Answer:
Stoma represents the interface between the environment and the plant, helping to obtain the necessary CO2.
Explanation:
Stoma are groups of two or more specialized epidermal cells whose function is to regulate gas exchange and perspiration.
The frequency or density varies widely from a few tens to thousands per mm2, due to the influence of environmental factors, leaf morphology and genetic composition.
The 78 kg climber here is supported in the "chimney" by the friction forces exerted on her shoes and back. The static coefficients of friction between her shoes and the wall, and between her back and the wall, are 0.88 and 0.63, respectively. What is the minimum normal force he must exert? Assume the walls are vertical and that friction forces are both at a maximum.
Answer:
N = 516.56 N
Explanation:
By the means of a sum of forces on the x-axis:
[tex]N_b-N-f=0[/tex]
Where [tex]N_b[/tex] is the force on her back and [tex]N_f[/tex] is the force on her feet:
[tex]N_b=N-f = N[/tex]
On the y-axis:
[tex]Ff_b+Ff_f-m*g=0[/tex]
[tex]\mu_b*N_b+\mu_f*N_f-m*g=0[/tex]
[tex]\mu_b*N+\mu_f*N=m*g[/tex]
[tex](\mu_b+\mu_f)*N=m*g[/tex]
[tex]N=\frac{m*g}{\mu_b+\mu_f}[/tex] using [tex]g=10m/s^2[/tex]
N = 516.56N
A 600kg car is moving at 5 m/s to the right and elastically collided with a stationary 900 kg car. What is the velocity of the 900 kg car after the collision if the 600 kg car moves left at .714 m/s?
Answer:
[tex]\mathrm{v}_{2} \text { velocity after the collision is } 3.3 \mathrm{m} / \mathrm{s}[/tex]
Explanation:
It says “Momentum before the collision is equal to momentum after the collision.” Elastic Collision formula is applied to calculate the mass or velocity of the elastic bodies.
[tex]m_{1} v_{1}=m_{2} v_{2}[/tex]
[tex]\mathrm{m}_{1} \text { and } \mathrm{m}_{2} \text { are masses of the object }[/tex]
[tex]\mathrm{v}_{1} \text { velocity before the collision }[/tex]
[tex]\mathrm{v}_{2} \text { velocity after the collision }[/tex]
[tex]\mathrm{m}_{1}=600 \mathrm{kg}[/tex]
[tex]\mathrm{m}_{2}=900 \mathrm{kg}[/tex]
[tex]\text { Velocity before the collision } v_{1}=5 \mathrm{m} / \mathrm{s}[/tex]
[tex]600 \times 5=900 \times v_{2}[/tex]
[tex]3000=900 \times v_{2}[/tex]
[tex]\mathrm{v}_{2}=\frac{3000}{900}[/tex]
[tex]\mathrm{v}_{2}=3.3 \mathrm{m} / \mathrm{s}[/tex]
[tex]\mathrm{v}_{2} \text { velocity after the collision is } 3.3 \mathrm{m} / \mathrm{s}[/tex]
Final answer:
After calculating with the conservation of momentum for an elastic collision, the 900 kg car will have a velocity of 3.81 m/s to the right after it is hit by the 600 kg car.
Explanation:
The question asks about the post-collision velocity of a 900 kg car that was initially stationary and was hit by a 600 kg car moving at 5 m/s which, after the collision, moved left at 0.714 m/s. Using the principle of conservation of momentum for elastic collisions, we can set up the equation as follows:
Initial momentum = Final momentum
(600 kg × 5 m/s) + (900 kg × 0 m/s) = (600 kg × -0.714 m/s) + (900 kg × v)
Solving for v (the velocity of the 900 kg car after the collision), we obtain:
3000 kg·m/s = -428.4 kg·m/s + 900 kg × v
v = (3000 kg·m/s + 428.4 kg·m/s) / 900 kg
v = 3.81 m/s
Thus, the velocity of the 900 kg car after the collision is 3.81 m/s to the right.
For lunch you and your friends decide to stop at the nearest deli and have a sandwich made fresh for you with 0.100kg{\rm kg} of turkey. The slices of turkey are weighed on a plate of mass 0.400kg{\rm kg} placed atop a vertical spring of negligible mass and force constant of 200N/m{\rm N/m} . The slices of turkey are dropped on the plate all at the same time from a height of 0.250m{\rm m} . They make a totally inelastic collision with the plate and set the scale into vertical simple harmonic motion (SHM). You may assume that the collision time is extremely small.What is the amplitude of oscillations A of the scale after the slices of turkey land on the plate?
Answer:
0.02268 m
Explanation:
[tex]m_1[/tex] = Mass of turkey slices = 0.1 kg
[tex]m_2[/tex] = Mass of plate = 0.4 kg
[tex]u_1[/tex] = Initial Velocity of turkey slices = 0 m/s
[tex]u_2[/tex] = Initial Velocity of plate = 0 m/s
[tex]v_1[/tex] = Final Velocity of turkey slices
[tex]v_2[/tex] = Final Velocity of plate
k = Spring constant = 200 N/m
x = Compression of spring
g = Acceleration due to gravity = 9.81 m/s²
Equation of motion
[tex]v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times 9.81\times 0.25+0^2}\\\Rightarrow v=2.21472\ m/s[/tex]
The final velocity of the turkey slice is 2.21472 m/s = v₁
For the spring
[tex]x=\frac{m_1g}{k}\\\Rightarrow x=\frac{0.1\times 9.81}{200}\\\Rightarrow x=0.004905\ m[/tex]
As the linear momentum is conserved
[tex]m_1v_1=(m_1+m_2)v_2\\\Rightarrow v_2=\frac{m_1v_1}{m_1+m_2}\\\Rightarrow v_2=\frac{0.1\times 2.21472}{0.1+0.4}\\\Rightarrow v_2=0.442944\ m/s[/tex]
Here the kinetic and potential energy of the system is conserved
[tex]\frac{1}{2}(m_1+m_2)v_2^2+\frac{1}{2}kx^2=\frac{1}{2}kA^2\\\Rightarrow A=\sqrt{\frac{(m_1+m_2)v_2^2+kx^2}{k}}\\\Rightarrow A=\sqrt{\frac{(0.1+0.4)0.442944^2+200\times 0.004905^2}{200}}\\\Rightarrow A=0.02268\ m[/tex]
The amplitude of oscillations is 0.02268 m
A jet aircraft is traveling at 201 m/s in horizontal flight. The engine takes in air at a rate of 44.1 kg/s and burns fuel at a rate of 3.28 kg/s. The exhaust gases are ejected at 669 m/s relative to the aircraft. Find the thrust of the jet engine. Answer in units of N.
