
= 63.43 NE
Therefore, the correct option is A) Displacement: 6.71 m, Direction: 63.4 degrees north of east
A centripetal force of [tex]F_c[/tex] acts on a car going around a curve. If another car goes around the same curve but with twice the speed, the centripetal force acting on that car is ___________.
Answer:
The centripetal on the car will become 4 times when the velocity gets twice.
Explanation:
As we know that centripetal force on the car of mass m and moving with constant speed v given as
[tex]F_c=\dfrac{mv^2}{r}[/tex]
m=mass
v=velocity
r=radius of the circular arc
We are assuming that the mass of the both the car is same.
If the velocity of the car gets twice 2 v
The new centripetal force on the car
[tex]F_c'=\dfrac{m(2v)^2}{r}[/tex]
[tex]F_c'=4\dfrac{mv^2}{r}[/tex]
[tex]F_c'=4\ F_c[/tex]
Therefore we can say that centripetal on the car will become 4 times when the velocity gets twice.
A coin is placed 32 cm from the center of a horizontal turntable, initially at rest. The turntable then begins to rotate. When the speed of the coin is 110 cm/s (rotating at a constant rate), the coin just begins to slip. The acceleration of gravity is 980 cm/s^2 . What is the coefficient of static friction between the coin and the turntable?
Answer:
0.39
Explanation:
distance from the center (r) = 32 cm = 0.32 m
speed of the coin (v) = 110 cm/s = 1.1 m/s
acceleration due to gravity (g) = 980 m/s^{2} = 9.8 m/s^{2}
find the coefficient of static friction (k) between the coin and the turn table
frictional force = kmg
before the table begins to move, the frictional force balances the centripetal force ([tex]\frac{mv^{2} }{r}[/tex])
therefore
frictional force = centripetal force
kmg = [tex]\frac{mv^{2} }{r}[/tex]
kg = [tex]\frac{v^{2} }{r}[/tex]
k = [tex]\frac{v^{2} }{r}[/tex] ÷ g
k = [tex]\frac{1.1^{2} }{0.32}[/tex] ÷ 9.8 = 0.39
Rocks on either side of a strike-slip fault move past each other without much upward or downward movement is called___________.
Answer:
strike-slip fault
Explanation:
A fault is a crack on the crust of the Earth where forces on the rocks causes the displacement of the rocks.
The movement of the crust determines the type of fault.
When rocks slip pass horizontally with respect to each other then it is called a strike slip fault. The crust moves towards each other causing horizontal compression. The displacement of the rock takes place parallel to the horizontal force.
Hence, the statement here is referring to the strike-slip fault.
Which of the following types of light cannot be studied with telescopes on the ground?a. Visible light b. X-rays c. Radio waves d. All of the above
Answer:
Which of the following types of light cannot be studied with telescopes on the ground?
The Answer is X-rays
Explanation:
Earth’s atmosphere blocks most of the radiation from space preventing some electromagnetic spectrum from reaching the Earth because they are absorbed or reflected by the Earth's atmosphere. Visible light and radio waves get through to telescopes on the ground, while X-rays are absorbed by most molecules in the Earth’s atmosphere making it visible only from above the atmosphere.
Some Terms Explained:
Telescope: is an optical instrument that makes distant objects appear magnified. The purposes of a telescope are to gather light using either a lens or a mirror and resolve detail. There are two basic types of telescopes, refractors and reflectors.
Visible light: covers the range of wavelengths from 400–700 nanometers.
X-rays: range in wavelength from 0.001–10 nanometers. They are shorter in wavelength than Ultraviolet rays and longer than gamma rays. They will pass through most substances, and this makes them useful to see inside things.
Radio waves: the longest waves, longer than 1 meter they have the lowest energy. It is used for long distance communication.
Final answer:
X-rays cannot be studied with telescopes on the ground due to Earth's atmosphere blocking them. Space telescopes are needed for such observations.
Explanation:
The type of light that cannot be studied with telescopes on the ground is X-rays. Ground-based telescopes can effectively observe visible light and radio waves, but X-rays are mostly blocked by Earth's atmosphere. To study X-ray emissions, astronomers use space telescopes. Other ranges of the electromagnetic spectrum, such as ultraviolet and gamma rays, are similarly studied using telescopes positioned in space to bypass atmospheric interference.
One long wire lies along an x axis and carries a current of 43 A in the positive x direction. A second long wire is perpendicular to the xy plane, passes through the point (0, 5.9 m, 0), and carries a current of 41 A in the positive z direction. What is the magnitude of the resulting magnetic field at the point (0, 1.7 m, 0)?
Answer:
Bnet=1.006*10^-6T
Explanation:
One long wire lies along an x axis and carries a current of 43 A in the positive x direction. A second long wire is perpendicular to the xy plane, passes through the point (0, 5.9 m, 0), and carries a current of 41 A in the positive z direction. What is the magnitude of the resulting magnetic field at the point (0, 1.7 m, 0)?
the magnetic field Bnet=[tex]\sqrt{b1^2+b2^2}[/tex]
the magnetic field due this long wire is given by
B1=∨I1/[tex](2\pi *R1)[/tex]..............................1
B2=∨I2/[tex](2\pi *R2)[/tex]............................2
Bnet=[tex]\sqrt{(vI1/2*pi*R1)^2+(vI2/2*pi*R2)^2}[/tex].......................3
Bnet=v/2*pi[tex]\sqrt{(I1/R1)^2+(i2/R2)^2}[/tex]
Bnet=4*pi*10^-7/(2[tex]\pi[/tex])[tex]\sqrt{(43/1.7)^2+(41/29.5)^2}[/tex]
Bnet=0.0000002*(641.72)^.5
Bnet=1.006*10^-6T
Calculate the moment of inertia of a skater given the following information. (a) The 60.0-kg skater is approximated as a cylinder that has a 0.110-m radius. (b) The skater with arms extended is approximated by a cylinder that is 52.5 kg, has a 0.110-m radius, and has two 0.900-m-long arms which are 3.75 kg each and extend straight out from the cylinder like rods rotated about their ends.
Answer:
a) I = 0.363 kg*[tex]m^{2}[/tex]
b) [tex]I_{T}[/tex] = 0.82385 kg*[tex]m^{2}[/tex]
Explanation:
a) If we approximate the skater how a cylinder his moment of inertia is:
I = [tex]\frac{mr^{2} }{2}[/tex]
I = [tex]\frac{(60)(0.110)^{2} }{2}[/tex]
I = 0.363 kg*[tex]m^{2}[/tex]
b) If the skater has his arms extended then:
[tex]I_{T} = I_{B} + I_{A}[/tex]
where [tex]I_{B}[/tex]: Body’s moment of inertia
[tex]I_{A}[/tex]: Moment of inertia of the arms
[tex]I_{B}[/tex] = [tex]\frac{mr^{2} }{2}[/tex]
[tex]I_{B}[/tex] = [tex]\frac{(52.5)(0.110)^{2} }{2}[/tex] = 0.3176 kg*[tex]m^{2}[/tex]
[tex]I_{A}[/tex] = 2 * [tex]\frac{mL^{2} }{12}[/tex]
[tex]I_{A}[/tex] = 2 * [tex]\frac{(3.75)(0.9)^{2} }{12}[/tex] = 0.50625 kg*[tex]m^{2}[/tex]
[tex]I_{T}[/tex] = 0.3176 + 0.50625 = 0.82385 kg*[tex]m^{2}[/tex]
The moment of inertia for a skater can be calculated using the mass and radius in the formula for the moment of inertia of a cylinder. If the skater extends their arms, the moment of inertia increases, and this is calculated by adding the moment of inertia of the skater to the moments of inertia of the arms.
Explanation:The moment of inertia of a body is a measure of its resistance to rotational motion. It depends on the mass and how that mass is distributed relative to the axis of rotation. In the case of the skater (approximated as a cylinder), you calculate this using the formula for the moment of inertia of a cylinder, which is I=0.5MR², where M is the mass and R is the radius.
(a) For the skater without extended arms, substitute the given mass (60 kg) and radius (0.11 m) into the formula to get: I = 0.5 * 60 kg * (0.11 m)² = 0.363 kg-m².
(b) For the skater with extended arms, first, calculate the skater's body moment of inertia as before, but now with 52.5 kg mass. Secondly, calculate the moment of inertia of each arm (approximated as a rod rotating about one end) with the formula I=1/3mL², where m is the mass of an arm and L is the length. Add those values to get the total moment of inertia.
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Anti-lock brakes (ABS) __________
A. prevent skidding and allows drivers to steer during an emergency braking situation.
B. sense a skid and automatically apply the brakes.
C. stop the vehicle faster than normal brakes.
D. all of the above
Answer:
A. prevent skidding and allows drivers to steer during an emergency braking situation.
Explanation:
In an urgent stopping case, anti-lock brakes avoid skidding and allow drivers to steer. ABS can also help improve automobile balance (evitating spinouts), steering (that is directing the vehicle where the driver needs it to go) and braking (distance needed to stop the vehicle). Hence, the correct answer is A.
Anti-lock brakes (ABS) primarily prevent skidding and allow drivers to steer during an emergency braking situation. They do not necessarily stop the vehicle faster. The main benefit of ABS is that it enables steering while braking.
Explanation:The correct answer to your question is A. Anti-lock brakes (ABS) primarily prevent skidding and allow drivers to maintain steering control during an emergency braking situation. When the system senses a wheel is about to lock up and skid, it automatically modulates brake pressure to that wheel. This process repeats rapidly, often several times per second.
Note that ABS does not necessarily stop the vehicle faster. In fact, in certain conditions like gravel or snow, stopping distances can be longer. However, the crucial advantage is that ABS allows you to steer while braking, potentially avoiding a collision.
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A client experiences difficulty in performing the prone iso-abs exercise. Which of the following should be suggested as an regression?
a. Quadruped arm and opposite leg raise
b. Cable rotation
c. Rolling active resistance row
d. Cable chop
Answer:
a. Quadruped arm and opposite leg raise
Explanation:
Quadruped arm and opposite leg lift
- Kneel on the floor, lean forward and place your hands down.
- Keep your knees in line with your hips and hands directly under your shoulders.
- Simultaneously raise one arm and extend the opposite leg, so that they are in line with the spine.
- Go back to the starting position.
This method is usually used as an alternative to iso-abs exercise or also known as a bridge, which allows you to exercise the abdominal and spinal area at the same time.
It is also used together with other exercises for the treatment of hyperlordosis.
The suitable regression for a prone iso-abs exercise is the Quadruped arm and opposite leg raise.
The question asks for a regressed exercise alternative to the prone iso-abs exercise, which typically involves the individual maintaining a prone plank position, activating their abdominal core muscles without additional movement. When a client experiences difficulty with this exercise, an appropriate regression should reduce the demand on the core muscles while still enabling them to engage effectively.
The correct regression exercise from the options provided is a. Quadruped arm and opposite leg raise. This exercise involves the individual starting on their hands and knees (a quadruped position), then extending one arm and the opposite leg to create a straight line from fingertips to toes. This movement stabilises the core and has a reduced intensity compared to the prone iso-abs exercise.
The other options, b. Cable rotation, c. Rolling active resistance row, and d. Cable chop, are more dynamic exercises involving rotation and should not be confused with static core stabilisation exercises.