A rectangular loop of wire with dimensions 1.80 cm by 9.00 cm and resistance 0.800 Ω is being pulled to the right out of a region of uniform magnetic field. The magnetic field has magnitude 2.60 T and is directed into the plane of (Figure 1) .
a) At the instant when the speed of the loop is 3.00 m/s and it is still partially in the field region, what is the magnitude of the force that the magnetic field exerts on the loop?
b) What is the direction of the force that the magnetic field exerts on the loop?

C. The opportunity cost of a student who is staying up all night to study for an exam that he has to take in the early morning is sleep or rest.

The law of increasing opportunity cost is an economic principle that describes how opportunity costs increase as resources are applied.

As the student gives up his sleep or night rest in the place of his exam preparation, we say that the opportunity cost is the sleep or rest.

Thus, the opportunity cost of a student who is staying up all night to study for an exam that he has to take in the early morning is sleep or rest.

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Answer:a)45cos40 b)45sin40 c) 2.95 d)102

Explanation:

a) [tex]v_{x} = 45 cos40[/tex]

b) [tex]v_{y}=45sin40[/tex]

c) Since air resistance is negligible, the only force acting on the golf ball is gravity. Thus, its vertical acceleration is -g. We know the final velocity must be 0 m/s, because this will be when the golf ball reaches the maximum height and starts to change direction (it falls back to the ground). We also know initial velocity in the vertical direction (see part b). Thus, we can use this equation: [tex]v_{f} = v_{i} + at[/tex].

[tex]0 m/s = 45sin40 + (-9.8m/s^2)t\\t = 2.95s[/tex]

d) The horizontal distance traveled is dependent on (1) how long the ball is in the air and (2) what the horizontal velocity is. (1) was found in part c, and (2) was found in part a.

[tex]x=vt\\x=45cos40(2.95) =102m[/tex]

The confidence interval is given as lower interval is -0.77 while The upper interval is 1.67

Multivitamin treatment

n1 is 10

σ1 = 1.8

x21 = 5

vitamin C treatment

n2 = 10

σ2 = 1.5

x2 = 4.55

zα/2 = 1.645

The formula for the confidence interval is given as

(x1 - x2) ± zα/2[tex]\sqrt{\frac{sd1^2}{n1}+\frac{sd2^2}{n2} }[/tex]

When we input the values we have above we would have

CI = (-0.7688550 , 1.668855 )

The the lower interval is -0.77

The upper interval is 1.67

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A random selection of volunteers at a research institute has been exposed to a weak flu virus. After the volunteers began to have flu symptoms, 10 of them were given multivitamin tablets daily that contained 1 gram of vitamin C and 3 grams of various other vitamins and minerals. The remaining 10 volunteers were given tablets containing 4 grams of vitamin C only. For each individual, the length of time taken to recover from the flu was recorded. At the end of the experiment the following data were obtained:

Treated with multivitamin

Days to recover from flu

2.4, 6.4, 9.1, 4.1, 4.6, 6.4, 6.4, 3.2, 6.9, 0.5

Treated with Vitamin C

5.2, 3, 3.6, 5.5, 7.5, 6.7, 1.3, 1.9, 5.3, 5.5

Suppose that it is known that the population standard deviation of recovery time from the flu is 1.8 days when treated with multivitamins and that the

population standard deviation of recovery time from the flu is 1.5 days when treated with vitamin C tablets. Suppose also that both populations are

approximately normally distributed. Construct a 90% confidence interval for the difference µµ₂ between the mean recovery time when treated with

multivitamins (μ,) and the mean recovery time when treated with vitamin C only (H2). Then complete the table below.

Carry your intermediate computations to at least three decimal places. Round your answers to at least two decimal places. (If necessary, consult a list of formulas.)

What is the lower limit of the 90% confidence interval?

What is the upper limit of the 90% confidence interval?

Answer:

d

Explanation:

Gravitational force is a type of force that pulls objects to the Earth's surface.

The thermal energy that is generated due to friction is 344J.

Now we know that the total mechanical energy in the system is constant. The loss in energy is given by the loss in energy.

