Answer:
In this problem basically we will use directions to find displacement and distance .
(Keep in mind that displacement is vector while distance is scalar quantity)
And then we will use displacement and distance to find average velocity and average speed.
Explanation:
A thin piece of semiconducting silicon will be used to fabricate an electrical device. This layer is 0.10 cm thick and cut into a strip 0.50 cm wide by 1.50 cm long. Electrical contacts are placed at opposite ends of its length. The intrinsic carrier concentration of the silicon at room temperature (300K) is 1.0x1010/cm3 and the bandgap energy is 1.12 eV.
Required:
a. If the application of 1.0 volt to the contacts results in a current of 0.019 amps, what is the resistivity in (ohm-cm) of the material?
b. If the material's conductivity is due to doping with aluminum to a level of [Al]= 1x10^17 atoms/cm^3, what is the resulting conductivity "type" and what is the mobility of these "majority" carriers in this material (assuming that the aluminum is fully ionized - i.e. all Al atoms donated electrons).
We have that for the Question "a)what is the resistivity in (ohm-cm) of the material? b) what is the resulting conductivity "type" and what is the mobility of these "majority" carriers in this material"
Answer:
Resistivity = [tex]1.754 ohm-cm[/tex]Conductivity = [tex]6.25*10^{25} cm^3/V-s[/tex]
From the question we are told
This layer is 0.10 cm thick and cut into a strip 0.50 cm wide by 1.50 cm long. The intrinsic carrier concentration of the silicon at room temperature (300K) is 1.0x1010/cm3 and the bandgap energy is 1.12 eV.
A) Resistivity is given as,
[tex]p = \frac{RA}{l}[/tex]
where,
[tex]R = \frac{V}{I}[/tex]
Therefore,
[tex]p = \frac{VA}{Il}\\\\p = \frac{1*(0.1*0.5)}{0.019*1.5}\\\\p = 1.754 ohm-cm[/tex]
B) Conductivity is given as,
[tex]U = \frac{\rho}{pe}\\\\U = \frac{10^{17}}{10^{10}*1.6*10^{-19}}\\\\U = 6.25*10^{25} cm^3/V-s[/tex]
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At which type of boundary is new oceanic crust created?
A. a convergent plate boundary
B. a divergent plate boundary
C. a subduction plate boundary
D. a transform plate boundary
Answer:
c.
Explanation:
If the two plates that meet at a convergent plate boundary both are of oceanic crust, the older, denser plate will subduct beneath the less dense plate. The older plate subducts into a trench, resulting in earthquakes. Melting of mantle material creates volcanoes at the subduction zone.
When two oceanic plates converge, the denser plate will end up sinking below the less dense plate, leading to the formation of an oceanic subduction zone. Old, dense crust tends to be subducted back into the earth. An example of a subduction zone formed from a convergent boundary is the Chile-Peru trench….
Answer:
a divergent plate boundary
A student that is running in a gym at a speed of 3.5m/s grabs the rope hanging from the ceiling and swings on it.
a. how high will he swing? [63cm]
b. How high will he be when his speed reduced to half of its initial value? [16cm, ¼ of the initial value]
Can someone explain the logic behind the second part of the question (why is it 1/4 the initial value)?
a. Assuming all energy involved is conserved, at the lowest point of the swing (which includes the moment the student grabs the rope), the student only has kinetic energy,
K = 1/2 m (3.5 m/s)²
and at the highest point of the swing, the student only has potential energy
P = mgh
The energies at the bottom and top of the swing must be equal, so
1/2 m (3.5 m/s)² = mgh
h = (3.5 m/s)² / (2g)
h = 0.625 m ≈ 63 cm
b. In part (a), we found the relationship
h = v²/(2g)
If we cut the speed v in half, we replace v in the equation above with v/2 :
h = (v/2)²/(2g)
and simplifying this gives
h = (v²/4)/(2g) = 1/4 • v²/(2g)
The factor of 1/4 tells you that reducing the speed by a factor of 1/2 reduces the height by a factor of 1/4. So he can swing as high as
1/4 (3.5 m/s)²/(2g) = 0.15625 m ≈ 16 cm
1.25 is the closest to 1.04 or not I want to answer please. I think it's true, but I want to prove it scientifically, please.
