Answer:
The order is Star 3 < Star 2 < Star 1 < Star 4
Explanation:
Lets name the stars with their written order.
Star 1: 2.7 x 10^18 km3
Star 2: 6.9 x 10^12 km3
Star 3: 2.2 x 10^12 km3
Star 4: 4.9 x 10^21 km3
Star with lowest power of 10 has the least volume. Therefore, star 2 and star 3 would be the stars with least volume. Star 2's coefficient is better than Star 3. Thus, the order will be Star 3 < Star 2 < Star 1 < Star 4.
In the manufacturing of computer chips, cylinders of silicon are cut into thin wafers that are 3.00 in. in diameter and have a mass of 1.50 g of silicon. How thick, in millimeters, is each wafer if silicon has a density of?
The thickness of a silicon wafer can be found using the volume of a cylinder and the density formula. Given the diameter and the mass, rearranging to solve for height will give the thickness of the wafer in millimeters.
Explanation:The thickness of the silicon wafer can be found by using the formula for the volume of a cylinder (Volume = pi * radius2 * height) and the definition of density (Density = mass/volume). Given the silicon wafer has a diameter of 3 inches (or a radius of 1.5 inches), and a mass of 1.5g, we can determine the volume of the wafer from the given density. Rearranging the equation for the volume of a cylinder to solve for height (or thickness, in this case) gives: Thickness = Volume / (pi * radius2). Assuming measurements are converted correctly for consistent units, this calculation will give the thickness of the wafer in millimeters.
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A Ping-Pong ball has a diameter of 1.99 cm and average density of 0.121 g/cm3 . What force would be required to hold it completely submerged under water? The acceleration of gravity is 9.8 m/s 2 . Answer in units of N.
Answer:
0.035 N
Explanation:
Density = mass/volume
D = m/v
m = D× v ..................... Equation 1
Where D = density of the ping-pong, v = volume of the ping-pong, m = mass of the ping-pong
Note: The ping-pong is spherical in shape.
v = 4/3πr²
Where r = radius, π = pie
d = 1.99 cm, π = 3.14
v = 4/3(1.99/2)²(3.14)
v = 4.12 cm³
Also D = 0.121 g/cm³
Therefore,
m = 0.121(4.12)
m = 0.499 g
W = mg
Where W = weight of the ping-pong
W = (0.499/1000)×9.81
W = 0.005 N.
From Archimedes principle,
Upthrust = density of water × volume of water displaced × acceleration due to gravity.
U = D'vg/1000...................... Equation 2
Note: The volume of water displaced is equal to the volume of the ping-pong.
given: v = 4.12 cm³, g = 9.81 m/s², D' = 1 g/cm³
Substitute into equation 2
U = 1(4.12)(9.81)/1000
U = 0.04 N
The force required to hold the ball completely submerged under water = U-W
= 0.04-0.005 = 0.035 N
You are given a vector in the xy plane that has a magnitude of 84.0 units and a y-component of -67.0 units.
(a) What are the two possibilities for its x-component?
(b) Assuming the x-component is known to be positive, specify the magnitude of the vector which, if you add it to the original one, would give a resultant vector that is 80.0 units long and points entirely in the negative x-direction.
(c) Specify the direction of the vector.
Answer:
Explanation:
a)Magnitude = [tex]\sqrt{(x1-y2)^{2} + (x1-x2)^{2} }[/tex]
84=[tex]\sqrt{(0- (-67))^{2} + (x-0)^{2} }[/tex]
x= +50.67 or -50.67 units
b) We are given that the resultant is entirely in the -ve x direction which means that the y-component of the resultant is 0; It means that the y-component of the next vector = -ve of the y component of the initial vector i.e 67.
To make the magnitude 80 units in the negative x direction where the y component is 0, the x component must be -130.67(-50.67 - 80) as the x component is + 50.67units.
Magnitude = [tex]\sqrt{(0- (67))^{2} + (-130.67)^{2} }[/tex] = 146.85 units
c) The direction vector = 67/146.85 i - 130.67/146.85 j where i corresponds to the vector in y direction and j corresponds to the vector in x direction. Or this vector is at an angle of 180 - [tex]Tan^{-1}(67/130.67)degrees[/tex] i.e 152.85 degrees from the +ve x-axis.
A) The two possibilities for the x-component are; +50.67 units or -50.67 units
B) The magnitude of the vector added to the original one is; 146.85 units
C) The direction of the vector is; θ = 207.15°
A) We are given;
Magnitude of vector = 84 units
Y-component of the vector = -67 units
We know that the formula for for 2 vectors like this in x and y direction is;
A = xi^ + yj^
Where A is the magnitude of the resultant
x is the value of the x-component
y is the value of the y-component
Thus;
A = √(x² + y²)
84 = √(x² + (-67)²)
84² = x² + 4489
7056 = x² + 4489
x = ±√(7056 - 4489)
x = ±50.67 units
B) From A above, let us take the positive value of the x-component and as such our original vector will be;
A = 50.67i^ - 67j^
We want to add another vector to this that would make the resultant to be -80 units in the x direction. Thus, A = -80i and if the new additional vector is V^, then we have;
-80i^ = (50.67i^ - 67j^) + V^
V^ = -(80 + 50.67)i^ + 67j^
V = -130.67i^ + 67j^
The magnitude of vector V is;
V = √(x² + y²)
V = √(-130.67)² + 67²)
V = 146.85 units
C) The direction of the vector V is;
θ = tan^(-1) (y/x)
θ = tan^(-1) (67/-130.67)
θ = -27.15°
Since it points entirely in the negative x axis, then the angle is;
180 - (-27.15) = 207.15°
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Suppose two wagons, one with twice as much mass as the other, fly apart when a compressed spring that joins them is released. The heavier wagon rolls ____ as fast as the lighter wagon.
