Both of these questions are the same but their answers in the answer key are different. Why?

This question is incomplete, the complete question is;

A magnetic field is passing through a loop of wire whose area is 0.015 m2. The direction of the magnetic field is parallel to the normal to the loop, and the magnitude of the field is increasing at the rate of 0.20 T/s.

(a) Determine the magnitude of the emf induced in the loop.

(b) Suppose the area of the loop can be enlarged or shrunk. If the magnetic field is increasing as in part (a), at what rate (in m^2/s) should the area be changed at the instant when B = 1.5 T, if the induced emf is to be zero? (Give the magnitude of the rate of change of the area.) (m2/s)

Answer:

a) the magnitude of the emf induced in the loop is 0.003 V

b) dA/dt = 0.002 m²/s

Explanation:

Area of the loop wire A = 0.015 m²

magnitude of the field is increasing dB/dt = 0.20 T/s

a)

Determine the magnitude of the emf induced in the loop.

V = A( dB/dt )

we substitute

V = 0.015 m² × 0.20 T/s

V = 0.003 V

Therefore,  the magnitude of the emf induced in the loop is 0.003 V

b)  the induced emf is;

V = B( dA/dt ) + A( dB/dt )

given that; induced emf is 0, B = 1.5

so we substitute

0 = [ 1.5T × ( dA/dt ) ] + [ 0.015 m² × 0.20 T/s ]

-[ 1.5T × ( dA/dt )] = 0.003 m²T/s

dA/dt = -[ 0.003 m²T/s / 1.5T ]

dA/dt = -0.002 m²/s

the negative shows that the area is decreasing

hence, dA/dt = 0.002 m²/s

Answer:

D

Explanation:

Anyone could be leaning forward toward the speaker but be distracted and I believe if you're paying attention to the speaker, you would ask questions to make sure you're understanding what they are speaking

Answer:

A

.............................

Answer:

I am serious about that

Explanation:

.......

Explanation:

Hi Linda,

How's it going?

Sorry I haven't been in touch for such a long time but I've had exams so I've been studying every free minute. Anyway, I'd love to hear all your news and I'm hoping we can get together soon to catch up. We just moved to a bigger flat so maybe you can come and visit one weekend?

How's the new job?

Looking forward to hearing from you!

Helga

Answer:

hhhhhhhjjjkkllkcftkbgfjknhglncg

Answer:

23s

Explanation:

s=ut+1/2at^2

the distance (s) is 400, initial velocity (u) is 0, acceleration (a) is 1.5 therefore

400=0t+1/2(1.5)t^2

400/0.75=0.75t^2/0.75

t^2=√533.33

t=23s

I hope this helps and sorry if it's wrong

Answer:

v=26.23*105 m/s

or 2.6 × 106 m/s

Explanation:

Force generated by magnetic field will only provide centripetal acceleration thus the entering speed will be same as the exit speed

so,

.5mv2=eV potential differnce*charge= kinetic energy

.5*35*1.66*10-27*v2= 1.6*10-19*5*250000

v2=68.84*1011

v=26.23*105 m/s

or 2.6 × 106 m/s

Answer:poop

Explanation:

poop

Answer:

λ₂ = 357.3 nm

Explanation:

The expression for double-slit interference is

          d sin θ = m λ                 constructive interference

          d sin θ = (m + ½) λ        destructive interference.

The initial data corresponds to a constructive interference, they indicate that we are in the fourth order (m = 4), let's look for the separation of the slits

         d sin θ = m λ₁

now ask for destructive interference for m = 4

        d sin θ = (m + ½) λ₂

we match these two expressions

         m λ₁ = (m + ½) λ₂

         λ₂ = ( m / m + ½) λλ₁  

let's calculate

         λ₂ =[tex]\frac{4}{(4.000 +0.5) \ 401}[/tex]

        λ₂ = 357.3 nm

Answer:

u=20 m/s, T=4s

Explanation:

Given final velocity v= 0 m/s and displacement h= 20 m; acceleration due to gravity = 10 m/ s 2

From equation of motion 

v2=u2+2gs−u2=−2(10).20u=20m/s

and time t can be determined by the formula

t=gv−u=−10−20=2s

total time = 2× time of ascend=2×2=4s

it is helpful for you

Answer:

ω₁ = 2.2 rev/s

Explanation:

Conservation of angular momentum

moment of inertia uniform disk is ½mR²

moment of inertia uniform rod about an end mL²/3

We can think of our rod as two rods of mass m/2 and length R

L = ½mR²ω₀

L = (½mR² + 2(m/2)R²/3)ω₁

ω₁ = ω₀(½mR² / (½mR² + mR²/3))

ω₁ = ω₀(½  / (½  + 1/3))

ω₁ = 0.6ω₀

ω₁ = 2.16

Answer:

2.3  ×  [tex]10^{-1}[/tex]  

Explanation:

1 kg = 1000 g.