Answer:
in general context yes it is closest to 1.04
Explanation:
theres no right or wrong way to scientifically prove this though.
Overall in scale its closest to 1.04 hope that helped
What is the angle of incidence when incident ray is reflected backwards along the same path
Answer:
The angle of incidence = The angle of reflection
Explanation:
For instance, the formula is <i = <r which means if the surface is smooth then the reflacted anglw will be equal according to the normal (dotted 90°) line.
Upon being reflected backwards the angle of incidence is simply the same, except from the other side
3) A 60. kg person is in an elevator. The elevator starts from rest and then accelerates upwards at 2.0 m/s^2 for 4.0 seconds. Calculate the work done by the normal force on the person. *
Answer:
WD = 960 J
Explanation:
WD = work done (J)
F = force (N)
s = displacement (m)
m = mass (kg) = 60
a = acceleration (m/s²) = 2
t = time (s) = 4
u = initial velocity (m/s) = 0
The formulas or equations that are relevant ate:
WD = F × s
F = m × a
s = u + at
We want to find WD, so we need to now the force and the displacement (or distance);
We calculate force, in Newtons, with the formula F = ma:
F = 60 × 2
F = 120 N
We also need displacement, which get with the formula s = u + at:
s = 0 + 2(4)
s = 8 m
Now we have F and s, we can calculate WD:
WD = 120 × 8
WD = 960 J
Methodology:
Starting with what you want to find, in this case WD, list the formula/s you could use;
Then, identify the information you need for the formula and whether or not you are given that information;
Next, list the formulas for the information you don't have and once again, identify whether the information you are given is sufficient to use those formulas;
Once you can calculate all necessary information, then proceed to calculate the values and finally, the answer;
I suggest also keeping a list of all the variables as I've done at the top of my working so it is clear for you to see and use.
please help me
please help me
please help me
Answer:
do it got a picture
on the edge
Explanation:
An object is travels 50 m in 4 s. It had no initial velocity and experiences constant acceleration. What is the magnitude of the acceleration?
Free-fall Acceleration is -10 m/s^2
I also need the Formula
Answer:
Explanation:
s = s₀ + v₀t + ½at²
50 = 0 + 0(4) + ½a(4²)
50 = 8a
a = 50/8 = 6.25 m/s²
The amount of work done in example B is:
Answer:
Explanation:
20 n is an unknown amount
If that is supposed to be 20 N(ewtons)
then W = Fd = 20(15) = 300 J
Answer: it will be 300 newton meters
Explanation:
A car is driving 12m/sec, has to stop suddenly because a pedestrian dashes out in front of the car. If the coefficient of kinetic friction between the tires and parking lot is ∪=60
what is the time, after the breaks are applied, before the car comes to a stop? Sketch the velocity time graph for the car's motion from the instant the breaks are applied until the car comes to a stop.
Answer:
Approximately [tex]2\; \rm s[/tex], assuming that the floor of this parking lot is level, [tex]\mu_{\rm k} = 0.60[/tex], and [tex]g = 9.81\; \rm m\cdot s^{-2}[/tex].
Explanation:
Let [tex]m[/tex] denote the mass of this vehicle. Weight of this vehicle: [tex]m\, g[/tex].
If the floor of this parking lot is level, the normal force on this vehicle would be equal to its weight: [tex]N = m \, g[/tex].
Given that [tex]\mu_{\rm k}[/tex], the kinetic friction between this vehicle and the ground would be consistently [tex]\mu_{\rm k} \, N = \mu_{\rm k} \, m \, g[/tex] until the vehicle comes to a stop.
Assuming that all forces on this vehicle other than friction are balanced. The net force of this vehicle during braking would be [tex](-\mu_{\rm k} \, m \, g)[/tex] (negative because this force is opposite to the direction of the motion.)
