Answer:
There are [tex]9.053\times 10^{14}[/tex] charge carriers.
Explanation:
Given that,
Current density (current per unit time), [tex]\dfrac{I}{A}=1.55\ A/cm^2=0.000155\ A/m^2[/tex]
The drift velocity of the electron, [tex]v_d=107\ cm/s=1.07\ m/s[/tex]
We need to find the no charge carriers present. The formula for drift velocity is given by :
[tex]v_d=\dfrac{I}{neA}\\\\\text{n is no of charge carriers}\\\\n=\dfrac{I}{eAv_d}\\\\n=\dfrac{0.000155}{1.6\times 10^{-19}\times 1.07}\\\\n=9.053\times 10^{14}[/tex]
So, there are [tex]9.053\times 10^{14}[/tex] charge carriers.
Besides being asleep and being awake list the different states of consciousness
Answer:
bias.
Consciousness.
Hypnosis.
Priming.
Sleep.
Trance.
Explanation:
An 800 N table is slid across the kitchen floor by pushing with a force of 100 N. If the table moves at a constant speed, what is the friction force with the floor?
A. 100 N
B. 8 N
C. 800 N
D. over 800 N
Answer:
A. 100 N
Explanation:
Given
Table = 800N
Pushing Force = 100N
Required
Determine the friction force with the floor
From the question, we understand that the table moves at a constant speed.
In this kind of situation, the friction force equals to the pushing force.
Hence:
Friction Force = Pushing Force
Friction Force = 100N
If the first stage provides a thrust of 5.25 mega-newtons [MN] and the space shuttle has a mass of 4,470,000 pound-mass [lbm], what is the acceleration of the spacecraft in miles per hour squared [mi/h2]
Answer:
20,861.65 mi/h²
Explanation:
We convert 4,470,000 pound-mass [lbm], to kg. Since 2.205 lbm = 1 kg, then 4,470,000 lbm = 4,470,000 lbm × 1 kg/2.205 = 2,027,210.88 kg
Since Force , F = ma where m = mass and a = acceleration, and our force of thrust , F = 5.25 MN = 5,250,000 N and or mass = mass of spacecraft = 2,027,210.88 kg, we then find the acceleration, a.
a = F/m = 5,250,000 N/2,027,210.88 kg = 2.59 m/s².
We now convert this acceleration into miles per hour. Since 1 mile = 1609 meters and 60 × 60 s = 1 hour ⇒ 3600 s = 1 hour, Our conversion factor for meter to mile is 1 mile/1609 m and that for second to hour is 3600 s/1 hour. We square the conversion factor for the time so we have (3600 s/1 hour)².
Multiplying both conversion factors with our acceleration, we have
a = 2.59 m/s²
= 2.59 m/s² × 1 mile/1609 m × (3600 s/1 hour)²
= 33566440/1609 miles/hour²
= 20,861.65 mi/h²
= 20,861.65 miles per hour squared
Which answer is it
0.32 m/s
3.2 m/s
9.8 m/s
40 m/s
Answer:
3.2 m/s
Explanation:
you take the amount of blocks height wise, and the number of blocks with wise, and divide. height over with, so in this case 16/5. 16 divided by 5 is 3.2
Protons can be accelerated to speeds near the speed of light in particle accelerators. Calculate the wavelength of a proton moving at 2.90 x 108 m/s if the proton was a mass of 1.673 x 10-24 g. 2. Give the values for the quantum numbers associated with the following orbitals a) 2p b) 3s c) 5d
Answer:
1. λ = 1.4x10⁻¹⁵ m
2. a) n=2, l=1, [tex]m_{l}[/tex]= -1, 0, +1, [tex]m_{s}[/tex] = +/- (1/2)
b) n=3, l=0, [tex]m_{l}[/tex]= 0, [tex]m_{s}[/tex] = +/- (1/2)
c) n=5, l=2, [tex]m_{l}[/tex]= -2, -1, 0, +1, +2, [tex]m_{s}[/tex] = +/- (1/2)
Explanation:
1. The proton's wavelength can be found using the Broglie equation:
[tex] \lambda = \frac{h}{mv} [/tex]
Where:
h: is the Planck's constant = 6.62x10⁻³⁴ J.s
m: is the proton's mass = 1.673x10⁻²⁴ g = 1.673x10⁻²⁷ kg
v: is the speed of the proton = 2.90x10⁸ m/s
The wavelength is:
[tex] \lambda = \frac{h}{mv} = \frac{6.62 \cdot 10^{-34} J.s}{1.673 \cdot 10 ^{-27} kg*2.90 \cdot 10^{8} m/s} = 1.4 \cdot 10^{-15} m [/tex]
2. a) 2p
We have:
n: principal quantum number = 2
l: angular momentum quantum number = 1 (since is "p")
[tex]m_{l}[/tex]: magnetic quantum number = {-l,... 0 ... +l}
Since l = 1 → [tex]m_{l} = -1, 0, +1[/tex]
[tex]m_{s}[/tex]: is the spin quantum number = +/- (1/2)
b) 3s:
n = 3
l = 0 (since is "s")
[tex]m_{l}[/tex] = 0
[tex]m_{s}[/tex] = +/- (1/2)
c) 5d:
n = 5
l = 2 (since is "d")
[tex]m_{l}[/tex] = -2, -1, 0, +1, +2
[tex]m_{s}[/tex] = +/- (1/2)
I hope it helps you!
Answer:
(1) The wavelength of the proton is 1.366 x 10⁻¹⁵ m
(2) 2p( l = 1, ml = -1,0,+1)
3s( n = 3, l = 0, ml = 0)
5d ( l = 2, ml = -2,-1,0,+1,+2)
Explanation:
Given;
mass of the proton; m = 1.673 x 10⁻²⁴ g = 1.673 x 10⁻²⁷ kg
velocity of the proton, v = 2.9 x 10⁸ m/s
The wavelength of the proton is calculated by applying De Broglie's equation;
[tex]\lambda = \frac{h}{mv}[/tex]
where;
h is Planck's constant = 6.626 x 10⁻³⁴ J/s
Substitute the given values and solve for wavelength of the proton;
[tex]\lambda = \frac{h}{mv}\\\\ \lambda = \frac{(6.626*10^{-34})}{(1.673*10^{-27})(2.9*10^8)}\\\\\lambda = 1.366 *10^{-15} \ m[/tex]
(2) the values for the quantum numbers associated with the following orbitals is given by;
n, which represents Principal Quantum number
[tex]l,[/tex] which represents Azimuthal Quantum number
[tex]m_l,[/tex] which represents Magnetic Quantum number
(a) 2p (number of orbital = 3):
[tex]l= 1\\m_l = -1,0,+1[/tex]
(b) 3s (number of orbital = 1):
[tex]n= 3\\l=0\\m_l= 0[/tex]
(c) 5d (number of orbital = 5)
[tex]l=2\\m_l = 2, -1, 0, +1, +2[/tex]
A golf ball rolls up a hill toward a miniature-golf hole. Assume the direction toward the hole is positive. If the golf ball starts with a speed of 2.0 m/s and slows at a constant rate of 0.50 m/s2. What is the golf ball’s velocity if the constant acceleration continues for 6.0 s? *
Answer:
0.2533333334
Explanation:
It is simple, just do the math
An animation is being created with a thousand tennis balls falling out of a truck, bouncing off the ground and each other. The animator wants to make the scene realistic but also wants to be efficient in the program. Which of these characteristics could be ignored when establishing boundary conditions for the system?
aerodynamic properties
gravitational forces
Newton's laws of motion
the diameter of the tennis balls
Answer:
aerodynamic properties
Explanation:
For this test the correct answers are:
1. Initial conditions
2. the speed of a character at the moment of contact
3. the hardness of the sidewalk where the mug landed
4. the location where potential energy is zero
5. aerodynamic properties
I just took this test and these were the correct answers.
What is the answer to this question?
Answer:buttt
It mean b
Explanation:
what vehicle are runaway ramps designed for
Answer:truck
Explanation:
truck
if a cross country runner covers 347 meters in 134 seconds what is her speed
A dart is loaded into a compressed spring. Starting from rest the spring is released and shoots the dart up into the air. From the instant the dart is released (before it starts to move) until it reaches its maximum height, what is the Net work done on the dart
Answer:
Wnet = 0
Explanation:
The work-energy theorem says that the total net work done on an object, is equal to the change in the kinetic energy of that object.In this case, the total change in kinetic energy of the dart is zero, due to the initial status is at rest (loaded into a compressed spring) and the final is also at rest (when it reaches to its maximum height).The total work done on the dart can be written as follows:[tex]W_{net} = \frac{1}{2} k* \Delta x^{2} - m*g*h_{max} (1)[/tex]
Now, in absence of friction, all the elastic potential energy becomes gravitational potential energy, so the following is true:[tex]\frac{1}{2} k* \Delta x^{2} = m*g*h_{max} (2)[/tex]
From (1) and (2) we conclude that Wnet = 0.A 0.26 kg rock is thrown vertically upward from the top of a cliff that is 32 m high. When it hits the ground at the base of the cliff, the rock has a speed of 29 m/s. Assuming that air resistance can be ignore and using conservation of mechanical energy, find (a) the initial speed of the rock and (b) the greatest height of the rock as measured from the base of the cliff.
Answer:
a) The initial speed of the rock is approximately 14.607 meters per second.
b) The greatest height of the rock from the base of the cliff is 42.878 meters.
Explanation:
a) The rock experiments a free-fall motion, that is a vertical uniform accelerated motion due to gravity, in which air friction and effects of Earth's rotation. By Principle of Energy Conservation we have the following model:
[tex]U_{g,1}+K_{1} = U_{g,2}+K_{2}[/tex] (Eq. 1)
Where:
[tex]U_{g,1}[/tex], [tex]U_{g,2}[/tex] - Initial and final gravitational potential energies, measured in joules.
[tex]K_{1}[/tex], [tex]K_{2}[/tex] - Initial and final translational kinetic energies, measured in joules.
By definitions of gravitational potential and translational kinetic energies we expand and simplify the equation above:
[tex]m\cdot g\cdot (y_{1}-y_{2})= \frac{1}{2}\cdot m\cdot (v_{2}^{2}-v_{1}^{2})[/tex]
[tex]g\cdot (y_{1}-y_{2}) = \frac{1}{2}\cdot (v_{2}^{2}-v_{1}^{2})[/tex] (Eq. 2)
Where:
[tex]g[/tex] - Gravitational acceleration, measured in meters per square second.
[tex]y_{1}[/tex], [tex]y_{2}[/tex] - Initial and final height, measured in meters.
[tex]v_{1}[/tex], [tex]v_{2}[/tex] - Initial and final speed of the rock, measured in meters per second.
If we know that [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], [tex]y_{1} = 32\,m[/tex], [tex]y_{2} = 0\,m[/tex] and [tex]v_{2} = 29\,\frac{m}{s}[/tex], then the equation is:
[tex]\left(9.807\,\frac{m}{s^{2}} \right)\cdot (32\,m-0\,m) = \frac{1}{2}\cdot \left[\left(29\,\frac{m}{s} \right)^{2}-v_{1}^{2}\right][/tex]
[tex]313.824 = 420.5-0.5\cdot v_{1}^{2}[/tex]
[tex]0.5\cdot v_{1}^{2} = 106.676[/tex]
[tex]v_{1} \approx 14.607\,\frac{m}{s}[/tex]
The initial speed of the rock is approximately 14.607 meters per second.
b) We use (Eq. 1) once again and if we know that [tex]g =9.807\,\frac{m}{s^{2}}[/tex], [tex]y_{1} = 32\,m[/tex], [tex]v_{1} \approx 14.607\,\frac{m}{s}[/tex] and [tex]v_{2} = 0\,\frac{m}{s}[/tex], then the equation is:
[tex]\left(9.807\,\frac{m}{s^{2}} \right)\cdot (32\,m-y_{2}) = \frac{1}{2}\cdot \left[\left(0\,\frac{m}{s} \right)^{2}-\left(14.607\,\frac{m}{s} \right)^{2}\right][/tex]
[tex]313.824-9.807\cdot y_{2} = -106.682[/tex]
[tex]9.807\cdot y_{2} = 420.506[/tex]
[tex]y_{2} = 42.878\,m[/tex]
The greatest height of the rock from the base of the cliff is 42.878 meters.
A bus accelerates from rest at 2.25 m/s2 for 12.3 seconds.
Answer:
6.15m for 12.3 seconds
Explanation:
You are given two vectors A⃗ =−3.00i^+7.00j^ and B⃗ =5.00i^+2.00j^. Let counterclockwise angles be positive.
1- What angle θA, where 0∘≤θA<360∘, does A⃗ make with the +x-axis?
2-What angle θB, where 0∘≤θB<360∘, does B⃗ make with the +x-axis?
Express your answer in degrees.
3-Vector C⃗ is the sum of A⃗ and B⃗ , so C⃗ =A⃗ +B⃗ . What angle θC, where 0∘≤θC<360∘, does C⃗ make with the +x-axis?
Express your answer in degrees
Answer:
Ax = -3 By= 7 tan phi = -7 / 3 phi = -66.8
Since theta is in the second quadrant theta = 180 - 66.8 = 113.2 deg
Bx = 5 By = 2 tan theta = .4 theta = 21.8 deg
Cx = Ax + Bx = 2 Cy = Ay + By = 9
tan theta = 9 / 2 = 4.5 theta = 77.5 deg
1) The solution is [tex]\theta_{1} \approx 113.141^{\circ}[/tex].
2) The solution is [tex]\theta_{1} \approx 21.874^{\circ}[/tex].
3) The solution is [tex]\theta_{1} \approx 77.467^{\circ}[/tex].
1) In this case, we can determine the angle, measured in sexagesimal degrees, with respect to the x-axis by definition of dot point:
[tex]\cos \theta = \frac{\vec A\,\bullet \, \vec u}{\|\vec A\|\cdot \|\vec u\|}[/tex] (1)
Where [tex]\|\vec A\|[/tex] and [tex]\|\vec u\|[/tex] are the norms of [tex]\vec A[/tex] and [tex]\vec u[/tex].
If we know that [tex]\vec A = (-3, 7)[/tex] and [tex]\vec u = (1, 0)[/tex], then the angle that [tex]\vec A[/tex] make with the x-axis is:
[tex]\|\vec A\| = \sqrt{(-3)^{2}+7^{2}}[/tex]
[tex]\|\vec A\| = \sqrt{58}[/tex]
[tex]\|\vec u\| = 1[/tex]
[tex]\cos \theta = \frac{(-3)\cdot (1)}{\sqrt{58} \cdot 1}[/tex]
[tex]\cos \theta \approx -0.393[/tex]
The solution is [tex]\theta_{1} \approx 113.141^{\circ}[/tex].
2) In this case, we can determine the angle, measured in sexagesimal degrees, with respect to the x-axis by definition of dot point:
[tex]\cos \theta = \frac{\vec B\,\bullet \, \vec u}{\|\vec B\|\cdot \|\vec u\|}[/tex] (1)
Where [tex]\|\vec B\|[/tex] and [tex]\|\vec u\|[/tex] are the norms of [tex]\vec B[/tex] and [tex]\vec u[/tex].
If we know that [tex]\vec B = (5, 2)[/tex] and [tex]\vec u = (1, 0)[/tex], then the angle that [tex]\vec A[/tex] make with the x-axis is:
[tex]\|\vec B\| = \sqrt{5^{2}+2^{2}}[/tex]
[tex]\|\vec B\| = \sqrt{29}[/tex]
[tex]\|\vec u\| = 1[/tex]
[tex]\cos \theta = \frac{(5)\cdot (1)}{\sqrt{29} \cdot 1}[/tex]
[tex]\cos \theta \approx 0.928[/tex]
The solution is [tex]\theta_{1} \approx 21.874^{\circ}[/tex].
3) In this case, we need to find the value of [tex]\vec C[/tex] by vectorial sum before calculating the angles with respect to x-axis:
[tex]\vec C = \vec A + \vec B = (-3, 7) + (5, 2)[/tex]
[tex]\vec C = (2, 9)[/tex]
In this case, we can determine the angle, measured in sexagesimal degrees, with respect to the x-axis by definition of dot point:
[tex]\cos \theta = \frac{\vec C\,\bullet \, \vec u}{\|\vec C\|\cdot \|\vec u\|}[/tex] (1)
Where [tex]\|\vec C\|[/tex] and [tex]\|\vec u\|[/tex] are the norms of [tex]\vec C[/tex] and [tex]\vec u[/tex].
If we know that [tex]\vec B = (2, 9)[/tex] and [tex]\vec u = (1, 0)[/tex], then the angle that [tex]\vec A[/tex] make with the x-axis is:
[tex]\|\vec C\| = \sqrt{2^{2}+9^{2}}[/tex]
[tex]\|\vec C\| = \sqrt{85}[/tex]
[tex]\|\vec u\| = 1[/tex]
[tex]\cos \theta = \frac{(2)\cdot (1)}{\sqrt{85} \cdot 1}[/tex]
[tex]\cos \theta \approx 0.217[/tex]
The solution is [tex]\theta_{1} \approx 77.467^{\circ}[/tex].
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how much tension (fm) must be supplied at the triceps to stabalize the arm against an external force (fe) of 200 N given dm
The missing part of the question is :
given dm-2 cm and de 25 cm
Answer:
The correct answer is = 2500 N
Explanation:
Given:
External force (fe) = 200 N
dm- 2 cm = 0.02 m
de = 25 cm = 0.25 m
Balancing torque about elbow
Fm*dm = Fe*de
By putting value
fm*0.02 = 200*0.25
fm= 200*0.25/0.02
fm = 50/0.02
fm = 2500 N
Thus, the correct answer is = 2500 N
All of the following would motivate someone toward attaining a career goal except
A expectation of money
B. low self-esteem
C. belief in success
D. financial stability
All would motivate someone toward attaining a career goal except low self-esteem. The correct option is B.
What is self-esteem?We tend to feel better about ourselves and about life in general when we have healthy self-esteem.
It improves our ability to deal with life's ups and downs. When we have low self-esteem, we tend to view ourselves and our lives in a more negative and critical light.
People's choices and decisions are heavily influenced by their self-esteem. In other words, self-esteem motivates people by making it more or less likely that they will take care of themselves and reach their full potential.
Except for low self-esteem, everyone would not motivate somebody to pursue a career goal.
Thus, the correct option is B as low self esteem cannot motivate anyone for career goal.
For more details regarding self-esteem, visit:
https://brainly.com/question/28768312
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A 5.00 gg bullet is fired horizontally into a 1.20 kgkg wooden block resting on a horizontal surface. The coefficient of kinetic friction between block and surface is 0.20. The bullet remains embedded in the block, which is observed to slide 0.390 mm along the surface before stopping.
Answer:
V= 295.2m/s
Explanation:
Given that mass of bullet = 5.0g = 0.005kg
Mass of wooden block = 1.20kg
The coefficient of kinetic friction between block and surface = 0.20
The force of friction of the block
Fk = (μ) mg
Fk = 0.2×1.2×9.8m/s
F = 2.352N
= Force × distance
The work done = 2.352× 0.390
= 0.91728J
The initial velocity of the block can be calculated
1/2mv^2
0.5×1.2×v^2 = 0.917J
0.6V^2 = 0.917J
V^2 = 0.917J/0.6
V^2 = 1.52
V = √1.52
V = 1.23m/s
Now we shall use conservation of momentum to find the velocity of the bullet
0.005kg×v = 1.2kg × 1.23m/s
0.005v = 1.476
V= 1.476/0.005
V= 295.2m/s
A block with mass 0.50 kg is forced against a horizontal spring of negligible mass, compressing the spring a distance of 0.20 m . When released, the block moves on a horizontal tabletop for 1.00 m before coming to rest. The spring constant k is 100 N/m. What is the coefficient of kinetic friction between the block and the tabletop
Answer:
μ = 0.41
Explanation:
[tex]\frac{1}{2} kx^2 = \mu mgd[/tex]
[tex]\mu = (\frac{1}{2} kx^2)/mgd[/tex]
[tex]\mu = (\frac{1}{2} (100)(0.2)^2)/(0.5)(9.81)(1)[/tex]
[tex]\mu = 0.41[/tex]
The coefficient of kinetic friction between the block and the tabletop is 0.41.
The given parameters;
mass of the block, m = 0.5 kgextension of the spring, x = 0.2 mspring constant, K = 100 N/mThe coefficient of kinetic friction between the block and the tabletop by applying the principle of conservation of energy as shown below.
Fd = Uₓ
μmgd = ¹/₂kx²
where;
μ is the coefficient of kinetic frictionThe coefficient of kinetic friction between the block and the tabletop;
[tex]\mu_k = \frac{kx^2}{mgd} \\\\\mu_k = \frac{100\times 0.2^2}{2\times 0.5\times9.8 \times 1 } \\\\\mu_k = 0.41[/tex]
Thus, the coefficient of kinetic friction between the block and the tabletop is 0.41.
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What is the weight, on Earth, of a book with a mass of 1.5 kg?
0 1,5 N
0 6,5 N
O 11.3N
0 14 7 N
Answer:
[tex]\boxed {\tt 14.7 \ Newtons}[/tex]
Explanation:
Weight is force due to gravity.
The formula for weight is:
[tex]W=m*g[/tex]
where m is the mass and g is the acceleration of gravity.
On Earth, the acceleration due to gravity is 9.8 meters per seconds squared.
The mass of the book is 1.5 kilograms. Therefore:
[tex]m= 1.5 \ kg \\g=9.8 \ m/s^2[/tex]
Substitute the values into the formula.
[tex]W= 1.5 \ kg *9.8 \ m/s^2[/tex]
Multiply.
[tex]W=14.7 \ kg*m/s^2[/tex]
1 kg meter per second squared is equal to 1 Newton, so our answer of 14.7 kg*m/s² is equal to 14.7 Newtons.
[tex]W=14.7 \ N[/tex]
The weight of a 1.5 kilogram book on Earth is 14.7 Newtons.
Answer:
Answer:
Explanation:
Weight is force due to gravity.
The formula for weight is:
where m is the mass and g is the acceleration of gravity.
On Earth, the acceleration due to gravity is 9.8 meters per seconds squared.
The mass of the book is 1.5 kilograms. Therefore:
Substitute the values into the formula.
Multiply.
1 kg meter per second squared is equal to 1 Newton, so our answer of 14.7 kg*m/s² is equal to 14.7 Newtons.
The weight of a 1.5 kilogram book on Earth is 14.7 Newtons.
Explanation:
9. A 0.25-kg mass is attached to a string and swung in a vertical circle whose radius is 0.75 m. At the bottom of the circle, the mass is observed to have a speed of 10 m/s. What is the magnitude of the tension in the string at that point
Answer:
T = 33.34 N
Explanation:
Given that,
Mass attached to the spring, m = 0.25 kg
Radius of the vertical circle, r = 0.75 m
Speed of the mass, v = 10 m/s
We need to find the magnitude of the tension in the string at that point. We know that the tension in the string is equal to the centripetal force acting on it. It can be given by
[tex]T=\dfrac{mv^2}{r}\\\\T=\dfrac{0.25\times (10)^2}{0.75}\\\\T=33.34\ N[/tex]
So, the tension in the string is 33.34 N.
A 80-kg person stands at rest on a scale while pulling vertically downwards on a rope that is above them. Use g = 9.80 m/s2.
What is the critical magnitude of tension that the rope is pulling on the person when the person just begins to lift off the scale?
Answer:
784 N
Explanation:
Given that a 80-kg person stands at rest on a scale while pulling vertically downwards on a rope that is above them. Use g = 9.80 m/s2.
The weight of the person = mg
The weight = 80 × 9.8
The weight = 784 N
The critical magnitude of tension that the rope is pulling on the person when the person just begins to lift off the scale will be equal to the weight of the person which is 784 N
A solenoid with 2,781 turns has a radius of 74.7 mm and is 38.4 cm long. If this solenoid carries a current of 68.4 A, what is the magnitude of the magnetic field near the center of the solenoid
Answer:
B = 0.62 T
Explanation:
Assuming that the solenoid can be treated as of infinite length (as its radius is much smaller than the length) we can consider that the magnetic field created by the current, near the center of it, is constant, and that it's almost zero outside it.Applying Ampere's law to a closed surface that encloses the wire carrying the current, we get:[tex]B= \mu_{o} *n*I (1)[/tex]
where μ₀ = 4*π*10⁻⁷ T*m/A, n = turns per unit length = 2,781/0.384 m and I = 68.4 AReplacing by the givens in (1) we get:B= 4*π*10⁻⁷ T*m/A* 7242.2 (1 /m) * 68.4 A = 0.62 TB = 0.62 TA current of 4sin4t A flows through a 5-F capacitor. Find the voltage v(t) across the capacitor given that v(0)
Answer:
[tex]v(t) = \frac{-cos4t}{5} + \frac{6}{5} Volts[/tex]
Explanation:
Given
current I = 4sin4t
capacitance C = 5F
v(0) = 1
The voltage across the capacitor is expressed as:
[tex]v(t) = \frac{1}{C} \int\limits^t_{t_0} {i(t)} \, dt + v(t_0)[/tex]
Given v(0) = 1, means at t = 0, v(t0) = 1
Substitute
[tex]v(t) = \frac{1}{5} \int\limits^t_{0} {4sin4t} \, dt + 1\\v(t) = \frac{1}{5} [\frac{-4cos4t}{4} ] + 1\\\\v(t) = \frac{1}{5} (-cos4t)+1 + \frac{1}{5}\\v(t) = \frac{-cos4t}{5} + \frac{6}{5}[/tex]
a mass of .4 kg is raised by a vertical distance of .450 m in the earth's gravitational field. what is the change in its gravitational potential energy
Answer:
E = 1.76 J
Explanation:
Given that,
Mass of an object, m = 0.4 kg
It moves by a vertical distance of 0.45 m in the Earth's gravitational field.
We need to find the change in its gravitational potential energy. It can be given by the formula as follow :
[tex]E=mgh\\\\E=0.4\times 9.8\times 0.45\\\\E=1.76\ J[/tex]
So, the change in its gravitational potential energy is 1.76 J.
A pencil is dropped from rest from a height of 1.8 meters above the ground. How much time does it take the pencil to hit the ground?
Answer:
0.60s
Explanation:
Using the motion equation as follows:
S = ut + 1/2gt²
Where;
S = distance travelled (m)
u = initial velocity (m/s)
t = time (s)
g = acceleration due to gravity (m/s²)
Based on the information provided, S = 1.8m, u = 0m/s, t = ?
S = ut + 1/2gt²
1.8 = (0×t) + 1/2 (9.8t²)
1.8 = 0 + 4.9t²
1.8 = 4.9t²
t² = 1.8/4.9
t² = 0.3673
t = √0.3673
t = 0.6060
t = 0.60s
a man in a gym is holding an 8.0 kg weight at arm's length, a distance of 0.55 m from his shoulder joint What is the torque about his shoulder joint due to the weight if his arm is horizontal
Answer:
[tex]the \: torque= 44 \: Nm.[/tex]
Explanation:
[tex] torque \: \boxed{t} \: is \: the \: product \: of \:one \: of \: the \: force \: - raduis \: \\ and \: sine \: of \: the \: angle \: of \: inclination : \\ torque \: \boxed{t} = fr \sin( \alpha ) \\ f = w = mg = 80. \times 10 = 80 \\ r = 0.55 \\ \alpha = 90 \\ \sin( \alpha ) = 1 \\ therefore \: the \: torque \: \boxed{t} \: is : \\ 80 \times 0.55 \times 1 = 44 \: Nm.[/tex]
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A man is holding a weight at arm's length, a distance apart from his shoulder joint, it will experience torque. The torque about his shoulder joint due to the weight if his arm is horizontal is 43.12 N.m .
What is torque?Torque is the rotation caused about a point when a force is applied at a distance from the point.
The mass of man is 8.0 kg and the distance from the shoulder joint is 0.55 m. The torque applied will be the product of force and distance.
τ =F x r
and Force = weight of the body = mass x acceleration due to gravity
Substitute the values to get the torque.
τ = 8 kg x 9.8 m/s² x 0.55 m
τ = 43.12 N.m
Hence, the torque about his shoulder joint due to the weight if his arm is horizontal is 43.12 N.m
Learn more about torque.
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QUESTION 1 During the time of Pangaea most of the dry land on Earth was joined into one huge landmass that covered nearly a third of the planet's surface. The giant ocean that surrounded the continent is known as what?
QUESTION 2Pangaea existed during what geological time periods, which were times of great change?
QUESTION 3 Most distributions of rocks within Earth's crust, including minerals, fossil fuels, and energy resources, are a direct result of the history of plate motions and collisions and the corresponding changes in the configurations of the continents and ocean basins. Research and explain how plate tectonics could account for these resources exactly.
PLS ANSWeR
Answer:
Pangea existed between about 299 million years ago (at the start of the Permian Period of geological time) to about 180 million years ago (during the Jurassic Period). It remained in its fully assembled state for some 100 million years before it began to break up.
Explanation:
Help please it’s urgent
A fisherman notices that one wave passes the bow of his anchored boat
every 3 seconds. He measured the wavelength to be 8.5 meters. How fast
are the waves traveling?
1.)2.83m/s
2.)0.283m/s
3.)28.3m/s
What average mechanical power (in W) must a 69.0 kg mountain climber generate to climb to the summit of a hill of height 305 m in 48.0 min
Answer: Power= 71.61W
Explanation:
POWER = Work done / Time
But Work done = Gravitational Potential energy
= Mgh
Therefore Power = Mgh/ t
g= 9.8m/s
Where t in mins to secs becomes 48.0x 60=2880 secs
Power= 69x 9.8 x 305/ 2880
=206,241/2880
71.61W
the difference between mechanical and chemical weathering
Answer:
Chemical weathering demands chemical reactions with minerals inside the rock and causes changes in rock composition. Sometimes this process will produce a different kind of product due to the reaction. Mechanical weathering only involves the physical breakage of rocks to smaller pieces of fragments.
Explanation: