Suppose you stand with one foot on ceramic flooring and one on a wool carpet, making contact over a 77.2 cm2 area with each foot. Both the ceramic and the carpet are 2.60 cm thick and are 10.0°C on their bottoms. At what rate must each foot supply heat to keep the top of the ceramic and carpet at 33.0°C? The thermal conductivity of ceramic is 0.84 J/(s · m · °C) and that of wool is

Answer: Their final relative velocity is -0.412 m/s.

Explanation:

According to the law of conservation,

      [tex]m_{1}v_{1} + m_{2}v_{2} = (m_{1} + m_{2})v[/tex]

Putting the given values into the above formula as follows.

      [tex]m_{1}v_{1} + m_{2}v_{2} = (m_{1} + m_{2})v[/tex]

     [tex]2.50 \times 10^{3} kg \times 0 m/s + 7.50 \times 10^{3} kg \times -0.550 m/s = (2.50 \times 10^{3} kg + 7.50 \times 10^{3} kg)v[/tex]

           [tex]-4.12 \times 10^{3} kg m/s = (10^{4} kg) v[/tex]

                   v = [tex]\frac{-4.12 \times 10^{3} kg m/s}{10^{4} kg}[/tex]

                      = -0.412 m/s

Thus, we can conclude that their final relative velocity is -0.412 m/s.

Answer:

29.61 rpm.

Explanation:

Given,

student arm length, l = 67 cm

distance of the bucket, r = 35 m

Minimum angular speed of the bucket so, the water not fall can be calculated by equating centrifugal force with weight.

Now,

[tex]mg = m r \omega^2[/tex]

[tex]\omega = \sqrt{\dfrac{g}{R}}[/tex]

R = 67 + 35 = 102 cm = 1.02 m

[tex]\omega = \sqrt{\dfrac{9.81}{1.02}}[/tex]

[tex]\omega = 3.101\ rad/s[/tex]

[tex]\omega = \dfrac{3.101}{2\pi} = 0.494\ rev/s[/tex]

[tex]\omega = 0.494 \times 60 = 29.61\ rpm[/tex]

minimum angular velocity is equal to 29.61 rpm.

Answer:

29.6 rpm

Explanation:

length of arm = 67 cm

distance of handle to the bottom = 35 cm

radius of rotation, R = 67 + 35 = 102 cm = 1.02 m

The centripetal force acting on the bucket is balanced by the weight of the bucket.

mRω² = mg

R x ω² = g

[tex]\omega = \sqrt\frac{g}{R}[/tex]

[tex]\omega = \sqrt\frac{9.8}{1.02}[/tex]

ω = 3.1 rad/s

Let f is the frequency in rps

ω = 2 x 3.14 x f

3.1 = 2 x 3.14 xf

f = 0.495 rps

f = 29.6 rpm

Answer:

True.

The day a huge asteroid strikes the earth will likely be a weekday.

Explanation:

Mathematically, there are 5 weekdays in the week and there are only 2 weekend days.

Total number of days in a week = 7

Since, the asteroids can hit on any day,

The probability of the asteroid hitting on a weekend day = (2/7) = 0.2857

The probability that an asteroid hitting on a weekday = (5/7) = 0.7143.

So, by the virtue of the probability of the asteroid hitting on a weekday or on a weekend day, the asteroid is more likely to hit on a weekday as

P(weekday) > P(weekend)

0.7143 > 0.2857.

Hope this Helps!!!

Answer:

Explanation:

According to  energy conservation principle;

[tex]\delta \ K.E = \delta \ P.E[/tex]

[tex]8.00*10^{-19} \ J = (1.6*10^{-19} \ C ) |V_b-V_a|[/tex]

[tex]|V_b-V_a|= \frac {8.00*10^{-19} }{(1.6*10^{-19} ) }[/tex]

[tex]|V_b-V_a|= 5 \ V[/tex]

The potential is positive hence the electric fielf=d is negative along the x-axis.

We can then say that the movement of the particle goes to the left through a potential difference of  [tex]|V_b-V_a|= 5 \ V[/tex].

There will be no significant change in the y-direction of the potential energy when the particle moves from point b to point c in the y-direction.