Answer:
The heavier wagon rolls 1/2 as fast as the lighter wagon.
Explanation:
When the compressed spring that joins them is released then the force acts on both wagons will be of equal magnitude but in the opposite direction. However as the mass of one wagon is twice that of other, so the acceleration will become half of the heavier wagon in comparison with lighter one.
Each of the following diagrams shows a spaceship somewhere along the way between Earth and the Moon (not to scale); the midpoint of the distance is marked to make it easier to see how the locations compare. Assume the spaceship has the same mass throughout the trip (that is, it is not burning any fuel). Rank the five positions of the spaceship from left to right based on the strength of the gravitational force that Earth exerts on the spaceship, from strongest to weakest.
The five positions of the spaceship from left to right are based on the strength of the gravitational force that Earth exerts on the spaceship, from strongest to weakest is [tex]5, 1, 2, 4, 3[/tex]
Gravity, or gravitation, is a natural phenomenon by which all things with mass or energy—including planets, stars, galaxies, and even light—are brought toward one another.
The gravitation force that Earth exerts on the spaceship will be:[tex]F_{ES}=(Gm_1m_E)/r^2[/tex]
Where [tex]F_{ES}[/tex] the force exerted on the spaceship by Earth [tex]m_1\\\\[/tex] is the mass of the spaceship and r is the distance between the.
[tex]F_{ES}\ \alpha\ 1/r^2[/tex]
This indicates larger the distance smaller will be the force. The correct order is [tex]5, 1, 2, 4, 3[/tex].
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The strength of the gravitational force that Earth exerts on a spaceship varies depending on the distance between them. The force is strongest when the spaceship is closest to Earth and weakest when it is closest to the Moon.
Explanation:Position 1: The spaceship is closest to Earth, so the gravitational force is strongest here. Position 2: The spaceship is moving away from Earth, so the gravitational force is slightly weaker than at Position 1 but stronger than at the other positions. Position 3: The spaceship is at the midpoint between Earth and the Moon, so the gravitational force is weaker than at Positions 1 and 2 but still stronger than at Positions 4 and 5. Position 4: The spaceship is closer to the Moon than to Earth, so the gravitational force from the Moon is stronger than the force from Earth. Position 5: The spaceship is closest to the Moon, so the gravitational force from the Moon is strongest here, and the force from Earth is weakest. Learn more about Gravitational force here:
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A 96.0 kg hoop rolls along a horizontal floor so that its center of mass has a speed of 0.240 m/s. How much work must be done on the hoop to stop it
Answer:
The work done by the hoop is equal to 5.529 Joules.
Explanation:
Given that,
Mass of the hoop, m = 96 kg
The speed of the center of mass, v = 0.24 m/s
To find,
The work done by the hoop.
Solution,
The initial energy of the hoop is given by the sum of linear kinetic energy and the rotational kinetic energy. So,
[tex]K_i=\dfrac{1}{2}mv^2+\dfrac{1}{2}I\omega^2[/tex]
I is the moment of inertia, [tex]I=mr^2[/tex]
Since, [tex]\omega=\dfrac{v}{r}[/tex]
[tex]K_i=mv^2[/tex]
[tex]K_i=96\times (0.24)^2=5.529\ J[/tex]
Finally it stops, so the final energy of the hoop will be, [tex]K_f=0[/tex]
The work done by the hoop is equal to the change in kinetic energy as :
[tex]W=K_f-K_i[/tex]
W = -5.529 Joules
So, the work done by the hoop is equal to 5.529 Joules. Therefore, this is the required solution.
A ball has a speed of 15 m/s. Only one external force acts on the ball. After this force acts, the speed of the ball is 7 m/s. Has the force done positive, zero, or negative work on the ball?
Answer:
Negative work
Explanation:
The work-energy theorem states that the work of the resultant forces acting on a particle modifies its kinetic energy:
[tex]W=\Delta K\\W=\frac{mv_f^2}{2}-\frac{mv_0^2}{2}\\W=\frac{m}{2}(v_f^2-v_0^2)\\W=\frac{m}{2}((7\frac{m}{s})^2-(15\frac{m}{s})^2)\\W=\frac{m}{2}(-178\frac{m^2}{s^2})[/tex]
Since the mass of the ball has to be positive, the work is negative.
The external force did negative work on the ball because the ball's speed decreased from 15 m/s to 7 m/s, indicating a decrease in kinetic energy.
