3. A microscope is focused on a black dot. When a 1.30 cm -thick piece of plastic is placed over the dot, the microscope objective has to be raised 0.410 cm to bring the dot back into focus. What is the index of refraction of the plastic

Answer:

simultaneously

Time taken to reach the ground depends on the vertical component of velocity, not horizontal component of velocity.

Answer:

It will travel Vx * t = 30 m/s * 2 s = 60 m

Answer:

a) Fnet = mg - Fb - Fr

b) 8.67 secs

Explanation:

mass of object = 80 kg

Buoyancy force = 1/50 * weight ( 80 * 9.81 ) = 15.696

Proportionality constant = 10 N-sec/m

a) Calculate  equation of motion of the object

Force of resistance on object  due to water = Fr ∝ V

                                                                         = Fr = Kv = 10 V

Given that : Fb( due to buoyancy ) , Fr ( Force of resistance ) acts in the positive y-direction on the object  while mg ( weight ) acts in the negative y - direction on the object.

Fnet = mg - Fb - Fr

∴ Equation of motion of the object ( Ma = mg - Fb - Fr )

b) Calculate how long before velocity of the object hits 40 m/s

Ma = mg - Fb - Fr

a = 9.81 - 0.1962 - 0.125 V = 9.6138 - 0.125 V

V = u + at ---- ( 1 )

u = 0

V = 40 m/s

a = 9.6138 - 0.125 V

back to equation 1

40 = 0 + ( 9.6138 - 0.125 (40) ) t

40 = 4.6138 t

∴ t = 40 / 4.6138 = 8.67 secs

Answer:

The rise in temperature is 0.06 K.

Explanation:

mass of bullet, m = 15 g

initial speed, u = 865 m/s

final speed, v = 534 m/s

mass of water, M = 13.5 kg

specific heat of water, c = 4200 J/kg K

The change in kinetic energy

[tex]K = 0.5 m(u^2 - v^2)\\\\K = 0.5\times 0.015\times (865^2-534^2)\\\\K = 3473 J[/tex]

According to the conservation of energy, the change in kinetic energy is used to heat the water.

K = m c T

where, T is the rise in temperature.

3473 = 13.5 x 4200 x T

T = 0.06 K

Answer:

T = 1/f = 1/50(s)

ω = 2πf = 100π (rad/s)

(vote 5 sao nhó :3 )

Answer:

1.84 hours

I hope it's helps you

Answer:

About 37.70 radians.

Explanation:

1 revolution = 2[tex]\pi[/tex] radians

∴ 6 revolutions = (6)(2[tex]\pi[/tex] radians)

6 revolutions = 37.6991 or ≈ 37.70 radians

Answer:The radioactive waste

Explanation:Fission is the splitting of a heavy unstable nucleus into two Lighter nuclei

A roller coaster uses 800 000 J of energy to get to the top of the first hill. During this climb, it gains 500 000 J of potential energy and pauses for a fraction of a second at the very top before heading down the other side. At the top of the hill total, the kinetic energy of the roller coaster would be zero as the velocity is zero at the top of the hill, therefore the total mechanical energy is only because of potential energy.

Mechanical energy is the combination of all the energy in motion represented by total kinetic energy and the total stored energy in the system which is represented by total potential energy.

The expression for total mechanical energy is as follows

ME= KE+PE

As total mechanical energy is the sum of all the kinetic as well as potential energy stored in the system.As given in the problem a roller coaster uses 800000 J of energy to get to the top of the first hill. During this climb, it gains 500 000 J of potential energy which means 300000 J of energy is lost in the frictional energy while climbing the hill,

Thus at the top of the hill, the total energy of the roller coasters is only due to the potential energy.

Learn more about mechanical energy from here brainly.com/question/12319302

#SPJ2

Answer:

5 m/s I hope it will help you

Explanation:

mark me as a brainlist answer

Answer:

the initial velocity of the car is 12.04 m/s

Explanation:

Given;

force applied by the break, f = 1,398 N

distance moved by the car before stopping, d = 25 m

weight of the car, W = 4,729 N

The mass of the car is calculated as;

W = mg

m = W/g

m = (4,729) / (9.81)

m = 482.06 kg

The deceleration of the car when the force was applied;

-F = ma

a = -F/m

a = -1,398 / 482.06

a = -2.9 m/s²

The initial velocity of the car is calculated as;

v² = u² + 2ad

where;

v is the final velocity of the car at the point it stops = 0

u is the initial velocity of the car before the break was applied

0 = u² + 2(-a)d

0 = u² - 2ad

u² = 2ad

u = √2ad

u = √(2 x 2.9 x 25)

u =√(145)

u = 12.04 m/s

Therefore, the initial velocity of the car is 12.04 m/s

Answer:

The smallest and largest areas could be 6400 m and 7500 m, respectively.

Explanation:

The area of a rectangle is given by:

[tex] A = l*w [/tex]

Where:

l: is the length = 100 m

w: is the width

We can calculate the smallest area with the lower value of the width.

[tex] A_{s} = 100 m*64 m = 6400 m^{2} [/tex]                            

And the largest area is:

[tex] A_{l} = 100 m*75 m = 7500 m^{2} [/tex]  

Therefore, the smallest and largest areas could be 6400 m and 7500 m, respectively.            

I hope it helps you!                        

Answer:

the largest areas that the field could be is [tex]A_l[/tex]=7587.75 m

the smallest areas that the field could be is [tex]A_s[/tex]=6318.25 m

Explanation:

to the find the largest and the smallest area of the field measurement error is to be considered.

we have to find the greatest possible error, since the measurement was made nearest whole mile, the greatest possible error is half of 1 mile and that is 0.5m.

therefore to find the largest possible area we add the error in the mix of the formular for finding the perimeter with the largest width as shown below:

[tex]A_l[/tex]= (L+0.5)(W+0.5)

(100+0.5)(75+0.5) = (100.5)(75.5) = 7587.75 m

To find the smallest length we will have to subtract instead of adding the error factor value of 0.5 as shown below:

[tex]A_s[/tex]= (L-0.5)(W-0.5)

(100-0.5)(64-0.5) = (99.5)(63.5) = 6318.25 m

Answer:

The constant minimum deceleration required to stop the car in time to avoid pileup is 6.32 m/s²

Explanation:

From the question, the car is traveling at 118 km/h, that is the initial velocity, u = 118km/h

The distance between the car and the accident at the moment when the driver sees the accident is 85 m, that is s = 85 ,

Since the driver slams on the brakes and the car will come to a stop, then the final velocity, v = 0 km/h = 0 m/s

First, convert 118 km/h to m/s

118 km/h = (118 × 1000) /3600 = 32.7778 m/s

∴ u = 32.7778 m/s

Now, to determine the deceleration, a, required to stop,

From one of the equations of motion for linear motion,

v² = u² + 2as

Then

0² = (32.7778)² + 2×a×85

0 = 1074.3841 + 170a

∴ 170a = - 1074.3841

a = - 1074.3841 / 170

a = - 6.3199

a ≅ - 6.32 m/s²

Hence, the constant minimum deceleration required to stop the car in time to avoid pileup is 6.32 m/s²

Answer:

233.33°F

Explanation:

(385K - 273.15) * 9/5 + 32 = 233.33°F

Answer:

wegkwe fhkrbhefdb

Explanation:B

Answer:

prove that Sin^6 ϴ-cos^6ϴ=(2Sin^2ϴ-1)(cos^2ϴ+sin^4ϴ)

please sove step by step with language it is opt maths question

Explanation:

Energy utilization focuses on technologies that can lead to new and potentially more efficient ways of using electricity in residential, commercial and industrial settings—as well as in the transportation sector

Answer:

"the ratio of increase in the volume of a solid per degree rise of temperature to its initial volume" -web

Explanation:

tbh up above ✅

Answer:

cubic meter

Explanation:

Increase in volume of a body on heating is referred to as volumetric expansion or cubical expansion

Answer:

approaches infinity

Explanation:

There are two momentums, the classical momentum which is equal to the product of mass and velocity, and the relativistic momentum, the one we should look at when we work with high speeds, and this happens because massive objects have a speed limit, in this case, we are approaching the speed of light, so we need to work with the relativistic momentum instead of the classical momentum.

The relativistic momentum can be written as:

[tex]p = \frac{1}{\sqrt{1 - \frac{u^2}{c^2} } } *m*u[/tex]

where

u = speed of the object relative to the observer, in this case we have that u tends to c, the speed of light.

m = mass of the object

c = speed of light.

So, as u tends to c, we will have:

[tex]\lim_{u \to c} p = \frac{1}{\sqrt{1 - \frac{u^2}{c^2} } } *m*u[/tex]

Notice that when u tends to c, the denominator on the first term tends to zero, thus, the relativistic momentum of the object will tend to infinity.

Then the correct option is infinity, as the particle speed approaches the speed of light, the relativistic momentum of the particle tends to infinity.

Answer:

15.88

is the correct answer

Answer:

the change in the kinetic energy of the system is -42.47 J

Explanation:

Given;

mass A, Ma = 2 kg

initial velocity of mass A, Ua = 15 m/s

Mass M, Mm = 4 kg

initial velocity of mass M, Um = 7 m/s

Let the common velocity of the two masses after collision = V

Apply the principle of conservation of linear momentum, to determine the final velocity of the two masses;

[tex]M_aU_a + M_mU_m = V(M_a + M_m)\\\\(2\times 15 )+ (4\times 7) = V(2+4)\\\\58 = 6V\\\\V = \frac{58}{6} = 9.67 \ m/s[/tex]

The initial kinetic of the two masses;

[tex]K.E_i = \frac{1}{2} M_aU_a^2 \ + \ \frac{1}{2} M_mU_m^2\\\\K.E_i = (0.5 \times 2\times 15^2) \ + \ (0.5 \times 4\times 7^2)\\\\K.E_i = 323 \ J[/tex]

The final kinetic energy of the two masses;

[tex]K.E_f = \frac{1}{2} M_aV^2 \ + \ \frac{1}{2} M_mV^2\\\\K.E_f = \frac{1}{2} V^2(M_a + M_m)\\\\K.E_f = \frac{1}{2} \times 9.67^2(2+ 4)\\\\K.E_f = 280.53 \ J[/tex]

The change in kinetic energy is calculated as;

[tex]\Delta K.E = K.E_f \ - \ K.E_i\\\\\Delta K.E = 280.53 \ J \ - \ 323 \ J\\\\\Delta K.E = -42.47 \ J[/tex]

Therefore, the change in the kinetic energy of the system is -42.47 J

Answer:

test her prototype and collect data about its flight

Answer:

The correct answer is:

(a) 27.5 Joules

(b) 141.5 Joules

Explanation:

Given:

Energy,

[tex]Q_c = 110 \ J[/tex]

Coefficient of performance refrigerator,

[tex]Cop(refrig)=4[/tex]

(a)

As we know,

⇒ [tex]Cop(refrig) = \frac{Q_c}{Work}[/tex]

or,

⇒ [tex]Work=\frac{Q_c}{Cop(refrig)}[/tex]

              [tex]=\frac{110}{4}[/tex]

              [tex]=27.5 \ Joules[/tex]

(b)

⇒ [tex]Heat \ expelled = Heat \ removed +Work \ done[/tex]

or,

⇒ [tex]Q_h = Q_c+Work[/tex]

         [tex]=114+27.5[/tex]

         [tex]=141.5 \ Joules[/tex]

The corresponding frequency of this sinewave, in kHz, expressed to 3 significant figures is: 155 kHz.

Given the following data:

Note: μs represents microseconds.

Conversion:

1 μs = [tex]1[/tex] × [tex]10^-6[/tex] seconds

645 μs = [tex]645[/tex] × [tex]10^-6[/tex] seconds

To find corresponding frequency of this sinewave, in kHz;

Mathematically, the frequency of a waveform is calculated by using the formula;

[tex]Frequency = \frac{1}{Period}[/tex]

Substituting the value into the formula, we have;

[tex]Frequency = \frac{1}{645 * 10^-6}[/tex]

Frequency = 1550.39 Hz

Next, we would convert the value of frequency in hertz (Hz) to Kilohertz (kHz);

Conversion:

1 hertz = 0.001 kilohertz

1550.39 hertz = X kilohertz

Cross-multiplying, we have;

X = [tex]0.001[/tex] × [tex]1550.39[/tex]

X = 155039 kHz

To 3 significant figures;

Frequency = 155 kHz

Therefore, the corresponding frequency of this sinewave, in kHz is 155.

Find more information: brainly.com/question/23460034

Answer:

The pressure of water progressively increases as the depth of the water increases. The pressure increases as the depth of a point in a liquid increases. The walls of the vessel in which liquids are held are likewise subjected to pressure. The sideways pressure exerted by liquids increases as the liquid depth increases.

i'm pretty sure the answer is A wider

Electric lines of force where intensity of electric field is maximum when its wider.

The physical field that surrounds electrically charged particles and exerts force on all other charged particles in the field, either attracting or repelling them, is known as an electric field (also known as an E-field. It can also refer to a system of charged particles' physical field.

Electric charges and time-varying electric currents are the building blocks of electric fields. The electromagnetic field, one of the four fundamental interactions (also known as forces) of nature, manifests itself in both electric and magnetic fields.

Electrical technology makes use of electric fields, which are significant in many branches of physics. For instance, in atomic physics and chemistry, the electric field acts as an attracting force to hold atoms' atomic nuclei and electrons together.

Therefore, Electric lines of force where intensity of electric field is maximum when its wider.

To learn more about electric field, refer to the link:

https://brainly.com/question/1443103

#SPJ2

This question is incomplete, the complete question is;

A solenoid has a length 11.34 cm , a radius 1.85 cm , and 1627 turns. The solenoid has a net resistance of 144.9 Ω . A circular loop with radius of 3.77 cm is placed around the solenoid, such that it lies in a plane whose normal is aligned with the solenoid axis, and the center of the outer loop lies on the solenoid axis. The outer loop has a resistance of 1651.6 Ω. At a time of 0 s , the solenoid is connected to a battery that supplies a potential 34.95 V. At a time 2.58 μs , what current flows through the outer loop?

Answer:

the current flows through the outer loop is 1.3 × 10⁻⁵ A

Explanation:

Given the data in the question;

Length [tex]l[/tex] = 11.34 cm = 0.1134 m

radius a = 1.85 cm = 0.0185 m

turns N = 1627

Net resistance [tex]R_{sol[/tex] = 144.9 Ω

radius b = 3.77 cm = 0.0377 m

[tex]R_o[/tex] = 1651.6 Ω

ε = 34.95 V

t = 2.58 μs = 2.58 × 10⁻⁶ s

Now, Inductance; L = μ₀N²πa² / [tex]l[/tex]

so

L = [ ( 4π × 10⁻⁷ ) × ( 1627 )² × π( 0.0185 )² ] / 0.1134

L = 0.003576665 / 0.1134

L = 0.03154

Now,

ε = d∅/dt = [tex]\frac{d}{dt}[/tex]( BA ) =  [tex]\frac{d}{dt}[/tex][ (μ₀In)πa² ]

so

ε = μ₀n [tex]\frac{dI}{dt}[/tex]( πa² )

ε = [ μ₀Nπa² / [tex]l[/tex] ] [tex]\frac{dI}{dt}[/tex]

ε = [ μ₀Nπa² / [tex]l[/tex] ] [ (ε/L)e^( -t/[tex]R_{sol[/tex]) ]

I = ε/[tex]R_o[/tex] = [ μ₀Nπa² / [tex]R_o[/tex][tex]l[/tex] ] [ (ε/L)e^( -t/[tex]R_{sol[/tex]) ]

so we substitute in our values;

I = [ (( 4π × 10⁻⁷ ) × 1627 × π(0.0185)²) / (1651.6 ×0.1134) ] [ ( 34.95 / 0.03154)e^( -2.58 × 10⁻⁶ / 144.9 ) ]

I = [ 2.198319 × 10⁻⁶ / 187.29144 ] [ 1108.116677 × e^( -1.7805 × 10⁻⁸ )

I = [ 1.17374 × 10⁻⁸ ] × [ 1108.116677 × 0.99999998 ]

I = [ 1.17374 × 10⁻⁸ ] × [ 1108.11665 ]

I = 1.3 × 10⁻⁵ A

Therefore, the current flows through the outer loop is 1.3 × 10⁻⁵ A

Answer:

1.28 *10^-5 A

Explanation:

Same work as above answer. Needs to be more precise

Explanation:

Hope it Will help he hsuejwoamxgehanwpalasmbwfwfqoqlmdbehendalmZbgevzuxwllw. yeh we pabdvddxhspapalw. X

The angular separation (in degrees) between the second and the third orders on the same side of the central bright fringe if the wavelength is 500 nm and A diffraction grating has 6000 lines per centimeter ruled on it, is  27.29°.

Waves spreading outward around obstructions are known as diffraction. Sound, electromagnetic radiation like light, X-rays, and gamma rays, as well as very small moving particles like atoms, neutrons, and electrons that exhibit wavelike qualities all exhibit diffraction.

Given:

The number of lines = 6000 per cm,

The Wavelength, λ = 500 nm = 500 × 10 ⁻⁹ m

Calculate the diffraction grating,

[tex]d = 1 / no\ of\ lines[/tex]

d = 10⁻² / 6000 m,

Calculate the second-order maxima angle and third-order maxima angle by the formula given below,

[tex]dsin\theta_1 = n_1 \lambda[/tex]

[tex]sin\theta_1 = n_1\lambda / d[/tex]

[tex]\theta _1 = sin^{-1}[2\times 500\times 10 ^{-9}/10^{-2}\times 6000][/tex]

θ₁ = sin⁻¹(0.6)

θ₁ = 36.87°

Similarly, for θ₂,

θ₂ = sin⁻¹(3 × 500 × 10 ⁻⁹ / 10⁻² × 6000)

θ₂ = sin⁻¹(0.9)

θ₂ = 64.16°

Calculate the separation as follows,

θ₂ - θ₁ = 64.16° - 36.87°

θ₂ - θ₁ =  27.29°

Therefore, the angular separation (in degrees) between the second and the third orders on the same side of the central bright fringe if the wavelength is 500 nm and A diffraction grating has 6000 lines per centimeter ruled on it, is  27.29°.

To know more about diffraction:

https://brainly.com/question/12290582

#SPJ2

Answer:

Explanation:

The law of reflection says that the reflected angle (measured from a vertical line to the surface  called the normal) is equal to the reflected angle measured from the same normal line.

All other properties of reflection flow from this one statement.