Thus, the kinetic energy is given as;

KE = 0.5 * mv^2 =0.5 * 15.0-kg * (1.10 m/s)^2 = 9.1 J

PE = mgh = 15.0-kg * 9.8 m/s^2 *  2.40 m = 352.8 J

The thermal energy is; 352.8 J - 9.1 J = 344J

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The force required to pull the two hemispheres apart is 4.2×10⁴ N and 29 number of horses are needed to pull these hemispheres apart.

=955×100 N/m²= 9.55×10⁴ N/m²

No. of horses required to separate the hemispheres = 4.2×10⁴/1450 = 29

Thus, we can conclude that the 29 horses are needed to pull the two hemispheres with a force of 4.2×10⁴ N.

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The magnitude of the tension in the string marked A is 39.5 N.

The tension in A is determined thus:

The angle at A, θ = tan⁻¹(3/8) = 20.56

When extrapolated below negative x, the angle at B, α = tan⁻¹(5/4) = 51.34

When extrapolated below negative x, the angle at C, β = tan⁻¹(1/6) = 9.46

Taking the horizontal components of tension;

56.3cos(9.46) = A * cos(20.56) + B * cos(51.34)

0.6247B= 55.53 - 0.936A

B = (55.53 - 0.936A)/0.6247 ----(1)

Taking the vertical components of tension;

56.3 * sin(9.46) + A * sin(20.6) = B * sin(51.3)

9.25 + 0.35A = 0.78B  ---- (2)

substitute the value (1)  in (2)

9.25 + 0.35A = 0.78{(55.53 - 0.936A)/0.6247}

(9.25 + 0.35A) * 0.6247 = 43.31 - 0.73A

0.22A + 0.73A = 43.31 - 5.78

0.93A = 37.53

A = 39.5 N

In conclusion, the tension in A is determined by solving for the vertical and horizontal components of tension.

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Answer:

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The distance from A to B is 39.375 Km.

The total distance travelled by the cyclist is given by the formula below:

Total distance = (15 + 20)/2 * 4.5

Total distance = 78.75 Km

Thus, distance from A to B = 78.75/2 = 39.375 Km

In conclusion, the total distance travelled is determined from the average speed and total time taken.

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(a) The initial vertical velocity of the stone is 16.85 m/s and the initial horizontal velocity is 14.14 m/s.

(b) The final vertical velocity is 0 and the final horizontal velocity is 14.14 m/s.

The initial vertical velocity of the stone is calculated as follows;

Vi = Vsinθ

Vi = 22 x sin(50)

Vi = 16.85 m/s

Vxi = V cosθ

Vxi = 22 x cos(50)

Vxi = 14.14 m/s

Vf = Vi - gt

where;

Vfx = Vxi = 14.14 m/s

Thus, the initial vertical velocity of the stone is 16.85 m/s and the initial horizontal velocity is 14.14 m/s.

The final vertical velocity is 0 and the final horizontal velocity is 14.14 m/s.

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The ratio of the speed of the lighter bullet to the speed of the heavier will be four times the heavier bullet.

Both can do same amount of work.

Kinetic energy of a body is the energy due to the motion of the body.

Kinetic energy = mv²/2

where m is mass and v is velocity of the object.

Since both objects have the same kinetic energy but one bullet has twice the mass of the other, the ratio of the speed of the lighter bullet to the speed of the heavier will be four times the heavier bullet.

The kinetic energy of both bullets is the same. Hence, they can do equal amount of work.

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There is a 446N force in the horizontal massless cable CD. The vertical component of the pivot A's force on the aluminum pole, which is 328.3N. The force that the pivot A applies to the aluminum pole has a horizontal component of 446N.

We need to be aware of the force in order to discover the solution.

where, H=3.8m, l=7.6m, m=10kg, M=23.5 kg and alpha= 37 degree,

                      [tex]F_V-mg-Mg=0\\F_v=328.2N\\[/tex]

                  [tex]F_H=T=446N[/tex]

In light of this, we may say that the tension in the horizontal massless cable CD and the horizontal component of the force applied by the pivot A to the aluminum pole are identical and each exert 466N of force, whereas the vertical component of that force is 328.3N.