Answer:

The work done on the jet by the catapult is 1.53 x  10⁷ J

Explanation:

Given;

force of thrust engines, F =  2.3 x 10⁵ N

distance moved by the thrust engines, d =  90 m

kinetic energy of the plane, K.E = 3.60 x 10⁷ J

Based on work-energy theorem;

the net work done on the plane by catapult and the thrust engine is given as;

[tex]W_n = K.E_f = 3.6 *10^7 \ J[/tex]

work done by the thrust engine on the jet;

[tex]W_T_._e_n = fd\\\\W_T_._e_n = 2.3*10^5 * 90\\\\W_T_._e_n = 2.07*10^7 \ J[/tex]

Work done on the jet by the catapult;

[tex]W_c_p= W_n - W_T_._e_g\\\\W_c_p= (3.6*10^7 - 2.07*10^7) J\\\\W_c_p= 1.53*10^7 \ J[/tex]

Therefore, the work done on the jet by the catapult is 1.53 x  10⁷ J

Answer:

(a) 11.8692‬ ohm

(b) 12.447 A

(c) 17.6 A

Explanation:

a)  inductive reactance Z = L Ω

    = L x 2π x F

    = 45.0 x 10⁻³ x 2(3.14) x 42

    = 11.8692‬ ohm

b) rms current

    = 100 / 8.034

    = 12.447 A

c) maximum current in the circuit

    = I eff x rac2

    = 12.447 x 1.414

    = 17.6 A

Answer:

Explanation:

velocity of proton  v = 1.5 x 10³ i  m /s

charge on proton e = 1.6 x 10⁻¹⁹ C

Let the magnetic field be B = Bx i + Bz k

force on charged particle ( proton )

F = e ( v x B )

2.06  x10⁻¹⁶ j = 1.6 x 10⁻¹⁹ [ 1.5 x 10³ i x ( Bx i + Bz k) ]

2.06  x10⁻¹⁶ j = - 1.6 x 10⁻¹⁹ x 1.5 x 10³  Bz j) ]

2.06  x10⁻¹⁶ = - 1.6 x 10⁻¹⁹ x 1.5 x 10³  Bz

Bz = -  .8583  

force on charged particle ( electron )

F = e ( v x B )

8.40  x10⁻¹⁶ j = -1.6 x 10⁻¹⁹ [ - 4.4 x 10³ k  x ( Bx i + Bz k) ]

8.4  x10⁻¹⁶ j =  1.6 x 10⁻¹⁹ x 4.4 x 10³  Bx j ]

- 8.4  x10⁻¹⁶ =  1.6 x 10⁻¹⁹ x 4.4 x 10³  Bx

Bx =  - 1.19

Magnetic field = - 1.19 i - .8583 k

magnitude = √ (1.19² + .8583²)

= 1.467 T

If it is making angle θ with x - axis in x -z plane

Tanθ = (.8583 / 1.19 )

36⁰ .

C )

v = - 3.7 x 10³j m /s

e = - 1.6 x 10⁻¹⁶ C

Force = F = e ( v x B )

= -1.6 x 10⁻¹⁹ [ -3.7 x 10³ j  x ( Bx i + Bz k) ]

=  - 1.6 x 10⁻¹⁹ x 3.7 x 10³  Bx  k -1.6 x 10⁻¹⁹ x 3.7 x 10³Bzi ]

=    5.08 i - 7.04 k

Tanθ = 54 ° .

Answer:

A)P = 1.2 × 10⁵Pa

B)F = 3.2 × 10⁷N

C) P = 1.2 × 10⁵Pa

Explanation:

Part A)

The relative pressure at the bottom of a column of fluid is given by

[tex]p_r = \rho g h[/tex]

where

[tex]\rho[/tex] is the fluid density

g is the gravitational acceleration  

h is the height of the column of fluid

At the bottom of the swimming pool, h=1.9 m, and the water density is  

[tex]\rho[/tex] = 1000 kg/m^3, therefore the relative pressure is

[tex]p_r = (1000 kg/m^3)(9.81 m/s^2)(1.9 m)=1.86 \cdot 10^4 Pa[/tex]

To find the absolute pressure, we must add to this the atmospheric pressure, [tex]p_a[/tex] :

[tex]p= p_r + p_a\\= 1.86 \cdot 10^4 Pa + 1.01 \cdot 10^5 Pa \\=1.2 \times 10^5 Pa[/tex]

part B

Total force acting on the bottom

force = pressure * area

area of pool = 30.0 m × 8.9 m

= 267m²

Force F =

1.2 × 10⁵ * 267m² N

= 32040000 N

F = 3.2 × 10⁷N

Part C

The pressure acting on the side wall will be

now the pressure at the side of the pool at the bottom is simply equal to absolute pressure as they are at same level

P = 1.2 × 10⁵

Answer:

a. Air fuel Ratio = 19.76 kg air/kg fuel

b. % Theoretical air used = 131%

c. Amount of H2O that condenses as the products are cooled to 25°C at 100kPa = 6.59 kmol

Explanation:

Answer:

18.8m

Explanation:

Let the distance traveled before stopping be 'd' m.

Given:

Mass of the skater (m) = 90 kg

Initial velocity of the skater (u) = 12.0 m/s

Final velocity of the skater (v) = 0 m/s (Stops finally)

Coefficient of kinetic friction (μ) = 0.390

Acceleration due to gravity (g) = 9.8 m/s²

Now, we know  from work-energy theorem, that the work done by the net force on a body is equal to the change in its kinetic energy.

Hence, the net force acting on the skater is only frictional force which acts in the direction opposite to motion.

Frictional force is given as:

[tex]f=\mu N[/tex]

Where, 'N' is the normal force acting on the skater. As there is no vertical motion, N=mg

[tex]f=\mu mg\\=0.390\times 90\times 9.8\\=343.98\ N[/tex]

work done by friction is a negative work as friction and displacement are in opposite direction and is given as

[tex]W=-fd=-343.98d[/tex]

Now, change in kinetic energy is given as:

[tex]\Delta K=\frac{1}{2}m(v^2-u^2)\\\\\Delta K=\frac{1}{2}\times 90(0-12^2)\\\\\Delta K=45\times (-144)=-6480\ J[/tex]

Therefore, from work-energy theorem,

[tex]W=\Delta K\\\\-343.98d=6480\\\\d=\frac{6480}{343.98}\\\\d=18.8m[/tex]

Hence, the skater covers a distance of 18.8 m before stopping.

Answer:

18.84m

Explanation:

We are given;

Mass; m = 90 Kg

Initial velocity; u = 12 m/s

Coefficient of friction; μ = 0.390

Let,the combined mass of the ice skater and the skate be M

Thus, So,if the coefficient of frictional force is μ,then frictional force acting is; μN. N= Mg. Thus F_friction = μMg

Now, the deceleration due to friction will be, F_friction/M

Thus, deceleration = μMg/M

M will cancel out and we have; μg

Now, deceleration means negative acceleration. Thus acceleration (a) =

-μg

Now, to find the distance, let's use equation of motion which is;

v² = u² + 2as

Putting -μg for a, we have;

v² = u² + 2(-μg)s

v² = u² — 2μgs

We want to know the distance covered before coming to rest. Thus, v = 0m/s

So,

0² = 12² - (2 x 0.39 x 9.8 x s)

0 = 144 - 7.644s

Thus,

7.644s = 144

Thus, s = 144/7.644 = 18.84m

Answer:

0.002638 C or 2.6388*10^-3 C

Explanation:

Given that

Quantity of the charge, q = -0.0005 C

Force on the charge of magnitude, F1 = 19 N

Distance from the second charge, r = 25 m

Magnitude of force of the second charge, q2 = ? N

F = (kq1q2) / r², where

k = 9*10^9

19 = (9*10^9 * 0.0005 * q2) / 25²

19 * 625 = 4.5*10^6 * q2

q2 = 11875 / 4.5*10^6

q2 = 0.002638 C or 2.6388*10^-3 C

Thus, the magnitude of the second charge is 0.002638 C or 2.6388*10^-3 C

Answer:

[tex]6.53\times10^-^1^7N[/tex]

Explanation:

The magnet of the magnetic field is 53 cm = 0.53m from wire is

[tex]B = \frac{\mu_0 I}{2\pi d}[/tex]

[tex]= \frac{(4\pi \times 10^-^7)(40)}{2 \pi (0.53)} \\\\= \frac{5.0265\times 10^-^5}{3.33} \\\\= 1.5095 \times 10^-^5[/tex]

the magnetic force exerted by the wire on the electron is

[tex]F = Bqv \sin \theta\\\\= 1.5095 \times 10^-^5 \times1.602\times10^-^1^9\times2.7\times10^7\\\\= 6.53\times10^-^1^7N[/tex]

From the right hand rule the direction of the force is parallel to the current (since the particle is electron)

Answer: f = 6.52*10^-16 N

Explanation:

if we assume that the force is directed at the y positive direction, then

B = μi / 2πr, where

μ = 4π*10^-7

B= (4π*10^-7 * 40) / 2 * π * 5.3*10^-2

B = 5.027*10^-5 / 0.333

B = 1.51*10^-4 T

Since v and B are perpendicular, then,

F = qvB

F = 1.6*10^-19 * 2.7*10^7 * 1.51*10^-4

F = 2.416*10^-23 * 2.7*10^7

F = 6.52*10^-16 N

Therefore, the magnitude of the force is, F = 6.52*10^-16 N and it moves in the i negative direction

Answer:

Explanation:

A police car's radar gun emits microwaves with frequency f. 1. =36 GHz. The beam reflects from a car that speeds away from the cruiser with 43 m/s. The receiver ...