0.00023 kg x 1000 g = 0.23 grams

Answer:

0.23×10⁴

Explanation:

kilogram to gram ÷ 1000

0.00023kg ÷ 1000

=0.23g

scientific notation=0.23×10⁴

Answer:

uninstell this apps nobody will give u ans its happening to me alsoi was in exam hall i thought this app will give answer but no

Answer:

[tex]dF=2.5Hz[/tex]

Explanation:

From the question we are told that:

Time [tex]T=0.2sec[/tex]

Generally the Period is given as

 [tex]T= 2 * 0.2 = 0.4[/tex]

Therefore difference in frequency dF

 [tex]dF=\frac{1}{T}[/tex]

 [tex]dF=\frac{1}{0.4}[/tex]

 [tex]dF=2.5Hz[/tex]

Answer:

[tex]v=0.9833\ c[/tex]

Explanation:

The density changes means that the length in the direction of the motion is changed.

Therefore,

[tex]$\text{Density} = \frac{m}{lwh}$[/tex]

Given :

Side,  b = h = 0.13 m

Mass, m = 3.3 kg

Density = 8100 [tex]kg/m^3[/tex]

So,

[tex]$8100=\frac{3.3}{l \times 0.13 \times 0.13}$[/tex]

[tex]$l=\frac{3.3}{8100 \times 0.13 \times 0.13}$[/tex]

l = 0.024 m

Then for relativistic length contraction,

[tex]$l= l' \sqrt{1-\frac{v^2}{c^2}}$[/tex]

[tex]$0.024= 0.13 \sqrt{1-\frac{v^2}{c^2}}$[/tex]

[tex]$0.184= \sqrt{1-\frac{v^2}{c^2}}$[/tex]

[tex]$0.033= 1-\frac{v^2}{c^2}}$[/tex]

[tex]$\frac{v^2}{c^2}= 0.967$[/tex]

[tex]$\frac{v}{c}=0.9833$[/tex]

[tex]v=0.9833\ c[/tex]

Therefore, the speed of the observer relative to the cube is 0.9833 c (in the units of c).

Explanation:

bro I have no idea fam......

Answer:

R = 0.0015Ω

Explanation:

The formula for calculating the resistivity of a material is expressed as;

ρ = RA/l

R is the resistance

ρ is the resistivity

A is the area of the wire

l is the length of the wire

Given

l = 85cm = 0.85m

A = πr²

A = 3.14*0.0018²

A = 0.0000101736m²

ρ = 1.75 × 10-8Ωm.

Substitute into the formula

1.75 × 10-8 = 0.0000101736R/0.85

1.4875× 10-8 = 0.0000101736R

R = 1.4875× 10-8/0.0000101736

R = 0.0015Ω

Answer: D. 0 m

Explanation:

Concept:

Here, we need to know the concept of displacement.

Displacement is defined to be the change in position of an object.

The difference between displacement and distance is the total movement of an object without any regard to direction, while displacement is the pure change of position.

If you are still confused, please refer to the attachment below for a graphical explanation.

Solve:

STEP ONE: the boy walks from point C to point D (a distance of 50 m)

C ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ D

                                              50 m

STEP TWO: the boy walks from point D to point C (a distance of 50 m)

D ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ C

                                               50 m

STEP THREE: find the displacement

The boy started with point C

The boy ended with point C

He did not change his position throughout the journey.

Therefore, his displacement is 0 m.

Hope this helps!! :)

Please let me know if you have any questions

Answer:

The speed of proton is 2.1 x 10^5 m/s .

Explanation:

potential difference, V = 234 V

let the initial speed of the proton is v.

The kinetic energy of proton is

KE = q V

[tex]0.5 mv^2 = e V \\\\0.5\times 1.67\times 10^{-27} v^2 = 1.6\times 10^{-19} \times 234\\\\v=2.1\times 10^5 m/s[/tex]

Answer:

a politician is a person active in party politic or person holding os seeking an elected seat in government.politicans purpose, support,and create laws that govern the land and by extension ,it's people.

Explanation:

mass=kilogram,temperature=Klevin,power=watt,density=kilogram per cubic metre

Explanation:

the unit of mass is kg , temperature is kelvin ,power is watt and density is kilogram per cubic meter.

Answer:ew

Explanation:

qeeqw

Answer:

1. The image is real

2. 5.85

3. h' = 3.05 mm

4. The image is upright

Explanation:

1. Start with the first lens and apply 1/f = 1/p + 1/q

1/5.01 = 1/13.7 + 1/q

q = 7.90 cm

Since that distance is behind the first lens, and the second lens is 62.5 cm behind the first lens, that distance is 62.5 - 7.90 = 54.6 cm in front of the second lens, and becomes the object for that lens, thus,

1/25.9 = 1/54.6 + 1/q

q = 49.3 cm behind the second lens

Using that information, since q is positive, the image is real

2. Also, using that information, you have the second answer, which is 49.3 cm

The height can be found from the two magnifications.

m = -q/p

m1 = -7.9/13.7 = -.577

m2 = -49.3/54.6 = -.903

Net m = (-.577)(-.903) = .521

Then, m = h'/h

.521 = h'/5.85

3. h' = 3.05 mm

4. For the fourth answer, since the overall magnification is positive, the final image is upright

Block on the incline:

• net force parallel to the incline

∑ F (para.) = m₁ g sin(38°) - T = m₁ a

where T is the magnitude of tension in the cord.

