Answer:
a) x = 0.098
b) x = 2.72 m
Explanation:
(a) To find the stretch of the fire net when the same person is lying in it, you can assume that the net is like a spring with constant spring k. It is necessary to find k.
When the person is falling down he acquires a kinetic energy K, this energy is equal to the elastic potential energy of the net when it is max stretched.
Then, you have:
[tex]K=U\\\\\frac{1}{2}mv^2=\frac{1}{2}kx^2[/tex] (1)
m: mass of the person = 62kg
k: spring constant = ?
v: velocity of the person just when he touches the fire net = ?
x: elongation of the fire net = 1.4 m
Before the calculation of the spring constant, you calculate the final velocity of the person by using the following formula:
[tex]v^2=v_o^2+2gy[/tex]
vo: initial velocity = 0 m/s
g: gravitational acceleration = 9.8 m/s^2
y: height from the person jumps = 20.0m
[tex]v=\sqrt{2gy}=\sqrt{2(9.8m/s^2)(20.0m)}=14\frac{m}{s}[/tex]
With this value you can find the spring constant k from the equation (1):
[tex]mv^2=kx^2\\\\k=\frac{mv^2}{x^2}=\frac{(62kg)(14m/s)^2}{(1.4m)^2}=6200\frac{N}{m}[/tex]
When the person is lying on the fire net the weight of the person is equal to the elastic force of the fire net:
[tex]W=F_e\\\\mg=kx[/tex]
you solve the last expression for x:
[tex]x=\frac{mg}{k}=\frac{(62kg)(9.8m/s^2)}{6200N/m}=0.098m[/tex]
When the person is lying on the fire net the elongation of the fire net is 0.098m
b) To find how much would the net stretch, If the person jumps from 38 m, you first calculate the final velocity of the person again:
[tex]v=\sqrt{2gy}=\sqrt{2(9.8m/s^2)(38m)}=27.29\frac{m}{s}[/tex]
Next, you calculate x from the equation (1):
[tex]x=\sqrt{\frac{mv^2}{k}}=\sqrt{\frac{(62kg)(27.29m/s)^2}{6200N/m}}\\\\x=2.72m[/tex]
The net fire is stretched 2.72 m
A) If the person is lying on the net, the net would stretch by : 0.0679 m
B) If the person jumped from 38m, the net would stretch by : 1.61 m
Given data :
Mass of person (m) = 62 kg
Initial height ( H₁ ) = 20 m
Net stretch ( H[tex]_{f}[/tex] ) = -1.2 m
Initial potential energy before jump ( PE₁ ) = mgH₁ = 62 * 9.81 * 20 = 12164.4J
Potential energy after the jump ( PE₂ ) = mgH[tex]_{f}[/tex] = 62 * 9.81 * ( -1.2 ) = -729.86J
The potential spring constant ( SE ) = 1/2 kx²
= 1/2 * k * ( 1.2 )²
Applying the principle of energy conservation
PE₁ = PE₂ + SE
12164.4 = - 729.86 + 0.5 * k * 1.44
∴ k = ( 12164.4 + 729.86 ) / ( 0.5 * 1.44 )
= 17908.69
A) Calculate the amount of stretch if same person lies on the net1/2 kx² = mgx ---- ( 1 )
Where x = stretch
equation ( 1 ) becomes equation ( 2 )
1/2 kx = mg ----- ( 2 )
x ( amount of stretch ) = ( 2mg ) / k
= ( 2 * 62 * 9.81 ) / 17908.69
= 0.0679 m
B ) Calculate How much the net would stretch if Height = 38 mHi = 38 m
MgHi = mgH[tex]_{f}[/tex] + 1/2 kx² ----- ( 3 )
where ; H[tex]_{f}[/tex] = -x
Back to equation ( 3 )
62 * 9.81 * 38 = 62 * 9.81 * (-x) + 1/2 ( 17908.69 x² )
23112.36 + 608.22x - [tex]\frac{17908.69 x^{2} }{2}[/tex] = 0
8954.345 x² - 608.22x - 23112.36 = 0 ----- ( 4 )
Resolving the quadratic equation ( 4 )
x = ± 1.60623 ≈ 1.61 m
Therefore we can conclude that the net would stretch 1.61 m if the person jumped from 38 m
Hence we can conclude that If the person is lying on the net, the net would stretch by : 0.0679 m, and If the person jumped from 38m the net would stretch by : 1.61 m
Learn more about simple spring : https://brainly.com/question/15727787
The resistivity of a metal increases slightly with increased temperature. This can be expressed as rho=rho0[1+α(T−T0)], where T0 is a reference temperature, usually 20∘C, and α is the temperature coefficient of resistivity. Part A First find an expression for the current I through a wire of length L, cross-section area A, and temperature T when connected across the terminals of an ideal battery with terminal voltage ΔV. Then, because the change in resistance is small, use the binomial approximation to simplify your expression. Your final expression should have the temperature coefficient α in the numerator. Express your answer in terms of L, A, T, T0, ΔV, rho0, and α.
Answer:
I = ΔVA[1 - α (T₀ - T)]/Lρ₀
Explanation:
We have the following data:
ΔV = Battery Terminal Voltage
I = Current through wire
L = Length of wire
A = Cross-sectional area of wire
T = Temperature of wire, when connected across battery
T₀ = Reference temperature
ρ = Resistivity of wire at temperature T
ρ₀ = Resistivity of wire at reference temperature
α = Temperature Coefficient of Resistance
From OHM'S LAW we know that;
ΔV = IR
I = ΔV/R
but, R = ρL/A (For Wire)
Therefore,
I = ΔV/(ρL/A)
I = ΔVA/ρL
but, ρ = ρ₀[1 + α (T₀ - T)]
Therefore,
I = ΔVA/Lρ₀[1 + α (T₀ - T)]
I = [ΔVA/Lρ₀] [1 + α (T₀ - T)]⁻¹
using Binomial Theorem:
(1 +x)⁻¹ = 1 - x + x² - x³ + ...
In case of [1 + α (T₀ - T)]⁻¹, x = α (T₀ - T).
Since, α generally has very low value. Thus, its higher powers can easily be neglected.
Therefore, using this Binomial Approximation, we can write:
[1 + α (T₀ - T)]⁻¹ = [1 - α (T₀ - T)]
Thus, the equation becomes:
I = ΔVA[1 - α (T₀ - T)]/Lρ₀
50 POINTS
What happens to kinetic energy of a moving object of the velocity decreases? (Look at photo multiple choice)
Answer: A Answers. Assuming that the terminal velocity doesn't change during the fall, then the kinetic energy would remain constant. However the terminal velocity decreases during the fall since the air becomes denser at lower altitudes.
Explanation:
What happens to the KE of an object when it slows down and heats up? - Quora. The kinetic energy goes down and the loss of the kinetic energy is through the production of heat energy. In real world this is due to friction, or an opposing force that decelerates the object, or a combination of both.
Answer:
Kinetic energy stays the same
a block (mass 0.6kg) is released from rest at point A at the top of a ramp inclined at 36.9 degress above the horizontal. The block moves a distance of 4m along the ramp to point b at the bottom of the ramp. How much work is done by gravity as the block goes from a to b
The motion of a free falling body is an example of __________ motion
Answer:
uniformly accelerated motion
Explanation:
The motion of the body where the acceleration is constant is known as uniformly accelerated motion. The value of the acceleration does not change with the function of time.
How much would it cost to cover the entire land area of United States with dollar bills? To answer this question, you may find the following information useful: The contiguous United States (all of the states minus Alaska and Hawaii) can be approximated as a rectangle that measures 1000 miles north to south and 3000 miles east to west, while Alaska has about one-fifth the area of the contiguous states and Hawaii is small enough to be ignored for this rough calculation. Also, note that a dollar bill measures around 6.5cm×15.5cm6.5cm×15.5cm.
Answer:
It would cost approximately $925,455,484 million to cover continental US and Alaska with $1 bills.
Explanation:
the area of a one dollar bill = 6.5 cm x 15.5 cm = 100.75 sq cm
the approximate area of continental US + Alaska = (1,000 miles x 3,000 miles) x 1.2 = 3,600,000 sq miles
each sq mile is roughly 2.58999 sq km, so the total area in sq km = 9,323,964 sq km
1 sq km = 1,000,000 sq meters
each sq meter = 10,000 sq cm
1 sq m = 10,000 / 100.75 = 99.25558 bills
1 sq km = 99,255,583.13 bills
9,323,964 sq km = 925,455,483,900,000 bills
If the US includes the area of the Hawaii and the Alaska. It would be costing approx. $925,455,484 mn to cover continental with the $1 bills.
As the area of 1 dollar bill = 6.5 cm x 15.5 cm = 100.75 cm2 The area of US + Alaska = (1,000 miles x 3,000 miles) x 1.2 = 3,600,000 sq Thus the total area in km2 = 9,323,964 sq km 1 km2 equals to 1,000,000 sq meters each sq meter = 10,000 sq cm 1 sq m = 10,000 / 100.75 = 99.25558 bills 1 km2 equals to 99,255,583.13 bills 9,323,964 sq km = 925,455,483,900,000 bills .Learn more about the would it cost to cover the entire land area.
brainly.com/question/17333335.
In order for an external force to do work on a system, Question 9 options: a) there must be a component of the force perpendicular to the motion. b) there must be a component of force parallel to the motion of the object. c) the force can be at any angle relative to the motion of the object.
Answer:
b) there must be a component of force parallel to the motion of the object.
Explanation:
We know that work done on a body by an external force is calculated by the formula given below:
W = F.d = Fd Cos θ
where,
W = Work Done by the force on the body
F = Magnitude of force
d = displacement of the body
θ = The angle between the direction of motion of the body and the force applied
It is clear from the formula of the work done, that "F Cosθ" represents the component of the force, that is acting in the direction of motion of the object or parallel to the direction of motion of the object. So, if there is no component of force parallel to motion of object, then this factor will become zero. As a result, the work done will also be zero.
Therefore, the correct option will be:
b) there must be a component of force parallel to the motion of object.
Having successfully ransomed a nearby city with her rail gun, Prof. Marcia Grail is looking for new and exciting ways to wreck science based havok for profit. She constructs a device capable of creating a constant intensity of electromagnetic radiation of any frequency over a circular space of radius 50cm. In order to do the most damage to biological systems (e.g. make those filthy normal humans pay), what type of electromagnetic wave does she need to set the device for
Answer:
Gamma radiation
Explanation:
Electromagnetic radiations are waves that do not require material medium for their propagation, and they travel at the same velocity. When these waves are arranged with respect to increasing frequency or decreasing wavelength, it forms a spectrum.
Since Prof. Marcia Grail's device can create an electromagnetic radiation of any frequency, the appropriate radiation to achieve her goal would be gamma radiation.
Gamma radiation is a high penetrating radiation with lethal effects on exposed biological molecules, causing induction of cancer and genetic mutations even at low level.
If the molar mass of helium is 4.0 g/mol and the molar mass of neon is 20.2 g/mol, then
a. All the atoms have exactly the same velocity.
b. All the atoms have the same average speed.
c.The average speed of the helium atoms is greater than the average speed of the neon atoms.
d. The average speed of the neon atoms is greater than the average speed of the helium atoms.
e. The atoms diffuse from high temperature to low temperature.
Three Small Identical Balls Have Charges -3 Times 10^-12 C, 8 Times 10^12 C And 4 Times 10^-12 C Respectively. They Are Brought In Contact And Then Separated. Calculate Charge On Each Ball.
Answer:
The charge in each ball will be 3 * 10^-12 C
Explanation:
(Assuming the correct charge of the second ball is 8 * 10^-12)
When the balls are brought in contact, all the charges are split evenly among then.
So first we need to find the total charge combined:
(-3 * 10^-12) + (8 * 10^-12) + (4 * 10^-12) = 9 * 10^-12 C
Then, when the balls are separated, each ball will have one third of the total charge, so in the end they will have the same charge:
(9 * 10^-12) / 3 = 3 * 10^-12 C
So the charge in each ball will be 3 * 10^-12 C
A particle moves with a velocity v in a circle of radius r, then its angular velocity is equal
to………….. and acts along the…………..
Answer:
Given that,
Speed = v
Radius = r
We have to ascertain the precise speed
Utilizing equation of speed
Where, v = velocity/speed
r = radius
= precise speed/ angular velocity
Angular Velocity :
The precise speed is characterized by the speed of turn.
The precise speed is straightforwardly corresponding to the direct speed and contrarily relative to the range of the molecule.
Subsequently, The precise speed is v/r
The dial of a scale looks like this: 00.0kg. A physicist placed a spring on it. The dial read 00.6kg. He then placed a metal chain on the scale, it read 02.1kg. The physicist, then decided to test Einstein’s equation, and compressed the spring and tied it with the chain and placed it on the scale. It read 02.7kg. Which of the following conclusions is the most likely the physicist will come to?
a. Einstein's equation has an error
b. The scale is broken
c. Compressing the spring didn't add energy
d. The scale's resolution is too low to read the change in mass
Answer:
d. The scale's resolution is too low to read the change in mass
Explanation:
If we want to find the change in energy of the spring, we will have to use the Hooke's Law. Hooke's Law states that:
F = kx
since,
w = Fd
dw = Fdx
integrating and using value of F, we get:
ΔE = (0.5)kx²
where,
ΔE = Energy added to spring
k = spring constant
x = displacement
The spring constant is typically in range of 4900 to 29400 N/m.
So if we take the extreme case of 29400 N/m and lets say we assume an unusually, extreme case of 1 m compression, we get the value of energy added to be:
ΔE = (0.5)(29400 N/m)(1 m)²
ΔE = 1.47 x 10⁴ J
Now, if we convert this energy to mass from Einstein's equation, we get:
ΔE = Δmc²
Δm = ΔE/c²
Δm = (1.47 x 10⁴ J)/(3 x 10⁸ m/s)²
Δm = 4.9 x 10⁻¹³ kg
As, you can see from the answer that even for the most extreme cases the value of mass associated with the additional energy is of very low magnitude.
Since, the scale only gives the mass value upto 1 decimal place.
Thus, it can not determine such a small change. So, the correct option is:
d. The scale's resolution is too low to read the change in mass
Suppose the kicker launches the ball at 60∘ instead of 30∘. Assuming that the goal is 4.55 m high and 40 m away, what minimum initial speed v0 would the ball need to have in order to just clear the goal?
Answer:22 m/s
Explanation:
Given
launch angle [tex]\theta =60^{\circ}[/tex]
height of goal [tex]h=4.55\ m[/tex]
and horizontal distance [tex]x=40\ m[/tex]
Suppose initial speed is [tex]u[/tex]
Trajectory of a Projectile is given by
[tex]y=x\tan \theta -\frac{1}{2}\frac{gx^2}{u^2\cos ^2\theta }[/tex]
substituting the values we get
[tex]4.55=40\tan (60)-0.5\times \frac{9.8\times (40)^2}{u^2\cdot \cos ^260 }[/tex]
[tex]4.55=69.28-0.5\times \frac{15,680}{u^2\cdot 0.25}[/tex]
[tex]\frac{31,360}{u^2}=69.28-4.55[/tex]
[tex]\frac{31,360}{64.73}=u^2[/tex]
[tex]u^2=484.47[/tex]
[tex]u=22.01\ m/s[/tex]
So, initial launch speed is [tex]22\ m/s[/tex]
A substance that produces H+ ions in solution is a.
A)soap
B)acid
C)salt
D)base
The speed of light changes when it goes from ethyl alcohol (n = 1.361) to carbon tetrachloride (n = 1.461). The ratio of the speed of light in carbon tetrachloride to the speed in ethyl alcohol is
Answer:
The ratio of the speed of light in carbon tetrachloride to the speed in ethyl alcohol is 0.93
Explanation:
The formula for the refractive index of a medium is given as:
n = c/v
where,
n = refractive index of the medium
c = speed of light in vacuum
v = speed of light in that medium
FOR CARBON TETRACHLORIDE:
n = n₁ = 1.461
v = v₁
Therefore,
1.461 = c/v₁
v₁ = c/1.461 ----- equation (1)
FOR ETHYL ALCOHOL:
n = n₂ = 1.361
v = v₂
Therefore,
1.361 = c/v₂
v₂ = c/1.361 ----- equation (2)
Now, dividing equation (1) by equation (2), we get:
v₁/v₂ = (c/1.461) / (c/1.361)
v₁/v₂ = 1.361/1.461
v₁/v₂ = 0.93
Hence, the ratio of the speed of light in carbon tetrachloride to the speed in ethyl alcohol is 0.93
Your friend says your body is made up of more than 99.9999% empty space. What do you think
Answer:false
Explanation:
A boat crosses a 200m wide river at 3m/s, north relative to water. The river flows at 1m/s as shown. What is the velocity of the boat as observed by a stationary observer on the river back from the boat departed?
Answer:
The resultant velocity = 3.16 m/s
Explanation:
Since the boat is moving North of the direction of the riverflow, the river would either be flowing westward or Eastward. The two motions form a right angle triangle with the resultant velocity being the hypotenuse of the traingle.
The resultant velocity will be given as ;
R² = B² + r²
Where B is the velocity of the boat and r is the velocity of the river
R² = 3² + 1²
R² = 10
R = √10 = 3.16 m/s
Therefore, the resultant velocity = 3.16 m/s
The frames for a pair of eyeglasses have a radius of 2.14 cm at 20.0°C. Lenses with radius of 2.16 cm have to be inserted into these frames. To what temperature must the technician heat the frames to accommodate the lenses? The frames are made of a material whose thermal expansion coefficient is 1.30 10-4/°C.
Answer:
The technician must heat the frame to a temperature of T₂ = 0.0072°C
Explanation:
Change in one-dimension of an object upon heating is given by the following equation:
ΔL = α L ΔT
For the radius this equation can be re-written as follows:
ΔR = α R ΔT
where,
ΔR = Change in Radius = 2.16 cm - 2.14 cm = 0.02 cm
α = Thermal Expansion Coefficient = 1.3 x 10⁻⁴ /°C
R = Original Radius = 2.14 cm
ΔT = Change in Temperature = T₂ - 20°C
Therefore,
0.02 cm = (1.3 x 10⁻⁴ /°C)(2.14 cm)(T₂ - 20°C)
T₂ - 20°C = (0.02 cm)/(1.3 x 10⁻⁴ /°C)(2.14 cm)
T₂ = 0.0072°C + 20°C
T₂ = 20.0072°C
Consider the interference/diffraction pattern from a double-slit arrangement of slit separation d = 6.60 um and slit width a. The wavelength of the monochromatic light incident normally upon the slits is 2n = d/10 (in air). There is a filter (of negligible thickness) placed on slit 2, so that the magnitude of the EM wave emitted from it is half of that emitted from slit 1. The space between the slits and the screen is filled with water, whose index of refraction is n = 1.33 (you can take noir as 1.00).
(a) What is the wavelength 2 of the light in water?
(b) If a << 1, What is the phase difference between the waves from slits 1 and 2?
(c) For a << 2, Derive an expression for the intensity / as a function of O and other relevant parameters, including the intensity at the center of the screen (where 0 = 0).
(d) Now suppose a = d/3. Redo part (b) above. How many interference maxima are present within the central diffraction peak? (Do not count the "clipped" maxima, if any.) (4) — E -2 d E ) в /Б/ = 1/5 | TT
Answer:
(a) λ = 0.496 um (b) S =2π Δ d sinθ/ λ (c) I =gI₀ (d) For the central diffraction peak, a total of 5 interference maxima are present or available.
Note: find an attached copy of a part of the solution to the given question below.
Explanation:
Solution
Recall that:
d = 6.6 um
λ₀ =d/10
λ₀ = 6.6 um
Now,
(a) We find the wavelength λ of the light in water.
Thus,
λ water = (λ₀ )/n
= 0.66/1.33
So,
λ water = λ = 0.496 um
(b) We find the phase difference between the waves from slit 1 and 2
Now,
if a <<d and a<<λ
Then the path difference between the rays will be
Δ S₂N = Δ d sinθ
Thus, the phase difference becomes,
S = 2π Δ/λ is S= 2π Δ d sinθ/ λ
(c) The next step is to derive an expression for the intensity I as function of O and other relevant parameters.
Now,
Let p be the point where these two rays interfere with each other.
Thus,
The electric field vector coming out from slot and and slot 2 is
E₁= E₀₁ cos (ks₁ p - wt) i
E₂ = E₀₂ cos (ks₂ p - wt) i
Note: Kindly find an attached copy of a part of the solution to the given question below.
An early submersible craft for deep-sea exploration was raised and lowered by a cable from a ship. When the craft was stationary, the tension in the cable was 6000 N. When the craft was lowered or raised at a steady rate, the motion through the water added an 1800 N drag force. What was the tension in the cable when the craft was being lowered to the seafloor?
Answer:
T = 4200N
Explanation:
When the submersible craft is at rest, the tension in the cable is 6000N.
With this information you can calculate the weight of the craft by summing the forces (the summation of the force is zero because the craft is at rest):
[tex]T-W=0\\\\W=T=6000N[/tex]
When the craft is going down with a constant speed, there is a drag force of 1800N. Then, by using the second Newton law you have:
[tex]T-W+F_d=0[/tex] (1)
Fd: drag force
The summation of the forces is zero because the craft moves with constant velocity, that is, there is no acceleration.
You calculate the new tension on the cable by solving the equation (1) for T:
[tex]T=W-Fd=6000N-1800N=4200N[/tex]
hence, the tension is 4200N
-
is a side effect of tobacco use.
A. Lightheadedness
B. Irritability
C. Dizziness
D. All of the above
Please select the best answer from the choices provided.
Answer:
D. all of the above
Explanation:
Whale sharks swim forward while ascending or descending. They swim along a straight-line path at a shallow angle as they move from the surface to deep water or from the depths to the surface. In one recorded dive, a shark started 50 m below the surface and swam at 0.85 m/s along a path tipped at a 13° angle above the horizontal until reaching the surface.
Required:
a. What was the horizontal distance between the shark's starting and ending positions?
b. What was the total distance that the shark swam?
c. How much time did this motion take?
Answer:
a)217m
b)222m
c)261s
Explanation:
Considering the figure in the attachment
depth 'h'=50m
speed 'v'=0.85m/s
angle 'θ'=13°
a) we have
[tex]\frac{h}{x}[/tex]=tanθ
x = h / tanθ => 50/tan13°
x=216.57≈217m
b)we have
[tex]\frac{h}{d}[/tex]=sinθ
d=h / sinθ => 50 / sin13°
d=222.27≈222m
c)Time needed = d / v = 222/0.85
261.18≈261s
Based on the information in the table, which two elements are most likely in the same group, and why?
A table with 5 columns and 5 rows labeled facts about 4 elements. The first column labeled element has entries bismuth (B i), nitrogen (N), oxygen (O), sodium (N a), thallium (T l). The second column labeled atomic mass (a m u) has entries 209, 14, 16, 23, 204. The third column labeled total electrons has entries 83, 7, 8, 11, 81. The fourth column labeled valence electrons has entries 5, 5, 6, 1, 3. The fifth column labeled year isolated has entries 1753, 1772, 1772, 1807, 1861.
bismuth and thallium, because their atomic masses are very similar
nitrogen and oxygen, because they were both first isolated in the same year
sodium and thallium, because their names both end in the same suffix: -ium
bismuth and nitrogen, because they have the same number of valence electrons
Answer: bismuth and nitrogen, because they have the same number of valence electrons
Explanation:
Elements are distributed in groups and periods in a periodic table.
Elements that belong to same groups will show similar chemical properties because they have same number of valence electrons.
The number of valence electrons in Bismuth and nitrogen are 5 and thus thus they will show similar chemical properties and thus belong to the same group.
The atomic masses of elements in a group will differ drastically.
The group number has got nothing to be the isolation year.
Thus bismuth and nitrogen belong to same group because they have the same number of valence electrons
Answer:
D
Explanation:
i got it right in my test
A p-type Si sample is used in the Haynes-Shockley experiment. The length of the sample is 2 cm, and two probes are separated by 1.8 cm. Voltage applied at the two ends is 5 V. A pulse arrives at the collection point at 0.608 ms, and the separation of the pulse is 180 sec. Calculate mobility and diffusion coefficient for minority carriers. Verify it from the Einstein relation.
Answer:
Mobility of the minority carriers, [tex]\mu_{n} =1184.21 cm^{2} /V-sec[/tex]
Diffusion coefficient for minority carriers,[tex]D_{n} = 29.20 cm^2 /s[/tex]
Verified from Einstein relation as [tex]\frac{D_{n} }{\mu_{n} } = 25 mV[/tex]
Explanation:
Length of sample, [tex]l_{s} = 2 cm[/tex]
Separation between the two probes, L = 1.8 cm
Drift time, [tex]t_{d} = 0.608 ms[/tex]
Applied voltage, V = 5 V
Mobility of the minority carriers ( electrons), [tex]\mu_{n} = \frac{V_{d} }{E}[/tex]
Where the drift velocity, [tex]V_{d} = \frac{L}{t_{d} }[/tex]
[tex]V_{d} = \frac{1.8}{0.608 * 10^{-3} } \\V_{d} = 2960.53 cm/s[/tex]
and the Electric field strength, [tex]E = \frac{V}{l_{s} }[/tex]
E = 5/2
E = 2.5 V/cm
Mobility of the minority carriers:
[tex]\mu_{n} = 2960.53/2.5\\\mu_{n} =1184.21 cm^{2} /V-sec[/tex]
The electron diffusion coefficient, [tex]D_{n} = \frac{(\triangle x)^{2} }{16 t_{d} }[/tex]
[tex]\triangle x = (\triangle t )V_{d}[/tex], where Δt = separation of pulse seen in an oscilloscope in time( it should be in micro second range)
[tex]\triangle x = \frac{(\triangle t) L}{t_{d} } \\\triangle x = \frac{180*10^{-6} * 1.8}{0.608*10^{-3} }\\\triangle x =0.533 cm[/tex]
[tex]D_{n} = \frac{0.533^{2} }{16 * 0.608 * 10^{-3} }\\D_{n} = 29.20 cm^2 /s[/tex]
For the Einstein equation to be satisfied, [tex]\frac{D_{n} }{\mu_{n} } = \frac{KT}{q} = 0.025 V[/tex]
[tex]\frac{D_{n} }{\mu_{n} } = \frac{29.20}{1184.21} \\\frac{D_{n} }{\mu_{n} } = 0.025 = 25 mV[/tex]
Verified.
Two fledglings leave a nest which is 2.50 m off the ground. One drops from rest and then 0.10 s later the second charges out of the nest with a velocity with horizontal and
downward components of 1.00 and 1.50 m/s, respectively.
1. Which fledgling hits the ground first (first or second)?
2. With what speed, in m/s, does the first fledgling hit the ground?
3. With what speed, in m/s, does the second fledgling hit the ground?
4. Which fledgling travels the greater displacement?
5. How far does a parent bird have to travel on the ground between the two fledglings, in m, to make sure they are alright?
t Answer:
1) the time of the pigeon 1 is less, so it comes first
2) v = - 6,997 m / s , 3) v = 10.15 m / s ,
4) the displacement of the second point in greater
5) x = 0.883 m
Explanation:
For this exercise we will use the kinematics equations
1) ask which chick reaches the ground first
we calculate for the first chick that has zero initial velocity
y = y₀ + v₀ t - ½ g t²
0 = yo - ½ g t²
t = √ 2 y₀ / g
let's calculate
t = √ (2 2.50 / 9.8)
t = 0.714 s
We calculate the time it takes for the second chick that has velocity v = (1 i ^ - 1.5 j⁾ m / s
y = y₀ + v₀t - ½ g t²
0 = 2.5 - 1.5 t - ½ 9.8 t²
4.9 t² + 1.5 t - 2.5 = 0
t² + 0.306 t - 0.510 = 0
we solve the quadratic equation
t = [0.306 ± √ (0.306² - 4 (-0.510))] / 2
t = [0.306 ± 1.46] / 2
The results are
t₁ = -0.577 s
t₂ = 0.883 m / s
we take positive time as correct
the time of the pigeon 1 is less, so it comes first
2) the speed of the first chick is
v = v₀ - g t
we can see that
v = -gt
v = - 9.8 0.714
v = - 6,997 m / s
the negative sign indicates that the speed is down
3) the speed of the other bird is
v = -1.5 - 9.8 0.883
v = 10.15 m / s
4) which chick has the greatest displacement. The first point falls vertically and its displacement is y₀
The second point describes a parabola and its displacement is
d = √ (x² + y₀²)
therefore we see that the displacement of the second point in greater
5) calculate the horizontal displacement of the second point
x = vx t
x = 1 0.883
x = 0.883 m
A television weighs 8.50 pounds. How many grams is this? (Hint: You need to
use two unit conversion fractions. 1 pound equals about 0.454 kg.)
Answer:
3859 grams
Explanation:
Given: Weight of a television = 8.50 pounds
To find: Weight of a television in grams
Solution:
1 pound = 0.454 kg and 1 kg = 1000 g
So,
1 pound = 0.454 × 1000 = 454 grams
8.50 pounds = 8.50 × 454 = 3859 grams
Therefore,
Weight of television in grams = 3859 grams
4. An electric iron has a
power rating of 750W
a. How many joules of
electric energy does it
change into heat energy
every second?
b. How many joules of
work can it do in 3
seconds
c. How long does it take
the iron to do 1500J of
work?
5. Use the kinetic particle
theory to explain why a
solid has a definite shape
and liquid has none.
Answer:
4.
a) W = 750 J
b) W = 2250 J
c) t = 2 sec
5. Answered in explanation
Explanation:
4.
The formula of power is given as:
P = W/t
where,
P = Power
W = Work Done
t = Time Taken
a)
Here,
P = 750 W
t = 1 sec
W = ?
Therefore,
750 W = W/1 sec
W = 750 J
b)
Here,
P = 750 W
t = 3 sec
W = ?
Therefore,
750 W = W/3 sec
W = (750 W)(3 sec)
W = 2250 J
c)
Here,
P = 750 W
t = ?
W = 1500 J
Therefore,
750 W = 1500 J/t
t = 1500 J/750 W
t = 2 sec
5.
According to Kinetic Particle Theory, the molecules are tightly packed with each other, by strong inter-molecular forces and they can only vibrate at their position. While, molecules or particles in liquids have lesser attractive forces among them. They can move in layers and can take the shape of any container. This is the reason why solid has a definite shape and liquid has none.
Helppppp please thanks
Answer:
It represents battery
Answer: I think #1 represent lights and #2 represent the battery
You are working in a biology lab and learning to use a new ultracentrifuge for blood ts. The specifications for the centrifuge say that a red blood cell rotating in the ultracentrifuge moves at 470 m/s and has a radial acceleration of 150,000 g's (that is, 150,000 times 9.8 m/s2). The radius of the centrifuge is 0.15 m. You wonder if this claim is correct.
What is the radial acceleration of the ultra-centrifuge using calculations ?
Answer:
The radial acceleration is [tex]a_r = 1472667 \ m/s^2[/tex]
Explanation:
From the question we are told that
The velocity of the red blood cell is [tex]v_r = 470 \ m/s[/tex]
The radial acceleration of the red blood cell is [tex]a_r = 150000g = 150000*9.8 = 1470000 \ m/s^2[/tex]
The radius of the centrifuge is [tex]r = 0.15 \ m[/tex]
Generally radial acceleration is mathematically represented as
[tex]a_r = \frac{v^2}{r}[/tex]
substituting values
[tex]a_r = \frac{470^2}{0.15}[/tex]
[tex]a_r = 1472667 \ m/s^2[/tex]
Convert 45m/s to mph. Choose the best answer to two significant figures.
101mph
45mph
200mph
24mph
Explanation:
45 m/s × (3.28 ft/m) × (1 mi / 5280 ft) × (3600 s/h) = 101 mph
This is actually 3 significant figures. To write in 2 significant figures, you must use scientific notation:
1.0×10² mph
standing on a hill that is 3.5 m high above level ground. A wall that is on level ground is 15 m away from the student. The student decides to throw a ball horizontally at a combined height of 4.5 m, at the wall. The student measures the time it takes to make an impact and finds it to be 0.65 s. At what height, from the ground, did the student hit the side of the wal
Answer:
The height will be "2.42 m".
Explanation:
The given values are:
time, t = 0.65 seconds
∴ g = 9.8
As we know,
⇒ [tex]x=vt+\frac{1}{2}gt^2[/tex]
∴ v = 0
On putting the estimated values, we get
⇒ [tex]=0+\frac{1}{2}\times 9.8\times (0.65)^2[/tex]
⇒ [tex]=\frac{1}{2}\times 9.8\times 0.4225[/tex]
⇒ [tex]=\frac{1}{2}\times 4.1405[/tex]
⇒ [tex]=2.07025 \ m[/tex]
Now,
Height, [tex]h=4.5-2.07025[/tex]
⇒ [tex]=2.42 \ m[/tex]