By Newton's second law of motion, the acceleration of this vehicle would be:
[tex]\begin{aligned}a &= \frac{F_\text{net}}{m} \\ &= \frac{-\mu_{\rm k} \, m \, g}{m} \\ &= -\mu_{\rm k}\, g \\ &= -0.60 \times 9.81\; \rm m\cdot s^{-2} \\ &= -5.886\; \rm m\cdot s^{-2}\end{aligned}[/tex].
In other words, braking would reduce the velocity of this vehicle by a constant [tex]5.886\; \rm m\cdot s^{-1}[/tex] every second until the vehicle comes to a stop. Calculate the time it would take to reduce the velocity of this vehicle from [tex]v_{0} = 12\; \rm m\cdot s^{-1}[/tex] to [tex]v_{1} = 0\; \rm m\cdot s^{-1}[/tex]:
[tex]\begin{aligned}t &= \frac{v_{1} - v_{0}}{a} \\ &= \frac{0\; \rm m\cdot s^{-1} - 12\; \rm m\cdot s^{-1}}{-5.886\; \rm m \cdot s^{-2}} \\ &\approx 2.0\; \rm s \end{aligned}[/tex].
Acceleration is the slope of the velocity-time graph. Since the acceleration here is constant, the velocity-time graph of this vehicle would be a line with a negative slope.
The symbol or variable to used find initial velocity is
Answer:
v down exponenet 1 brainlest
Explanation:
Answer:
v0 [vee nought] is the initial velocity when time=0
The center of gravity of a loaded truck depends on how the truck is packed. If it is 4.0 m high and 2.4 m wide, and its CG is 2.2 m above the ground, how steep a slope can the truck be parked on without tipping over
The slope of the road can be given as the ratio of the change in vertical
distance per unit change in horizontal distance.
The maximum steepness of the slope where the truck can be parked without tipping over is approximately 54.55 %.Reasons:
Width of the truck = 2.4 meters
Height of the truck = 4.0 meters
Height of the center of gravity = 2.2 meters
Required:
The allowable steepness of the slope the truck can be parked without tipping over.
Solution:
Let, C represent the Center of Gravity, CG
At the tipping point, the angle of elevation of the slope = θ
Where;
[tex]tan\left(\theta \right) = \dfrac{\overline{AM}}{\overline{CM}}[/tex]
The steepness of the slope is therefore;
[tex]\mathrm{The \ steepness \ of \ the \ slope}= \dfrac{\overline{AM}}{\overline{CM}} \times 100[/tex]
Where;
[tex]\overline{AM}[/tex] = Half the width of the truck = [tex]\dfrac{2.4 \, m}{2}[/tex] = 1.2 m
[tex]\overline{CM}[/tex] = The elevation of the center of gravity above the ground = 2.2 m
[tex]\mathrm{The \ steepness \ of \ the \ slope}= \dfrac{1.2}{2.2} \times 100 \approx 54.55\%[/tex]
[tex]tan\left(\theta \right) = \mathbf{\dfrac{2.2}{1.2}} = \dfrac{11}{6}[/tex]
[tex]Elevation \ of \ the \ road \ \theta = arctan\left( \dfrac{6}{11} \right) \approx 28.6 ^{\circ}[/tex]
The maximum steepness of the slope where the truck can be parked is 54.55 %.
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you are standing ata known distance from the statue of liberty describe how you could determine its height using only a meter stick
Use the meter stick to measure your height to the level of your eyes, then use trigonometry ratio formula to calculate the height from your eyes' level and above of the statue of liberty with angle of elevation, then, add the two heights.
If you are standing at a known distance from the statue of liberty, a meter stick can be used to measure your known distance away from the statue of liberty.
To determine its height using only a meter stick, the angle at which you look at the peak of statue of liberty must be measured or known. The height of the statue of liberty can be calculated if you know the angle of elevation at which you look at the peak of the statue, and the availability of the meter stick.
Use the meter stick to measure your height to the level of your eyes, then use trigonometry ratio formula to calculate the height from your eyes' level and above of the statue of liberty. That is,
Tan Ф = opposite / adjacent
Tan Ф = H/d
H = d x TanФ
Where
H = calculated height from the eyes level and above
Ф = angle of elevation
d = known distance away from the statue
Let h = Your measured height of your body to the eyes level
Then,
The height of the statue of liberty = H + h
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Some students conduct an experiment to prove conservation of momentum. They use two objects that collide Measurements
are taken before and after the collision.
Which two quantities will the students multiply together before and after the collision?
A. mass and velocity
B. distance and time
C. mass and acceleration
D. velocity and time
This question involves the concepts of the law of conservation of momentum, velocity, and mass.
The two quantities, the students should multiply before and after the collision are "A. mass and velocity".
According to the law of conservation of momentum, In an isolated system, the total momentum of the system before the collision is always equal to the total momentum of the system after the collision.
To prove the law of conservation of momentum, consider two balls of masses ‘m₁’ and ‘m₂’, moving with velocities ‘u₁’ and ‘u₂’, respectively, such that u₁ is greater than u₂. After some time, these balls collide with each other and their velocities become ‘v₁’ and ‘v₂’, respectively.
This situation is illustrated in the attached picture.
So, according to the law of conservation of momentum:
Total Momentum Before Collision = Total Momentum After Collision
m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂
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Sorry this is a year late, but here it is for those of you who are stuck on the same thing.
======================================Proving Conservation of Momentum Quick Check - 5/5NOTE: Please Check and Confirm That You Are On The Same Assignment with The Same Questions and Number of Questions. Thank You and Good Luck!
=======================================1. Mass & velocity
2. The total momentum after the collision is the same as the total momentum before the collision.
3. 0.54 kg⋅m/s
4. The system has external forces, such as friction and air resistance, acting on it.
5. 3.0 m/s
an observer sees two spaceships flying apart with speed .99c. What is the speed of one spaceship as viewed by the other
Answer:
V2 = (V1 - u) / (1 - V1 u / c^2)
V1 = speed of ship in observer frame = .99 c to right
u = speed of frame 2 = -.99 c to left relative to observer
V2 = speed of V1 relative to V2
V2 = (.99 - (-.99 ) / (1 - .99 (-.99)) c
V2 = 1.98 / (1 + .99^2) c = .99995 c
A Pump discharges water at 1MPaa and 165 deg C. Determine the specific volume and internal energy of the water at the discharge point.
This problem is describing a pump from which water is discharged at 1 MPa and 165 °C and is asking for the specific volume and internal energy at those conditions, thus, we can use the steam tables for resolving this requirement.
First of all, we need to remember that water can be a saturated liquid, vapour or liquid-vapour mixture, and this is determined for the temperature and pressure it is at.
In this case, we find that at 165 °C the saturation pressure is about 0.6178 MPa; this means we are referring to a saturated liquid so that both the specific volume and internal energy can be simply read from the steam tables as vf and uf as follows:
[tex]v=0.001127\frac{m^3}{kg}\\\\u=761.67 \frac{kJ}{kg}[/tex]
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https://brainly.com/question/13177371https://brainly.com/question/15298076Rachel drops a ball from a hot–air balloon while her friend Lisa is watching her from the ground. Which statement about the ball's motion is true from Lisa's point of view?
Assume that there is no air resistance and the hot–air balloon is moving horizontally.
A. The ball drops to the ground along a straight–line path.
B.When the ball lands, the hot–air balloon will be ahead of it.
C. When the ball lands, the hot–air balloon will be behind it.
D. When the ball lands, the hot–air balloon will be directly above it.
Answer:
According Lisa, both the ball and the balloon have the same forward velocity of Vx.
(D) is correct
what does stimlus mean
Answer:
a thing or event that evokes a specific functional reaction in an organ or tissue.
or
a thing that rouses activity or energy in someone or something; a spur or incentive.
identify the following prefixes:
1) Di-
2) Tetra-
3) Deca-
4) Hepta-
Explanation:
Di -. 2
Tetra. -3
deca. -. 10
Hepta. -- 7
The elevation at the base of a ski hill is 350 m above sea level. A ski lift raises a skier (total mass=72 kg, including equipment) to the top of the hill. If the skier's gravitational potential energy relative to the base of the hill is now 9.2 x 105 J, what is the elevation at the top of the hill?