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From the calculation, the atomic weight of the boron atom is 10.812.

We can find the atomic weight of boron from;

(19.78/100 * 10.0129) + ( 80.22/100 * 11.00931)

= 1.981 + 8.831

= 10.812

The number of protons, neutrons, and electrons in the following isotopes are;

N - 7 electrons, 7 protons and 7 neutrons

Cl - 17 electrons, 17 protons and 18 neutrons

Ca - 20 protons, 20 electrons and 28 neutrons

Cu - 29 protons, 29 electrons and 3 neutrons

U - 92 protons, 92 electrons and 138 neutrons

The number of outer shell electrons in the atoms are;

Na - 1

P - 5

Br - 7

I - 7

Te - 6

The elements are classified as follows;

Radon - nonmetal

Francium - metal

Arsenic - semiconductor

phosphorus - nonmetal

Hafnium - metal

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The wave interaction that is shown in the photo is refraction as light moves from air to water.

Refraction refers to the change in the frequency of a wave and the direction of the wave as it moves from one medium to another. We know that waves makes a body under water to look slightly different than when it is in air.

Thus, the wave interaction that is shown in the photo is refraction as light moves from air to water.

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Answer:

In simple words, thermal equilibrium means that the two systems are at the same temperature.

Answer:

36 cm

Explanation:

Mass of stick; m1 = 0.20kg at midpoint.

Total length; L=1.0 m

Pivot at 0.40m

Atached mass m2 = 0.50kg

Applying rotational equilibrium we have;

Ʈnet = 0

(m1g) • r1 = (m2g) • r2

(0.2) (0.1m) = (0.5)(x)

x = 0.04m =4cm

measured away from 40cm mark gives a position on the stick of; 40cm - 4cm = 36 cm

The deceleration of the bullet over 0.2 second, given the data from the question is –2500 m/s²

This is defined as the rate of change of velocity which time. It is expressed as

a = (v – u) / t

Where

a is the acceleration

v is the final velocity

u is the initial velocity

t is the time

NOTE: Deceleration is the opposite of acceleration

With the above equation for acceleration, we can obtain the deceleration of the bullet. Details below:

a = (v – u) / t

a = (0 – 500) / 0.2

a = –500 / 0.2

a = –2500 m/s²

Thus, the deceleration of the bullet is –2500 m/s²

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The speed of the roller coaster at point 2 is 28 m/s, at point 3 is 16 m/s and at point 4 is 22 m/s.

The speed of the roller coaster at any position is calculated by applying the principle of conservation of energy.

K.E = P.E

¹/₂mv² = mgh

v = √2gh

where;

v(2) = √[(2 x 9.8)(39 - 0)]

v(2) = 28 m/s

v(3) = √[(2 x 9.8)(39 -  26)]

v(3) = 16 m/s

v(4) = √[(2 x 9.8)(39 - 14)]

v(4) = 22 m/s

Thus, the speed of the roller coaster at point 2 is 28 m/s, at point 3 is 16 m/s and at point 4 is 22 m/s.

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A. During long jump athlete runs before taking the jump by doing so he provides himself a larger inertia.

Inertia is the reluctance of an object to stop moving once in motion or start moving when it is at rest.

When an athlete runs before taking the jump, he is trying to increase his inertia, that is his reluctance to stop, thereby increasing his forward motion or jump.

Thus, during long jump athlete runs before taking the jump by doing so he provides himself a larger inertia.

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D. The canister with the fastest moving particle is Z due to concentration of particle.

The speed of the particles depend on the mean distance of the particles.

The canister with the largest concentration per particle will contain particles with the greatest speed.

If the particle concentration is increasing from W to Z, then Z will have fastest moving particle.

Thus, the canister with the fastest moving particle is Z due to concentration of particle.

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Answer:

d. Z

Explanation:

If all the canisters had the same amount of particles, which canister would have the fastest moving particles?

