Answer: The temperature increases on the stratosphere with the altitude, given that absorption of the ultraviolet rays by the ozone.
Explanation: On the stratosphere, the water vapor and the umidity are almost nonexistents and, in view of the absorption of ultraviolet rays by the ozon, the temperature increases, reaching 35,6º Fahrenheit.
The ozone is a unusual type of oxygen molecule. In the stratosphere, the ozone appears on a large scale and warms it up by the absorption of the ultraviolet rays energy.
The temperature in the stratosphere increases with altitude due to the absorption of UV radiation by ozone, which leads to exothermic reactions that generate heat, especially at higher altitudes where ozone concentration is greater.
Explanation:The temperature increases with height in the stratosphere primarily because of the absorption of ultraviolet (UV) radiation by ozone. Ozone, a form of oxygen molecule, is very good at absorbing UV radiation from the Sun. When UV radiation is absorbed by ozone, it leads to exothermic chemical reactions which generate heat, thus raising the temperature of the surrounding atmosphere.
At higher altitudes in the stratosphere, the concentration of ozone is greater and, consequently, more UV radiation is absorbed, leading to a higher temperature in this region compared to the lower part of the stratosphere. It is this temperature inversion that distinguishes the stratosphere from the troposphere below, where the temperature normally decreases with altitude. The absorption of UV radiation and its conversion into heat makes the top of the stratosphere hotter as it is closer to the source of shortwave radiation, and the amount of absorbed shortwave radiation diminishes towards the bottom of the stratosphere.
A particle moving along a straight line is subjected to a deceleration ???? = (−2???? 3 ) m/???? 2 , where v is in m/s. If it has a velocity v = 10 m/s and a position s = 15 m when t = 0, determine its velocity and position when t = 25 s.
Answer:
(a). The velocity is 0.099 m/s.
(b). The position is 19.75 m.
Explanation:
Given that,
The deceleration is
[tex]a=(-2v^3)\ m/s^2[/tex]
We need to calculate the velocity at t = 25 s
The acceleration is the first derivative of velocity of the particle.
[tex]a=\dfrac{dv}{dt}[/tex]
[tex]\dfrac{dv}{dt}=-2v^3[/tex]
[tex]\dfrac{dv}{-v^3}=2dt[/tex]
On integrating
[tex]int{\dfrac{dv}{-v^3}}=\int{2dt}[/tex]
[tex]\dfrac{1}{2v^2}=2t+C[/tex]
[tex]v^2=\dfrac{1}{4t+2C}[/tex]....(I)
At t = 0, v = 10 m/s
[tex]10^2=\dfrac{1}{4\times0+2C}[/tex]
[tex]C=\dfrac{1}{200}[/tex]
Put the value of C in equation (I)
[tex]v^2=\dfrac{1}{4\times25+2\times\dfrac{1}{200}}[/tex]
[tex]v=\sqrt{\dfrac{1}{4\times25+2\times\dfrac{1}{200}}}[/tex]
[tex]v=0.099\ m/s[/tex]
The velocity is 0.099 m/s.
(b). We need to calculate the position at t = 25 sec
The velocity is the first derivative of position of the particle.
[tex]\dfrac{ds}{dt}=v[/tex]
On integrating
[tex]\int{ds}=\int(\sqrt{\dfrac{200}{800t+1}})dt[/tex]
[tex]s=\dfrac{\sqrt{200}\times2\sqrt{800t+1}}{800}+C'[/tex]
At t = 0, s = 15 m
[tex]15=\dfrac{200}{800}+C'[/tex]
[tex]C'=15-\dfrac{200}{800}[/tex]
[tex]C'=14.75[/tex]
Put the value in the equation
[tex]s=\dfrac{\sqrt{200}\times2\sqrt{800\times25+1}}{800}+14.75[/tex]
[tex]s=19.75\ m[/tex]
The position is 19.75 m.
Hence, (a). The velocity is 0.099 m/s.
(b). The position is 19.75 m.
The velocity of an object depends on its position.
Part A: The velocity of the moving particle at t = 25 s is 0.099 m/s.
Part B: The position of the particle at t = 25 s is 15.475 m.
What is velocity?Velocity is defined as the rate at which the position of an object changes.
Given that the deacceleration of the particle is [tex]a = (-2v^3)[/tex]. The velocity of the particle v = 10 m/s and a position s = 15 m when t = 0.
The acceleration of the particle is given below.
[tex]a = \dfrac {dv}{dt}[/tex]
[tex]-2v^3 = \dfrac {dv}{dt}[/tex]
[tex]- 2 dt = \dfrac {1}{v^3}dv[/tex]
Integrating the above equation, we get,
[tex]\int (-2 dt) = \int \dfrac {1}{v^3} dv[/tex]
[tex]2t + C = \dfrac {1}{2v^2}[/tex]
[tex]v^2 = \dfrac {1}{4t + 2C}[/tex]
Putting t=0 s, v = 10 m/s
[tex]100 = \dfrac {1}{4\times 0 + 2C}[/tex]
[tex]100 = \dfrac {1}{2C}[/tex]
[tex]C = 5 \times 10^{-3}[/tex]
Part A: Velocity
The velocity of the particle at t = 25 s is given as below.
[tex]v^2 = \dfrac { 1}{(4\times 25) + (2\times 5\times 10^{-3})}[/tex]
[tex]v^2 = \dfrac{1}{100.01}[/tex]
[tex]v = 0.099 \;\rm m/s[/tex]
Hence the velocity of the moving particle at t = 25 s is 0.099 m/s.
Part B: Position
The velocity of the particle is given as,
[tex]v = \dfrac {ds}{dt}[/tex]
[tex]v dt = ds[/tex]
Integrating the above equation, we get,
[tex]\int vdt = \int ds[/tex]
[tex]\int \sqrt{\dfrac {1}{4t + (\dfrac {1}{100})}} dt = s[/tex]
[tex]\int\sqrt{ {\dfrac {100}{400 t + 1}}}dt = s[/tex]
[tex]s =\dfrac {\sqrt{100} \times 2 \sqrt{400 t + 1}}{400} + C'[/tex]
Put t = 0 s, and s = 10 m,
[tex]15=\dfrac {\sqrt{100} \times 2 \sqrt{400 \times 0 + 1}}{400} + C'[/tex]
[tex]15 = \dfrac {\sqrt{100}\times 2}{400} + C'[/tex]
[tex]C' = 14.95[/tex]
Now the position at t = 25 s is,
[tex]s =\dfrac {\sqrt{100} \times 2 \sqrt{400 \times 25 + 1}}{400} + 14.95[/tex]
[tex]s =15.475 \;\rm m[/tex]
Hence the position of the particle at t = 25 s is 15.475 m.
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Suppose that a sound has initial intensity β1 measured in decibels. This sound now increases in intensity by a factor f. What is the new level of sound β2?
The new level of sound β2 = β1 + 10 log10(f).
Suppose that a sound has initial intensity level β1 measured in decibels.
This sound now increases in intensity level by a factor f. What is the new level of sound β2?
Formula to calculate the new sound intensity β2 = β1 + 10 log10(f)
The new level of sound β2 can be calculated using the formula β2 = β1 + 10 log10(f)
Here,β1 is the initial sound intensity levelβ2 is the new sound intensity level if is the factor by which the sound intensity level is increased by.
Sound intensity, also known as acoustic intensity, is defined as the power carried by sound waves per unit area in a direction perpendicular to that area.
The SI unit of intensity, which includes sound intensity, is the watt per square meter (W/m2)
Hence, The new level of sound β2 = β1 + 10 log10(f).
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The new level of the sound, β2, can be found by using the property of logarithms. The formula to calculate the new level is β2 = β1 + 10 log10(f). If one sound is twice as intense as another, it has a sound level about 3 dB higher.
Explanation:The new level of the sound, β2, can be found by using the property of logarithms. When the intensity of a sound increases by a factor of f, the sound level increases by 10 times the logarithm (base 10) of f. So, in this case, if the initial sound has intensity β1, the new level of sound β2 can be obtained by:
β2 = β1 + 10 log10(f)
For example, if the initial sound has intensity β1 = 60 dB and it increases by a factor of f = 2, then:
β2 = 60 + 10 log10(2) = 60 + 3 = 63 dB
What happens to the wavelength of a wave on a string when the frequency is doubled
Answer:
[tex]\lambda'=\dfrac{1}{2}\times \lambda[/tex]
Explanation:
The number of oscillation per unit time is called frequency of a wave while the distance between two consecutive crests and the trough is called its wavelength.
The wavelength of a wave is given by :
[tex]\lambda=\dfrac{v}{f}[/tex]
If frequency is doubled, f' = 2f
[tex]\lambda'=\dfrac{v}{f'}[/tex]
[tex]\lambda'=\dfrac{v}{2f}[/tex]
[tex]\lambda'=\dfrac{1}{2}\times \dfrac{v}{f}[/tex]
[tex]\lambda'=\dfrac{1}{2}\times \lambda[/tex]
So, when the frequency is doubled the wavelength of a wave on the string becomes half.
In afternoon, the decibel level of a busy freeway is 100 dB with 100 cars passing a given point every minute. Late at night, the traffic flow is only 20 cars per minute. What is the late-night decibel level? A. 20 dB b. 50 dB c. 80 dB d. 88 dB e. 93 dB
Answer:
option E
Explanation:
given,
decibel level of a busy freeway = 100 dB
when flow is = 100 car/minutes
traffic flow= 20 car/minute
Intensity factor =[tex]\dfrac{I}{I_0}[/tex]
=[tex]\dfrac{20}{100}[/tex]
=[tex]\dfrac{1}{5}[/tex]
[tex]\Delta \beta = 10 log\dfrac{I}{I_0}[/tex]
[tex]\Delta \beta = 10 log\dfrac{1}{5}[/tex]
[tex]\Delta \beta = - 6.989\ dB[/tex]
The late night decibel level
= 100 - 6.989
= 93
The correct answer option E
Two uniform spheres,each of mass 0.260kg are fixed at points A and B
A.)Find the magnitude of the initial acceleration of a uniform sphere with mass 0.010 kg if released from rest at point P and acted on only by forces of gravitational attraction of the spheres at A and B.
B.) Find the direction of the initial acceleration of a uniform sphere with mass 0.010 kg.
Complete Question:
Two uniform spheres,each of mass 0.260kg are fixed at points A and B
A.)Find the magnitude of the initial acceleration of a uniform sphere with mass 0.010 kg if released from rest at point P and acted on only by forces of gravitational attraction of the spheres at A and B.
B.) Find the direction of the initial acceleration of a uniform sphere with mass 0.010 kg.
Thunderstorms are often produced ahead of this type of front
Answer:
Cold front
Explanation:
The occurrence of cold front takes place when colder air at the surface pushes up the warmer into the atmosphere.
The movement of these cold fronts takes place at a faster pace as compared to the other fronts and are these are linked with the most violent types of weather such as severe thunderstorms are produced.
You watch distant Sally Homemaker driving nails into a front porch at a regular rate of 1 stroke per second. You hear the sound of the blows exactly synchronized with the blows you see. And then you hear one more blow after you see the hammering stop. Calculate the distance of Sally from you.
Answer:
343m
Explanation:
From the question Sally hits one blow every second
However, the sound is delayed by that same amount of time (1 sec)
If the speed of sound is 343m/s
Speed = Distance/Time
Speed = 343m/s
Time = 1s
Distance = Speed x Time
= 343 x 1 = 343m
An arsenic surface emits electrons after shining with a light source of the frequency of 5.32×1023 Hz. Calculate the maximum kinetic energy of the ejected electron if the work function of arsenic is 7.67×10.
Answer:
The maximum kinetic energy of the ejected electron=[tex]3.53\times 10^{-10} J[/tex]
Explanation:
We are given that
Frequency of light source=[tex]5.32\times 10^{23}[/tex] Hz
Work function of arsenic=[tex]7.67\times 10[/tex] eV
We have to find the maximum kinetic energy of ejected electron.
We know that the maximum kinetic energy of ejected electron
[tex]K.E=h\nu-w_o[/tex]
Where h=Plank's constant=[tex]6.63\times 10^{-34} J.s[/tex]
[tex]\nu[/tex] =Frequency of light source
[tex]w_o[/tex]=Work function
Substitute the values in the given formula
Then, the maximum kinetic energy of ejected electron
[tex]K.E=6.63\times 10^{-34}\times 5.32\times 10^{23}-7.67\times 10\times 1.6\times 10^{-19}[/tex]
Because 1 e V=[tex]1.6\times 10^{-19} J[/tex]
[tex]K.E=35.2716\times 10^{-11}-12.272\times 10^{-18}[/tex]
[tex]K.E=3.53\times 10^{-10}[/tex] J
Hence, the maximum kinetic energy of the ejected electron=[tex]3.53\times 10^{-10} J[/tex]
What factors determine whether a collision between two reactant molecules will result in a reaction?
Answer:
Minimum required energy
The correct orientation
Explanation:
A collision is described as successful or effective if the particles:
Collide with the right amount of energy. This is why, generally increasing the temperature of a reaction leads to an increased number of successful collisions. Higher temperature means a higher average kinetic energy. It is this minimum energy requirement (sometimes called activation energy) that must be met to achieve a successful collisionCollide in the correct orientation. Meeting the minimum energy requirement is not enough, particles must also collide in the correct orientation.The success of a collision between reactant molecules in causing a reaction depends on the molecules colliding with enough energy to overcome the activation energy, proper orientation during the collision, and a higher collision frequency influenced by reactant concentration.
Explanation:To answer your question about the factors that determine whether a collision between two reactant molecules results in a reaction, we need to consider several key factors. First, reactant molecules must collide with sufficient energy; this energy must be equal to or higher than the activation energy of the reaction. Second, the molecules must have proper orientation during the collision to allow the necessary rearrangement of electrons and atoms. Lastly, the collision frequency is also essential, which is influenced by the concentration of the reactants; higher concentration generally leads to more collisions and an increased likelihood of reaction.
The deformation of molecules during a collision can also be a contributing factor, as more violent collisions tend to lead to greater deformation and enhance the probability of reaction. Additionally, external factors such as temperature and the presence of catalysts can increase the kinetic energy of the molecules or reduce the activation energy, both of which can lead to a higher rate of successful collisions and reactions.
What will a spring scale read for the weight of a 58.0-kg woman in an elevator that moves in free fall?
The answer is 0N but could someone pls explain why?
Answer:
0 N
Explanation:
The elevator is under free fall. So, when a body is under free fall, the acceleration is only due to gravity. So, the acceleration of the elevator or the woman inside it, is acceleration due to gravity in the downward direction.
The spring scale gives the value of the normal force acting on the woman and doesn't give the exact weight of the woman. Under normal conditions, when the spring scale is at rest, then the upward normal force equals the weight and hence weight of a body is equal to the normal force acting on the body.
But, here, the body is not at rest. Weight\tex](mg)[/tex] acts in the downward direction and normal force[tex](N)[/tex] acts in the upward direction. The woman is moving down with acceleration equal to acceleration due to gravity[tex](g)[/tex]
So, we apply Newton's second law on the woman.