Answer:22,833.12 N
Explanation:
Given
Velocity of Jet [tex]v=201 m/s[/tex]
Velocity of Exhaust gases [tex]u=669 m/s[/tex]
Rate of intake of air [tex]\frac{\mathrm{d} M_a}{\mathrm{d} t}=44.1 kg/s[/tex]
Rate at which Fuel is burned is [tex]\frac{\mathrm{d} M_f}{\mathrm{d} t}=3.28 kg/s[/tex]
Rate of change of mass in Rocket [tex]\frac{\mathrm{d} M}{\mathrm{d} t}=\frac{\mathrm{d} M_a}{\mathrm{d} t}+\frac{\mathrm{d} M_f}{\mathrm{d} t}=44.1+3.28=47.38 kg/s[/tex]
Thrust on Rocket is given by
[tex]F=\frac{\mathrm{d} M}{\mathrm{d} t}u-\frac{\mathrm{d} M_a}{\mathrm{d} t}v[/tex]
[tex]F=47.38\times 669-44.1\times 201[/tex]
[tex]F=31,697.22-8864.1=22,833.12 N[/tex]
Two football players collide head-on in midair while trying to catch a thrown football and cling together. The first player is 94 kg and has an initial velocity of 6 m/s, while the second player is 116 kg and has an initial velocity of -5 m/sn the next three parts, (d) through (f), justify that friction could be ignored compared to the forces of collision by considering the change in momentum of the first player. Let's assume that the collision lasts for 10ms. Calculate the force on the first player by the second player during the collision
Final answer:
The combined velocity of the two football players just after they collide and cling together is calculated to be 0.80 m/s in the direction of the first player's original motion, using the law of conservation of momentum.
Explanation:
To determine the velocity just after impact when two football players collide and cling together, we can use the law of conservation of momentum. The initial momentum of the system is the sum of the momentum of each player before they collide. For the first player with a mass of 95.0 kg and velocity of 6.00 m/s, and the second player with a mass of 115 kg and velocity of -3.50 m/s, the total initial momentum is:
(95.0 kg × 6.00 m/s) + (115 kg × -3.50 m/s) = 570 kg·m/s - 402.5 kg·m/s = 167.5 kg·m/s.
After the collision, the two players cling together and thus have a combined mass of 95.0 kg + 115 kg = 210 kg. The final velocity of the two players clinging together can be found by dividing the total initial momentum by the combined mass:
Final velocity = Total initial momentum / Combined mass = 167.5 kg·m/s / 210 kg = 0.80 m/s.
Therefore, the combined velocity of the two football players just after the collision is 0.80 m/s in the direction of the first player's initial motion.
A thermosensory neuron in the skin converts heat energy to nerve impulses via a conversion called
Answer:
Sensory transduction
Explanation:
The term sensory transduction refers to the conversion process where the sensory energy is converted in order to change the potential of a membrane.
In other words, it can defined as the process of energy conversion such that stimulus can be transmitted or received by the sensory receptors and the nervous system may initiate with the sensory receptors.
Transduction takes in all of the five receptors of the body. Thus skin is also one of the receptors and hence conversion of heat energy into impulses takes place with the help of thermo-sensory neuron.
Part A What is the function of the nuclear pore complex found in eukaryotes? What is the function of the nuclear pore complex found in eukaryotes? It synthesizes the proteins required to copy DNA and make mRNA. It assembles ribosomes from raw materials that are synthesized in the nucleus. It selectively transports molecules out of the nucleus but prevents all inbound molecules from entering the nucleus. It regulates the movement of proteins and RNAs into and out of the nucleus.
Answer:
It regulates the movement of proteins and RNAs into and out of the nucleus
Explanation:
The nuclear pore complex are protein channels connecting the outer membrane of the nucleus to the inner membrane of the nucleus. They securely regulates the almost all of the transport of protein and RNAs into and out of the nucleus.
The nuclear pore complex in eukaryotes is a protein complex that regulates the transportation of molecules between the nucleus and cytoplasm. It selectively allows passage of specific molecules, including ions, proteins, and RNA. It is integral to cellular functioning and to maintaining the cell’s genetic stability.
Explanation:The nuclear pore complex found in eukaryotes regulates the traffic of molecules between the nucleus and cytoplasm. This rosette-shaped complex, composed of multiple proteins, allows for the selective passage of molecules such as ions, RNA, and proteins. Hence, it is crucial for the cell's functioning. For instance, RNA, which is created and spliced within the nucleus, needs to be transported to the cytoplasm for translation, and this transportation occurs through the nuclear pore complex. Moreover, the nuclear pore complex helps maintain the structure of the nucleus by allowing the passage of ions and molecules, ensuring the proper functioning of the nucleoplasm.
The composition of the nuclear pore complex makes it an efficient system for the transfer and regulation of certain molecules. Besides, it secures the cell's nucleus against unwanted, larger, or harmful substances that could potentially penetrate and damage the nucleus. Consequently, the nuclear pore complex contributes significantly to maintaining the cell's health and its genetic stability.
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You open the refrigerator in your room and put in a case of room-temperature root beer. After an hour, the root beer is ice cold. If your room air did not exchange any heat with the outdoor air during that time, the room air will be________.
Answer:
warmer
Explanation:
The law of conservation of energy tells us that energy cannot be created or destroyed, it can be transferred or converted from one from to another. In this question when the beer that is at room temperature is put in the fridge, it loses some heat energy. This heat energy is not destroyed, the fridge through multiple processes eventually releases this heat to the room through pipes at the back which is why they are normally warm. the heat from the food inside is expelled to the room. It is not lost.
Which wavelength produces fluorescence? Why do you think this wavelength produces fluorescence while the other does not?
Answer:
Long wavelength
Explanation:
Wavelengths that corresponds to the bands of blue and red are strongly absorbed whereas the wavelengths that lie in the mid-range corresponds to green light that are absorbed weakly.
Fluorescence produced is always directed towards longer wavelengths of the spectra as compared to the corresponding spectra for absorption.
A bird is flying with a speed of 18.6 m/s over water when it accidentally drops a 2.30 kg fish. The acceleration of gravity is 9.81 m/s 2 . If the altitude of the bird is 5.50 m and air resistance is disregarded, what is the speed of the fish when it hits the water?
Answer:21.3 m/s
Explanation:
Given
speed [tex]u=18.6 m/s[/tex]
mass of fish [tex]m_f=2.30 kg[/tex]
Altitude [tex]h=5.50 m[/tex]
Time taken to cover h
[tex]h=ut+\frac{at^2}{2}[/tex]
[tex]5.5=\frac{9.8\times t^2}{2}[/tex]
[tex]t^2=1.122[/tex]
[tex]t=1.05 s[/tex]
Vertical velocity after [tex]t=1.05 s[/tex]
[tex]v_y=0+gt[/tex]
[tex]v_y=9.8\times 1.05=10.38 m/s[/tex]
Horizontal velocity will remain same [tex]u=18.6 m/s[/tex]
Net velocity [tex]v_{net}=\sqrt{u^2+v_y^2}[/tex]
[tex]v_{net}=\sqrt{18.6^2+10.38^2}[/tex]
[tex]v_{net}=\sqrt{453.76}=21.30 m/s[/tex]
If this energy were used to vaporize water at 100.0 ∘C, how much water (in liters) could be vaporized? The enthalpy of vaporization of water at 100.0 ∘C is 40.7 kJ⋅mol−1. (Assume the density of water is 1.00 g/mL.)
Answer:
0.429 L of water
Explanation:
First to all, you are not putting the value of the energy given to vaporize water, so, to explain better this problem, I will assume a value of energy that I took in a similar exercise before, which is 970 kJ.
Now, assuming that the water density is 1 g/mL, this is the same as saying that 1 g of water = 1 mL of water
If this is true, then, we can assume that 1 kg of water = 1 L of water.
Knowing this, we have to use the expression to get energy which is:
Q = m * ΔH
Solving for m:
m = Q / ΔH
Now "m" is the mass, but in this case, the mass of water is the same as the volume, so it's not neccesary to do a unit conversion.
Before we begin with the calculation, we need to put the enthalpy of vaporization in the correct units, which would be in grams. To do that, we need the molar mass of water:
MM = 18 g/mol
The enthalpy in mass:
ΔH = 40.7 kJ/mol / 18 g/mol = 2.261 kJ/g
Finally, solving for m:
m = 970 / 2.261 = 429 g
Converting this into volume:
429 g = 429 mL
429 / 1000 = 0.429 L of water
When we are not putting the value of the energy given to vaporize water I will assume a value of energy that I took in a similar exercise before, which is 970 kJ.Now, we are assuming that the water density is 1 g/mL, this is the same as saying that 1 g of water = 1 mL of waterAlthough when If this is true, then, we can assume that 1 kg of water = 1 L of water.
Now, we have to use the expression to get energy which is:Then Q = m * ΔHSolving for is m:Now m = Q / ΔH Now after that "m" is the mass, but in this case, the mass of water is the same as the volume, also that it's not necessary to do unit conversion.Before that we begin with the calculation, then we need to put the enthalpy of vaporization in the correct units, also that which would be in grams. To do that, then we need the molar mass of water:
Then MM = 18 g/molThe enthalpy in mass:After that ΔH = 40.7 kJ/mol / 18 g/mol = 2.261 kJ/gFinally, solving for m:Then m = 970 / 2.261 = 429 gThen Converting this into volume:Now, 429 g = 429 mLThus, 429 / 1000 = 0.429 L of water
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Abnormal protrusion of the eye out of the orbit is known as
Answer:
Exophthalmos
Explanation:
Exophthalmos is a disorder which can be either bilateral or unilateral. Sometimes it is also known by other names like Exophthalmus, Excophthamia, Exobitism.
It is basically the bulging of eye anterior out of orbit which if left unattended may result in eye openings even while sleeping consequently resulting in comeal dryness and damage which ultimately may lead to blindness.
It is commonly caused by trauma or swelling of eye surrounding tissues resulting from trauma.
A crate is sitting in the center of a flatbed truck. As the truck accelerates to the east, the crate moves with it, not sliding on the bed of the truck. In what direction is the friction force exerted by the bed of the truck on the crate?
Answer:East
Explanation:
Given
The truck is accelerating towards east along with crate and crate is not sliding.
Friction Force on the crate will act towards the east as friction Force always opposes the motion of an object. Also in this case, if friction force is absent then crate would have moved backward.
Thus static Friction will help the crate to move with truck.
A car moves horizontally with a constant acceleration of 3 m/s2. A ball is suspended by astring from the ceiling of the car. The ball does not swing, being at rest with respect to thecar. What angle does the string make with the vertical?
Answer:
β = 16.7°
Explanation:
The sum of forces on the x-axis are:
[tex]T*sin\beta=m*a[/tex]
The sum of forces on the y-axis are:
[tex]T*cos\beta=m*g[/tex]
By dividing x-axis by the y-axis equation:
[tex]tan\beta=a/g[/tex]
Solving for β:
[tex]\beta=atan(a/g)[/tex]
β = 16.7°
You are standing 2.5 m directly in front of one of the two loudspeakers shown in the figure. They are 3.0 m apart and both are playing a 686 Hz tone in phase. Part A As you begin to walk directly away from the speaker, at what distances from the speaker do you hear a minimum sound intensity? The room temperature is 20 degrees C. Express your answer numerically using two significant figures. If there is more than one answer, enter your answers in ascending order separated by commas
Answer:
L = 3.8 m
Explanation:
As we know that the frequency of sound is given as
[tex]f = 686 Hz[/tex]
speed of the sound is given as
[tex]v = 332 + 0.6 t[/tex]
[tex]v = 332 + (0.6 \times 20)[/tex]
[tex]v = 344 m/s[/tex]
now we have wavelength of sound is given as
[tex]\lambda = \frac{v}{f}[/tex]
[tex]\lambda = \frac{344}{686}[/tex]
[tex]\lambda = 0.50 m[/tex]
now we have path difference at initial position given as
[tex]\Delta L = \sqrt{L^2 + d^2} - L[/tex]
[tex]\Delta L = \sqrt{3^2 + 2.5^2} - 2.5[/tex]
[tex]\Delta L = 3.9 - 2.5 = 1.4 m[/tex]
now we know that for minimum sound intensity we have
[tex]\Delta L = \frac{2N + 1}{2}\lambda[/tex]
[tex]\Delta L = \frac{2N + 1}{2}(0.50)[/tex]
so we have
N = 2
[tex]\Delta L = 1.25 m[/tex]
so we have
[tex]\sqrt{2.5^2 + L^2} - L = 1.25[/tex]
[tex]2.5^2 + L^2 = L^2 + 1.25^2 + 2.5L[/tex]
[tex]L = 1.875 m[/tex]
Now for N = 1
[tex]\Delta L = 0.75 m[/tex]
so we have
[tex]\sqrt{2.5^2 + L^2} - L = 0.75[/tex]
[tex]2.5^2 + L^2 = L^2 + 0.75^2 + 1.5L[/tex]
[tex]L = 3.8 m[/tex]
so the next minimum intensity will be at L = 3.8 m
Two speakers emit the same sound wave, identical frequency, wavelength, and amplitude. What other quantity would be necessary to determine if constructive or destructive interference occurs at a particular point some distance from the speakers?
Answer:
Phase Difference
Explanation:
When the sound waves have same wavelength, frequency and amplitude we just need the phase difference between them at a particular location to determine whether the waves are in constructive interference or destructive interference.
Interference is a phenomenon in which there is superposition of two coherent waves at a particular location in the medium of propagation.
When the waves are in constructive interference then we get a resultant wave of maximum amplitude and vice-versa in case of destructive interference.
For constructive interference the waves must have either no phase difference or a phase difference of nλ, where n is any natural number.For destructive interference the waves must have a phase difference of n×0.5λ, where n is any odd number.a person throws a rock at 3 M/s down over the edge of a very tall cliff on Earth how far will the rock have fallen in 4 seconds if the rock never hit the bottom?
The rock will be at 90.4 m from the top of the cliff.
Explanation:
The rock is thrown with the “initial velocity” 3 m/s. We need to find how much distance does the rock traveled in 4 seconds (t).
From the “kinematic equations” take
[tex]s=u t+\frac{1}{2} a t^{2}[/tex]
Where, “s” is distance traveled, “u” initial velocity of the object, “t” time the object traveled and “a” acceleration due to gravity is [tex]9.8 \mathrm{m} / \mathrm{s}^{2}.[/tex]
Substitute the given values in the above formula,
[tex]s=3 \times 4+\frac{1}{2} \times 9.8 \times 4^{2}[/tex]
[tex]s=12+\frac{1}{2} \times 9.8 \times 16[/tex]
[tex]s=12+\frac{1}{2} \times 156.8[/tex]
[tex]s=12+78.4[/tex]
[tex]s=90.4[/tex]
The rock is at height of 90.4 m from the top of the cliff.
The door is 3.00 m tall and 1.25 m wide, and it weighs 750 N . You can ignore the friction at the hinges. If Exena applies a force of 220 N at the edge of the door and perpendicular to it, how much time does it take her to close the door?
Answer:
0.674 s = t
Explanation:
Assuming that the door is completely open, exena need to rotate the door 90°.
Now, using the next equation:
T = I∝
Where T is the torque, I is the moment of inertia and ∝ is the angular aceleration.
Also, the torque could be calculated by:
T = Fd
where F is the force and d is the lever arm.
so:
T = 220N*1.25m
T = 275 N*m
Addittionaly, the moment of inertia of the door is calculated as:
I = [tex]\frac{1}{3}Ma^2[/tex]
where M is the mass of the door and a is the wide.
I =[tex]\frac{1}{3}(750/9.8)(1.25)^2[/tex]
I = 39.85 kg*m^2
Replacing in the first equation and solving for ∝, we get::
T = I∝
275 = 39.85∝
∝ = 6.9 rad/s
Now, the next equation give as a relation between θ (the angle that exena need to rotate) ∝ (the angular aceleration) and t (the time):
θ = [tex]\frac{1}{2}[/tex]∝[tex]t^2[/tex]
Replacing the values of θ and ∝ and solving for t, we get:
[tex]\sqrt{\frac{2(\pi/2)}{6.9 rad/s}}[/tex] = t
0.674 s = t
A block with mass m = 7.4 kg is attached to two springs with spring constants kleft = 31 N/m and kright = 53 N/m. The block is pulled a distance x = 0.27 m to the left of its equilibrium position and released from rest.
1)What is the magnitude of the net force on the block (the moment it is released)?
N
2)What is the effective spring constant of the two springs?
N/m
3)What is the period of oscillation of the block?
s
4)How long does it take the block to return to equilibrium for the first time?
s
5)What is the speed of the block as it passes through the equilibrium position?
m/s
6)What is the magnitude of the acceleration of the block as it passes through equilibrium?
m/s2
7)Where is the block located, relative to equilibrium, at a time 1.06 s after it is released? (if the block is left of equilibrium give the answer as a negative value; if the block is right of equilibrium give the answer as a positive value)
m
8)What is the net force on the block at this time 1.06 s? (a negative force is to the left; a positive force is to the right)
N
9)What is the total energy stored in the system?
J
10)If the block had been given an initial push, how would the period of oscillation change?
the period would increase
the period would decrease
the period would not change
I need help with this question please
The correct answers are as follows:
1) The magnitude of the net force on the block at the moment it is released is given by Hooke's Law for springs in parallel, which states that the net force is the sum of the forces exerted by each spring. Since the block is pulled to the left, the force exerted by the left spring is to the right, and the force exerted by the right spring is to the left. Thus, the net force[tex]\( F \)[/tex] is:
[tex]\[ F = k_{\text{left}} \cdot x + k_{\text{right}} \cdot x \] \[ F = (31 \, \text{N/m} + 53 \, \text{N/m}) \cdot 0.27 \, \text{m} \] \[ F = 84 \, \text{N/m} \cdot 0.27 \, \text{m} \] \[ F = 22.68 \, \text{N} \][/tex]2) The effective spring constant [tex]\( k_{\text{eff}} \)[/tex] of the two springs in parallel is the sum of the individual spring constants:
[tex]\[ k_{\text{eff}} = k_{\text{left}} + k_{\text{right}} \] \[ k_{\text{eff}} = 31 \, \text{N/m} + 53 \, \text{N/m} \] \[ k_{\text{eff}} = 84 \, \text{N/m} \][/tex]
3) The period of oscillation [tex]\( T \)[/tex] for a mass-spring system is given by:
[tex]\[ T = 2\pi \sqrt{\frac{m}{k_{\text{eff}}}} \] \[ T = 2\pi \sqrt{\frac{7.4 \, \text{kg}}{84 \, \text{N/m}}} \] \[ T = 2\pi \sqrt{\frac{7.4}{84}} \] \[ T = 2\pi \sqrt{0.0881} \] \[ T = 2\pi \cdot 0.2968 \] \[ T \approx 1.86 \, \text{s} \][/tex]
4) The time it takes for the block to return to equilibrium for the first time is half of the period of oscillation:
[tex]\[ t = \frac{T}{2} \] \[ t = \frac{1.86 \, \text{s}}{2} \] \[ t \approx 0.93 \, \text{s} \][/tex]
5) The speed of the block as it passes through the equilibrium position can be found using the conservation of energy. The total mechanical energy is constant, so the potential energy at the release point is converted into kinetic energy at the equilibrium position:
[tex]\[ \frac{1}{2} k_{\text{eff}} x^2 = \frac{1}{2} m v^2 \] \[ k_{\text{eff}} x^2 = m v^2 \] \[ v^2 = \frac{k_{\text{eff}} x^2}{m} \] \[ v = \sqrt{\frac{k_{\text{eff}} x^2}{m}} \] \[ v = \sqrt{\frac{84 \, \text{N/m} \cdot (0.27 \, \text{m})^2}{7.4 \, \text{kg}}} \] \[ v = \sqrt{\frac{84 \cdot 0.0729}{7.4}} \] \[ v = \sqrt{\frac{6.1296}{7.4}} \] \[ v \approx \sqrt{0.8284} \] \[ v \approx 0.909 \, \text{m/s} \][/tex]
6) The magnitude of the acceleration[tex]\( a \)[/tex]of the block as it passes through equilibrium is given by Newton's second law:
[tex]\[ F = m \cdot a \] \[ a = \frac{F}{m} \] \[ a = \frac{22.68 \, \text{N}}{7.4 \, \text{kg}} \] \[ a \approx 3.065 \, \text{m/s}^2 \][/tex]
7) The position[tex]\( x(t) \) of the block at a time \( t \)[/tex] after it is released can be found using the equation for simple harmonic motion: [tex]\[ x(t) = A \cos(2\pi f t) \] \[ x(t) = 0.27 \cos\left(\frac{2\pi}{1.86} \cdot 1.06\right) \] \[ x(t) = 0.27 \cos(3.61\pi) \] \[ x(t) \approx 0.27 \cdot (-1) \] \[ x(t) \approx -0.27 \, \text{m} \][/tex]
8) The net force on the block at time [tex]\( t \)[/tex] is given by Hooke's Law, taking into account the position of the block:
[tex]\[ F(t) = k_{\text{eff}} \cdot x(t) \] \[ F(t) = 84 \, \text{N/m} \cdot (-0.27 \, \text{m}) \] \[ F(t) \approx -22.68 \, \text{N} \][/tex]
9) The total energy stored in the system[tex]\( E \)[/tex] is the potential energy at the maximum displacement, which is equal to the kinetic energy at the equilibrium position:
[tex]\[ E = \frac{1}{2} k_{\text{eff}} x^2 \] \[ E = \frac{1}{2} \cdot 84 \, \text{N/m} \cdot (0.27 \, \text{m})^2 \] \[ E = 42 \, \text{N/m} \cdot 0.0729 \, \text{m}^2 \] \[ E \approx 3.065 \, \text{J} \][/tex]
10) The period of oscillation is independent of the amplitude of the motion and depends only on the mass and the spring constant. Therefore, if the block had been given an initial push, the period of oscillation would not change. The correct answer is: the period would not change.
You charge a parallel-plate capacitor, remove it from the battery, and prevent the wires connected to the plates from touching each other. When you pull the plates apart to a larger separation, do the following quantities increase, decrease, or stay the same?
a. C
b. Q
c. E between the plates
d. delta-V
Answer:
a. C will decrease
b. Q will remain the same
c. E will decrease
d. Delta-V will increase
Explanation:
Justification for C:
As we know that for parallel plate capacitors, capacitance is calculated using:
C = (ϵ_r * ϵ_o * A) / d - Say it Equation 1
Where:
ϵ_r - is the permittivity of the dielectric material between two plates
ϵ_o - Electric Constant
A - Area of capacitor's plates
d - distance between capacitor plates
From equation 1 it is clear that capacitance will decrease if distance between the plates will be increased.
Justification of Q
As charge will not be able to travel across the plates, therefore it will remain the same
Justification of E
As we know that E = Delta-V / Delta-d, thus considering Delta-V is increasing on increasing Delta-d (As justified below) as both of these are directly proportional to each other, therefore Electric field (E) will remain constant as capacitors' plates are being separated.
Moreover, as the E depends on charge density which remains same while plates of capacitor are being separated therefore E will remain the same.
Justification of Delta-V
As we know that Q = C * V, therefore considering charge remains the same on increasing distance between plates, voltage must increase to satisfy the equation.
f R = 12 cm, M = 360 g, and m = 70 g (below), find the speed of the block after it has descended 50 cm starting from rest. Solve the problem using energy conservation principles. (Treat the pulley as a uniform disk.) Solve the problem using energy conservation principles. (b) Repeat (a) with R = 5.0 cm.
Answer:
a)v= 1.6573 m/s
Explanation:
a) Considering center of the disc as our reference point. The potential energy as well as the kinetic energy are both zero.
let initially the block is at a distance h from the reference point.So its potential energy is -mgh as its initial KE is zero.
let the block descends from h to h'
During this descend
PE of the block = -mgh' {- sign indicates that the block is descending}
KE= 1/2 mv^2
rotation KE of the disc= 1/2Iω^2
Now applying the law of conservation of energy we have
[tex]-mgh = \frac{1}{2}mv^2+\frac{1}{2}I\omega^2-mgh'[/tex]
[tex]mg(h'-h) = \frac{1}{2}mv^2+\frac{1}{2}I\omega^2[/tex] ................i
Rotational inertia of the disc = [tex]\frac{1}{2}MR^2[/tex]
Angular speed ω =[tex]\frac{v}{R}[/tex]
by putting vales of ω and I we get
so, [tex]\frac{1}{2}I\omega^2= \frac{1}{4}Mv^2[/tex]
Now, put this value of rotational KE in the equation i
[tex]mg(h'-h) = \frac{1}{4}(2m+M)v^2[/tex]
⇒[tex]v= \sqrt{\frac{4mg(h'-h)}{2m+M} }[/tex]
Given that (h'-h)= 0.5 m M= 360 g m= 70 g
[tex]v= \sqrt{\frac{4\times70\times 9.81\times 0.5}{140+360} }[/tex]
v= 1.6573 m/s\
b) The rotational Kinetic energy of the disc is independent of its radius hence on changing the radius there is no change in speed of the block.
Answer:
The speed of the block is 1.65 m/s.
Explanation:
Given that,
Radius = 12 cm
Mass of pulley= 360 g
Mass of block = 70 g
Distance = 50 cm
(a). We need to calculate the speed
Using energy conservation
[tex]P.E=K.E[/tex]
[tex]P.E=mgh[/tex]
[tex]K.E=\dfrac{1}{2}mv^2+\dfrac{1}{2}I\omega^2[/tex]
[tex]K.E=\dfrac{1}{2}mv^2+\dfrac{1}{2}I\times(\dfrac{v}{r})^2[/tex]
[tex]K.E=\dfrac{1}{2}mv^2+\dfrac{1}{2}\times0.5Mr^2\times(\dfrac{v}{r})^2[/tex]
[tex]K.E=\dfrac{1}{2}mv^2+\dfrac{1}{2}\times0.5M\times v^2[/tex]
[tex] K.E=\dfrac{1}{2}v^2(m+0.5M)[/tex]
Put the value into the formula
[tex]mgh=\dfrac{1}{2}v^2(m+0.5M)[/tex]
[tex]v^2=\dfrac{2mgh}{m+0.5M}[/tex]
[tex]v=\sqrt{\dfrac{2mgh}{m+0.5M}}[/tex]
[tex]v=\sqrt{\dfrac{2\times70\times10^{-3}\times9.8\times50\times10^{-2}}{70\times10^{-3}+0.5\times360\times10^{-3}}}[/tex]
[tex]v=1.65\ m/s[/tex]
(b), We need to calculate the speed of the block
When r = 5.0 cm
Here, The speed of the block is independent of radius of pulley.
Hence, The speed of the block is 1.65 m/s.
An ideal refrigerator extracts 500 joules of heat from a reservoir at 295 K and rejects heat to a reservoir at 493 K. What is the ideal coefficient of performance and how much work is done in each cycle?
Answer:
C.O.P = 1.49
W = 335.57 joules
Explanation:
C.O.P = coefficient of performance = (benefit/cost) = Qc/W ...equ 1 where C.O.P is coefficient of performance, Qc is heat from cold reservoir, w is work done on refrigerator.
Qh = Qc + W...equ 2
W = Qh - Qc ...equ 3 where What is heat entering hot reservoir.
Substituting for W in equ 1
Qh/(Qh - Qc) = 1/((Qh /Qc) -1) ..equ 4
Since the second law states that entropy dumped into hot reservoir must be already as much as entropy absorbed from cold reservoir which gives us
(Qh/Th)>= (Qc/Tc)..equ 5
Cross multiple equ 5 to get
(Qh/Qc) = (Th/Tc)...equ 6
Sub equ 6 into equation 4
C.O.P = 1/((Th/Tc) -1)...equ7
Where Th is temp of hot reservoir = 493k and Tc is temp of cold reservoir = 295k
C.O.P = 1/((493/295) - 1)
C.O.P = 1.49
To solve for W= work done on every cycle
We substitute C.O.P into equ 1
Where Qc = 500 joules
1.49 = 500/W
W = 500/1.49
W = 335.57 joules
A wind turbine is initially spinning at a constant angular speed. As the wind's strength gradually increases, the turbine experiences a constant angular acceleration of . After making 2870 revolutions, its angular speed is 133 rad/s. (a) What is the initial angular velocity of the turbine? (b) How much time elapses while the turbine is speeding up?
:Answer:
a. the initial angular velocity = 166.5rad/s
b. t= 135.7seconds
Explanation:
a. v = rw
angular velocity for 2870 rev, w = (2πN)/60 =2*3.142*2870rev)/60 =300.5rad/sec
w₁-w₂ = 300.5 - 133= 166.5rad/s
the initial angular velocity = 166.5rad/s
acceleration = change in velocity / time
b. 1 revolution = 2π
1 rad = 0.159rev
? = 2870 rev
=2870*1)/0.159 = 18050.3rad
the final angular speed = rev/time = rad/time
133 rad/s = 18050.3/t
133t = 18050.3
t= 135.7seconds
A potter's wheel has the shape of a solid uniform disk of mass 7 kg and radius 0.65 m. It spins about an axis perpendicular to the disk at its center. A small 2.1 kg lump of very dense clay is dropped onto the wheel at a distance 0.41 m from the axis.
What is the moment of inertia of the system about the axis of spin?
Answer:
1.832 kgm^2
Explanation:
mass of potter's wheel, M = 7 kg
radius of wheel, R = 0.65 m
mass of clay, m = 2.1 kg
distance of clay from centre, r = 0.41 m
Moment of inertia = Moment of inertia of disc + moment f inertia of the clay
I = 1/2 MR^2 + mr^2
I = 0.5 x 7 x 0.65 x 0.65 + 2.1 x 0.41 x 0.41
I = 1.47875 + 0.353
I = 1.832 kgm^2
Thus, the moment of inertia is 1.832 kgm^2.
The moment of inertia of the system about the axis of spin is mathematically given as
I = 1.832 kgm^2
What is the moment of inertia of the system about the axis of spin?Question Parameter(s):
A potter's wheel has the shape of a solid uniform disk of mass of 7 kg
and a radius of 0.65 m
A small 2.1 kg lump of very dense clay
the wheel at a distance of 0.41 m from the axis.
Generally, the equation for the moment of inertia is mathematically given as
I = 1/2 MR^2 + mr^2
I = 0.5 x 7 (0.65)^2 + 2.1 (0.41)^2
I = 1.47875 + 0.353
I = 1.832 kgm^2
In conclusion moment of inertia is
I = 1.832 kgm^2
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An infinite plane of charge has a surface charge density of 5 µC/m2 . How far apart are the equipotential surfaces whose potentials differ by 105 V? The permittivity of free space is 8.85 × 10−12 C 2 /N · m2 . Answer in units of mm.
Answer:
Distance in mm will be 0.3718 mm
Explanation:
We have given charge surface charge density [tex]\rho _s=5\mu c/m^2=5\times 10^{-6}\mu c/m^2[/tex]
We know that electric field due to surface charge density is given by
[tex]E=\frac{\rho _S}{2\epsilon _0}=\frac{5\times 10^{-6}}{2\times 8.85\times 10^{-12}}=2.824\times 10^5Volt/m[/tex]
We have given potential difference V = 105 volt
We know that potential difference is given by [tex]V=Ed[/tex]
So [tex]105=2.824\times 10^5\times d[/tex]
[tex]d=37.181\times 10^{-5}m=0.3718mm[/tex]
Where does the heat come from that drives this convection current in the mantle
Answer:
Earth's interior (Core)
Explanation:
The earth is comprised of 3 distinct layers namely the Core, the Mantle and the Crust, which are divided based on their composition as well as density.
The core of the earth is extremely very hot where the inner core remains solid and outer core acts a liquid. It is mainly comprised of iron, nickel and other siderophile elements.
A large amount of heat (energy) is radiated from this core region towards the surface of the earth. Due to this, the mantle rocks forms magma that creates the convection currents, where the hot and less dense magma rises upward and the cool and denser magma sinks to the bottom. This occurs continuously, as a result of which the lithospheric plates are forced to move over the less dense layer of asthenosphere.
Thus, the heat energy that drives the convection current in the mantle is provided from the interior (core) of the earth.
Answer:
earths core
Explanation:
A 3.00-kg rifle fires a 0.00500-kg bullet at a speed of 300 m/s. Which force is greater in magnitude:(i) the force that the rifleexerts on the bullet; or (ii) the force that the bulletexerts on the rifle?A. the force that the rifle exerts on the bulletB. the force that the bullet exerts on the rifleC. both forces have the same magnitudeD. not enough information given to decide
Answer:
C. both forces have the same magnitude
Explanation:
Here the action force is equal to the reaction force in accordance with the Newton's third law of motion.
Also when we apply the conservation of momentum so that the momentum bullet and the momentum of the gun are equal and according to the second law of motion by Newton, we have force equal to the rate of change in momentum.
We have the equation for momentum as:
[tex]p=m.v[/tex]
Newton's second law is Mathematically given as:
[tex]F=\frac{dp}{dt}[/tex]
Momentum is constant and the reaction time is equal, so the force exerted will also be equal.
The correct answer is C. both forces have the same magnitude.
According to Newton's third law of motion, for every action, there is an equal and opposite reaction. This means that when the rifle fires the bullet, the force exerted by the rifle on the bullet is equal in magnitude to the force exerted by the bullet on the rifle.
Let's denote the force exerted by the rifle on the bullet as [tex]\( F_{rb} \)[/tex]and the force exerted by the bullet on the rifle as[tex]\( F_{br} \).[/tex] According to Newton's third law:
[tex]\[ F_{rb} = -F_{br} \][/tex]
The magnitudes of these forces are equal, even though they are in opposite directions.
To calculate the magnitude of these forces, we can use the impulse-momentum theorem, which states that the change in momentum of an object is equal to the impulse applied to it.
Before the rifle is fired, both the bullet and the rifle are at rest, so their initial momenta are zero. After the rifle is fired, the bullet has a velocity of 300 m/s, and we can assume the rifle has a much smaller recoil velocity due to its much larger mass.
The change in momentum of the bullet is:
[tex]\[ \Delta p_{bullet} = m_{bullet} \times v_{bullet} \][/tex]
[tex]\[ \Delta p_{bullet} = 0.00500 \, \text{kg} \times 300 \, \text{m/s} \][/tex]
[tex]\[ \Delta p_{bullet} = 1.5 \, \text{kg} \cdot \text{m/s} \][/tex]
The change in momentum of the rifle is equal in magnitude and opposite in direction to the change in momentum of the bullet, assuming no other forces are acting on the system (like air resistance or friction):
[tex]\[ \Delta p_{rifle} = -\Delta p_{bullet} \][/tex]
[tex]\[ \Delta p_{rifle} = -1.5 \, \text{kg} \cdot \text{m/s} \][/tex]
Since the time interval [tex]\( \Delta t \)[/tex] over which the forces are applied is the same for both the bullet and the rifle, the forces can be calculated using the impulse-momentum theorem:
[tex]\[ F_{rb} = \frac{\Delta p_{bullet}}{\Delta t} \][/tex]
[tex]\[ F_{br} = \frac{\Delta p_{rifle}}{\Delta t} \][/tex]
Since [tex]\( \Delta p_{bullet} = -\Delta p_{rifle} \)[/tex], it follows that:
[tex]\[ F_{rb} = -F_{br} \][/tex]
Therefore, the forces are equal in magnitude and opposite in direction, which is consistent with Newton's third law. The correct choice is C, both forces have the same magnitude.
Determine the COP of a heat pump that supplies energyto a house at a rate of 8000 kJ/h for each kW of electric power it draws. Also, determine the rate of energy absorption from the outdoor air.
The Coefficient of Performance (COP) of the heat pump is 2.22 and the rate at which it absorbs energy from the outdoor air is 1222 Watts.
Explanation:The quality of a heat pump is judged by how much energy is transferred by heat into the warm space compared with how much input work is required. This measure is referred to as the Coefficient of Performance (COP). To calculate the COP of a heat pump which supplies energy at a rate of 8000 kJ/h for each kW of electrical power it draws, we have to convert all the units to the same base, which in this case will be watts (W).
1 kW = 1000 W and 8000 kJ/h = (8000*1000) J/3600 s = 2222 W
Hence, using the formula for the COP of the heat pump: COPhp = Qh/W, we substitute the given values and we get: COPhp = 2222W/1000W = 2.22. This means that for every 1 Watt of electricity the heat pump uses, it generates 2.22 Watts of heat for the house.
Additionally, the rate of energy absorption from the outdoor air is the difference between the rate of heat supply to the house and the electric power drawn, which is 2222W - 1000W = 1222W.
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The COP of the heat pump is 2.22, and the rate of energy absorption from the outdoor air is 1.22 kJ/s.
Determining the COP of a Heat Pump
The Coefficient of Performance (COP) of a heat pump is defined as the ratio of heat energy delivered to the heated space (Qh) to the energy input (W). In this case, we are given that the heat pump supplies 8000 kJ/h for each kW of electric power it draws.
Firstly, convert the supplied energy and power input to consistent units. Since 1 kW = 1 kJ/s, we have:
Energy supplied, Qh = 8000 kJ/h = 8000 / 3600 kJ/s = 2.22 kJ/s
Power input, W = 1 kW = 1 kJ/s
Apply the COP formula:
COP = Qh / W = 2.22 kJ/s / 1 kJ/s = 2.22
To determine the rate of energy absorption from the outdoor air (Qc), use the energy balance equation:
Qh = Qc + W
Solving for Qc:
Qc = Qh - W = 2.22 kJ/s - 1 kJ/s = 1.22 kJ/s
Thus, the COP of the heat pump is 2.22 and the rate of energy absorption from the outdoor air is 1.22 kJ/s.
All of the following statements about the pyramid of biomass are correct EXCEPT: a. Biomass is the total dry mass of the organisms presentb. The base of the pyramid generally represents primary consumersc The amount of biomass at a particular level of the pyramid depends on the amount of energy availabled. Certain toxins tend to become concentrated at the upper levels of the pyramide. Biomass pyramids tend to vary for different ecosystems.
The correct answer is B. The base of the pyramid generally represents primary consumers.
Explanation
A biomass pyramid is a graphic representation of the biomass present in a unit area of various trophic levels, this graphic representation shows the relationship between biomass and the trophic level that quantifies the biomass available at each trophic level of an energy community at a specific time. In general, an ecosystem is represented in a pyramid in which the primary producers occupy the base of the pyramid because they have more biomass in a unitary area. An example of the organization of a biomass pyramid is an ecosystem in which caterpillars feed on oak trees; In turn, the caterpillars are consumed by a bluebird, which is consumed by a sparrowhawk. In this example, the oak tree is at the base of the biomass pyramid, because it feeds dozens of caterpillars, thanks to its massive biomass, and the sparrowhawk occupies the highest level of the pyramid. Therefore, it is incorrect to affirm that the base of the pyramid generally represents primary consumers, because the primary producers are generally at the base of the pyramid. So, the correct answer is B. The base of the pyramid generally represents primary consumers.
A 120-kg object and a 420-kg object are separated by 3.00 m At what position (other than an infinitely remote one) can the 51.0-kg object be placed so as to experience a net force of zero from the other two objects?
Answer:
1.045 m from 120 kg
Explanation:
m1 = 120 kg
m2 = 420 kg
m = 51 kg
d = 3 m
Let m is placed at a distance y from 120 kg so that the net force on 51 kg is zero.
By use of the gravitational force
Force on m due to m1 is equal to the force on m due to m2.
[tex]\frac{Gm_{1}m}{y^{2}}=\frac{Gm_{2}m}{\left ( d-y \right )^{2}}[/tex]
[tex]\frac{m_{1}}{y^{2}}=\frac{m_{2}}{\left ( d-y \right )^{2}}[/tex]
[tex]\frac{3-y}{y}=\sqrt{\frac{7}{2}}[/tex]
3 - y = 1.87 y
3 = 2.87 y
y = 1.045 m
Thus, the net force on 51 kg is zero if it is placed at a distance of 1.045 m from 120 kg.