Z

The tension in the rope B is determined as 10.9 N.

tanθ = (6 - 4)/(5 - 0)

tan θ = (2)/(5)

tan θ = 0.4

θ = arc tan(0.4) = 21.8 ⁰

θ = 21.8 ⁰ + 21.8 ⁰ = 43.6⁰

Apply cosine rule to determine the tension in rope B;

A² = B² + C² - 2BC(cos A)

B = C

A² = B² + B² - (2B²)(cos A)

A² = 2B² - 2B²(cos 43.6)

A² = 0.55B²

B² = A²/0.55

B² = 65.3/0.55

B² = 118.73

B = √(118.73)

B = 10.9 N

Thus, the tension in the rope B is determined as 10.9 N.

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Hello!

This is an example of a force summation in the vertical direction.

We have the tension of rope A upward (+), and the equal vertical components of the tensions of rope B and C downward (-).

These forces sum to zero, since the knot is stationary.

[tex]\Sigma F = T_A - T_{By} - T_{Cy} \\\\0 = T_A - T_{By} - T_{Cy}[/tex]

Ropes 'B' and 'C' form equivalent angles from the vertical. (If you were to draw a line from rope A down). We can use right-triangle trig to determine the angle:

[tex]tan^{-1}(\frac{O}{A}) = \theta[/tex]

The ropes are 5 m long and 2 m tall, which are the opposite and adjacent sides respectively:
[tex]tan^{-1}(\frac{5}{2}) = 68.2^o[/tex]

The vertical components are the adjacent sides from this angle, so, we would use cosine.

[tex]0 = T_A - T_Bcos\theta - T_Ccos\theta[/tex]

Rope 'B' and 'C' have the same tensions since they form the same angle with the vertical and are the same length, so we can call them 'T'.

[tex]0 = T_A - 2Tcos\theta[/tex]

Solving for 'T':

[tex]2Tcos\theta = T_A \\\\T = \frac{T_A}{2cos\theta}\\\\T = \frac{65.3}{2cos(68.2)} = \boxed{87.92 N}[/tex]

The angular speed of the merry-go-round just after the child has jumped onto it is 0.696 rad/s.

The principle of conservation of angular momentum states that the sum of initial angular momentum is equal to final angular momentum.

Li = Lf

Li = Ii ωi  + Ic ωc

Li = Iiωi  +  MR²(V/R)

Li =  Iiωi + MRV ----- (1)

Lf = If ωf

ωf = Lf/If

If = Im  + MR²

ωf = Lf / ( Im  + MR²)

Recall, Lf = Li

ωf = (Iiωi + MRV) / ( Im  + MR²)

where;

ωf = (Iiωi + MRV) / ( Im  + MR²)

ωf = (0 + 40 x 1.5 x 8) / (600 + 40(1.5)²)

ωf = (480) / (690)

ωf = 0.696 rad/s

Thus, the angular speed of the merry-go-round just after the child has jumped onto it is 0.696 rad/s.

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The magnitude of the acceleration of block m2  is 0.09 m/s² and the tension in the cord is 1.8 x 10⁻³ N.

The acceleration of the blocks is calculated from the net force on the blocks.

∑F = ma

a = ∑F/m

a = (F) / (m₁ + m₂)

where;

a = (0.03)/(0.04 + 0.02)

a = 0.09 m/s²

T = m₂a

T = (0.02 x 0.09) = 1.8 x 10⁻³ N

Thus, the magnitude of the acceleration of block m2  is 0.09 m/s² and the tension in the cord is 1.8 x 10⁻³ N.

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1.06 is the maximum refractive index that the liquid may have for the light to be totally reflected.

Only when a light source passes from a denser to a rarer medium can it completely reflect.

When the angle of incidence surpasses a specific critical value, specular reflection occurs in the more highly refractive of the two mediums at their interface, and this reflection is known as total reflection.

sin [tex]i_{c}[/tex] = μ[tex]_{r}[/tex] / μ[tex]_{d[/tex]

From the diagram

Angle of incidence = 60°

sin60°  ≥ sin[tex]i_{c}[/tex] = μ[tex]_{r}[/tex]/μ[tex]_{d}[/tex]

μ[tex]_{r}[/tex] ≤ μ[tex]_{d}[/tex] sin60°

μ[tex]_{r}[/tex] ≤ √1.5 × √3/2

   = 1.06

Hence, the maximum index that the liquid may have for the light to be totally reflected is 1.06

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Answer: A candle 4.75 cm tall is 40.0 cm to the left of a plane mirror. Then, the image formed by the mirror will be to the right of the mirror and the height of this image is same as that of the object.