[tex]\textrm{Net force} = \textrm{mass}\times \textrm{acceleration}\\F_{net}=ma[/tex]
Net force is equal to the difference of the downward force and upward force.
[tex]F_{net}=mg-N[/tex]
Now, [tex]F_{net}=ma[/tex]
[tex]mg-N=mg\\N=mg-mg=0[/tex]
Therefore, the reading on the spring scale is 0 N.
At the top of a frictionless inclined plane, a 0.50-kilogram block of ice possesses 5.0 joules of gravitational poteRtial energy with respect to the bottom of the incline. After sliding halfway down the plane, the block's gravitational potential energy is _______.
The block of ice would have lost half of its gravitational potential energy after sliding halfway down the frictionless inclined plane. Therefore, its gravitational potential energy will be 2.5 Joules.
Explanation:The question is about the concept of energy conservation in physics, specifically the gravitational potential energy. The gravitational potential energy (U) of an object is given by the formula U = mgh, where 'm' is the mass of the object, 'g' is the acceleration due to gravity, and 'h' is the height from the reference point.
In this case, we know that the block of ice has a gravitational potential energy of 5.0 Joules when it's at the top of the incline. When the block of ice has slid halfway down the incline, it will have lost half of its height. Therefore, it would have lost half of its gravitational potential energy. Hence, after sliding halfway down the incline, the block's gravitational potential energy will be 2.5 Joules.
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A metal disk of radius 4.0 cm is mounted on a frictionless axle. Current can flow through the axle out along the disk, to a sliding contact on the rim of the disk. A uniform magnetic field B= 1.25 T is parallel to the axle of the disk. When the current is 5.5 A, the disk rotates with constant angular velocity. What's the frictional force at the rim between the stationary electrical contact and t?
Answer:
Friction force is 0.1375 N
Solution:
As per the question:
Radius of the metal disc, R = 4.0 cm = 0.04 m
Magnetic field, B = 1.25 T
Current, I = 5.5 A
Force on a current carrying conductor in a magnetic field is given by considering it on a differential element:
[tex]dF = IB\times dR[/tex]
Integrating the above eqn:
[tex]\int dF = IB\int_{0}^{R}dr[/tex]
[tex]\tau = IB\times \frac{R^{2}}{2} = \frac{1}{2}IBR^{2}[/tex] (1)
Now the torque is given by:
[tex]\tau = F\times R[/tex] (2)
From eqn (1) and (2):
[tex]F\times R = \frac{1}{2}IBR^{2}[/tex]
Thus the Frictional force is given by:
[tex]F = \frac{1}{2}\times 5.5\times 1.25\times 0.04 = 0.1375\ N[/tex]
A gas, behaving ideally, has a pressure P1 and at a volume V1. The pressure of the gas is changed to P2. Using Avogadro’s, Charles’, or Boyle’s law, develop an expression that would solve for the new volume V2.
Answer:
Boyle's Law
[tex]\therefore P_1.V_1=P_2.V_2[/tex]
Explanation:
Given that:
initially:
pressure of gas, [tex]= P_1[/tex]
volume of gas, [tex]= V_1[/tex]
finally:
pressure of gas, [tex]= P_2[/tex]
volume of gas, [tex]= V_2[/tex]
To solve for final volume [tex]V_2[/tex]
According to Avogadro’s law the volume of an ideal gas is directly proportional to the no. of moles of the gas under a constant temperature and pressure.
According to the Charles' law, at constant pressure the volume of a given mass of an ideal gas is directly proportional to its temperature.
But here we have a change in the pressure of the Gas so we cannot apply Avogadro’s law and Charles' law.
Here nothing is said about the temperature, so we consider the Boyle's Law which states that at constant temperature the volume of a given mass of an ideal gas is inversely proportional to its pressure.
Mathematically:
[tex]P_1\propto \frac{1}{V_1}[/tex]
[tex]\Rightarrow P_1.V_1=k\ \rm(constant)[/tex]
[tex]\therefore P_1.V_1=P_2.V_2[/tex]
Final answer:
To calculate the new volume V2 in Boyle's Law, use the formula V2 = P1V1/P2, ensuring that P1 and P2 are in the same pressure units and that V1 and V2 are in the same volume units.
Explanation:
To determine the new volume V2 after a change in pressure of an ideal gas, one can utilize Boyle's Law, which states that at a constant temperature, the pressure of a gas is inversely proportional to its volume. The formula representing Boyle's Law is P1V1 = P2V2. In this equation, P1 and V1 are the initial pressure and volume, and P2 and V2 are the final pressure and volume. To solve for V2, you simply rearrange the equation to V2 = P1V1/P2.
Remember that in order for Boyle's Law to be applied correctly, both volumes should be in the same unit. If you solve for V2 and find it to be greater than V1, it indicates that the pressure has decreased, consistent with the inverse proportionality.
What two kinds of crust are involved in a subduction zone
Answer:
Oceanic crust and continental crust
Explanation:
A subduction zone is normally between oceanic crust which is made of basalt and continental crust which is made of granite. Oceanic crust is denser than continental crust. So when oceanic crust collides with continental crusts, it subsducts underneath the continental crust since it is denser.