The elevation at the top of the hill is 1,653.85 m.
The given parameters;
initial height of the skier, h₁ = 350 mlet the final height of the skier at the hill top, = h₂total mass, m = 72 kggravitational potential energy of the skier, P.E = 9.2 x 10⁵ JThe elevation at the top of the hill is calculated as follows;
[tex]P.E = mg\Delta h\\\\P.E = mg(h_2 -h_1)\\\\h_2 -h_1 = \frac{P.E}{mg} \\\\h_2 = \frac{P.E}{mg} + h_1\\\\h_2 = \frac{9.2 \times 10^5 }{72 \times 9.8} \ + \ 350 \ m\\\\h_2 = 1,653.85 \ m[/tex]
Thus, the elevation at the top of the hill is 1,653.85 m.
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The average normal body temperature measured in the mouth is 310 K. What would Celsius thermometers read for this temperature?
54.1°C
23.8°C
36.9°C
42.7°C
________
= 310 - 273
= 37°C
Actually,310 Kelvin is same with 37°C, and as you see, there is no 37°C
So, The Nearest Number To 37°C is 36,9°C
Answer:
36.9
Explanation:
Plato
After passing point 2 the hill becomes frictionless and the ring's rotational velocity remains constant. What is the linear velocity of the ring at point 3 in m/s
The energy in the system is given by the initial potential energy at the point 1.
The linear velocity at point 3, is approximately 33.59 m/s.
Reasons:
The parameters are;
Height at point 1, h₁ = 83 m
Radius of the ring = 8 cm
Mass of the ring, M = 8 kg
Height at point 2, h₂ = 32 m
At point 2, we have;
Change in potential energy = Kinetic energy
Which gives;
(83 - 32) × 9.81 × 8 = 0.5 × 8 × v² + 0.5 × 8 × 0.08² × (v/0.08)²
Which gives;
v ≈ 22.37 m/s
At point 3, the rotational kinetic energy remains constant while the
translational kinetic energy increases as follows;
K.E. at point 3 = Initial kinetic energy + Change in potential energy
Which gives;
K.E. at point 3 = 0.5 × 8 × v₃³ ≈ 0.5×8×22.37² + 32×9.81×8
[tex]v_3^2 = \dfrac{0.5 \times 8 \times 22.37^2 + 32 \times 9.81 \times 8}{0.5 \times 8} = 1128.15[/tex]
v₃ ≈ √(1128.15) ≈ 33.59
The linear velocity at point 3, v₃ ≈ 33.59 m/s
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The probable question parameters as obtained from a similar question online are;
Height at point 1, h₁ = 83 m
Radius of the ring = 8 cm
Mass of the ring, M = 8 kg
Height at point 2, h₂ = 32 m
When tightening a bolt, you push perpendicularly on a wrench with a force of 185 N at a distance of 0.11 m from the center of the bolt. How much torque are you exerting relative to the center of the bolt
Explanation:
The torque exerted on the wrench is
[tex]\tau = Fd\sin{\theta} = Fd[/tex]
since the force is applied perpendicular to the wrench, in which sin90 = 1. The torque then is
[tex]\tau = (185\:\text{N})(0.11\:\text{m}) = 20.35\:\text{N-m}[/tex]
I need help with this equation. 4 tutors so far on the math side are unable to help me solve the problem.
How do I resolve moments about the point P?
Answer:
By applying the definition of torques ( [tex]\vec \tau = \vec r \times \vec F[/tex] ) and them remembering a few tricks.
Namely: if you wrap your RIGHT hand fingers around something and stick your thumb out, the direction your finger wraps gives you the verse of rotation and the thumb the orientation of the torque. Bottom force (4N) will give a counterclockwise rotation, torque is pointing up; top force (3N) will give a clockwise rotation and its torque its pointing down (read up and down as if the sheet the image is printed on is on your table).
In terms of magnitude the trick is easy: You want to multiply the intensity of the force (3N and 4N) by the distance between the point and the line the force it is applied to (that is, you don't care about the length of r itself, but the distance at a right angle, which is 0.9 and 0.8m respectively.
At this point, assuming "upwards" (relative to the plane of the sheet that is) torques positive, the 3N force gives you a torque of [tex]- 3N \times 0.9m = - 2.7N\cdot m[/tex] and the 4N force provides [tex]+4N\times 0.8 m = +3.2 N\cdot m[/tex]
If the penny is thrown horizontally at 25 m/s from the 170 meter building, how long will it take for the penny to hit the ground?
9514 1404 393
Answer:
about 5.89 seconds
Explanation:
The penny will hit the ground at the same time it would if it were simply dropped. The equation for the vertical motion is ...
h(t) = -4.9t^2 +170 . . . . . where 170 is the initial height in meters
h(t) = 0 when ...
4.9t^2 = 170
t = √(170/4.9) ≈ 5.89
The penny will hit the ground in about 5.89 seconds.
A ball is dropped from an 80.0 m building. What is the ball's velocity after 3.00 s? Use an order-of-magnitude estimation to identify the correct choice.
A. -2.9 m/s
B. -29.4 m/s
C -8.8 m/s
D. -88.3 m/s
Answer:b
Explanation:
-29.4 m/s
The velocity of the ball dropped from 80 m if it reaches the ground within 3 seconds is 26.6 m/s. If it is in midway within this time, then the velocity will be 29.4 m/s.
What is velocity ?Velocity of a moving object is the measure of its distance travelled per unit time. Velocity is a vector quantity having both magnitude and direction. Acceleration is the rate of change in velocity.
Given that, height of the building = 80 m
the ball is moving downwards by acceleration due to gravity g = 9.8 m/s².
Then after 3 seconds, the velocity of the ball is calculated as follows:
velocity = acceleration × time
v = 9.8 m/s² × 3 s = -29.4 m/s
If the ball reaches the ground within the time of 3 s, then, the velocity is:
v = 80 m/3s = 26.6 m/s.
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100 J of work was done to lift a 10-N rock and set it at Position A near the edge of a cliff.
1. If the 100 Joules of work lifted the rock to the top of the cliff, how much potential energy did the rock gain?
2. At point C, the rock's potential energy will be
3. The rock's kinetic energy at point A is
4. At point B, some of the rock's potential energy will be changed to Kinetic energy
5. What is the mass of the rock?
6. What is the rock's velocity just before it hits the ground?
The rock to the right is sitting at the top of a ramp. I wonder how much work it required to get that rock up there.
Answer:
lol
Explanation:
Disk A, with a mass of 2.0 kg and a radius of 40 cm , rotates clockwise about a frictionless vertical axle at 50 rev/s . Disk B, also 2.0 kg but with a radius of 20 cm , rotates counterclockwise about that same axle, but at a greater height than disk A, at 50 rev/s . Disk B slides down the axle until it lands on top of disk A, after which they rotate together.
After the collision, what is magnitude of their common angular velocity (in rev/s)?
Hi there!
For this problem, we must use the conservation of angular momentum. This is an example of an inelastic "collision", so:
I₁w₁ + I₂w₂ = (I₁ + I₂)wf
We know that the moment of inertia of a disk is 1/2mR², so we can calculate the moments of inertia for both disks:
Disk 1: 1/2(2)(0.40²) = .16 kgm²/s
Disk 2: 1/2(2)(0.20²) = .04 kgm²/s
Plug in the values. Let counterclockwise be positive.
.16(-50) + .04(50) = (.16 + .04)wf
Solve:
wf = -30 rev/s
Read the sentence from the text. “They are as glossy as satin or sunlight reflecting off water!" What does the word glossy mean in the sentence? O A. pointed o B. shiny O C. small O D. strong
Answer:
b Shiny
Explanation:
Trust me it's right