Explanation: To find the answer, we need to know about the reflection at a plane mirror.

There are some points about the reflection at a plane surface.

Thus, we can conclude that, a candle 4.75 cm tall is 40.0 cm to the left of a plane mirror. Then, the image formed by the mirror will be to the right of the mirror and the height of this image is same as that of the object.

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In response, a plane mirror is 40.0 cm to the left of a candle that is 4.75 cm tall. The mirror's generated image will then be to the right of the mirror and have the same height as the object.

In order to determine the solution, we must understand how a plane mirror reflects light.

As a result, we may say that a flat mirror is 40.0 cm to the left of a candle that is 4.75 cm tall. The mirror's generated image will then be to the right of the mirror and have the same height as the object.

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An object is considered to be in a condition of equilibrium when it is balanced with regard to all external forces.

Equilibrium:

An object is considered to be in equilibrium if both its angular acceleration and the acceleration of its center of mass are equal to zero. In layman's terms: The item must either be at rest or moving at a constant speed if it is not accelerating because F = ma (force = mass x acceleration). Even in motion, a body can be in equilibrium. This kind of equilibrium is referred to as a dynamic equilibrium.

A weight suspended by a spring or a brick laying on a flat surface is an example. The equilibrium is unstable if the force with the smallest deviation tends to increase the displacement. As an example, imagine a ball bearing on the edge of a razor blade.

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The net current in the conductors and the value of the line integral

This is further explained below.

Generally,

To put it another way, the total current In flowing across a surface S (contained by C) is proportional to the line integral of the magnetic B-field (in tesla, T).

[tex]\oint_C \mathbf{B} \cdot \mathrm{d}\boldsymbol{\ell} = \mu_0 \iint_S \mathbf{J} \cdot \mathrm{d}\mathbf{S} = \mu_0I_\mathrm{enc}[/tex]

[tex]I=\frac{3.2\cdot 10^{-4}}{4\pi \cdot 10^{-7}}=254.77\, A[/tex]

B)

In conclusion, It is possible for the line integral to go around the loop in either direction (clockwise or counterclockwise), the vector area dS to point in either of the two normal directions and Ienc, which is the net current passing through the surface S, to be positive in either direction—but both directions can be chosen as positive in this example. The right-hand rule solves these ambiguities.

The resultant remains the same at 3.2 *10^4 Tm

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The bigger object will move some distance because the initial and final momentum of the colliding particles are not zero.

The term momentum has to do with the product of mass and velocity. We know that during a collision, momentum is conserved. This implies that the momentum before collision is equal to the momentum after collision. Thus, the total momentum of the system is constant.

Given the fact that the initial and final momentum of the colliding particles are not zero, the  the big object is not going to stop immediately but must move some distance towards the smaller object.

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The  energy she has at each position shown are:

Velocity is known to be a term that connote the direction of any kind of a moving body or an object.

Note that the Speed is known to be a a scalar quantity and as such,  Velocity is said to be a vector quantity.

Note also that from the question given,  Jessica's height of the swing 3 meters above the ground, therefore:

Jessica Position at maximum height :

PE = mgh

PE = 44kg x  9.8m/s² x 3

PE = 1,294 J

Jessica  Position at minimum height:

PE = 0 J

Jessica  Position at maximum velocity:

KE = 1/2 x mv²

KE = 1/2 x 44kg x (5m/s²)²

KE = 550 J

Jessica  Position at minimum velocity:

KE = 0 J

Therefore, The  energy she has at each position shown are:

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The primary means by which heat is transferred through fluids is convection currents (option D).

Convection is the transmission of heat in a fluid by the circulation of currents.

Heat can be transferred by different methods depending on the medium. Fluids like gases and liquids transfer heat through the process of convection.

Therefore, the primary means by which heat is transferred through fluids is convection currents.

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