Answer:
Approximately [tex]25\; \rm cm[/tex], assuming that the permeability between the two wires is constant, and that the two wires are of infinite lengths.
Explanation:
Note that [tex]y_1 < y_2[/tex], which means that the first wire (with current [tex]I_1[/tex]) is underneath the second wire (with current [tex]I_2[/tex].) Let [tex]y[/tex] denote the [tex]y[/tex]-coordinate (in [tex]\rm cm[/tex]) of a point of interest. The two wires partition this region into three parts:
The region under the first wire, where [tex]y < y_1[/tex];The region between the first and the second wire, where [tex]y_1 < y < y_2[/tex]; andThe region above the second wire, where [tex]y > y_2[/tex].Apply the right-hand rule to find the direction of the magnetic field in each of the three regions. Assume that [tex]I_1[/tex] points to the left while [tex]I_2[/tex] points to the right. Let [tex]B_1[/tex] and [tex]B_2[/tex] denote the magnetic field due to [tex]I_1[/tex] and [tex]I_2[/tex], respectively.
In the region below the first wire, [tex]B_1[/tex] points out of the [tex]x y[/tex]-plane while [tex]B_2[/tex] points into the [tex]x y[/tex]-plane.In the region between the two wires, both [tex]B_1[/tex] and [tex]B_2[/tex] point into the [tex]x y[/tex]-plane.In the region above the second wire, [tex]B_1[/tex] points into the [tex]x y[/tex]-plane while [tex]B_2[/tex] points out of the [tex]x y[/tex]-plane.The (net) magnetic field on this plane would be zero only in regions where [tex]B_1[/tex] and[tex]B_2[/tex] points in opposite directions. That rules out the region between the two wires.
At a distance of [tex]R[/tex] away from a wire with current [tex]I[/tex] and infinite length, the formula for the magnitude of the magnetic field [tex]B[/tex] due to that wire is:
[tex]\displaystyle B = \frac{\mu\, I}{2\pi\, R}[/tex],
where [tex]\mu[/tex] is the permeability of the space between the wire and the point of interest (should be constant.) The exact value of [tex]\mu[/tex] does not affect the answer to this question, as long as it is constant throughout this region.
Note that the value of [tex]R[/tex] in this formula is supposed to be positive. Let [tex]R_1[/tex] and [tex]R_2[/tex] denote the distance between the point of interest and the two wires, respectively.
In the region under the first wire, [tex]R_1 = y_1 - y = 9 - y[/tex], while [tex]R_2 = y_2 - y = 15 - y[/tex]. In the region above the second wire, [tex]R_1 = y - 9[/tex], while [tex]R_2 = y - 15[/tex].Make sure that given the corresponding range of [tex]y[/tex], these distances are all positive.
Strength of the magnetic field due to the first wire: [tex]\displaystyle B_1 = \frac{\mu\, I_1}{4\pi\, R_1}[/tex].Strength of the magnetic field due to the second wire: [tex]\displaystyle B_1 = \frac{\mu\, I_2}{4\pi\, R_2}[/tex].For the net magnetic field to be zero at a certain [tex]y[/tex]-value, the strength of the two magnetic fields at that point should match. That is:
[tex]B_1 = B_2[/tex].
[tex]\displaystyle \frac{\mu\, I_1}{4\pi\, R_1} = \frac{\mu\, I_2}{4\pi\, R_2}[/tex].
Simplify this equation:
[tex]\displaystyle \frac{I_1}{R_1} = \frac{I_2}{R_2}[/tex].
In the region under the first wire (where [tex]y < 9[/tex],) this equation becomes [tex]\displaystyle \frac{47}{9 - y} = \frac{29}{15 - y}[/tex].In the region above the second wire (where [tex]y > 15[/tex],) this equation becomes [tex]\displaystyle \frac{47}{y - 9} = \frac{29}{y - 15}[/tex].These two equations give the same result: [tex]y = 25[/tex]. However, based on the respective assumptions on the value of [tex]y[/tex], this value corresponds to the region above the second wire.
⦁ Consider an atom. Which contributes most to the mass of the atom?
Answer:
protons
Explanation:
A 64.7 cm long straight section of wire is located entirely inside a uniform magnetic field of |B| = 0.370 T. The wire is perpendicular to the direction of the magnetic field. When a current runs through the wire a magnetic force of |F| = 0.110 N acts on that section of wire. Calculate the size of the current.
Answer:
The current is [tex]I = 0.4595 \ A[/tex]
Explanation:
From the question we are told that
The length of the wire is [tex]L = 64.7 \ cm = 0.647 \ m[/tex]
The magnetic field is [tex]B = 0.370 \ T[/tex]
The magnetic force is [tex]F = 0.110 \ N[/tex]
Generally the magnetic force exerted by the field is mathematically represented as
[tex]F = IL B sin \theta[/tex]
Making I the subject we have
[tex]I = \frac{F}{LB \ sin\theta}[/tex]
The angle here is 90° since the wire is perpendicular to the direction of the magnetic field
Substituting values
[tex]I = \frac{0.110}{ (0.370) * 0.647 \ sin(90)}[/tex]
[tex]I = 0.4595 \ A[/tex]
1. Who is Nancy Lieberman?
2. What were some of her accomplishments throughout her life?
3. How could her accomplishments motivate other women in their lives?
4. Describe how you think she felt at the peak of her success?
5. How did her success contribute to women’s sports?
who can help me with that 5 questions please, I need help
Answer:
Nancy Elizabeth Lieberman (nacido el 1 de julio de, 1958), apodado "Señora mágica", ] es un ex profesional jugador de baloncesto y entrenador de la WNBA (WNBA) que es actualmente un organismo de radiodifusión para los New Orleans Pelicans de la Nacional La Asociación de Baloncesto (NBA) y la entrenadora principal de Power , un equipo en el BIG3 que lideró en su Campeonato 0.]bLieerman es considerada como una de las figuras más importantes del baloncesto femenino estadounidense.
Explanation:
A woman on a bicycle travels at 9 m/s relative to the ground as she passes a little boy on a tricycle going in the opposite direction. If the boy is traveling at 1 m/s relative to the ground, how fast does the boy appear to be moving relative to the woman
Answer:
Useless
Explanation:
Question
WILL VOTE MOST BRAINLIEST
Which of the following explains why the inner planets are different than the outer planets?
A) Due to the greater inertia of the dust particles, the dust remained close to the sun to form the inner planets leaving gases to form the outer planets.
B) The lighter gases boiled off of the protoplanets closest to the sun, leaving dust and metals behind to form the inner planets.
C) Due to the greater momentum of the gas particles, the gas particles flew farther away from the sun than dust and metals to form the outer planets, leaving dust and metals to form the inner planets.
Answer:
B) The lighter gases boiled off of the protoplanets closest to the sun, leaving dust and metals behind to form the inner planets.
Explanation:
Inner planets are smaller and rockier than outer gas planets,outer planets are larger,because of their lower gravity they don't attract extreme amounts of gas in their planets.The four outer planets were so far from the Sun that its winds could not blow away their ice and gases
On his way to deliver presents Santa has a minor accident. If the sleigh (1200 kg) was traveling at 322 m/s and the jet(4800 kg) was traveling at 680 m/s and they collided head on, what was the final velocity of the two objects after the collision
Answer:
608.4m/s
Explanation:
We are given that
Mass of Sleigh,M=1200 kg
Speed of Sleigh,u=322 m/s
Speed of jet,u'=680 m/s
Mass of jet,m=4800 kg
Total mass=M+m=1200+4800=6000 kg
We have to find the final velocity of the two objects after the collision.
The collision is inelastic .
By using law of conservation of momentum
[tex]Mu+mu'=(m+M)v[/tex]
Using the formula
[tex]1200\times 322+4800\times 680=6000v[/tex]
[tex]6000v=3650400[/tex]
[tex]v=\frac{3650400}{6000}[/tex]
[tex]v=608.4m/s[/tex]
Hence, the final velocity of two objects after the collision=608.4m/s
A loop of wire carrying a steady current I is initially at rest perpendicular to a uniform magnetic field of magnitude B, as shown above. The loop is then rotated about a diameter at a constant rate. The torque on the loop is maximum when the loop has rotated, with respect to its initial position, through an angle of:__________.
(A) 30°
(B) 45°
(C) 90°
(D) 180°
(E) 360°
your answer is b to be presided
A gas at pressure 100 atm and volume 3.7 L has the pressure reduced to 32 atm what is the new volume of the gas
Answer:
The new volume is [tex]11.6L[/tex]
Explanation:
This problem is on the application of Boyle's law which states that" the volume of a given mass of gas at constant temperature is inversely proportional to the pressure"[tex]pressure \alpha \frac{1}{volume}[/tex]
Given data
initial pressure p1= 100atm
initial volume v1= 3.7L
final pressure p2= 32 atm
final volume v2= ?
Apply Boyle's law we have
[tex]p1v1= p2v2\\[/tex]
[tex]v2= \frac{p1v1}{p2}[/tex]
Substituting our data we have
[tex]v2= \frac{100*3.7}{32} \\v2=\frac{370}{32} \\v2= 11.6 L[/tex]
The new volume is [tex]11.6L[/tex]
17. A ranger needs to capture a monkey hanging on a tree branch. The ranger aims his dart gun directly at the monkey and fires the tranquilizer dart. However, the monkey lets go of the branch at exactly the same time as the ranger fires the dart. Will the monkey get hit or will it avoid the dart?
Answer:
The monkey will get hit by the dart. This is because when the dart is fired the force of gravity makes it fall from its original position. The monkey also has the same fall from the branches which means they will both be falling at the same gravitational rate.
This will eventually make the monkey to get hit by the dart.
What is a limitation of the electron cloud model theory that
a law about electrons would not have?
Answer:
Electron cloud model describes the region in an atom which negatively charged and which has probability to find an electron. according to this model, an electron can move closer or away from the nucleus that is it can be inside the nucleus but according to Bohr's model, an electron is always at a fixed distance from the nucleus. Thus, it is the limitation of the electron cloud model but still it is a widely accepted model.