Final answer:
In subduction zones, the denser oceanic crust is pushed beneath another plate, which may be either oceanic or continental. This process results in the formation of ocean trenches and volcanic activity. Subduction zones play a crucial role in the cycle of plate tectonics.
Explanation:
The two kinds of crust involved in a subduction zone are the oceanic crust and the continental crust. In a subduction zone, one tectonic plate, usually the denser oceanic plate, is forced beneath another plate, which can be either oceanic or continental. Often it is the older and colder oceanic plate that subducts beneath the younger and warmer one. The subduction process is marked by features such as ocean trenches, volcanic island arcs, and is associated with powerful earthquakes and volcanic eruptions.
Subduction zones are integral to the Earth's plate tectonics cycle, balancing the creation of new crust at mid-ocean ridges by destroying the old crust as it descends into the mantle. Such areas of crust destruction are most commonly found around the edges of the Pacific Ocean. The descending plate undergoes changes due to high pressures and temperatures, leading to the release of water and the melting of the crust, forming magma that can result in volcanic activity.
Dan is gliding on his skateboard at 4.00 m/s . He suddenly jumps backward off the skateboard, kicking the skateboard forward at 8.00 m/s . Dan's mass is 50.0 kg and the skateboard's mass is 7.00 kg .
Answer:
v₂ = 3.44 m/s
Explanation:
given,
Dan is gliding on skateboard at = 4 m/s
speed of skateboard = 8 m/s
mass of Dan = 50 Kg
mass of skateboard = 7 Kg
Speed of Dan when his feet hit the ground = ?
Using conservation of momentum
PE(initial) = PE(final)
(M+m)V = m v₁ + M v₂
(50+7)x 4 = 7 x 8 + 50 x v₂
50 x v₂ =172
v₂ = 3.44 m/s
speed of Dan when his feet hit the ground = v₂ = 3.44 m/s
An escalator carries you from one level to the next in an airport terminal. The upper level is 4.4m above the lower level, and the length of the escalator is 7.3m . Assume that your mass is 63kg .Part AHow much work does the up escalator do on you when you ride it from the lower level to the upper level?Express your answer to two significant figures and include the appropriate units.W =___________.Part BHow much work does the down escalator do on you when you ride it from the upper level to the lower level?Express your answer to two significant figures and include the appropriate units.W =___________.
Answer:
Part a)
[tex]W = 2716.5 J[/tex]
Part B)
[tex]W = -2716.5 J[/tex]
Explanation:
Part A)
While escalator is moving up work done to move the person upwards is given as
[tex]W = mgh[/tex]
here we know that
m = 63 kg
h = 4.4 m
now we have
[tex]W = 63 \times 9.8 \times 4.4[/tex]
[tex]W = 2716.5 J[/tex]
Part B)
Work done while we move from up to down
So we have
[tex]W = - mgh[/tex]
so we have
[tex]W = -2716.5 J[/tex]
What experiments did thomas edison conduct with electricity
Answer:
Vacuum lamp experiment
Explanation:
Edison began his work on electricity the year 1878, where he made use of platinum wire to make filaments for his lamp as the melting point of Platinum is.
However, later he concluded that Platinum pores absorbed air while heating thus reducing its melting point.
So he came up with the idea of vacuum lamp and developed quite a good vacuum lamp with the help of carbon.
Being highly resistive in nature, carbon would end up burning at quite a rapid rate.
Along with his staff, in the year 1879, Edison conducted his first fruitful experiment by using vacuum lamp with carbon filament.
A steel plate has a hole in it with a diameter of exactly 1.0 cm when the plate is at a temperature of 20 degrees C. A steel ring has an inner diameter of exactly 1.0 cm at 20 degrees C. Both the plate and the ring are heated to 100 degrees C. Which statement is true?
a. the hole in the plate gets smaller, and the opening in the ring gets larger.
b. The opening in the ring gets larger, but we need the relative size of the plate and the hole to know what happens to the hole.
c. the hole in the plate and the opening in the ring get larger.
d. The hole in the plate and the opening in the ring get smaller.
e.The hole in the plate and the opening in the ring gets smaller.
Answer:
c. the hole in the plate and the opening in the ring get larger.
Explanation:
The hole in the plate and the ring both of them get enlarged.
The thermal expansion is very much similar to the optical expansion of the shadow of an object.
The change in length of an object is given by the relation:
[tex]\Delta l=l.\alpha.\Delta T[/tex]
where:
l = original length of the object
[tex]\alpha=[/tex] coefficient of linear expansion
ΔT = change in temperature
When heated from 20 degrees C to 100 degrees C, the diameter of the hole in a steel plate and the inner diameter of a steel ring both increase due to thermal expansion. So the correct option is C.
Explanation:The correct answer to the question about what happens to the diameter of a hole in a steel plate and the inner diameter of a steel ring when both are heated is that both will get larger. This is due to the thermal expansion of materials, which states that objects expand in all dimensions as their temperature increases. When the steel plate and ring are heated from 20 degrees C to 100 degrees C, the particles within the steel vibrate more and push each other farther apart, causing both the solid material and the empty space within, like the hole and the opening of the ring, to increase in size.