Explanation:To determine whether the force has done positive, negative, or zero work on the ball, we must consider the change in the ball's kinetic energy. Work done by a force is defined as the change in kinetic energy of an object. The formula for work done (W) is given by the change in kinetic energy:
W = ΔKE = ½mv2final - ½mv2initial
Since the speed of the ball decreased from 15 m/s to 7 m/s, the kinetic energy of the ball also decreased. A decrease in kinetic energy means that negative work was done on the ball by the external force.
A ________ is a device that converts digital signals from a computer into analog signals so that telephone lines may be used as a transmission medium to send and receive electronic data.
Answer:
analog-to-digital
Explanation:
An analog-to-digital converter, or ADC as it is more commonly called, is a device that converts analog signals into digital signals.
The force exerted on the small piston of a hydraulic lift is 780 N . If the area of the small piston is 0.0075 m2 and the area of the large piston is 0.13 m2, what is the force exerted by the large piston?
Answer:
13520 N
Explanation:
Pascal's Principle: The principle states that the pressure applied to an enclosed fluid is transmitted undiminished to every portion of the fluid and the walls of the containing vessel.
The operation of hydraulic press and car brake system is based on pascal's principle.
From pascal's principle,
F/A = f/a ........................... Equation 1
Where F = force exerted by the large piston, A = area of the large piston, f = force applied to the small piston, area of the small piston.
Making F the subject of the equation
F = A(f/a)......................... Equation 2
Given: A = 0.13 m², a = 0.0075 m², f = 780 N
Substituting into equation 2
F = 0.13(780/0.0075)
F = 13520 N.
Thus the force exerted by the large piston = 13520 N
the volume of an iron sphere is 3.00cm cubed after being heated from 20.0c to 600.0 c what was the initial volume of the iron sphere at 20.0c
Answer:
V = 2.94cm³
The initial volume at 20°C is 2.94cm³
Explanation:
As the temperature of the iron sphere increases the volume of the sphere also increase.
Using the equation for volumetric expansion:
∆V = VαΔT
where ;
V is the initial volume
α is the volumetric expansion coefficient
∆V is the change in Volume
∆T is the change in temperature
After Expansion the final volume can be written as:
Vf = V + ∆V
Vf = V + VαΔT
Vf = V(1 + αΔT)
making V the subject of formula;
V = Vf/(1+αΔT) .....1
Given:
Vf = 3.00cm³
ΔT= 600-20 = 580
And from the test book.
α = 35.5 × 10⁻⁶K⁻¹
Substituting the values into eqn 1
V = 3.00/(1+580× 35.5×10^-6)
V = 3.00/(1+0.021) = 3.00/1.021
V = 2.94cm³
The initial volume at 20°C is 2.94cm³
Answer: 1cm3
Explanation:
V1 =?
V2 = 3cm3
T1 = 20°C = 20 + 273 = 293K
T2 = 600°C = 600 + 273 = 873K
V1 /T1 = V2 /T2
V1 / 293 = 3 / 873
V1 = 293 x ( 3 / 873)
V1 = 1 cm3
What conclusion can be derived by comparing the central tendencies of the two data sets?
A: {7, 6, 3, 1, 6, 2, 4, 6, 3, 5}
B: {2, 2, 2, 3, 4, 5, 2, 8, 7, 6}
A.
The mean of set A is smaller than the mean of set B.
B.
The median of set A is greater than the median of set B.
C.
The median and the mean of set B are greater than those of set A.
D.
The mode of set B is greater than the mode of set A.
At standard temperature and pressure, a 0.50 mol sample of H2 gas and a separate 1.0 mol sample of O2 gas have the same A. average molecular knetic energyB. average molecular speedC. volumeD. effusion rateE. density
Answer:
The correct answer is option A.
Explanation:
The average kinetic energy of the gas particle only depends upon the temperature of the gas.
The formula for average kinetic energy is:
[tex]K.E=\frac{3}{2}kT[/tex]
where,
k = Boltzmann’s constant = [tex]1.38\times 10^{-23}J/K[/tex]
T = temperature
So, at standard temperature and pressure 0.50 moles of hydrogen and 1.0 mole of oxygen sample will have same value of average kinetic energy.
Where as in other option enlisted in question , molar masses of both gases will be involved which will give different answers for both the gases.
This question involves the concepts of the average kinetic energy of the molecules of a gas and temperature.
The gases will have the same "A. average molecular kinetic energy".
The average kinetic energy of gas molecules is given by the following formula:
[tex]K.E = \frac{3}{2}KT[/tex]
where,
K.E = average kinetic energy
K = Boltzmann constant
T = absolute temperature
Hence, the average kinetic energy depends upon the absolute temperature only. Since the temperature is the same for both gases. Hence, their average kinetic energy will also be the same.
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A spring with a force constant of 5.0 N/m has a relaxed length of 2.63 m. When a mass is attached to the end of the spring and allowed to come to rest, the vertical length of the spring is 3.93 m. Calculate the elastic potential ene
Answer:
4.225 J
Explanation:
Elastic Potential Energy: This is the potential energy stored in an elastic material.This also the energy required to stretch an elastic material. The S.I unit is Joules.
Mathematically it is expressed as
E = 1/2ke²....................... Equation 1
Where E =elastic potential Energy, k = spring constant, e = extension.
Given: k = 5.0 N/m, e = 3.93-2.63 = 1.3 m.
Substitute into equation 1
E = 1/2(5)(1.3)²
E = 8.45/2
E = 4.225 J.
Thus the Elastic potential Energy = 4.225 J.
When 1.60 × 10 5 J 1.60×105 J of heat transfer occurs into a meat pie initially at 17.5 °C , 17.5 °C, its entropy increases by 485 J / K . 485 J/K. Estimate the final temperature of the pie.
Answer:
Explanation:
Given
Heat transfer [tex]Q=1.6\times 10^5\ J[/tex]
initial Temperature [tex]T_i=17.5^{\circ}\approx 290.5\ K[/tex]
Entropy change [tex]dS=485\ J/K[/tex]
The expression for entropy is given by
[tex]dQ=TdS[/tex]
[tex]T=\frac{dQ}{dS}[/tex]
[tex]T=\frac{1.6\times 10^5}{485}[/tex]
[tex]T=329.89\ K[/tex]
Temperature can be written as average of initial and final temperature
[tex]T=\frac{T_i+T_f}{2}[/tex]
[tex]329.89=\frac{T_f+290.5}{2}[/tex]
[tex]T_f=659.78-290.5[/tex]
[tex]T_f=369.28\ K[/tex]
A factory has a solid copper sphere that needs to be drawn into a wire. The mass of the copper sphere is 76.5 kg. The copper needs to be drawn into a wire with a diameter of 9.50 mm. What length of wire, in meters, can be produced?
Answer:
120.125 m
Explanation:
Density = Mass/volume
D = m/v .............................. Equation 1.
Where D = Density of the solid copper sphere, m = mass of the solid copper sphere, v = volume of the solid copper sphere.
Making v the subject of the equation,
v = m/D............................... Equation 2
Given: m = 76.5 kg,
Constant: D = 8960 kg/m .
Substituting into equation 2
v = 76.5/8960
v = 0.0085379 m³
Since the copper sphere is to be drawn into wire,
Volume of the copper sphere = volume of the wire
v = volume of the wire
Volume of wire = πd²L/4
Where d = diameter of the wire, L = length of the wire.
Note: A wire takes the shape of a cylinder.
v = πd²L/4 ........................ equation 3.
making L the subject of the equation,
L = 4v/πd²..................... Equation 4
Given: v = 0.0085379 m³, d = 9.50 mm = 0.0095 and π = 3.14
Substitute into equation 4
L = 4×0.0085379/(3.15×0.0095²)
L = 0.0341516/0.0002843
L = 120.125 m.
L = 120.125 m
Thus the length of the wire produced = 120.125 m
Which layer in the Earth has a composition similar to the granite shown in this photograph?
Answer:
Earth crust and specifically the continental crust.
Explanation:
If we examine the earth crust there is mostly the granite and basalt and most of the granite is present in the continental crust part which is less thicker and denser. That's why we say that the continental crust has the composition similar to that of granite.
What soil conditions favor the use of belled caissons? What soil conditions favor piles over caissons? What type of piles are especially well suited to repair or improvement of existing foundations and why? List and explain some cost thresholds frequently encountered in foundation design.
Answer:
What soil conditions favor the use of belled caissons?
Answer:
- where the bell can be unearthed from a solid surface.
- where the supporting stratum below the bottom of the caisson is impermeable to water movement.
What soil conditions favor piles over caissons?
Answer:
- non-cosheal soils
- subterranean water or excessive depth of bearing strata make caisson unworkable
What type of piles are especially well suited to repair or improvement of existing foundations ?
Answer:
Without hammering, minipiles or helical piles are placed which escapes much of the vibration and noise associated with traditional pile installation. for working close to existing buildings or for improving the exiting foundations where excessive vibration could damage exiting structures or noise may interfere with ongoing activities these piles are good options.
Why?
Their slenderness involves little or no displacement of the soil, thus minimizing the risk of disturbance to nearby foundations.
List and explain some cost thresholds frequently encountered in foundation design.
Answer:
building below the water table- site dewatering must occur, strengthening of slopes supper system must be done and waterproofing of the foundation all of which entails money
building near existing building - this requires underpinning(The process of reinforcing the base of an existing building or other structure underpins it.)
increase in column/wall load- building height determines the foundation depth
The answer discusses the use of belled caissons, piles, and helical piles in different soil conditions, along with common cost thresholds in foundation design.
Belles caissons are favored in soil conditions with the potential for liquefaction, as driving deep piles or piers can strengthen the soil and reduce liquefaction risk. On the other hand, piles are preferred over caissons in soil conditions that are unsuitable for drilling due to hard rock or boulders. Helical piles are especially well-suited for repairing or improving existing foundations as they can be installed quickly with minimal noise and vibration, making them ideal for retrofitting projects. Some common cost thresholds in foundation design include budgets for materials, labor, and specialized equipment, all of which can impact the overall project cost.
The brakes on your automobile are capable of creating a deceleration of 4.9 m/s2. If you are going 149 km/h and suddenly see a state trooper, what is the minimum time in which you can get your car under the 100 km/h speed limit?
Answer:
You need at least 2.8 s to slow down your car to 100 km/h. If we add reaction time (≅0.3 s), you will need 3.1 s.
Explanation:
Hi there!
The equation of velocity for an object moving in a straight line is the following:
v = v0 + a · t
Where:
v = velocity at time t.
v0 = initial velocity.
a = acceleration.
t = time.
We have to find the time at which the velocity is 100 km/h with a decceleration of 4.9 m/s² and an initial velocity of 149 km/h. Let´s first convert km/h into m/s:
149 km/h · (1000 m / 1 km) · ( 1 h / 3600 s) = 41.4 m/s
100 km/h · (1000 m / 1 km) · ( 1 h / 3600 s) = 27.8 m/s
Now, let´s solve the equation of velocity for the time:
v = v0 + a · t
(v - v0) / a = t
Replacing with the data:
(27.8 m/s - 41.4 m/s) / -4.9 m/s² = t
Notice that the acceleration is negative because you are slowing down.
t = 2.8 s
You need at least 2.8 s to slow down your car to 100 km/h. If we add reaction time (≅0.3 s), you will need 3.1 s.
To get the car under the speed limit, it would take approximately 4.34 seconds.
Explanation:To calculate the minimum time it takes for the car to get under the 100 km/h speed limit, we need to find the deceleration required to slow down from 149 km/h to 100 km/h.
First, let's convert the speeds to m/s. 149 km/h is equal to 41.4 m/s and 100 km/h is equal to 27.8 m/s.
Using the equation v² = u² + 2as, where v is the final velocity, u is the initial velocity, a is the acceleration, and s is the distance, we can rearrange the equation to solve for acceleration:
a = (v² - u²) / (2s)
Plugging in the values, we get a = (27.8² - 41.4²) / (2(-30.5)).
Solving this equation, we find that the minimum deceleration required is approximately -3.13 m/s².
Finally, we can use the formula a = Δv / t to find the minimum time:
t = Δv / a = (41.4 - 27.8) / 3.13 = 4.34 seconds.
A simple pendulum is swinging back and forth through a small angle, its motion repeating every 1.06 s. How much longer should the pendulum be made in order to increase its period by 0.32 s?
Answer:
The pendulum should be made longer by 0.194m in order to increase its period by 0.32s
Explanation:
using the formula T= 2π[tex]\sqrt{\frac{L}{g} }[/tex]
rearranging the equation and making L subject of formula we have;
L=T²g/4π²
lets calculate the length when T=1.06s
g=9.8m/s² , π=3.124
[tex]L=\frac{1.06^{2}*9.8 }{4*3.142^{2} }[/tex]
L=0.279m
the new period after its increased by 0.32s = 1.06+0.32 =1.38s
[tex]L_{2}=\frac{1.38^{2}*9.8 }{4 *3.142^{2} }[/tex]
[tex]L_{2}=0.473m[/tex]
increase in length = 0.473 -0.279
=0.194m
The pendulum should be "0.194 m" longer.
According to the question,
Time,
[tex]T = 1.06 \ s[/tex]We know,
[tex]g = 9.8 \ m/s^2[/tex][tex]\pi = 3.124[/tex]By using the formula,
→ [tex]T = 2 \pi \sqrt{\frac{L}{g} }[/tex]
or,
→ [tex]L = \frac{T^2g}{4 \pi^2}[/tex]
By substituting the values, we get
→ [tex]= \frac{1.06^2\times 9.8}{4\times 3.142^2}[/tex]
→ [tex]= 0.279 \ m[/tex]
Now,
The new period after it increased by 0.32 s, we get
= [tex]1.06+0.32[/tex]
= [tex]1.38 \ s[/tex]
then,
→ [tex]L_2 = \frac{1.38^2\times 9.8}{4\times 3.142^2}[/tex]
[tex]= 0.473 \ m[/tex]
hence,
The increase in length will be:
= [tex]L_2-L[/tex]
= [tex]0.473-0.279[/tex]
= [tex]0.194 \ m[/tex]
Thus the answer above is right.
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For a particular scientific experiment, it is important to be completely isolated from any magnetic field, including the earth's field. The earth's field is approximately 50 μT, but at any particular location it may be a bit more or less than this. A 1.00-m-diameter current loop with 200 turns of wire is adjusted to carry a current of 0.203 A ; at this current, the coil's field at the center is exactly equal to the earth's field in magnitude but opposite in direction, so that the total field at the center of the coil is zero.
Answer:
51.019 μT
Explanation:
[tex]\mu_0[/tex] = Vacuum permeability = [tex]4\pi \times 10^{-7}\ H/m[/tex]
i = Current in wire = 0.203 A
N = Number of turns = 200
d = Diameter = 1 m
Magnetic field is given by
[tex]B=\dfrac{N\mu_0i}{d}\\\Rightarrow B=\dfrac{200\times 4\pi \times 10^{-7}\times 0.203}{1}\\\Rightarrow B=0.000051019\ T[/tex]
The strength of the magnetic field at this location would be 51.019 μT
The magnetic field at the given point is 51.019 μT.
Magnetic field:
It is given that the current in the coil is I = 0.203A
the diameter of the loop carrying the current is d = 2r = 1m
and the number of turns is N = 200.
The magnetic field to a current-carrying loop having N number of turns and carrying a current I with the radius of the loop being r is given by:
[tex]B=\frac{\mu_oNI}{2r} \\\\B=\frac{4\pi\times10^{-7}\times200\times0.203}{1}\\\\B=51.019\;\mu T[/tex]
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The removal, installation, and repair of landing gear tires by the holder of a private pilot certificate on an aircraft owned or operated is considered to be
Answer:
Preventive Maintenance.
Explanation:
Preventive maintenance is nothing but The removal, installation, and repair of landing gear tires by the holder of a private pilot certificate on an aircraft owned or operated. Preventive maintenance in an aircraft is performed by any person with holding at least a private pilot certificate. Only he/she can approve an aircraft to return to service after performing Preventive maintenance tests.
A person has been exposed to a particular antigen and now experiences a repeat exposure. What stimulates a quicker immune response?a) Memory T cellsb) immunityc) antibodiesd) macrophages
Answer:
a. Memory T cells
Explanation:
Memory T cells are actually the antigen-specific T cells that remain long-term after an infection has been eliminated. These memory T cells are quickly converted into large numbers of effector T cells upon reexposure to the specific invading antigen, thus providing a rapid response to past infection that has been experienced before
As shown in the figure below, a bullet is fired at and passes through a piece of target paper suspended by a massless string. The bullet has a mass m, a speed v before the collision with the target, and a speed (0.516)v after passing through the target. The collision is inelastic and during the collision, the amount of energy lost is equal to a fraction [(0.413)KEb BC] of the kinetic energy of the bullet before the collision. Determine the mass M of the target and the speed V of the target the instant after the collision in terms of the mass m of the bullet and speed v of the bullet before the collision. (Express your answers to at least 3 decimals.)
The mass (M) of the target and the final speed (V) of the bullet and target combined after an inelastic collision can be determined by setting the total momentum before the collision equal to the total momentum after the collision and solving for M and V using the given information.
Explanation:The question deals with the concept of conservation of momentum in an inelastic collision. Since the bullet and the target paper are involved in an inelastic collision, the total momentum before the collision is equal to the total momentum after the collision. This can be defined as m*v=(m+M)*V, where m and v are the mass and velocity of the bullet respectively, M is the mass of the target, and V is the final velocity of the bullet and target combined.
Given in the question is that after collision, the speed of the bullet becomes (0.516)*v, therefore the velocity V of the bullet and the target combined, the moment after the collision would be V = 0.516 * v. Solving these equations will give the required values for M and V in terms of the initial mass and velocity of the bullet.
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The question involves an inelastic collision between a bullet and a target. Using conservation of momentum, the target's mass is found to be 0.937 times the bullet's, and the target's speed immediately after collision is approximately 0.484 times the bullet's initial speed.
Explanation:Given that the collision is inelastic, the quantity of momentum before and immediately after the collision is conserved. Hence, we can express this conservation of momentum as: m*v = (m+M)V. From this we get the mass M of the target as: M = [m*(1 - 0.516)]/0.516 = 0.937m which means the mass of the target is approximately 0.937 times the mass of the bullet.
Subsequently, we can solve for the speed V of the target using the above conservation of momentum giving: V = (m*v)/(m+M) = v/(1+1.064) = 0.484v approximately.
In conclusion, the mass M of the target the instant after the collision is 0.937m and the speed V of the target is 0.484v in terms of the mass m of the bullet and initial speed v of the bullet.
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_____ is a disorder that results from damage to the brain's motor centers, causing difficulty with motor control so that speech and movements are impaired. Klinefelter syndrome Muscular dystrophy Cerebral palsy Crohn's disease.
Answer:
Cerebral palsy
Explanation:
Cerebral palsy - it is referred to that disorder which is related to damages that caused permanent disorder in the functioning of body parts. it affects the proper functioning of muscles thus cause the coordination problem.
it is caused due to abnormalities in the brain that result in the coordination of the body. As it is related to abnormalities in the brain thus it also causes a problem in vision, speaking, hearing, etc
A skier of mass 103 kg comes down a slopeof constant angle 32◦with the horizontal.What is the force on the skier parallel tothe slope? Neglect friction. The accelerationof gravity is 9.8 m/s2.Answer in units of N.
Final answer:
The force on the skier parallel to the slope is found by calculating the component of the skier's weight that acts along the slope. This force is the weight multiplied by the sine of the angle of the slope, which results in 534.8 Newtons for a 103 kg skier on a 32° slope.
Explanation:
To determine the force on the skier parallel to the slope, we can make use of the component of the gravitational force along the slope. Since we are neglecting friction, the only force acting on the skier in the direction parallel to the slope is the component of the skier's weight in that direction.
The weight of the skier can be calculated by multiplying the mass (m) by the acceleration due to gravity (g), which is W = m × g. The component of the weight parallel to the slope is Wparallel = W × sin(θ), where θ is the angle of the slope. Substituting the given values, we have W = 103 kg × 9.8 m/s² = 1009.4 N. The parallel component is then 1009.4 N × sin(32°).
To find the sine of 32°, we use a calculator and get sin(32°) ≈ 0.5299. Multiplying this by the weight gives the parallel force on the skier, which is 1009.4 N × 0.5299 ≈ 534.8 N. Therefore, the force on the skier parallel to the slope is 534.8 Newtons.
An unknown solid with a mass of 2.00 kilograms remains in the solid state while it absorbs 32.0 kilojoules of heat. Its temperature rises 4.00 degrees Celsius. What is the specific heat of the unknown solid?
Answer: The specific heat of the unknown solid is [tex]4.00J/g^0C[/tex]
Explanation:
As we know that,
[tex]q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})[/tex] (1)
where,
q = heat absorbed = 32.0 kJ = [tex]32.0\times 10^3J[/tex] J (1kg=1000g)
[tex]m[/tex] = mass of unknown solid= 2.00 kg = [tex]2.00\times 10^3g[/tex] (1kg=1000g)
[tex]T_{final}[/tex] = final temperature
[tex]T_{initial}[/tex] = initial temperature
[tex]\Delta T[/tex] =[tex]4.00^0C[/tex]
[tex]c[/tex] = specific heat of unknown solid = ?
Now put all the given values in equation (1), we get
[tex]32.0\times 10^3J=2.00\times 10^3g\times c\times (4.00^0C)][/tex]
[tex]c=4.00J/g^0C[/tex]
Therefore, the specific heat of the unknown solid is [tex]4.00J/g^0C[/tex]
The scientific heat of the unknown solid will be "4.00 J/g°C".
Specific heat:Given values are:
Heat absorbed, q = 32.0 kJ or, [tex]32.0\times 10^3[/tex] J
Mass, m = 2.00 kg or, [tex]2.00\times 10^3[/tex] g
Rise in temperature, ΔT = 4.00°C
We know the relation,
→ q = m×c×ΔT
or,
→ = m×c×([tex]T_{final} - T_{initial}[/tex])
By substituting the values,
[tex]32.0\times 10^3=2.00\times 10^3\times c\times 4.00[/tex]
[tex]c = 4.00[/tex] J/g°C
Thus the above answer is appropriate.
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What occurs in a nuclear power plant when Uranium-235 splits into two smaller isotopes? A nuclear fusion B nuclear fission C a chemical reaction D a neutralization reaction
B) Nuclear fission
Explanation:
Nuclear power plants work by using the process of nuclear fission.
Nuclear fission occurs when a heavy, unstable radioactive nuclei decays, breaking apart into two or more lighter nuclei, more stable. In the process, several neutrons are also released, alongside with energy.
In nuclear power plants, the nucleus used for the process is the Uranium-235. When an atom of uranium-235 absorbs a slow neutron, it becomes a very unstable nucleus of uranium-236, which quickly decays into a nucleus of Barium-141, Kripton-92 and 3 neutrons.
The uranium nuclei are located in the so-called fuel rods, which are placed in a moderator (usually water). The purpose of the moderator is to slow down the neutrons emitted in the reaction: this way, these neutrons can be absorbed by other nuclei of uranium-235, causing more fission reactions to occur.
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What is the difference between a continuous spectrum and a line spectrum
Answer:
Explanation:
The continuous spectrum is a band of visible colors of light. The continuous spectrum contains all the colors of all the visible wavelengths. Usually, most of the light is emitted from a single source. Whereas a line spectrum contains only a few colors and wavelengths of the visible spectrum with gaps of the discontinuity between them. The line spectrum is usually emitted by an excited electron of an atom that is going back to its ground state.A continuous spectrum shows all colors of the rainbow with no gaps, produced by a solid or very dense gas emitting radiation. In contrast, a line spectrum consists of only certain discrete wavelengths - either as an absorption spectrum with dark lines representing absorbed wavelengths, or as an emission spectrum showing bright lines for emitted wavelengths from excited gas atoms.
Explanation:The difference between a continuous spectrum and a line spectrum mainly lies in the type of light they represent. A continuous spectrum is formed when a solid or a very dense gas gives off radiation, showing an array of all wavelengths or colors of the rainbow. This can be seen when white light is passed through a prism as represented in Figure 5.10. It's like viewing a rainbow where all the colours blend into each other without any gaps.
On the other hand, a line spectrum, which could either be an absorption or an emission spectrum, consists of light in which only certain discrete wavelengths are present. Absorption spectrum appears as a pattern of dark lines or missing colors superimposed on the continuous spectrum, created when a cloud of gas absorbs certain wavelengths from the continuous spectrum behind it. Meanwhile, an emission spectrum appears as a series of bright lines when we examine an excited gas cloud, demonstrating that the gas is emitting light at only certain wavelengths, as showcased in Figure 5.12.
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A rigid tank contains nitrogen gas at 227 °C and 100 kPa gage. The gas is heated until the gage pressure reads 250 kPa. If the atmospheric pressure is 100 kPa, determine the final temperature of the gas in °C.
Answer:
T₂ =602 °C
Explanation:
Given that
T₁ = 227°C =227+273 K
T₁ =500 k
Gauge pressure at condition 1 given = 100 KPa
The absolute pressure at condition 1 will be
P₁ = 100 + 100 KPa
P₁ =200 KPa
Gauge pressure at condition 2 given = 250 KPa
The absolute pressure at condition 2 will be
P₂ = 250 + 100 KPa
P₂ =350 KPa
The temperature at condition 2 = T₂
We know that
[tex]\dfrac{T_2}{T_1}=\dfrac{P_2}{P_1}\\T_2=T_1\times \dfrac{P_2}{P_1}\\T_2=500\times \dfrac{350}{200}\ K\\[/tex]
T₂ = 875 K
T₂ =875- 273 °C
T₂ =602 °C
The problem involves using Charles's Law, a form of the ideal gas law, to find the final temperature of nitrogen after heating, by converting all pressures to absolute pressures and applying the law to relate initial and final states.
Explanation:The student's question involves determining the final temperature of nitrogen gas in a rigid tank after heating, given initial temperature and pressures, using the ideal gas law. To solve this problem, we assume the nitrogen behaves as an ideal gas and use the relation between pressure, volume, temperature, and the number of moles of gas, which is constant since the tank is rigid.
Firstly, convert all pressures into absolute pressures by adding atmospheric pressure to the gage pressures. Then, apply the ideal gas law in the form of Charles's Law (P1/T1 = P2/T2), which relates pressure and temperature at constant volume and number of moles. To find the final temperature (T2), rearrange the equation to T2 = (P2/P1) * T1, where P2 and P1 are the final and initial absolute pressures, respectively, and T1 is the initial temperature in Kelvin.
A 2.0 kg, 20-cm-diameter turntable rotates at 100 rpm ons tionless bearings. Two 500 g blocks fall from above, hit the tum ble simultaneously at opposite ends of a diameter, and stick. W is the turntable's angular velocity, in rpm, just after this event
There are mistakes in the question.The correct question is here
A 2.0 kg, 20-cm-diameter turntable rotates at 100 rpm on frictionless bearings. Two 500 g blocks fall from above, hit the turntable simultaneously at opposite ends of a diameter, and stick. What is the turntable’s angular velocity, in rpm, just after this event?
Answer:
w=50 rpm
Explanation:
Given data
The mass turntable M=2kg
Diameter of the turntable d=20 cm=0.2 m
Angular velocity ω=100 rpm= 100×(2π/60) =10.47 rad/s
Two blocks Mass m=500 g=0.5 kg
To find
Turntable angular velocity
Solution
We can find the angular velocity of the turntable as follow
Lets consider turntable to be disk shape and the blocks to be small as compared to turntable
[tex]I_{turntable}w=I_{block1}w^{i}+I_{turntable}w^{i}+I_{block2}w^{i}[/tex]
where I is moment of inertia
[tex]w^{i}=\frac{I_{turntable}w}{I_{block1}w^{i}+I_{block2}w^{i}+I_{turntable}w^{i}}\\ So\\I_{turntable}=M\frac{r^{2} }{2}\\I_{turntable}=2*(\frac{(0.2/2)}{2} )\\ I_{turntable}=0.01 \\And\\I_{block1}=I_{block2}=mr^{2}\\I_{block1}=I_{block2}=(0.5)*(0.2/2)^{2} \\ I_{block1}=I_{block2}==0.005\\so\\w^{i}=\frac{I_{turntable}w}{I_{block1}w^{i}+I_{block2}w^{i}+I_{turntable}w^{i}}\\w^{i}=\frac{0.01*(10.47)}{0.005+0.005+0.01} \\w^{i}=5.235 rad/s\\w^{i}=5.235*(60/2\pi )\\w^{i}=50 rpm[/tex]
The problem is a case of angular momentum conservation within the domain of rotational dynamics in physics. The turntable's initial angular momentum remains conserved despite the addition of the blocks. By accounting for the added moment of inertia from the blocks, the final angular velocity of the system can be calculated.
Explanation:The subject we're discussing here comes under the physics concept of rotational dynamics particularly focusing on the conservation of angular momentum.
Before the blocks hit the turntable, we know that the turntable is rotating with an angular velocity given in RPM (revolutions per minute), which we can convert to rad/s for our calculations. So, the initial angular momentum can be represented as Lim = (moment of inertia of the system) * (initial angular velocity).
Once the blocks fall onto the turntable, they contribute to the moment of inertia of the system, while the angular momentum of the system remains conserved. Thus resulting in a decreased angular velocity. The final angular momentum can be represented as Lfm = (moment of inertia including the blocks) * (final angular velocity).
Since the initial and final angular momenta need to be equal (Lfm = Lim), we can solve the resulting equation for the final angular velocity.
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