A wall clock has a second hand 22.0 cmcm long. For related problem-solving tips and strategies, you may want to view a Video Tutor Solution of Fast car, flat curve. Part A What is the radial acceleration of the tip of this hand
Answer:
Explanation:
The tip of the second hand moves on a circular path having radius equal to .22 m . Redial acceleration is given by the expression
ω²R where ω is angular velocity and R is radius of the circular path .
angular velocity of second hand = 2π / T where T is time period of circular motion . For second hand it is 60 s.
ω = 2π / T
= 2π / 60
= .1047
angular acceleration = .1047² x .22
= 2.41 x 10⁻³ rad / s² .
So, the required radial acceleration is [tex]a=0.21 cm/s[/tex]
Acceleration:The rate of change of velocity with respect to time.
Acceleration is a vector quantity.
The formula for the acceleration is,
[tex]a=\frac{V^2}{R}[/tex]
It is given that the radius is 22 cm
Now, substituting the given values into the above formula we get,
[tex]a=\frac{V^2}{22}[/tex]
Here, 1 second is equal to [tex]\frac{2\pi}{60}[/tex] then,
[tex]\\w=\frac{2\pi}{60}\\v=wR\\v=\frac{2\pi}{60}\times 19\\a=(\frac{2\pi}{60}\times 19)^2\times 19\\a=0.21 cm/s[/tex]
Learn more about the topic acceleration:
https://brainly.com/question/11021097
9. Santa brings a special baby a bouncy swing. The child bounces in a harness, with a spring constant k, and is suspended off of the ground. a. If the spring stretches x1 = 0.27 m from equilibrium while supporting an 7.15-kg child, what is its spring constant, in newtons per meter?
Answer:
259.52N
Explanation:
Force initiated on a spring that causes an extension is related by the expression below;
F= K×e
Where F is the Force
K is the spring constant
e is the extension caused by the spring.
By change of subject formula for K;
K = F/e
Now F is the same as weight and is given by mass× acceleration. In this case acceleration is g=9.8m/s2; this is because the child's mass is going to be under the influence of gravity as it swings up the harness}
Hence W =7.15×9.8=70.07N
Hence K = 70.07/0.27 =259.5185N/m
=259.52N/m to 2 decimal place.
A cosmic ray electron moves at 7.6 x 106 m/s perpendicular to the Earth's magnetic field at an altitude where the field strength is 1.02 x 10-5 T. What is the radius of the circular path the electron follows
Answer:
r = 4.23 m
Explanation:
To find the radius of the circular path of the electron you use the following formula:
[tex]r=\frac{mv}{qB}[/tex] (1)
This formula can be used because the motion of the electron is perpendicular to the direction of the magnetic field vector.
m: mass of the electron = 9.1*10^-31 kg
q: charge of the electron = 1.6*10^-19 C
B: magnitude of the magnetic field = 1.02*10^-5 T
v: velocity of the electron = 7.6*10^6 m/s
You replace the values of m, v, q and B in the equation (1):
[tex]r=\frac{(9.1*10^{-31}kg)(7.6*10^6 m/s)}{(1.6*10^{-19}C)(1.02*10^{-5}T)}\\\\r=4.23m[/tex]
hence, the raiuds of the orbit of the electron is 4.23m
A motor vehicle has a mass of 1.8 tonnes and its wheelbase is 3 m. The centre of gravity of the vehicle is situated in the central plane 0.9 m above the ground and 1.7 m behind the front axle. When moving on the level at 90 km/h the brakes applied and it comes to a rest in a distance of 50 m.
Calculate the normal reactions at the front and rear wheels during the braking period and the least coefficient of friction required between the tyres and the road. (Assume g = 10 m/s2)
Answer:
normal reaction, front: 11,175 Nnormal reaction, rear: 6,825 Nminimum coefficient of friction: 0.625Explanation:
The speed in meters per second is ...
(90 km/h)(1000 m/km)(1 h/(3600 s)) = 25 m/s
The braking acceleration can be found from ...
a = v²/(2d) = (25 m/s)²/(2×50 m) = 6.25 m/s²
Then the braking force is ...
F = Ma = (1800 kg)(6.25 m/s²) = 11,250 N
The torque on the center of gravity is ...
T = (11,250 N)(0.90 m) = 10,125 N·m
__
If we let x and y represent the normal forces on the front and rear wheels, respectively, then we have ...
x + y = (10 m/s²)(1800 kg) = 18000 . . . . . newtons
1.7x -1.3y = 10,125 . . . . . . . . . . . . . . . . . . . newton-meters
The latter equation balances the torque due to the wheel normal forces with the torque due to braking forces.
Multiplying the first equation by 1.3 and adding that to the second, we have ...
3.0x = (1.3)(18,000) + 10,125
x = 33,525/3 = 11,175 . . . . . . . . . . . newtons normal force on front tyres
y = 18000 -11175 = 6,825 . . . . . . . .newtons normal force on back tyres
The least coefficient of friction is the ratio of horizontal to vertical acceleration, 6.25/10 = 0.625.
After t hours a freight train is s(t) = 18t2 − 2t3 miles due north of its starting point (for 0 ≤ t ≤ 9). (a) Find its velocity at time t = 3 hours. 108 Incorrect: Your answer is incorrect. mi/hr (b) Find its velocity at time t = 7 hours. 546 Incorrect: Your answer is incorrect. mi/hr (c) Find its acceleration at time t = 1 hour. 42 Incorrect: Your answer is incorrect. mi/hr2
Answer:
Explanation:
Given the equation modelled by the height of the train given as:
s(t) = 18t²-2t³ for for 0 ≤ t ≤ 9
a) Velocity is the rate of change of displacement.
Velocity = dS(t)/dt
V = dS(t)/dt = 36t - 6t² miles
Velocity at t = 3hrs is determiner by substituting t = 3 into the velocity function.
V = 36(3) -6(3)²
V= 108 - 72
Velocity = 36mi/hr
b) for Velocity at time = 7hrs
V(7) = 36(7) - 6(7)²
V(7) = 252 - 294
V(7) = -42mi/hr
The velocity at t = 7hrs is -42mi/hr
c) Acceleration is the rate of change of velocity.
a(t) = dV(t)/dt
Given v(t) = 36t - 6t²
a(t) = 36 - 12t
Acceleration at t=1 is given as:
a(1) = 36 -12(1)
a(1) = 24mi/hr²
Choose the INCORRECT statement: A) Gauss' law can be derived from Coulomb's law B) Gauss' law states that the net number of lines crossing any closed surface in an outward direction is proportional to the net charge enclosed within the surface C) Coulomb's law can be derived from Gauss' law and symmetry D) Gauss' law applies to a closed surface of any shape E) If a closed surface encloses no charge, the electric field is zero everywhere on the surface
Answer:
A
Explanation:
A)
INCORRECT
No, Gauss's Law can not be derived from Coulomb's Law alone. Since, Coulomb's Law provides electric field produced by a single point charge. Therefore, it is important to assume principle of super position along with Coulomb's law to get the field due to all charges. So, Gauss's Law can be proved by the help of both Super Position Principle and Coulomb's law.
B)
CORRECT
Yes, Gauss' law states that the net number of lines crossing any closed surface in an outward direction is proportional to the net charge enclosed within the surface.
C)
CORRECT
Yes, Coulomb's law can be derived from Gauss' law and symmetry.
D)
CORRECT
Yes, Gauss's Law applies to a closed surface of any shape.
E)
CORRECT
Yes, if a closed surface encloses no charge, the electric field is zero everywhere on the surface.
An advertisement for a new fish food claims that lab studies show that fish grew three inches in three weeks while eating the food. Travis wondered if the food was worth the extra cost. To evaluate the claim in the advertisement, Travis should
Answer:
look at the data for the control group to see how much the fish grew without the new food.
Explanation:
i got it right
Two vectors are being added, one at an angle of 20.0 , and the other at 80.0. The only thing you know about the magnitudes is they are both positive. Will the equilibrant vector be in the:
a. first quadrant
b. second quadrant
c. third quadrant
d. fourth quadrant
e. you cannot tell which quadrant from the available information
Answer:a
Explanation: they are all positive
The equivalent vectors of both the vectors lie in the first quadrant. So, the right choice is a. first quadrant.
How the resultant of the vectors can be calculated?The resultant of the vectors can be calculated using the Parallelogram theorem of vector addition. In this theorem, a parallelogram is formed using the vectors, having the vectors as the adjacent side and the diagonal of this parallelogram is the resultant of both the vectors.
How the theorem is implemented to find out the resultant?The vectors A and B are drawn at the given angle, angles of 20° and 80° respectively. Using the vectors, A and B, the dotted parallelogram is drawn, and the vector C is the resultant.
Learn more about vectors here:
https://brainly.com/question/25705666
Suppose the electric field in problems 2 was caused by a point charge. The test charge is moved to a distance twice as far from the charge. What is the magnitude of the force that the field exerts on the test charge now ?
Answer:
it is reduced four times.
Explanation:
By definition, the electric field is the force per unit charge created by a charge distribution.
If the charge creating the field is a point charge, the force exerted by it on a test charge, must obey Coulomb´s Law, so, it must be inversely proportional to the square of the distance between the charges.
So, if the distance increases twice, as the force is inversely proportional to the square of the distance, and the square of 2 is 4, this means that the magnitude of the force exerted on the test charge must be 4 times smaller.
Three long parallel wires, each carrying 20 A in the same direction, are placed in the same plane with the spacing of 10 cm. What is the magnitude of net force per metre on central wire?
Answer:
F/L = 8*10^-4 N/m
Explanation:
To calculate the magnitude of the force per meter in the central wire, you take into account the contribution to the force of the others two wires:
[tex]F_N=F_{1,2}+F_{2,3}[/tex] (1)
F1,2 : force between first and second wire
F2,3 : force between second and third wire
The force per meter between two wires of the same length is given by:
[tex]\frac{F}{L}=\frac{\mu_oI_1I_2}{2\pi r}[/tex]
μo: magnetic permeability of vacuum = 4pi*10^-7 T/A
r: distance between wires
Then, you have in the equation (1):
[tex]\frac{F_N}{L}=\frac{\mu_oI_1I_2}{2\pi r}+\frac{\mu_oI_2I_3}{2\pi r}\\\\\frac{F_N}{L}=\frac{\mu_oI_1}{2\pi r}[I_2+I_3][/tex]
But
I1 = I2 = I3 = 10A
r = 10cm = 0.1m
You replace the values of the currents and the distance r and you obtain:
[tex]\frac{F_N}{L}=\frac{\mu_oI^2}{\pi r}\\\\\frac{F_N}{L}=\frac{(4\pi*10^{-7}T/A)(20A)^2}{2\pi (0.1m)}=8*10^{-4}\frac{N}{m}[/tex]
hence, the net force per meter is 8*10^-4 N/m
There is an old physics joke involving cows, and you will need to use its punchline to solve this problem
A cow is standing in the middle of an open, flat field. A plumb bob with a mass of 1 kg is suspended via an unstretchable string 10 meters long so that it is hanging down roughly 2 meters away from the center of mass of the cow. Making any reasonable assumptions you like or need to, estimate the angle of deflection of the plumb bob from vertical due to the gravitational field of the cow.
Answer:
The angle of deflection will be "1.07 × 10⁻⁷°".
Explanation:
The given values are:
Mass of a cow,
m = 1100 kg
Mass of bob,
mb = 1 kg
The total distance between a cow and bob will be,
d = 2 m
Let,
The tension be "t".
The angle with the verticles be "[tex]\theta[/tex]".
Now,
Vertically equating forces
⇒ [tex]T\times Cos \theta =mb\times g[/tex] ...(equation 1)
Horizontally equating forces
⇒ [tex]T\times Sin \theta = G\times M\times \frac{mb}{d^2}[/tex] ...(equation 2)
From equation 1 and equation 2, we get
⇒ [tex]tan \thata=\frac{G\times M}{g\times d^2}[/tex]
O putting the estimated values, will be
⇒ [tex]\theta = tan(\frac{6.674\times 10^{-11}\times 1100}{9.8\times 2^2} )[/tex]
⇒ [tex]\theta = 1.07\times 10^{-7}^{\circ}[/tex]
Beginning from rest when = 20°, a 35-kg child slides with negligible friction down the sliding board which is in the shape of a 2.5-m circular arc. Determine the tangential acceleration and speed of the child, and the normal force exerted on her (a) when = 30° and (b) when = 90°
Answer:
a) [tex]a_{T}=8.50m/s^{2}[/tex], [tex]v_{T}=2.78 m/s[/tex] and [tex]N=279.83N[/tex]
b) [tex]a_{T}=0m/s^{2}[/tex], [tex]v_{T}=5.68 m/s[/tex] and [tex]N=795.2N[/tex]
Explanation:
a)
In order to solve this problem we need to start by drawing a diagram of what the problem looks like: See attached picture. Next, we can start by finding the initial height of the child which will happen at an angle of 20°. We can find this by subtracting the distance from the highest point and the initial point from the radius of the circle so we get:
[tex]h_{0}=2.5m-h_{1}[/tex]
so we get:
[tex]h_{1}=2.5m(sin (20^{o}))[/tex]
[tex]h_{1}=0.855m[/tex]
so
[tex]h_{0}=2.5m-h_{1}[/tex]
[tex]h_{0}=2.5m-0.855m[/tex]
[tex]h_{0}=1.645m[/tex]
once we got this value, we can find the final height the same way. This time the angle is 30° so we get:
[tex]h_{f}=2.5m-h_{2}[/tex]
[tex]h_{f}=2.5m-2.5m(sin 30^{o}))[/tex]
[tex]h_{f}=1.25m[/tex]
Once we have these heights, we can go ahead and use an energy balance equation to find the velocity at 30° so we get:
[tex]U_{0}+K_{0}=U_{f}+K_{f}[/tex]
the initial kinetic energy is zero because its initial velocity is zero too, so the equation simplifies to:
[tex]U_{0}=U_{f}+K_{f}[/tex]
so now we substitute with the corresponding formulas:
[tex]mgh_{0}=mgh_{f}+\frac{1}{2}mv_{f}^{2}[/tex]
if we divided both sides of the equation by the mass, then the equation simplifies to:
[tex]gh_{0}=gh_{f}+\frac{1}{2}v_{f}^{2}[/tex]
and now we can solve for the final velocity so we get:
[tex]v_{f}=\sqrt{2g(h_{0}-h_{f}}[/tex]
and we can now substitute values:
[tex]v_{f}=\sqrt{2(9.81m/s^{2})(1.645m-1.25m)}[/tex]
which solves to:
[tex]v_{f}=2.78m/s[/tex]
which is our first answer. Once we got the velocity at 30° we can find the other data the problem is asking us for:
We can build a free body diagram (see attached picture) and do a balance of forces so we get:
[tex]\sum{F_{x}}=ma_{T}[/tex]
in this case we only have one x-force which is the x-component of the weight, so we get that:
[tex]w_{x}=ma_{T}[/tex]
so we get:
[tex]mgcos\theta=ma_{T}[/tex]
and solve for the tangential acceleration so we get:
[tex]a_{T}=gcos\theta[/tex]
[tex]a_{T}=9.81m/s^{2}*cos(30^{o})[/tex]
[tex]a_{t}=8.50m/s^{2}[/tex]
Now we can find the centripetal acceleration by using the formula:
[tex]a_{c}=\frac{V_{T}^{2}}{R}[/tex]
[tex]a_{c}=\frac{(2.78m/s)^{2}}{2.5m}[/tex]
so we get:
[tex]a_{c}=3.09m/s^{2}[/tex]
Next, we can find the normal force which is found by doing a sum of forces on y, so we get:
[tex]\sum{F_{y}}=ma_{c}[/tex]
so we get:
[tex]N-w_{y}=ma_{c}[/tex]
and we solve for the normal force so we get:
[tex]N=ma_{c}+w_{y}[/tex]
and substitute:
[tex]N=ma_{c}+mg sin\theta[/tex]
when factoring we get:
[tex]N=m(a_{c}+g sin\theta)[/tex]
and we substitute:
[tex]N=(35kg)(3.09m/s^{2}+9.81m/s^{2} sin30^{o})[/tex]
which yields:
N=279.83N
b)
The procedure for part b is mostly the same with some differences due to the angle. First:
[tex]h_{0}=1.645m[/tex]
[tex]h_{f}=0m[/tex]
so
[tex]U_{0}+K_{0}=U_{f}+K_{f}[/tex]
in this case the initial kinetic energy is zero because the initial velocity is zero and the final potential energy is zero because the final height is zero as well, so the equation simplifies to:
[tex]U_{0}=K_{f}[/tex]
so we get:
[tex]mgh_{0}=\frac{1}{2}mv_{f}^{2}[/tex]
so we solve for the final velocity so we get:
[tex]v_{f}=\sqrt{2gh_{0}}[/tex]
and we substitute:
[tex]v_{f}=\sqrt{2(9.81m/s^{2})(1.645m)}[/tex]
[tex]v_{f}=5.68m/s[/tex]
according to the free body diagram we get that:
[tex]a_{T}=gcos\theta[/tex]
[tex]a_{T}=9.81m/s^{2}(cos 90^{o})[/tex]
which yields:
[tex]a_{T}=0[/tex]
we can also find the centripetal acceleration, so we get:
[tex]a_{c}=\frac{V_{T}^{2}}{R}[/tex]
[tex]a_{c}=\frac{(5.68m/s)^{2}}{2.5m}[/tex]
so we get:
[tex]a_{c}=12.91m/s^{2}[/tex]
and we can do a sum of forces on y to find the normal force:
[tex]\sum{F_{y}}=ma_{c}[/tex]
so we get:
[tex]N-w_{y}=ma_{c}[/tex]
and we solve for the normal force so we get:
[tex]N=ma_{c}+w_{y}[/tex]
and substitute:
[tex]N=ma_{c}+mg [/tex]
when factoring we get:
[tex]N=m(a_{c}+g)[/tex]
and we substitute:
[tex]N=(35kg)(12.91m/s^{2}+9.81m/s^{2} sin30^{o})[/tex]
which yields:
N=795.2N
Following are the calculation to the angles:
For angle 30°:
consider the forces in tangential in direction:
[tex]\Sigma F_t= ma_t\\\\ W \cos \theta= m a_t \\\\m g \cos \theta = m a_t\\\\ a_t = g \cos \theta = 9.81 \cos 30^{\circ} \\\\a_t = 8.496 \ \frac{m}{s^2}\\\\[/tex]
calculate the speed of the child:
[tex]g\cos \theta = a_t\\\\ g\cos \theta = \frac{v.dv}{R d\theta} \\\\g R \cos \theta d \theta= V \ dv\\\\[/tex]
Integrating the equation:
[tex]\int_{20}^{30} g R \cos \theta d \theta= \int_{0}^{v} V \ dv\\\\g R (\sin \theta)_{20}^{30} = [\frac{V^2}{2}]_{0}^{v} \ dv\\\\g R (\sin 30-\sin 20) = \frac{V^2}{2}\\\\9.81 \times 2.5 (\sin 30-\sin 20) = \frac{V^2}{2}\\\\v=2.79 \ \frac{m}{s}\\\\[/tex]
consider the forces in the normal declaration:
[tex]\Sigma F_n = ma_n \\\\N-mg \sin \theta= \frac{mv^2}{R} \\\\N- (35\times 9.81\times \sin 30) = \frac{35 \times 2.78^2}{2.5}\\\\N= 279.87\ N[/tex]
For angle 90°:
consider the forces in tangential in direction:
[tex]\Sigma F_t= ma_t\\\\ W \cos \theta= m a_t \\\\m g \cos \theta = m a_t\\\\ a_t = g \cos \theta = 9.81 \cos 90^{\circ} \\\\a_t = 0\ \frac{m}{s^2}\\\\[/tex]
calculate the speed of the child:
[tex]g\cos \theta = a_t\\\\ g\cos \theta = \frac{v.dv}{R d\theta} \\\\g R \cos \theta d \theta= V \ dv\\\\[/tex]
Integrating the equation:
[tex]\int_{20}^{30} g R \cos \theta d \theta= \int_{0}^{v} V \ dv\\\\g R (\sin \theta)_{20}^{90} = [\frac{V^2}{2}]_{0}^{v} \ dv\\\\g R (\sin 90-\sin 20) = \frac{V^2}{2}\\\\9.81 \times 2.5 (\sin 90-\sin 20) = \frac{V^2}{2}\\\\V=\sqrt{\frac{9.81 \times 2.5 (\sin 90-\sin 20) }{2}}[/tex]
[tex]=\sqrt{\frac{9.81 \times 2.5 (1 -0.342)}{2}}\\\\=\sqrt{8.06}\\\\=2.83[/tex]
consider the forces in the normal declaration:
[tex]\Sigma F_n = ma_n \\\\N-mg \sin \theta= \frac{mv^2}{R} \\\\N- (35\times 9.81\times \sin 90) = \frac{35 \times 2.78^2}{2.5}\\\\N= \frac{35 \times 2.78^2}{2.5 \times 35\times 9.81\times 1}\\[/tex]
[tex]= \frac{270.494}{858.375}\\\\=0.315[/tex]
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brainly.com/question/13077544
Human reaction times are worsened by alcohol. How much further (in feet) would a drunk driver's car travel before he hits the brakes than a sober driver's car? Assume that both are initially traveling at 50.0 mi/h and their cars have the same acceleration while slowing down, and that the sober driver takes 0.33 s to hit the brakes in a crisis, while the drunk driver takes 1.0 s to do so. (5280 ft = 1 mi)
Answer:
A drunk driver's car travel 49.13 ft further than a sober driver's car, before it hits the brakes
Explanation:
Distance covered by the car after application of brakes, until it stops can be found by using 3rd equation of motion:
2as = Vf² - Vi²
s = (Vf² - Vi²)/2a
where,
Vf = Final Velocity of Car = 0 mi/h
Vi = Initial Velocity of Car = 50 mi/h
a = deceleration of car
s = distance covered
Vf, Vi and a for both drivers is same as per the question. Therefore, distance covered by both car after application of brakes will also be same.
So, the difference in distance covered occurs before application of brakes during response time. Since, the car is in uniform speed before applying brakes. Therefore, following equation shall be used:
s = vt
FOR SOBER DRIVER:
v = (50 mi/h)(1 h/ 3600 s)(5280 ft/mi) = 73.33 ft/s
t = 0.33 s
s = s₁
Therefore,
s₁ = (73.33 ft/s)(0.33 s)
s₁ = 24.2 ft
FOR DRUNK DRIVER:
v = (50 mi/h)(1 h/ 3600 s)(5280 ft/mi) = 73.33 ft/s
t = 1 s
s = s₂
Therefore,
s₂ = (73.33 ft/s)(1 s)
s₂ = 73.33 ft
Now, the distance traveled by drunk driver's car further than sober driver's car is given by:
ΔS = s₂ - s₁
ΔS = 73.33 ft - 24.2 ft
ΔS = 49.13 ft
Monochromatic light is incident on (and perpendicular to) two slits separated by 0.200 mm, which causes an interference pattern on a screen 613 cm away. The light has a wavelength of 656.3 nm. (a) What is the fraction of the maximum intensity at a distance of 0.600 cm from the central maximum of the interference pattern
Answer:
I = 0.636*Imax
Explanation:
(a) To find the fraction of the maximum intensity at a distance y from the central maximum you use the following formula:
[tex]I=I_{max}cos^2(\frac{\pi d}{\lambda L}y)[/tex] (1)
I: intensity of light
Imax: maximum intensity of light
d: separation between slits = 0.200mm = 0.200 *10^-3 m
L: distance from the screen = 613cm = 0.613 m
y: distance to the central peak of the interference pattern
λ: wavelength of light = 656.3 nm = 656.3 *10^-9 m
You replace the values of all variables in the equation (1):
[tex]I=I_{max}cos^2(\frac{\pi (0.200*10^{-3}m)}{(656.3*10^{-9}m)(0.613m)}0.600m)\\\\I=I_{max}cos^2(937.06)=0.636I_{max}[/tex]
Hence, the fraction of the maximum intensity is I = 0.636*Imax
A 50.0 Watt stereo emits sound waves isotropically at a wavelength of 0.700 meters. This stereo is stationary, but a person in a car is moving away from this stereo at a speed of 40.0 m/s. The frequency of sound waves that the car receives is ________. In addition, when the car is 70.0 meters away from the speaker, the car will hear sound waves with a sound intensity level of _________ .
Answer:
a) f' = 432 Hz
b) I = 8.12*10^-4 W/m^2
Explanation:
a) To calculate the frequency of sound waves that car receives, you take into account the Doppler effect. In this case (observer moves away of the source) you have the following formula:
[tex]f'=f(\frac{v-v_o}{v+v_s})[/tex] (1)
where
f: frequency of the source = ?
v: speed of sound = 343 m/s
vo: speed of the observer = 40.0 m/s
vs: speed of the source = 0 m/s (stationary)
You replace the values of all parameters in the equation (1):
To calculate f' you first calculate the frequency of the sound wave, by using the following formula:
[tex]v=\lambda f\\\\[/tex]
v: speed of sound
λ: wavelength = 0.700 m
[tex]f=\frac{v}{\lambda}=\frac{343m/s}{0.700m}=480Hz[/tex]
Next, you replace the values of all parameters in the equation (1):
[tex]f'=(490Hz)(\frac{343m/s-40.0m/s}{343m/s})=432Hz[/tex]
hence, the frequency perceived by the car is 432 Hz
b) To calculate the power of the sound wave, when the car is 70.0 maway from the speaker, you use the following formula:
[tex]I=\frac{P}{4\pi r^2}[/tex]
P: power of the source = 50.0 W
r: distance to the source = 70.0 m
[tex]I=\frac{50.0 W}{4\pi(70.0m)^2}=8.12*10^{-4}\frac{W}{m^2}[/tex]
hence, the intensity is 8.12*10^⁻4 W/m^2
If light energy to electric energy conversion using solar cells is 12 % efficient, how many square miles of land must be covered with solar cells to supply the electrical energy for 350000 houses? Assume there is no cloud cover.
Complete Question
The light energy that falls on a square meter of ground over the course of a typical sunny day is about 20 MJ . The average rate of electric energy consumption in one house is 1.0 kW .
If light energy to electric energy conversion using solar cells is 12 % efficient, how many square miles of land must be covered with solar cells to supply the electrical energy for 350000 houses? Assume there is no cloud cover.
Answer:
The area is [tex]A = 1.26 *10^{7} m^2[/tex]
Explanation:
From the question we are told that
The efficiency is [tex]\eta =[/tex]12%
The number of houses is [tex]N = 350000[/tex]
The light energy per day is [tex]E = 20 \ MJ[/tex]
The average rating of electric energy for a house is [tex]E_h = 1.0 \ k W = 1000W[/tex]
Generally the electric energy which the solar cells covering [tex]1 \ m^2[/tex] produces in a day is
[tex]E_s = \eta * E[/tex]
[tex]E_s = 0.12 * 20*10^{6}[/tex]
[tex]E_s = 2.4 MJ m^{-2}[/tex]
Energy for required by one house for one day is
[tex]E_H = E_h * 1 \ day[/tex]
[tex]E_H = 1000 * 24 * 3600[/tex]
[tex]E_H = 86.4 MJ[/tex]
Energy needed for 350000 house is
[tex]E_z = 86.4 *10^{6} * 350000[/tex]
[tex]E_z = 3.02 *10^{7} MJ[/tex]
The area covered is mathematically represented as
[tex]A = \frac{3.02*10^{7} \ MJ}{2.4 \ MJ m^{-2}}[/tex]
[tex]A = 1.26 *10^{7} m^2[/tex]
When we say that an object wants to maintain its state of motion, we’re talking about inertia. Which term determines the quantity of inertia for an object?
A.
mass
B.
velocity
C.
acceleration
D.
weight
Answer:
A. mass
Explanation:
Mass determines the quantity of inertia for an object. Mass is the quantity that depends upon the inertia of an object. The inertia that an object has is directly proportional to the mass of the object.
An object that has more mass has a greater tendency as compared to the object that has less mass to resist changes in its state of motion.
Can someone please help me with this question thank you!
A flat, rectangular coil consisting of 60 turns measures 23.0 cmcm by 34.0 cmcm . It is in a uniform, 1.40-TT, magnetic field, with the plane of the coil parallel to the field. In 0.230 ss , it is rotated so that the plane of the coil is perpendicular to the field.
(a) What is the change in the magnetic flux through the coil due to the rotation?
(b) What is the magnitude of the average emf induced in the coil during the rotation?
(c) What is the average current induced in the coil during the rotation?
Answer:
(a) ΔФ = -0.109W
(b) emf = 28.43V
(c) Iin = emf/R
Explanation:
(a) In order to calculate the magnetic flux you use the following formula:
[tex]\Delta\Phi_B=\Phi-\Phi_o=BAcos(90\°)-BAcos(0\°)[/tex] (1)
B: magnitude of the magnetic field = 1.40T
A: area of the rectangular coil = (0.23m)(0.34m)=0.078m^2
Where it has been taken into account that at the beginning the normal vector to the cross sectional area of the coil, and the magnetic field vector are parallel. When the coil is rotated the vectors are perpendicular.
Then, you obtain:
[tex]\Delta\Phi_B=(1.40T)(0.078m^2)=-0.109W[/tex]
The change in the magnetic flux is -0.109 W
(b) During the rotation of the coil the emf induced is given by:
[tex]emf=-N\frac{\Delta \Phi}{\Delta t}[/tex] (2)
N: turns of the coil = 60
ΔФ: change in the magnetic flux = 0.109W
Δt: lapse time of the rotation = 0.230s
You replace the values of the parameters in the equation (2):
[tex]emf=-(60)(\frac{-0.109W}{0.230s})=28.43V[/tex]
The induced emf is 28.43V
(c) The induced current in the coil is given by:
[tex]I_{in}=\frac{emf}{R}[/tex] (3)
R: resistance of the coil (it is necessary to have this value)
emf :induced emf = 28.43V
A gardener pushes a shovel into the ground with a force of 75 N. The angle of the shovel to the ground is 80 degrees. What is the downward force of the shovel? A. 13.0 N B. 73.9 N C. 75.2 N D. 80.0 N
Answer:
B
Explanation:
Resolve the 75N force into 2 components; horizontal and vertical. And remember that there is no acceleration in the downward direction, so apply Newton's second law and equate it to 0.