A high-wire walker always attempts to keep his center of mass (com) over the wire (or rope). He normally carries a long, heavy pole to help: If he leans, say, to his right (his com moves to the right) and is in danger of rotating around the wire, he moves the pole to his left (its com moves to the left) to slow the rotation and allow himself time to adjust his balance. Assume that the walker has a mass of 71.0 kg and a rotational inertia of 14.0 kg · m² about the wire. What is the magnitude of his angular acceleration about the wire if his com is 5.0 cm to the right of the wire under the following conditions?
(a) he carries no pole and
(b) the 14.0 kg pole he carries has its com 10 cm to the left of the wire?
Answer:
a)
2.5 rads⁻²
b)
1.5 rads⁻²
Explanation:
a)
[tex]R[/tex] = distance of the center of mass from the wire = 5 cm = 0.05 m
[tex]F_{g}[/tex] = Force of gravity on walker
[tex]M[/tex] = mass of walker = 71 kg
Force of gravity on walker is given as
[tex]F_{g} = Mg\\F_{g} = (71)(9.8)\\F_{g} = 695.8 N[/tex]
[tex]I[/tex] = Rotational inertia of walker = 14 kgm²
[tex]\alpha[/tex] = Angular acceleration about the wire
Torque equation is given as
[tex]F_{g} R = I \alpha \\(695.8) (0.05) = (14) \alpha \\\alpha = \frac{(695.8) (0.05)}{14} \\\alpha= 2.5 rads^{-2}[/tex]
b)
[tex]F_{p}[/tex] = Force of gravity on pole
[tex]m[/tex] = mass of walker = 14 kg
Force of gravity on pole is given as
[tex]F_{p} = mg\\F_{g} = (14)(9.8)\\F_{g} = 137.2 N[/tex]
[tex]r[/tex] = distance of the center of mass of pole from the wire = 10 cm = 0.10 m
Torque equation is given as
[tex]F_{g} R - F_{p} r = I \alpha \\(695.8) (0.05) - (137.2)(0.10) = (14) \alpha \\\alpha= \frac{21.07}{14} \\\alpha=1.5 rads^{-2}[/tex]
What temperature has the same value in both the fahrenheit and kelvin scales?
Answer:
574.25 °
Explanation:
To know this, we need to get the expressions to calculate °F and °K.
In the case of Kelvin:
°K = °C + 273
in fahrenheit:
°F = 9/5(°C + 32)
Now, in order to know the temperature which both Kelvin and Fahrenheit are the same, we need to use both above equations, and solve for °C.
°C = °K - 273 (1)
°C = 5/9(°F - 32) (2)
Using 1 and 2 into a same expression:
°K - 273 = 5/9(°F - 32)
With this, we need to know now the moment which K = F, so, all we need to do is replace the F for the K in the above expression. Doing this, we have:
°K - 273 = 5/9(°K - 32)
°K = 0.555(°K - 32) + 273
°K = 5/9K - 17.78 + 273
°K - 5/9K = 255.22
4/9°K = 255.22
°K = 255.22 * 9 / 4
°K = 574.25 °
And this is the temperature in which both Kelvin and Fahrenheit are the same.
An electroscope is a fork-shaped device commonly used to detect the presence of charge. The tin leaves of an electroscope will separate when charged negatively. Over time, however, these leaves will cease separating, even without being touched by another object. Explain why this may occur.
Answer:
All these is caused by the repulsion force.
Explanation:
The electroscope produces a series of electric charges that produce a repulsion force when is putted in contact with a electric charged object.
As the physics law mentions, two different forces are repealed, the electrocospe is charged negatively and the object positively, causing a repulsion force that avoids that both objects touch the other.
The height of the tide measured at a seaside community varies according to the number of hours t after midnight. If the height h, in feet, is currently given by the equation h=-1/2 t^2+6t-9, when will the tide first be at 6 ft?
Explanation:
Given that, the height of the tide measured at a seaside community varies according to the number of hours t after midnight. The height is given by the equation as :
[tex]h=-\dfrac{1}{2}t^2+6t-9[/tex]
When the tide first be at 6 ft, put h = 6 ft in above equation as :
[tex]-\dfrac{1}{2}t^2+6t-9=6[/tex]
[tex]-t^2+12t-18=0[/tex]
On solving the above equation to find the value of t. It is equal to :
t = 3.551 seconds
or
t = 8.449 seconds
So, the tide of 6 ft is at 3.551 seconds and 8.449 seconds. Hence, this is the required solution.
A bicycle tire is spinning counterclockwise at 2.70 rad/s. During a time period At 1.90 s, the tire is stopped and spun in the opposite (clockwise) direction also at 2.70 rad/s. Calculate the change in the tire's angular velocity Aw and the tire's average angular acceleration aav. (Indicate the direction with the signs of your answers.)
(a) the change in the tire's angular velocity Aco (in rad/s) rad/s
(b) the tire's average angular acceleration aav (in rad/s2) rad/s
The change in angular velocity of the bicycle tire is -5.40 rad/s, indicating a change in direction. The average angular acceleration during this time period was -2.84 rad/s2, also reflected by the negative sign representing a change in direction.
Explanation:This question pertains to physics, specifically focusing on angular velocity and angular acceleration. For the bicycle tire spinning counterclockwise at 2.70 rad/s to stop and spin in the clockwise direction at the same angular velocity, its total change in angular velocity (∆ω) is a combination of the initial and final angular velocities, given by ∆ω = ω_final - ω_initial. Since the tire goes from spinning counterclockwise (considered positive) to clockwise (considered negative), ∆ω = (-2.70 rad/s) - (2.70 rad/s) = -5.40 rad/s.
Next, to find the average angular acceleration (∆av), we use the formula ∆av = ∆ω / ∆t. Substituting in the time given (1.90 s) along with our calculated ∆ω, ∆av = -5.40 rad/s / 1.90 s = -2.84 rad/s2. Note that the negative sign indicates a change in direction.
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A force in the +x-direction with magnitude ????(x) = 18.0 N − (0.530 N/m)x is applied to a 6.00 kg box that is sitting on the horizontal, frictionless surface of a frozen lake. ????(x) is the only horizontal force on the box. If the box is initially at rest at x = 0, what is its speed after it has traveled 14.0 m?
Answer:
[tex]v_f=8.17\frac{m}{s}[/tex]
Explanation:
First, we calculate the work done by this force after the box traveled 14 m, which is given by:
[tex]W=\int\limits^{x_f}_{x_0} {F(x)} \, dx \\W=\int\limits^{14}_{0} ({18N-0.530\frac{N}{m}x}) \, dx\\W=[(18N)x-(0.530\frac{N}{m})\frac{x^2}{2}]^{14}_{0}\\W=(18N)14m-(0.530\frac{N}{m})\frac{(14m)^2}{2}-(18N)0+(0.530\frac{N}{m})\frac{0^2}{2}\\W=252N\cdot m-52N\cdot m\\W=200N\cdot m[/tex]
Since we have a frictionless surface, according to the the work–energy principle, the work done by all forces acting on a particle equals the change in the kinetic energy of the particle, that is:
[tex]W=\Delta K\\W=K_f-K_i\\W=\frac{mv_f^2}{2}-\frac{mv_i^2}{2}[/tex]
The box is initially at rest, so [tex]v_i=0[/tex]. Solving for [tex]v_f[/tex]:
[tex]v_f=\sqrt{\frac{2W}{m}}\\v_f=\sqrt{\frac{2(200N\cdot m)}{6kg}}\\v_f=\sqrt{66.67\frac{m^2}{s^2}}\\v_f=8.17\frac{m}{s}[/tex]
A hand pump is being used to inflate a bicycle tire that has a gauge pressure of 41.0 lb/in2. If the pump is a cylinder of length 16.8 in. with a cross-sectional area of 3.00 in.2, how far down must the piston be pushed before air will flow into the tire? Assume the temperature is constant.
Answer:
L - h = 12.3672 in
Explanation:
Given
P = 41.0 lb/in² = 41 P.S.I
L = 16.8 in
A = 3.00 in²
h = ?
In order that air flows into the tire, the pressure in the pump must be more than the tire pressure, 41.0 PSI.
We assume that air follows ideal gas equation, the temperature of the compressed air remains constant as the piston moves down. Taking one atmospheric pressure to be 14.6959 P.S.I , we can use the ideal gas equation
P*V = n*R*T
As number of moles of air do not change during its compression in the pump, n*R*T of the gas equation is constant. Therefore we have
P₁*V₁ = P₂*V₂ ⇒ V₂ = P₁*V₁ / P₂
where
1 and 2 are initial and final states respectively,
V₁ = A*L = (3.00 in²)*(16.8 in) ⇒ V₁ = 50.4 in³
P₁ = 14.6959 P.S.I
P₂ = P₁ + P = (14.6959 lb/in²) + (41.0 lb/in²) = 55.6959 lb/in²
Inserting various values we get
V₂ = (14.6959 P.S.I)*(50.4 in³) / (55.6959 lb/in²)
⇒ V₂ = 13.2985 in³
Length of pump, measured from bottom, this volume corresponds to is
h = V₂ / A = (13.2985 in³) / (3.00 in²)
⇒ h = 4.4328 in
Piston must be pushed down by more than
L - h = 16.8 in - 4.4328 in = 12.3672 in
Let r denote the distance between the center of the earth and the center of the moon. What is the magnitude of the acceleration ae of the earth due to the gravitational pull of the moon?
The magnitude of the acceleration (ae) of the Earth due to the gravitational pull of the Moon can be calculated using Newton's law of universal gravitation.
Explanation:The magnitude of the acceleration (ae) of the Earth due to the gravitational pull of the Moon can be calculated using Newton's law of universal gravitation. The formula is F = G × (m1 × m2) / r^2, where F is the gravitational force, G is the gravitational constant, m1 and m2 are the masses of the objects, and r is the distance between their centers. In this case, the Earth is the object experiencing the acceleration and the Moon is the object exerting the gravitational force.
Given the distance between the center of the Earth and the center of the Moon as r = 4700 km (or 4.7 x 10^6 m) and the gravitational constant F = G × (m1 × m2) / r^2[tex]F = G × (m1 × m2) / r^2[/tex], the formula becomes:
[tex]F = G × (m1 × m2) / r^2[/tex]
Since we know the mass of the Earth, we can calculate the acceleration by rearranging the formula to solve for ae:
ae = F / m1
Substituting the values, we get:
[tex]ae = (G × (m1 × m2) / r^2) / m1[/tex]
Simplifying the equation, we are left with:
[tex]ae = G × m2 / r^2[/tex]
Now, we can plug in the known values. The mass of the Moon (m2) is approximately 7.35 x 10^22 kg:
[tex]ae = (6.67 x 10^-11 N · m^2/kg^2) × (7.35 x 10^22 kg) / (4.7 x 10^6 m)^2[/tex]
Calculating this expression will give us the value for the magnitude of the acceleration of the Earth due to the gravitational pull of the Moon.
A mass is oscillating on the end of a spring. The distance, y, of the mass from its equilibrium point is given by the formula y=4zcos(8πwt) where y is in centimeters, t is time in seconds, and z and w are positive constants.
a. What is the furthest distance of the mass from its equilibrium point?
b. How many oscillations are completed in 1 second?
Answer:
(a.) 4z
(b.) 4w
Explanation:
From the equation y=4zcos(8πwt), where z and w are positive constants.
Comparing this equation to the equation of a wave y = Acos(Wt), where A is the amplitude (largest distance from equilibrium) and W is the angular frequency (W=2πf)
(a.) Comparing our wave equation with the given equation, we see that A = 4z in this case (furthest distance of the mass from equilibrium)
(b.) Comparing similarly we can see from our given equation that angular frequency W =8πw we also know that W = 2πf from our wave equation, therefore 2πf = 8πw
Solving for f we have f = 8πw÷2π
f = 4w (Proves our second answer because the frequency is the number of oscillations completed per second)
The angle between the two force of magnitude 20N and 15N is 60 degrees (20N force being horizontal) determine the resultant in magnitude and direction
Answer:
The magnitude of the resultant is 30.4 N.
The resultant angle direction is 25.3°.
Explanation:
To find the resultant of the magnitude and direction for given forces “P” and “Q” are 20 N and 15 N respectively, the angle (θ) between them is 60°.
We know that from triangle law of forces,
[tex]R=\sqrt{P^{2}+2 P Q \cos \theta+Q^{2}}[/tex]
Substitute the given values in the above formula,
[tex]R=\sqrt{20^{2}+2 (20)(15) Q \cos 60+15^{2}}[/tex]
[tex]R=\sqrt{400+600(0.5) + 225}[/tex]
[tex]R=\sqrt{400+300 + 225}[/tex]
[tex]R=\sqrt{925}[/tex]
R = 30.4 N
The magnitude of the resultant is 30.4 N.
To find the direction of the resultant we know that [tex]\text {Resultant angle}=\tan ^{-1} \frac{Q \sin \theta}{P+Q \cos \theta}[/tex]
Substitute the given values in the above formula,
[tex]\text {Resultant angle}=\tan ^{-1} \frac{15 \sin 60}{20+15 \cos 60}[/tex]
[tex]\text {Resultant angle}=\tan ^{-1} \frac{12.99}{20+7.5}[/tex]
[tex]\text {Resultant angle}=\tan ^{-1} \frac{12.99}{27.5}[/tex]
[tex]\text { Resultant angle }=\tan ^{-1} 0.472[/tex]
Resultant angle=25.3°
The resultant angle direction is 25.3°.
The intensity level is 65 db at a distance 5.00 m from a barking dog. What would be a reasonable estimate for the intensity level if two identical dogs very close to each other were barking? You can ignore any interference effects.
Answer:
68 db
Explanation:
Since now instead of one two dogs are barking simultaneously close to each other, therefore we take n =2.
Ignoring interference effects, the barking of two dogs result in a higher level of intensity which is given by,
β(db)=10×㏒(2)
=3 db
So, a reasonable estimate for the raised Intensity Level is: 65db+3db = 68db
The intensity level of two identical dogs barking together would be approximately 68 dB, as an increase in
sound intensity leads to an increase in the decibel level.
Explanation:The intensity level of a sound is determined by its amplitude and is measured in decibels (dB). The relationship between intensity and decibel level is logarithmic, which means that an increase of 10 dB corresponds to a tenfold increase in intensity. Hence, if two identical dogs are barking very close to each other their combined sound would be twice as intense as one dog barking alone. However, this does not mean that the decibel level would double. In fact, when the intensity of a sound is doubled, the increase in decibel level is about 3 dB. Thus, if a single dog barking has an intensity level of 65 dB, two dogs barking together would have an estimated intensity level of 68 dB.
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\If an object is placed at twice the focal length of a convex lens, where is the image located on the other side of the lens?
Question 2 options:
1 f
2 f
3 f
0.5 f
Answer:Image will be located at 2F on the other side of the lens.
Explanation:
Answer:
2 f
Explanation: