Answer:
13.9 kJ
Explanation:
Given that
Length of the side, l = 1.8 m
Drag coefficient, C(d) = 1.4
Distance of run, d = 200 m
Velocity of run, v = 5 m/s
Density, ρ = 1.23
Using the Aerodynamics Drag Force formula. We have
F(d) = 1/2.ρ.A.C(d).v²
The Area, A needed is 1.8 * 1.8 = 3.24 m². So that,
F(d) = 1/2 * 1.23 * 3.24 * 1.4 * 5²
F(d) = 139.482/2
F(d) = 69.74
recall that, energy =
W = F * d
W = 69.74 * 200
W= 13948
W = 13.9kJ
Therefore, the thermal energy added to the air by the drag force is 13.9kJ
How can socialism
impact populations?
Answer:
it represents a fundamental difference. (more info below)
Explanation:
Production is incessantly developing and expanding in socialist countries, and employment is guaranteed for the entire productive population. Consequently, the relative overpopulation problem has been eliminated. This represents the fundamental difference between socialism's demographic law and capitalism's law.
hope this helped!
A car traveling on a flat (unbanked), circular track accelerates uniformly from rest with a tangential acceleration of 1.90 m/s2. The car makes it one quarter of the way around the circle before it skids off the track. From these data, determine the coefficient of static friction between the car and track.
Required:
Determine the coefficient of static friction between the car and the track.
Answer:
Approximately [tex]0.608[/tex] (assuming that [tex]g = 9.81\; \rm N\cdot kg^{-1}[/tex].)
Explanation:
The question provided very little information about this motion. Therefore, replace these quantities with letters. These unknown quantities should not appear in the conclusion if this question is actually solvable.
Let [tex]m[/tex] represent the mass of this car.Let [tex]r[/tex] represent the radius of the circular track.This answer will approach this question in two steps:
Step one: determine the centripetal force when the car is about to skid.Step two: calculate the coefficient of static friction.For simplicity, let [tex]a_{T}[/tex] represent the tangential acceleration ([tex]1.90\; \rm m \cdot s^{-2}[/tex]) of this car.
Centripetal Force when the car is about to skidThe question gave no information about the distance that the car has travelled before it skidded. However, information about the angular displacement is indeed available: the car travelled (without skidding) one-quarter of a circle, which corresponds to [tex]90^\circ[/tex] or [tex]\displaystyle \frac{\pi}{2}[/tex] radians.
The angular acceleration of this car can be found as [tex]\displaystyle \alpha = \frac{a_{T}}{r}[/tex]. ([tex]a_T[/tex] is the tangential acceleration of the car, and [tex]r[/tex] is the radius of this circular track.)
Consider the SUVAT equation that relates initial and final (tangential) velocity ([tex]u[/tex] and [tex]v[/tex]) to (tangential) acceleration [tex]a_{T}[/tex] and displacement [tex]x[/tex]:
[tex]v^2 - u^2 = 2\, a_{T}\cdot x[/tex].
The idea is to solve for the final angular velocity using the angular analogy of that equation:
[tex]\left(\omega(\text{final})\right)^2 - \left(\omega(\text{initial})\right)^2 = 2\, \alpha\, \theta[/tex].
In this equation, [tex]\theta[/tex] represents angular displacement. For this motion in particular:
[tex]\omega(\text{initial}) = 0[/tex] since the car was initially not moving.[tex]\theta = \displaystyle \frac{\pi}{2}[/tex] since the car travelled one-quarter of the circle.Solve this equation for [tex]\omega(\text{final})[/tex] in terms of [tex]a_T[/tex] and [tex]r[/tex]:
[tex]\begin{aligned}\omega(\text{final}) &= \sqrt{2\cdot \frac{a_T}{r} \cdot \frac{\pi}{2}} = \sqrt{\frac{\pi\, a_T}{r}}\end{aligned}[/tex].
Let [tex]m[/tex] represent the mass of this car. The centripetal force at this moment would be:
[tex]\begin{aligned}F_C &= m\, \omega^2\, r \\ &=m\cdot \left(\frac{\pi\, a_T}{r}\right)\cdot r = \pi\, m\, a_T\end{aligned}[/tex].
Coefficient of static friction between the car and the trackSince the track is flat (not banked,) the only force on the car in the horizontal direction would be the static friction between the tires and the track. Also, the size of the normal force on the car should be equal to its weight, [tex]m\, g[/tex].
Note that even if the size of the normal force does not change, the size of the static friction between the surfaces can vary. However, when the car is just about to skid, the centripetal force at that very moment should be equal to the maximum static friction between these surfaces. It is the largest-possible static friction that depends on the coefficient of static friction.
Let [tex]\mu_s[/tex] denote the coefficient of static friction. The size of the largest-possible static friction between the car and the track would be:
[tex]F(\text{static, max}) = \mu_s\, N = \mu_s\, m\, g[/tex].
The size of this force should be equal to that of the centripetal force when the car is about to skid:
[tex]\mu_s\, m\, g = \pi\, m\, a_{T}[/tex].
Solve this equation for [tex]\mu_s[/tex]:
[tex]\mu_s = \displaystyle \frac{\pi\, a_T}{g}[/tex].
Indeed, the expression for [tex]\mu_s[/tex] does not include any unknown letter. Let [tex]g = 9.81\; \rm N\cdot kg^{-1}[/tex]. Evaluate this expression for [tex]a_T = 1.90\;\rm m \cdot s^{-2}[/tex]:
[tex]\mu_s = \displaystyle \frac{\pi\, a_T}{g} \approx 0.608[/tex].
(Three significant figures.)
what statement is true according to newton’s first law of motion?
a. in the absence of unbalanced force an object at rest will stay at rest and an object in motion will come to a stop.
b. in the absence of an unbalanced force, an object will start moving and an object in motion will come to a stop.
c. in the absence of an unbalanced force, an object at rest will stay at rest and an object in motion will stay in motion.
d. in the absence of an unbalanced force, an object will start moving and an object in motion will stay in motion.
Answer:
c. in the absence of an unbalanced force, an object at rest will stay at rest and an object in motion will stay in motion.
Explanation:
First law: things keep doing what they are doing, unless force is applied.
A gas in a closed container is heated with 12J of energy, causing the lid of the container to rise 3m with 5N of force. What is the total change in energy?
Answer:
27J
Explanation:
From conservation of Thermal energy, the total internal energy is the total sum of energy supplied or taken from the system plus work done for or on the system.
Now the change in internal energy would be the sum of the received energy substended in the gas plus the work done by the system which is workdone that it will sustend in pushing the lid. This is expressed mathematically as;
U = Q + (F×d);
U- change in internal energy
Q is the energy received by the system and is positive when energy is received by the system.
Fxd is the workdone and is positive since the gas pushes up the lid- the system does work.
U=12+(3×5)= 27J
Representar con una escala de 1cm = 10N dos fuerzas que tengan igual dirección, distinto sentido y sus intensidades son de 40n y 60n, respectivamente.
Alguien que me lo hagaaaaaaa
Answer:
To solve this problem we just need to graph two forces with same direction, pointing to different sides with intensities of 40 N and 60 N.
The image attached shows these forces.
Notice that the vectors are parallel, that's because they have the same direction, but they point to different sides, and their magnitudes have a difference of 20 N.
wheel rotates from rest with constant angular acceleration. Part A If it rotates through 8.00 revolutions in the first 2.50 s, how many more revolutions will it rotate through in the next 5.00 s?
Answer:
x2 = 64 revolutions.
it rotate through 64 revolutions in the next 5.00 s
Explanation:
Given;
wheel rotates from rest with constant angular acceleration.
Initial angular speed v = 0
Time t = 2.50
Distance x = 8 rev
Applying equation of motion;
x = vt +0.5at^2 ........1
Since v = 0
x = 0.5at^2
making a the subject of formula;
a = x/0.5t^2 = 2x/t^2
a = angular acceleration
t = time taken
x = angular distance
Substituting the values;
a = 2(8)/2.5^2
a = 2.56 rev/s^2
velocity at t = 2.50
v1 = a×t = 2.56×2.50 = 6.4 rev/s
Through the next 5 second;
t2 = 5 seconds
a2 = 2.56 rev/s^2
v2 = 6.4 rev/s
From equation 1;
x = vt +0.5at^2
Substituting the values;
x2 = 6.4(5) + 0.5×2.56×5^2
x2 = 64 revolutions.
it rotate through 64 revolutions in the next 5.00 s
Which of the following best describes the current age of the Sun?
A.) It is near the end of its lifespan.
B.) It is about halfway through its lifespan.
C.) It is early in its lifespan.
D.) We do not have a good understanding of the Sun's age.
Answer: Its b, The only problem with this is is there supposed to be a picture?
Explanation: NASA has used there fancy gadgets to figure this out but if there was a picture, this answer could be different.
a steel ball is dropped from a diving platform use the approximate value of g as 10 m/s^2 to solve the following problem what is the velocity of the ball 0.9 seconds after its released
Answer:
The final speed of the ball is 9 m/s.
Explanation:
We have,
A steel ball is dropped from a diving platform. It is required to find the velocity of the ball 0.9 seconds after its released. It will move under the action of gravity. Using equation of motion to find it as :
[tex]v=u+at[/tex]
u = 0 (at rest), a = g
[tex]v=gt\\\\v=10\times 0.9\\\\v=9\ m/s[/tex]
So, the final speed of the ball is 9 m/s.
1. An object with a mass of 15 kilograms is pushed by a force of 30 Newtons. How much does
it accelerate?
Answer: [tex]2m/s^2[/tex]
Explanation:
[tex]Formula: F=ma[/tex]
Where;
F = force
m = mass
a = acceleration
Solve for a;
[tex]a=\frac{F}{m}[/tex]
[tex]a=\frac{30N}{15kg}\\ a=2m/s^2[/tex]
On a brisk walk, a person burns about 331 Cal/h. If the brisk walk were done at 3.0 mi/h, how far would a person have to walk
to burn off 1 lb of body fat? (A pound of body fat stores an amount of chemical energy equivalent to 3,500 Cal.)
mi?
Answer:
32mi
Explanation:
If 1lb contains 3,500 Cal
It means the number of hours required to burn 3500cal would be;
3500/331 = 10.57hours
But a brisk walk is 3.0 mi/h,
It means a distance of 3.0 × 10.57 mi would be covered = 31.71 miles
32miles{ approximated to the nearest whole}
Note Distance = speed × time
A 72.0 kg swimmer jumps into the old swimming hole from a tree limb that is 3.90 m above the water.
A. Use energy conservation to find his speed just as he hits the water if he just holds his nose and drops in.
b) Use energy conservation to find his speed just he hits the water if he bravely jumps straight up (but just beyond the board!) at 2.90 m/s .
c) Use energy conservation to find his speed just he hits the water if he manages to jump downward at 2.90 m/s .
Answer:
Explanation:
The Law of Energy Conservation states that K1 + U1 = K2 + U2
m= 72.0 kg
h= 3.90 m
a)
K1 + U1 = K2 + U2
0 + mgh = 1/2mvf^2 + 0
mass cancels out so gh=1/2vf^2
(9.8 m/s^2)(3.9 m)=(.5)(vf^2)
vf= 8.74 m/s
b)
1/2mv^2 + mgh = 1/2mv^2 + 0
mass cancels again
(.5)(2.9^2 m/s) + (9.8 m/s^2)(3.9 m) = (.5)(vf^2)
vf= 9.21 m/s
c)
This would be the same as the past problem as the velocity gets squared so direction along the axis doesn't matter. Thus, vf= 9.21 m/s
Assuming 100% efficient energy conversion, how much water stored behind a 50
centimeter high hydroelectric dam would be required to charge the battery?
Answer:
Explanation:
The power rating of the battery isn't provided. But let us assume that it is one of the common batteries with ratings of 12 V and 50 A.h
Potential energy possessed by water at that height = mgh
m = mass of the water = ρV
ρ = density of water = 1000 kg/m³
V = volume of water = ?
g = acceleration due to gravity = 9.8 m/s²
h = height of water = 50 cm = 0.5 m
Potential energy = ρVgh = 1000 × V × 9.8 × 0.5 = (4900V) J
Energy of the battery = qV
q = 50 A.h = 50 × 3600 = 180,000 C
V = 12 V
qV = 180,000 × 12 = 2,160,000 J
Energy = 2,160,000 J
At a 100% conversion rate, the energy of the water totally powers the battery
(4900V) = (2,160,000)
4900V = 2,160,000
V = (2,160,000/4900)
V = 440.82 m³
Hence, with our assumed power ratings for the battery (12 V and 50 A.h), 440.82 m³ of water at the given height of 50 cm would power the battery.
Incase the power ratings of the battery in the complete question is different, this solution provides you with how to obtain the correct answer, given any battery power rating.
Hope this Helps!!!
Thana reminds Alston that because the electric field is uniform, a constant electric force is exerted on the electron. Alston recognizes that, in this case, they can use the kinematic equations to describe the motion of the charged particle while it is inside the region containing the electric field. He asks Thana to write down an equation they can use to calculate the acceleration of the particle while it is inside the region containing a uniform electric field. Which of These equations is correct?
Answer:
a = - e E / m
a = - 1,758 10¹¹ E
Explanation:
For this exercise we can use Newton's second law
F = m a
where the force is electric
the forces given by the product of the charge by the electric field
F = q E
in this case tell us that the charge is the charge of the electron
q = -e = - 1.6 10⁻¹⁹ C
we substitute
- e E = m a
a = - e E / m
we calculate
a = - 1.6 10⁻¹⁹ / 9.1 10⁻³¹ E
a = - 1,758 10¹¹ E
The negative sign indicates that the acceleration is in the opposite direction to the electric field
A mass of 5.00 kg pulls down vertically on a string that is wound around a rod of radius 0.100 m and negligible moment of inertia. The rod is fixed in the center of a disk. The disk has mass 125 kg and radius 0.2 m. They turn freely about a fixed axis through the center. What is the angular acceleration of the rod, in radians/s 2
Answer:
0.981 rad/sec^2
Explanation:
mass that pulls on string = 5 kg
weight due to mass = 5 x 9.81 = 49.05 N
radius of rod = 0.1 m
torque produced by this force on the rod = force x radius
torque = 49.05 x 0.1 = 4.905 N-m
mass of disk = 125 kg
radius of disk = 0.2 m
moment of inertia of the disk I = m[tex]r^{2}[/tex]
I = 125 x [tex]0.2^{2}[/tex] = 5 kg-m^2
from the equation, T = Iα
where T is torque
I is moment of inertia
α is angular acceleration
imputing values,
4.905 = 5α
α = 4.905/5 = 0.981 rad/sec^2
An experiment invilves three charges objects: A, B, and C. Object A repels object B and attracts onject C. object C ir repelled by ebonite charged with fur. What is the charge on the object?
Answer:
A and B is positive charge
C_negative
Explanation:
because when an ebonite is rubbed with fur produce negative charge due to law of electrostatic like charge repel and unlike attract
1. How is it possible to use pools to model apparent weightlessness, similar to what astronauts
experience on the Moon or on the space station? Explain
Answer:
by using it's buoyant or floating effect by Archimedes.
the buoyant force act on the astronauts body and make he/ she feels like in low gravity.
the buoyant force equation is
F = Density of liquid x earth gravitational field x volume of astronauts body and suit.
the Weight of astronauts in the pools will be less than in the land or air.
Weight in water = weight in air/land - buoyant force
so the astronauts will feel like in the outer space with low gravity.
4. A neutrally charged conductor has a negatively charged rod brought close to it, and thus has an induced positive charge on the surface closest to the rod. What can we say about the overall charge on the conductor
Answer:
Overall charge still remains zero on conductor until touched by charged rod.
Explanation:
Here, we want to know what has happened to the overall charge on the conductor.
Since the conductor is neutral, the overall charge on the conductor must remain zero because positive charge is induced on close end to rod then equal and negaitve charge is induced on far end to rod.
Thus, overall charge still remains zero on conductor until touched by charged rod.
A cheetah bites into its prey. One tooth exerts a force of 320 N. The area of the point of the tooth is 0.5 cm². The pressure of the tooth on the prey, in N/cm², is
a) 0.0013 N/cm²
b) 128 N/cm²
c) 320 N/cm²
d) 640 N/cm²
Answer:
640N/cm^2Answer D is correct
Explanation:
[tex]pressure = \frac{force}{area} \\ = \frac{320}{0.5} \\ = 640[/tex]
hope this helps
brainliest appreciated
good luck! have a nice day!
g Tether ball is a game children play in which a ball hangs from a rope attached to the top of a tall pole. The children hit the ball, causing it to swing around the pole. What is the total initial acceleration of a tether ball on a 2.0 m rope whose angular velocity changes from 13 rad/s to 7.0 rad/s in 15 s
Answer:
a = -0.8 m/s²
Here, negative sign indicates that the acceleration has opposite direction to the direction of motion.
Explanation:
First we find the angular acceleration of the ball from the following formula:
α = (ωf - ωi)/t
where,
α = angular acceleration = ?
ωf = final angular velocity = 7 rad/s
ωi = initial angular velocity = 13 rad/s
t = Time taken = 15 s
Therefore,
α = (7 rad/s - 13 rad/s)/15 s
α = - 0.4 rad/s
negative sign shows that acceleration is in opposite direction to the direction of motion.
Now, for the linear acceleration, we use the formula:
a = rα
where,
a = linear acceleration = ?
r = radius of circular path = length of rope = 2 m
therefore,
a = (2 m)(- 0.4 rad/s²)
a = -0.8 m/s²
Here, negative sign indicates that the acceleration has opposite direction to the direction of motion.
In which situation is chemical energy being converted to another form of energy?
Answer:
A burning candle. (chemical energy into energy of heat and light, i.e. thermal and wave)
Explanation:
An aluminium pot whose thermal conductivity is 237 W/m.K has a flat, circular bottom
with diameter 15 cm and thickness 0.4 cm. Heat is transferred steadily to boiling water in
the pot through its bottom at a rate of 1400 W. If the inner surface of the bottom of the pot
is at 105 °C, determine the temperature at the outer surface of the bottom of the pot
Answer:
T₁ = 378.33 k = 105.33°C
Explanation:
From Fourier's Law of heat conduction, we know that:
Q = - KAΔT/t
where,
Q = Heat Transfer Rate = 1400 W
K = Thermal Conductivity of Material (Aluminum) = 237 W/m.k
A =Surface Area through which heat transfer is taking place=circular bottom
A = π(radius)² = π(0.15 m)² = 0.0707 m²
ΔT = Difference in Temperature of both sides of surface = T₂ - T₁
T₁ = Temperature of outer surface = ?
T₂ = Temperature of inner surface = 105°C + 273 = 378 k
ΔT = 388 k - T₁
t = thickness of the surface (Bottom of Pot) = 0.4 cm = 0.004 m
Therefore,
1400 W = - (237 W/m.k)(0.0707 m²)(378 k - T₁)/0.004 m
(1400 W)/(4188.14 W/k) = - (378 k - T₁)
T₁ = 0.33 k + 378 k
T₁ = 378.33 k = 105.33°C
A conducting bar with mass m and length L slides over horizontal rails that are connected to a voltage source. The voltage source maintains a constant current I in the rails and bar, and a constant, uniform, vertical magnetic field B fills the region between the rails . Find the magnitude and direction of the net force on the conducting bar. Ignore friction, air resistance and electrical resistance :
A. v2m / ILB to yhe right
B. 3v2m /2 ILB to yhe left
C. 5v2m/ 2ILB to the right
D. v2m / 2ILB to the left
Answer:
F = ILB
Explanation:
To find the net force on the conducting bar you take into account the following expression:
[tex]\vec{F}=I( \vec{L}X \vec{B})[/tex]
I: current in the conducting bar
L: length of the bar
B: magnitude of the magnetic field
In this case the direction of the magnetic field and the motion of the bar are perpendicular between them. The direction of the bar is + i, and the magnetic field poits upward + k. The cross product of these vector give us the direction of the net force:
+i X +k = +j
The direction of the force is to the right and its magnitude is F = ILB
A 15.0-kg bucket of water is suspended by a very light rope wrapped around a solid uniform cylinder 0.300 m in diameter with mass 12.0 kg. The cylinder pivots on a frictionless axle through its center. The bucket is released from rest at the top of a well and falls 10.0 m to the water. (a) What is the tension in the rope while the bucket is falling? (b) With what speed does the bucket strike the water? (c) What is the time of fall? (d) While the bucket is falling, what is the force exerted on the cylinder by the axle?
Answer:
a. 42N
b. 11.8m/s
c. 1.69s
d. 160N
Explanation:
a) The tension of the rope is 130.66 N.
b) The speed of the bucket while strike the water = 4.64 m/s.
c) The time of fall is = 4.303 second.
d) While the bucket is falling, what is the force exerted on the cylinder by the axle is 130.66 N.
Mass of the water bucket; M = 15.0 kg
Mass of the cylinder; m = 12.0 kg
Height of the bucket; h = 10.0 m.
They are connected by a rope and a pivots.
So, acceleration of them is same and let it be a.
So equation of motion of both of them be:
Mg - T = Ma
and, T - mg = ma
Hence, a = g(M-m)/(M+m)
= 9.8(15-12)/(15+12)
= 1.08 m/s²
And, T = m(g+a)
= 12.0(9.8+1.08)
= 130.66 N.
a) so tension of the rope is 130.66 N.
b) speed of the bucket while strike the water = √2ah =√(2×1.08×10.0) m/s = 4.64 m/s.
c) The time of fall is = √2h/a = √(2×10/1.08) second = 4.303 second.
d) While the bucket is falling, what is the force exerted on the cylinder by the axle is tension of the rope, that is, 130.66 N.
Learn more about speed here:
https://brainly.com/question/28224010
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5.Which of the following does not affect rate of evaporation?
O Wind speed
O Surface area
O Temperature
O Insoluble heavy impurities
Answer:
D
Explanation:
Insoluble impurities would not change the constituent of the substance. Soluble would for example salt water takes longer time for the water to become vapour when subjected to the same temperature that normal water.
Wind would affect, the more windy the tendency for particles of the liquid to be moved into the atmosphere.
With an increase in surface area, the evaporation rate increase . Take a clue from water placed on the ground and exposed to the atmosphere and that same quantity of water is placed in a cup. That on the floor would evaporate faster.
Similarly the higher the temperature a substance is subjected to the easier is it's rate of evaporation. Take for instance water in a cup placed in the sun and that same placed in a room with mild temperatures than that of the sun.With time that in the sun decreases in volume faster than that in the room.
Can someone help me with this question
Answer:
hypothesis , hope it helps
Explanation:
Answer:
Inference
Explanation:
Inference is something you predict after testing that's a result after an hypothesis has been made. Hypothesis is an intelligent guess based on some observed phenomena which can be subjected to further testing.
Astronauts are testing the gravity on a new planet. A rock is dropped between two photogates that are 0.5 meters apart. The first photogate reads a velocity of 1.2 m/s and the the second photogate reads a velocity of 4.3 m/s . What is the acceleration of gravity on this new planet?
Answer:
a = 17 m / s²
Explanation:
For this experiment that the astronauts are carrying out, the value of the relation of gravity is cosecant, therefore we can use the kinematic relations
v² = v₀² + 2a y
They indicate the initial speed 1.2 m / s the final speed 4.3 m / s and the distance remembered 0.5 m
we clear
a = (v² - v₀²) / 2y
we calculate
a = (4.3² -1.2²) / 2 0.5
a = 17 m / s²
this is the gravity of the new planet
Constants Canada geese migrate essentially along a north-south direction for well over a thousand kilometers in some cases, traveling at speeds up to about 100 km/h. The one goose is flying at 100 km/h relative to the air but a 44 km/h wind is blowing from west to east.
1. At what angle relative to the north-south direction should this bird head so that it will be traveling directly southward relative to the ground?2. How long will it take the bird to cover a ground distance of 450 from north to south?
Answer:
a. 63.89° in the north-southward manner
b. 2.2 sec
Explanation:
The goose is flying at 100 km/h
Air from east to west is 44 km/h
angle relative to the north-south direction for the bird to travel south will be
cos∅ = 44/100 = 0.44
∅ = [tex]cos^{-1}[/tex]0.44 = 63.89° in the north-southward manner
Speed south relative to the ground will be v
Tan 63.89 = v/100
2.04 = v/100
v = 2.04 x 100 = 204 km/hr
to cover a distance of 450 m from north to south at this speed time will be
t = d/v = 450/204 = 2.2 sec
The current through an inductor of inductance L is given by I(t) = Imax sin(ωt).
(a) Derive an expression for the induced emf in the inductor as a function of time.
(b) At t = 0, is the current through the inductor increasing or decreasing?
(c) At t = 0, is the induced emf opposing or aiding the flow of the charge carriers? (Remember that the direction of a positive induced emf is the same as the current direction and the direction of a negative induced emf is opposite the current direction.)
(d) How are the answers to parts b and c consistent with the behavior of inductors discussed in the text?
Answer:
(a) [tex]emf_L=-LI_{max}\omega cos(\omega t)[/tex]
(b) neither increasing or decreasing
(c) opposite to the flow of charge carriers
Explanation:
The current through an inductor of inductance L is given by:
[tex]I(t)=I_{max}sin(\omega t)[/tex] (1)
(a) The induced emf is given by the following formula
[tex]emf_L=-L\frac{dI}{dt}[/tex] (2)
You derivative the expression (1) in the expression (2):
[tex]emf_L=-L\frac{d}{dt}(I_{max}sin(\omega t))\\\\emf_L=-LI_{max}\omega cos(\omega t)[/tex]
(b) At t=0 the current is zero
(c) At t = 0 the emf is:
[tex]emf_L=-\omega LI_{max}[/tex]
w, L and Imax have positive values, then the emf is negative. Hence, the induced emf is opposite to the flow of the charge carriers.
(d) read the text carefully
At t zero, the current through the inductor neither increasing nor decreasing because current is zero.
The current through an inductor of inductance L can be calculated by
[tex]\bold {I_t = I_m_a_x sin (\omega t)}[/tex].........1
(a) The induced emf can be calculated by
[tex]\bold {emf_L = - L \dfrac {dI}{dt}}[/tex]............2
Derivative the equation (1) in the equation (2)
[tex]\bold {emf _L= -L \dfrac {d (I _m_a_x sin (\omega t)} {dt}}\\\\\bold {emf _L= -L (I _m_a_x \omega cos( \omega t) }[/tex]
(b) At t=0 the current is zero,
(c) At t = 0 the emf is:
[tex]\bold {emf_L = -\omega LI _m_a_x}[/tex]
Therefore, at t zero, the current through the inductor neither increasing nor decreasing because current is zero.
To know more about inductance,
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Assuming 100% efficient energy conversion, how much water stored behind a 50 centimetre high hydroelectric dam would be required to charge battery
Complete question is;
Assuming 100% efficient energy conversion how much water stored behind a 50 centimeter high hydroelectric dam would be required to charge the battery with power rating, 12 V, 50 Ampere-minutes
Answer:
Amount of water required to charge the battery = 7.35 m³
Explanation:
The formula for Potential energy of the water at that height = mgh
Where;
m = mass of the water
g = acceleration due to gravity = 9.8 m/s²
H = height of water = 50 cm = 0.5 m
We know that in density, m = ρV
Where;
ρ = density of water = 1000 kg/m³
V = volume of water
So, potential energy is now given as;
Potential energy = ρVgH = 1000 × V × 9.8 × 0.5 = (4900V) J
Now, formula for energy of the battery is given as;
E = qV
We are given;
q = 50 A.min = 50 × 60 = 3,000 C
V = 12 V
Thus;
qV = 3,000 × 12 = 36,000 J
E = 36,000 J
At a 100% conversion rate, the energy of the water totally powers the battery.
Thus;
(4900V) = (36,000)
4900V = 36,000
V = 36,000/4900
V = 7.35 m³
As your bus rounds a flat curve at constant speed, a package with mass 0.900 kg , suspended from the luggage compartment of the bus by a string 50.0 cm long, is found to hang at rest relative to the bus, with the string making an angle of 30.0 â with the vertical. In this position, the package is 55.0 m from the center of curvature of the curve.
Required:
a. What is the radial acceleration of the bus?
b. What is the radius of the curve?
Answer:
a.[tex]5.66ms^{-2}[/tex]
b.55 m
Explanation:
We are given that
Mass ,m=0.9 kg
Length of string,l=50 cm=[tex]\frac{50}{100}=0.50 m[/tex]
1 m=100 cm
[tex]\theta=30^{\circ}[/tex]
R=55 m
a.Centripetal acceleration
[tex]a_c=gtan\theta[/tex]
[tex]a_c=9.8tan30^{\circ}[/tex]
[tex]a_c=5.66 m/s^2[/tex]
Hence, the radial acceleration of the bus=[tex]5.66m/s^2[/tex]
b. Radius of curve,R=55 m