Answer:

0.372 kg

Explanation:

The collision between the bullet and the block is inelastic, so only the total momentum of the system is conserved. So we can write:

[tex]mu=(M+m)v[/tex] (1)

where

[tex]m=11.9 g = 11.9\cdot 10^{-3}kg[/tex] is the mass of the bullet

[tex]u=261 m/s[/tex] is the initial velocity of the bullet

[tex]M[/tex] is the mass of the block

[tex]v[/tex] is the velocity at which the bullet and the block travels after the collision

We also know that the block is attached to a spring, and that the surface over which the block slides after the collision is frictionless. This means that the energy is conserved: so, the total kinetic energy of the block+bullet system just after the collision will entirely convert into elastic potential energy of the spring when the system comes to rest. So we can write

[tex]\frac{1}{2}(M+m)v^2 = \frac{1}{2}kx^2[/tex] (2)

where

k = 205 N/m is the spring constant

x = 35.0 cm = 0.35 m is the compression of the spring

From eq(1) we get

[tex]v=\frac{mu}{M+m}[/tex]

And substituting into eq(2), we can solve to find the mass of the block:

[tex](M+m) \frac{(mu)^2}{(M+m)^2}=kx^2\\\frac{(mu)^2}{M+m}=kx^2\\M+m=\frac{(mu)^2}{kx^2}\\M=\frac{(mu)^2}{kx^2}-m=\frac{(11.9\cdot 10^{-3}\cdot 261)^2}{(205)(0.35)^2}-11.9\cdot 10^{-3}=0.372 kg[/tex]

The mass of the wooden block is approximately 0.404 kg.

The initial kinetic energy of the bullet is given by:

KE_initial = 0.5 * mb * vb^2

where:

`m_b` is the mass of the bullet (11.9 g = 0.0119 kg)

`v_b` is the velocity of the bullet (261 m/s)

The potential energy stored in the compressed spring is given by:

PE_spring = 0.5 * k * x^2

where:

`k` is the spring constant (205 N/m)

`x` is the compression of the spring (0.35 m)

Since the system comes to a stop, the initial kinetic energy is converted to potential energy stored in the spring:

KE_initial = PE_spring

Solving for the mass of the wooden block, we get:

m_w = (2 * KE_initial) / (v_b^2 - (k * x) / m_b)

Plugging in the values, we get,

m_w = (2 * 0.5 * 0.0119 kg * (261 m/s)^2) / ((261 m/s)^2 - (205 N/m * 0.35 m) / 0.0119 kg)

m_w ≈ 0.404 kg

Therefore, the mass of the wooden block is approximately 0.404 kg.

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Answer:

The frequency of the sona-pulse reflected back is  [tex]f_b = 48109.22Hz[/tex]

Explanation:

From the question we are told that

     The speed of the bat is  [tex]v = 5.1 m/s[/tex]

      The speed of the insect is [tex]v_i = 1.1 m/s[/tex]

       The frequency emitted by the bat is [tex]f = 47000 \ Hz[/tex]

        The speed of sound is  [tex]v_s = 343 m/s[/tex]

Let look at this question in this manner

At the first instant the that the bat  emits the sonar pulse

     Let the bat be the source of sound

      Let the insect be the observer

This implies that the frequency of sound the the insect would receive is mathematically represented as

                    [tex]f_a = [\frac{v_s - v_i}{v_s - v}] f[/tex]

Substituting values

                 [tex]f_a = [\frac{343 - 1.1}{345 -5.1} ] * 47000[/tex]

                     [tex]f_a = 47556.4 Hz[/tex]

Now at the instant the sonar pules reaches the insect

            Let the bat be the observer

            Let the insect be the source of the sound

Here the sound wave is reflected back to the bat

This implies that the frequency of sound the the bat  would receive is mathematically represented as

                   [tex]f_b =[ \frac{x}{y} ] * f_a[/tex]

                   [tex]f_b =[ \frac{v_s + v}{v_s + v_i} ] * f_a[/tex]

                  [tex]f_b =[ \frac{343 + 5.1}{343 + 1.1} ] * 47556.4[/tex]

                  [tex]f_b = 48109.22Hz[/tex]

Final answer:

To calculate the frequency a bat hears from a sonar pulse reflected back by an insect, the Doppler effect formula is used. By substituting the given values into the formula, the result is approximately 47,544 Hz, which is the frequency at which the bat detects the echo from the insect.

Explanation:

To calculate the frequency at which a bat hears the sonar pulse reflected back from an insect, we can use the Doppler effect formula for sound in the same direction as the source of the sound is moving:

[tex]f' =\frac{f (v + v_d)}{(v + v_s)}[/tex]

Where:

f' is the frequency heard by the bat.

f is the emitted frequency of the bat's sonar pulse (47,000 Hz).

v is the speed of sound (343 m/s).

[tex]v_d[/tex] is the speed of the detector (bat), which is 5.1 m/s.

[tex]v_s[/tex] is the speed of the source (insect), which is 1.1 m/s.

Substituting the values into the equation:

[tex]f' = \frac{47000 Hz \times (343 m/s + 5.1 m/s)}{(343 m/s + 1.1 m/s)}[/tex]

[tex]f' = \frac{47000 Hz * 348.1 m/s}{344.1 m/s}[/tex]

[tex]f' = 47000 Hz \times 1.0116[/tex]

f' = 47544.2 Hz

Therefore, the bat hears the pulse reflected back from the insect at approximately 47,544 Hz.

The velocity that will be needed for the model and prototype to be similar is 108.97m/s  

Explanation:

length of Torpedo = 3m

diameter, [tex]d_{1} = 0.5m[/tex]

velocity of sea water, [tex]v_{1}[/tex]= 10m/s

dynamic viscosity of sea water, η[tex]_{1}[/tex] = 0.00097 Ns/m²

density of sea water, ρ[tex]_{1}[/tex] =  1023 kg/m³

Scale model = 1:15

[tex]\frac{d_{1} }{d_{2} }[/tex] = [tex]\frac{1}{15}[/tex]

Cross multiplying: d[tex]_{2}[/tex] = [tex]15d_{1} }[/tex] = 15 ×0.5 = 7.5m

Let:

velocity of air, [tex]v_{2}[/tex]

viscosity of air, η[tex]_{2}[/tex] =  0.000186Ns/m²

density of air, ρ[tex]_{2}[/tex] =  1.2 kg/m³

For the model and the prototype groups to be equal, Non-dimensional groups should be equal.

Reynold's number:   (ρ[tex]_{2}[/tex] ×[tex]v_{2}[/tex] ×d[tex]_{2}[/tex])/η[tex]_{2}[/tex] =  (ρ[tex]_{1}[/tex] ×[tex]v_{1}[/tex] ×d[tex]_{1}[/tex])/η[tex]_{1}[/tex]

[tex]v_{2}[/tex] = η[tex]_{2}[/tex]/(ρ[tex]_{2}[/tex] ×d[tex]_{2}[/tex])  ×  (ρ[tex]_{1}[/tex] ×[tex]v_{1}[/tex] ×d[tex]_{1}[/tex])/η[tex]_{1}[/tex]

[tex]v_{2}[/tex] =  [tex]\frac{0.000186}{1.2* 7.5}[/tex]×[tex]\frac{1023 *10*0.5}{0.00097}[/tex] , note: * means multiplication

[tex]v_{2}[/tex] = 108.97m/s  

velocity that will be needed for the model and prototype to be similar = 108.97m/s  

Answer:

1. Our ears can sort out the individual sine waves from a mixture of two or more sine waves, so we hear the pure tones that make up a complex tone.

Explanation:

A complex tone is a sound wave that consist of two or more forms of audible sound frequencies. Sound wave is a mechanical wave that is longitudinal, and could be represented by a sine wave because of it sinusoidal manner of propagation.

A Fourier analyzer can be used to differentiate individual sine waves from a combination of two or more of it; which is as the same function performed by human ear. To the human ear, a sound wave that consist of more than one sine wave will have perceptible harmonics which would be distorted and turn to a noise.

Thus, the human ear makes it possible to hear the pure tones that make up a complex tone.

Answer:

1. A Fourier analyzer sorts out the individual sine waves from a mixture of two or more sine waves, so we hear the pure tones that make up a complex tone.

Explanation:

Fourier analysis is a technique that is used to determine which sine waves constitute a given signal, i.e. to deconstruct the signal into its individual sine waves.  It is the process of decomposing a periodic function into its constituent sine or cosine waves.

What goes on inside our ears is a mathematical process called a Fourier transform. In the ear, there's a combination of different waves, Fourier analysis identifies contributions at different frequencies, allowing us to reconstruct the individual signals that go into it.

A complex tone perceived by the air is is an individual sine wave that the ear, by acting as a Fourier analyzer, decomposes to serious of sine waves that we hear as pure tones.

Answer:

Pabs = 247150 [Pa]

Explanation:

The pressure in the depth h can be calculated by the following expression.

Pabs = Po + (rho * g * h)

Where:

g = gravity = 9.81[m/s^2]

rho = density = 1000 [kg/m^3]

h = depth = 15 [m]

Po = 100000 [Pa]

Pabs = 100000 + (1000*9.81*15)

Pabs = 247150 [Pa]

Answer:

The answer is C.

Explanation:

Current can be represented in the symbol of I.

The symbol for voltage is V.

The symbol for power is W.

The symbol for resistance is R.

Answer:

a) [tex]t = 4.6\tau[/tex]

b) [tex]L = 0.0187 \: H[/tex]

Explanation:

The current flowing in a R-L series circuit is given by

[tex]I = I_{0} (1 - e^{\frac{-t}{\tau} })[/tex]

Where τ is the time constant and is given by

[tex]\tau = \frac{L}{R}[/tex]

Where L is the inductance and R is the resistance

Assuming the current has reached steady state when it is at 99% of its maximum value,

[tex]0.99I_{0} = I_{0} (1 - e^{\frac{-t}{\tau} })\\0.99 = (1 - e^{\frac{-t}{\tau} })\\1 - 0.99 = e^{\frac{-t}{\tau}}\\ln(0.01) = ln(e^{\frac{-t}{\tau}})\\-4.6 = \frac{-t}{\tau}\\t = 4.6\tau\\[/tex]

Therefore, it would take t = 4.6τ to reach the steady state.

(b) If an emergency power circuit needs to reach steady state within 1.2 ms of turning on and the circuit has a total resistance of 72 Ω, what values of the total inductance of the circuit are needed to satisfy the requirement?

[tex]t = 4.6\frac{L}{R}\\t = 4.6\frac{L}{R}\\0.0012 = 4.6\frac{L}{72}\\0.0864 = 4.6 L\\L = 0.0864/4.6\\L = 0.0187 \: H[/tex]

Therefore, an inductance of 0.0187 H is needed to satisfy the requirement.

Answer:

A sparrow flew faster. the sparrow flew 10 meters per second. The sparow flew 13.(3) meters per second

Explanation:

A sparrow flew faster than the hawk as it completes more distance in 60 seconds than that of hawk which is about 1200 meters. Speed is the distance travelled per unit time.

Speed is the measure of the distance travelled by an object per unit time taken. Speed is a vector quantity. It has both magnitude and direction.

Speed of an object can be calculated as: Distance travelled divided by time taken.

Speed of hawk is 600 meters/ 60 seconds

Speed of hawk = 10 m/s

Speed of sparrow is 400 meters/ 30 seconds

Speed of Sparrow = 13.33 m/s

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Final answer:

The calculated density ratio is 0.5.

The ratio of the densities of two blocks, A and B, is found by comparing the apparent weight reductions when they are submerged in a fluid. According to Archimedes' principle, this reduction is due to the buoyant force, which is equal to the weight of the fluid displaced.

Explanation:

The question revolves around calculating the ratio of the densities of two blocks, A and B, using the concept of buoyant force according to Archimedes' principle. The apparent weight of each block when submerged in a fluid gives us the buoyant force acting upon it, which is equal to the weight of the fluid displaced. Since both blocks are submerged in the same fluid, the difference in weights in air versus submerged gives us a measure of the volume of fluid displaced due to the respective densities of the blocks.

To determine the ratio of densities (A/B), we use the formula:

Ratio of densities = (Weight of A in air - Apparent weight of A) / (Weight of B in air - Apparent weight of B)

Substituting the given weights:

Ratio of densities = (12 N - 8 N) / (20 N - 12 N) = 4 N / 8 N = 0.5

Thus, the ratio of the densities of block A to block B is 0.5.

Answer:

Explanation:

Expression for times period of a satellite can be given as follows

Time period T = 1.8 x 60 x 60

= 6480

T² = [tex]\frac{4\times \pi^2\times r^3}{GM}[/tex] where T is time period , r is radius of orbit , G is gravitational constant and M is mass of the satellite.

6480² = 4 x 3.14² x 7.5³ x 10¹⁸ / GM

GM = 4 x 3.14² x 7.5³ x 10¹⁸ / 6480²

= 3.96 X 10¹⁴

Expression for acceleration due to gravity

g = GM / R² where R is radius of satellite

20 = 3.96 X 10¹⁴ / R²

R² = 3.96 X 10¹⁴ / 20

= 1.98 x 10¹³ m

R= 4.45 x 10⁶ m

Answer:

see explanation

Explanation:

Given that,

velocity of 1.50 km/s = 1.50 × 10³m/s

acceleration of 2.00 ✕ 1012 m/s2

electric field has a magnitude of strength of 18.0 N/C

[tex]\bar F= q[\bar E + \bar V \times \bar B]\\\\\bar F = [\bar E + \bar V \times ( B_x \hat i +B_y \hat j +B_z \hat z )]\\\\\\m \bar a = [\bar E + \bar V \times ( B_x \hat i +B_y \hat j +B_z \hat z )][/tex]

[tex]9.1 \times 10^-^3^1 \times 2\times 10^1^2 \hat k=-1.6\times10^-^1^9 \hat k [18\hat k+ 1.5\times 10^3 \hat i \times (B_x \hat i +B_y \hat j +B_z \hat k)][/tex][tex]42.2 \times 10^-^1^9 \hat k = -2.4 \times 10^1^6B_y \hat k + 2.4 \times 10 ^1^6 \hat j B_z\\[/tex]

[tex]B_x = undetermined[/tex]

[tex]B_y = \frac{42.2 \times 10^-^1^9}{-2.4 \times 10^-^1^6} \\\\= - 0.0176 T[/tex]

[tex]B_z = 0T[/tex]

Answer:

About eq (1)

[tex]mv^2/r = eVB[/tex]

when a charged particle (electron) enters into the magnetic field which is perpendicular to direction of motion than there will be magnetic force on particle and particle will travel in circular path in with constant speed.

So using force balance on charged particle:

[tex]F_{net} = Fc - Fm[/tex]

Since particle is traveling at constant speed, So acceleration is zero, and

[tex]F_{net} = 0[/tex]

[tex]Fc - Fm=0[/tex]

[tex]Fc = Fm[/tex]

Fc = centripetal force on particle [tex]= m*v^2/r[/tex]

Fm = magnetic force on electron = [tex]q*VxB = q*V*B*sin \theta[/tex]

q = charge on electron = e

since magnetic field is perpendicular to the velocity of particle, So theta = 90 deg

sin 90 deg = 1

So,

[tex]m*v^2/r = e*v*B[/tex]

About equation 2:

When this charged particle is released from rest in a potential difference V, and then it enters into above magnetic field, then using energy conservation on charge particle

[tex]KEi + PEi = KEf + PEf[/tex]

KEi = 0, since charged particle started from rest

[tex]PEi - PEf = q*dV[/tex]

[tex]PEi - PEf = eV[/tex]

KEf = final kinetic energy of particle when it leaves [tex]= (1/2)*m*v^2[/tex]

So,

[tex]0 + eV = (1/2)*m*v^2[/tex]

[tex]eV = (1/2)*m*v^2[/tex]

From above equation (1) and (2)

[tex]m*v^2/r = evB[/tex]

[tex]e/m = v/(r*B)[/tex]

Now

[tex]eV = (1/2)*m*v^2[/tex]

[tex]v = \sqrt{(2*e*V/m)}[/tex]

[tex]e/m = \sqrt{ (2*e*V/m)/(r*B)}[/tex]

[tex]\frac{e^2}{m^2} = \frac{2*e*V}{(m*r^2*B^2)}[/tex]

[tex]e/m = 2*V/(r^2*B^2)[/tex]

[tex]e/m = 2V/(Br)^2[/tex]

Answer: The length L of the wire used to construct the coil is 191.4m

Explanation: Please see the attachments below

Complete Question

A thin uniform film of refractive index 1.750 is placed on a sheet of glass with a refractive index 1.50. At room temperature ( 18.8 ∘C), this film is just thick enough for light with a wavelength 580.9 nm reflected off the top of the film to be canceled by light reflected from the top of the glass. After the glass is placed in an oven and slowly heated to 170 ∘C, you find that the film cancels reflected light with a wavelength 588.2 nm .

What is the coefficient of linear expansion of the film? (Ignore any changes in the refractive index of the film due to the temperature change.) Express your answer using two significant figures.

Answer:

the coefficient of linear expansion of the film is [tex]\alpha = 7.93 *10^{-5} / ^oC[/tex]

Explanation:

  From the question we are told that

     The refractive index of the film is  [tex]n_f = 1.750[/tex]

      The refractive index of the glass is [tex]n_g = 1. 50[/tex]

       The wavelength of light reflected at 18°C is [tex]\lambda _r = 580.9nm = 580.9*10^{-9}m[/tex]

      The wavelength of light reflected at 170°C is [tex]\lambda_h = 588.2 nm = 588.2 * 10^{-9}m[/tex]

For destructive interference the condition is  

         [tex]2t = \frac{m \lambda }{n_f}[/tex]

  Where m is the order of interference

              t is the thickness

For the smallest thickness is  when m= 1 and this is represented as

             [tex]t = \frac{\lambda }{2n_f }[/tex]

At 18°C  the thickness would be

              [tex]t_{r} = \frac{580.9 *10^{-9}}{2 * 1.750}[/tex]

              [tex]t_{r} = 166nm[/tex]\

At 170° the  thickness is  

               [tex]t_h = \frac{588.2 *10^{-9}}{2 * 1.750}[/tex]

               [tex]t_h = 168 nm[/tex]

The coefficient of linear expansion f the film is mathematically represented as

              [tex]\alpha = \frac{t_h - t_r}{t_r \Delta T}[/tex]

Substituting value

                 [tex]\alpha = \frac{168 *10^{-9} - 166 *10^{-9} }{166*10^{-9} * (170 -18)}[/tex]

                 [tex]\alpha = 7.93 *10^{-5} / ^oC[/tex]

Answer:

α=8.37 x [tex]10^{-5}[/tex] °[tex]C^{-1}[/tex]

Explanation:

Complete question : What is the coefficient of linear expansion of the film?

SOLUTION:

There is a net ( λ/2 ) phase change due to reflection for this film, therefore, destructive interference is given by

2t = m( λ/n)   where n=1.750

for smallest non-zero thickness

t= λ/2n

At 18.8°C, [tex]t_{o[/tex]=580.9 x [tex]10^{-9}[/tex]/(2 x 1.750)

[tex]t_{o[/tex]= 165.9nm

At 170°C, t= 588.2x [tex]10^{-9}[/tex]/(2x1.750)

t=168nm

t=[tex]t_{o[/tex](1 + αΔT)

=>α= (t-[tex]t_{o[/tex])/ ([tex]t_{o[/tex]ΔT)   [ΔT= 170-18.8 =151.2°C]

α= (168 x [tex]10^{-9}[/tex] - 165.9 x [tex]10^{-9}[/tex])/ (165.9 x [tex]10^{-9}[/tex] x 151.2)

α= 2.1 x [tex]10^{-9}[/tex]/ 2.508 x [tex]10^{-5}[/tex]

α=8.37 x [tex]10^{-5}[/tex] °[tex]C^{-1}[/tex]

Therefore, the coefficient of linear expansion of the film is 8.37 x [tex]10^{-5}[/tex] °[tex]C^{-1}[/tex]

Answer:

False

Explanation:

Friction is the resisting force of motion that converts kinetic energy into heat energy. Kinetic energy is the energy of movement. Removing kinetic energy makes the object slower.

Answer:

a) 22.06m/s

b)45°

Explanation:

Let 'm' be  mass of 1st and 2nd pieces

mass of 3rd piece=3m

[tex]v_{1}[/tex]=velocity of 1st piece = -47iˆ  m/s

[tex]v_{2}[/tex]=velocity of 2nd piece = -47jˆ  m/s

[tex]v_{3}[/tex]=velocity of 3rd piece=?

By considering conservation of linear momentum, we have

m[tex]v_{1}[/tex] + m[tex]v_{2}[/tex] + 3m[tex]v_{3}[/tex]=0

[tex]v_{1}[/tex] + [tex]v_{2}[/tex] + 3[tex]v_{3}[/tex]=0

3[tex]v_{3}[/tex]= - ([tex]v_{1}[/tex] + [tex]v_{2}[/tex] )

[tex]v_{3}[/tex] = -[tex]\frac{1}{3}[/tex] ([tex]v_{1}[/tex] + [tex]v_{2}[/tex] )

Substituting the values of [tex]v_{1}[/tex] and [tex]v_{2}[/tex] in above equation

[tex]v_{3}[/tex] = -[tex]\frac{1}{3}[/tex] (-47iˆ -47jˆ ) => 15.6iˆ + 15.6jˆ

(a)Magnitude of the velocity of  the third piece is given by

|[tex]v_{3}[/tex]| =√15.6²+15.6² => 22.06m/s

(b) its direction (as an angle relative to the +x axis)  'θ'

θ= [tex]tan^{-1} (\frac{15.6}{15.6} )[/tex] => [tex]tan^{-1} (1 )[/tex] =>45°

Answer:

 m_{p} = 0.3506 kg

Explanation:

For this exercise we use Newton's equilibrium equation

      B - Wc-Wp = 0

where B is the thrust of the water, Wc is the weight of the coins and Wp is the weight of the plastic block

       B = Wc + Wp

the state push for the Archimeas equation

      B = ρ_water g V

the volume of the water is the area of ​​the block times the submerged height h, which is

        h´ = 8 - h

where h is the height out of the water

      ρ_water g A h´ = [tex]m_{c}[/tex] g + [tex]m_{p}[/tex] g

      ρ_water A h´ = m_{c} + m_{p}

write this equation to make the graph

        h´= 1 /ρ_water A    (m_{c} +m_{p})

        h´ = 1 /ρ_waterA    (m_{c} + m_{p})

if we graph this expression, we get an equation of the line

        y = m x + b

where

        y = h´

        m = 1 /ρ_water A

        b = mp /ρ_water A

whereby

       m_{p} = b ρ_water A

       ρ_water = 1000 kg / m³

       b = 0.0312 m

       m = 0.0890 m / kg

we substitute the slope equation

       b = m_{p} / m

calculate

       m_{p}= 0.0312 / 0.0890

       m_{p} = 0.3506 kg