The first confirmed detections of extrasolar planets occurred in ____________. The first confirmed detections of extrasolar planets occurred in ____________. the mid-17th century the mid-20th century the 1990s 2009

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.

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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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.

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.

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.

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.

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

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

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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°

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.

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.

: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

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]

Answer:168 W

Explanation:

Given

Power needed [tex]P=42 W[/tex]

initial Launch velocity is v

Energy of projectile when it is launched [tex]E=\frac{1}{2}mv^2[/tex]

[tex]Power=\frac{Energy}{time}[/tex]

[tex]Power=\frac{E}{t}[/tex]

[tex]42=\frac{\frac{1}{2}mv^2}{t}--------1[/tex]

Power when it is launched with 2 v

[tex]E_2=\frac{1}{2}m(2v)^2=\frac{4}{2}mv^2[/tex]

[tex]P=\frac{2mv^2}{t}---------2[/tex]

Divide 1 & 2 we get

[tex]\frac{42}{P}=\frac{1}{2\times 2}[/tex]

[tex]P=42\times 4=168 W[/tex]    

To accelerate the projectile to twice its launch speed, four times the power is needed.

To find the power needed to accelerate the projectile from rest to a launch speed of 2v in a time of t, we need to recognize that power is directly proportional to the change in kinetic energy. The change in kinetic energy from rest to launch speed v is given by KE = (1/2)mv^2, and the change in kinetic energy from rest to launch speed 2v is given by KE' = (1/2)m(2v)^2 = 4(1/2)mv^2 = 4KE.

Since power is directly proportional to the change in kinetic energy, the power needed to accelerate the projectile to a launch speed of 2v is four times the power needed to accelerate it to a launch speed of v. Therefore, the power needed is 4(42.0 W) = 168.0 W.

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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

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.

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.

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.

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

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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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.

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

Answer:

Doug speed will be 65 km/hr

Explanation:

Let the Thor's speed is x km/hr

So Doug's speed = x+1 km/hr

We have given that Doug and Thor take same time to cover 390 km and 384 km respectively

We know that time is given by

[tex]time=\frac{distance}{speed}[/tex]

So time taken by Doug to cover the distance

[tex]time=\frac{390}{x+1}[/tex]

And time taken by Thor to cover the distance

[tex]time=\frac{384}{x}[/tex]

As both times are equal

So [tex]\frac{390}{x+1}=\frac{384}{x}[/tex]

[tex]6x=384[/tex]

[tex]x=64km/hr[/tex]

So Doug speed will be 64+1 = 65 km/hr

Answer:

A) 0.5cm  B) 1/30

Explanation:

The weight of the man = mass * acceleration due to gravity where the mass is 78kg and acceleration due to gravity is 9.81m/s^2

W = m * g = 78 * 9.81= 686.7 N

The force acting on the tendon is 8 times of the weight

Force = 8 * weight of the body = 8 * 686.7 = 5493.6 N

Young modulus of the tendon(e) = (F/A)/ (DL/L) where A is the cross sectional area in square meters, DL is the change in length of the tendon in meters and L is the original length of the tendon

e = (FL)/(ADL) cross multiply and make DL subject of the formula

DL = (FL) / (AL)

Convert the cross sectional area A into square meters and the length also

A = 110 / 1000000 since 1/1000 m = 1mm, 1/1000000 m^2 = 1 mm^2 and 1/100m = 1 cm

A = 0.00011 m ^2 and L = 0.15m

Substitute the values in the derived equation

DL = (5493.6 * 0.15)/ (1.5 * 10^ 9 * 1.1* 10^-4)

DL = 824.04 / 1.65 * 10^ 5

DL = 499.42 * 10^-5 = 499.42 *10^ -5 / 100 to convert it to meters

DL = 0.49942cm approx 0.5cm

B) fraction of the DL to L  = 0.5 / 15 = 1/30

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.

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.

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.

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

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]

Answer:

(a)20.65g

(b)0.19m

Explanation:

(a) The total mass would be it's mass per length multiplied by the total lenght

0.355(50 + 23*0.355) = 20.65 g

(b) The center of mass would be at point c where the mass on the left and on the right of c is the same

Hence the mass on the left side would be half of its total mass which is 20.65/2 = 10.32 g

[tex]c(50 + 23c) = 10.32[/tex]

[tex]23c^2 + 50c - 10.32 = 0 [/tex]

[tex]c \approx 0.19m[/tex]

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.

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.

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