Notice that we take down-the-incline to be the positive direction, so that if the 3.9-kg block pulls the 2.6-kg block upwards, then the acceleration of the system is positive.

Suspended block:

• net vertical force

∑ F (vert.) = T - m₂ g = m₂ a

Solve both equations for the acceleration a, set the results equal to one another, and solve for T :

a = g sin(38°) - T/m₁

a = T/m₂ - g

==>   g sin(38°) - T/m₁ = T/m₂ - g

==>   T (1/m₂ + 1/m₁) = g (sin(38°) + 1)

==>   T = g (sin(38°) + 1) / (1/m₂ + 1/m₁)

==>   T = (9.81 m/s²) (sin(38°) + 1) / (1/(2.6 kg) + 1/(3.9 kg)) ≈ 25 N

Temperature in kelvin scale=300k

We know

[tex]\boxed{\sf 0°C=273K}[/tex]

[tex] \\ \Large\sf\longmapsto 300K[/tex]

[tex] \\ \Large\sf\longmapsto 300-273[/tex]

[tex] \\ \Large\sf\longmapsto 27°C[/tex]

Answer:

The answer is "Option b, c, and a".

Explanation:

Here that the earth pulls on the phone, as it will accelerate towards Earth when we drop it.

We now understand the effects of gravity:

[tex]F \propto M\\\\F\propto \frac{1}{r^2}\\\\or\\\\F \propto \frac{M}{r^2}\\\\Sun (\frac{M}{r^2}) = \frac{10^{28}}{(10^9)^2} = 10^{10}[/tex]

The force of the sun is, therefore, [tex]10^{10}[/tex] times greater and the proper sequence, therefore, option steps are:

b. Pull-on phone from earth

c. Pull-on phone from sun

a. Pull phone from you

I'm not very good at this material. I'll try it, but if I were you, I wouldn't bet money on these answers.

"Isobaric" means constant pressure. So those are the horizontal lines, where every point on the line is at the same pressure. Those are the processes 1>2 and 3>4 .

I'm going around and around in my mind with the other labels, and I can't decide. So I'm afraid I can't answer any more of them ... they might be wrong.

Answer:

1 -> 2 & 3 -> 4: Isobaric process

4 -> 1: Work done BY the system

2 -> 3: Work done ON the system

W(total): W = 0J

Explanation:

The two horizontal lines (1 -> 2 & 3 -> 4) are "Isobaric" since isobaric processes take place at constant pressure. I believe 4 -> 1 is "Work done BY the system" since pressure increases when there is an increase of thermal energy, in other words, the system is absorbing heat. This is why the volume increases from 1 -> 2 after the system has absorbed heat in 4 -> 1. Following the directions of the arrows, 2 -> 3 would be "Work done ON the system" since pressure is DECREASING, meaning temperature is also exiting the system. That's why the next step (3 -> 4) shows a decrease in volume. This model depicts a process that has a W(total) of 0 J because this is a cycle.

I hope this helps :))

Answer:

1.24

Explanation:

The resistivity of copper[tex]\rho_1=2.65\times 10^{-8}\ \Omega-m[/tex]

The resistivity of Aluminum,[tex]\rho_2=1.72\times 10^{-8}\ \Omega-m[/tex]

The wires have same resistance per unit length.

The resistance of a wire is given by :

[tex]R=\rho \dfrac{l}{A}\\\\R=\rho \dfrac{l}{\pi (\dfrac{d}{2})^2}\\\\\dfrac{R}{l}=\rho \dfrac{1}{\pi (\dfrac{d}{2})^2}[/tex]

According to given condition,

[tex]\rho_1 \dfrac{1}{\pi (\dfrac{d_1}{2})^2}=\rho_2 \dfrac{1}{\pi (\dfrac{d_2}{2})^2}\\\\\rho_1 \dfrac{1}{{d_1}^2}=\rho_2 \dfrac{1}{{d_2}^2}\\\\(\dfrac{d_2}{d_1})^2=\dfrac{\rho_1}{\rho_2}\\\\\dfrac{d_2}{d_1}=\sqrt{\dfrac{\rho_1}{\rho_2}}\\\\\dfrac{d_2}{d_1}=\sqrt{\dfrac{2.65\times 10^{-8}}{1.72\times 10^{-8}}}\\\\=1.24[/tex]

So, the required ratio of the diameter of Aluminum to Copper wire is 1.24.