A long cylindrical rod of diameter 200 mm with thermal conductivity of 0.5 W/m⋅K experiences uniform volumetric heat generation of 24,000 W/m3. The rod is encapsulated by a circular seeve having an outer diameter of 400 mm and a thermal conductivity of 4 W/m⋅K. The outer surface of the sleeve is exposed to cross flow air at 27∘C with a convection coefficient of 25 W/m2⋅K.
(a) Find the temperature at the interface between the rod and sleeve and on the outer surface.
(b) What is the temperature at the center of the rod?

Answer:

2790 J/C

Explanation:

charge on sphere Q = 62 nC = [tex]62*10^{-9} C[/tex]

radius of the sphere r = 5.0 cm = 0.05 m

distance away from reference point d = 15.0 cm = 0.15 m

total distance of charge relative reference point R = r + d = 0.05 + 0.15 = 0.2 m

electric potential V is given as

[tex]V = \frac{kQ}{R}[/tex]

where k = Coulumb's constant = [tex]9*10^{9}[/tex] kg⋅m³⋅s⁻⁴⋅A⁻²

[tex]V = \frac{9*10^{9} * 62*10^{-9} }{0.2}[/tex] = [tex]\frac{9*62}{0.2}[/tex]

V = 2790 J/C

Answer:

If the particle is an electron [tex]E_y = 3.311 * 10^3 N/C[/tex]

If the particle is a proton, [tex]E_y = 6.08 * 10^6 N/C[/tex]

Explanation:

Initial speed at the origin, [tex]u = 3 * 10^6 m/s[/tex]

[tex]\theta = 38^0[/tex] to +ve x-axis

The particle crosses the x-axis at , x = 1.5 cm = 0.015 m

The particle can either be an electron or a proton:

Mass of an electron, [tex]m_e = 9.1 * 10^{-31} kg[/tex]

Mass of a proton, [tex]m_p = 1.67 * 10^{-27} kg[/tex]

The electric field intensity along the positive y axis [tex]E_y[/tex], can be given by the formula:

[tex]E_y = \frac{2 m u^2 sin \theta cos \theta}{qx} \\[/tex]

If the particle is an electron:

[tex]E_y = \frac{2 m_e u^2 sin \theta cos \theta}{qx} \\[/tex]

[tex]E_y = \frac{2 * 9.1 * 10^{-31} * (3*10^6)^2 *(sin38)( cos38)}{1.6*10^{-19} * 0.015} \\[/tex]

[tex]E_y = 3311.13 N/C\\E_y = 3.311 * 10^3 N/C[/tex]

If the particle is a proton:

[tex]E_y = \frac{2 m_p u^2 sin \theta cos \theta}{qx} \\[/tex]

[tex]E_y = \frac{2 * 1.67 * 10^{-27} * (3*10^6)^2 *(sin38)( cos38)}{1.6*10^{-19} * 0.015} \\[/tex]

[tex]E_y = 6.08 * 10^6 N/C[/tex]

Answer:

0.00185 °C

Explanation:

From the question,

The potential energy of the bird = heat gained by the water in the fish tank.

mgh = cm'(Δt)................... Equation 1

Where m = mass of the bird, g = acceleration due to gravity, h = height, c = specific heat capacity of water, m' = mass of water, Δt = rise in temperature of water.

make Δt the subject of the equation

Δt = mgh/cm'............... Equation 2

Given: m = 1 kg, h = 40 m, m' = 50.5 kg

constant: g = 9.8 m/s², c = 4200 J/kg.K

Substitute into equation 2

Δt = 1(40)(9.8)/(50.5×4200)

Δt = 392/212100

Δt = 0.00185 °C

Answer:

B = 0.024T positive z-direction

Explanation:

In this case you consider that the direction of the motion of the electron, and the direction of the magnetic field are perpendicular.

The magnitude of the magnetic force exerted on the electron is given by the following formula:

[tex]F=qvB[/tex]     (1)

q: charge of the electron = 1.6*10^-19 C

v: speed of the electron = 1.6*10^7 m/s

B: magnitude of the magnetic field = ?

By the Newton second law you also have that the magnetic force is equal to:

[tex]F=qvB=ma[/tex]       (2)

m: mass of the electron = 9.1*10^-31 kg

a: acceleration of the electron = 7.0*10^16 m/s^2

You solve for B from the equation (2):

[tex]B=\frac{ma}{qv}\\\\B=\frac{(9.1*10^{-31}kg)(7.0*10^{16}m/s^2)}{(1.6*10^{-19}C)(1.6*10^7m/s)}\\\\B=0.024T[/tex]

The direction of the magnetic field is found by using the right hand rule.

The electron moves upward (+^j). To obtain a magnetic forces points to the positive x-direction (+^i), the direction of the magnetic field has to be to the positive z-direction (^k). In fact, you have:

-^j X ^i = ^k

Where the minus sign of the ^j is because of the negative charge of the electron.

Then, the magnitude of the magnetic field is 0.024T and its direction is in the positive z-direction

Given that,

Height =1.5 m

Angle = 45°

We need to find the greater speed of the ball

Using conservation of energy

[tex]P.E_{i}+K.E_{f}=P.E_{f}+K.E_{f}[/tex]

[tex]mgh+\dfrac{1}{2}mv_{i}^2=mgh+\dfrac{1}{2}mv_{f}^2[/tex]

Here, initial velocity and final potential energy is zero.

[tex]mgh=\dfrac{1}{2}mv_{f}^2[/tex]

Put the value into the formula

[tex]9.8\times1.5=\dfrac{1}{2}v_{f}^2[/tex]

[tex]v_{f}^2=2\times9.8\times1.5[/tex]

[tex]v_{f}=\sqrt{2\times9.8\times1.5}[/tex]

[tex]v_{f}=5.42\ m/s[/tex]

Hence, the greater speed of the ball is 5.42 m/s.

Answer:

blue  θ₂ = 22.26º

red    θ₂ = 22.79º

Explanation:

When a light beam passes from one material medium to another, it undergoes a deviation from the path, described by the law of refraction

         n₁ sin θ₁ = n₂ sin θ₂

where n₁ and n₂ are the incident and transmitted media refractive indices and θ are the angles in the media

let's apply this equation to each wavelength

λ = blue

in this case n₁ = 1, n₂ = 1,645

       sin θ₂ = n₁/ n₂ sin₂ θ₁

let's calculate

       sin θ₂ = 1 / 1,645 sint 38.55

       sin θ₂ = 0.37884

       θ₂ = sin⁻¹ 0.37884

       θ₂ = 22.26º

λ = red

n₂ = 1,609

         sin θ₂ = 1 / 1,609 sin 38.55

         sin θ₂ = 0.3873

         θ₂ = sim⁻¹ 0.3873

         θ₂ = 22.79º

the refracted rays are between these two angles

Answer:

a) V =  140.8 m/s

b) T = 2027.52 N = 2.03 KN

Explanation:

a)

The formula for the speed of the wave is given as follows:

f₁ = V/2L

V = 2f₁L

where,

V = Speed of Wave = ?

f₁ = Fundamental Frequency = 80 Hz

L = Length of Wire = 88 cm = 0.88 m

Therefore,

V = (2)(80 Hz)(0.88 m)

V =  140.8 m/s

b)

Another formula for the speed of wave is:

V = √T/μ

V² = T/μ

T = V²μ

where,

T = Tension in String = ?

μ = Linear Mass Density of Wire = Mass of Wire/L = 0.09 kg/0.88 m

μ = 0.1 kg/m

Therefore,

T = (140.8 m/s)²(0.1 kg/m)

T = 2027.52 N = 2.03 KN

Answer:

9.6 Ns

Explanation:

Note: From newton's second law of motion,

Impulse = change in momentum

I = m(v-u).................. Equation 1

Where I = impulse, m = mass of the ball, v = final velocity, u = initial velocity.

Given: m = 2.4 kg, v = 2.5 m/s, u = -1.5 m/s (rebounds)

Substitute into equation 1

I = 2.4[2.5-(-1.5)]

I = 2.4(2.5+1.5)

I = 2.4(4)

I = 9.6 Ns

The magnitude of impulse will be "9.6 Ns".

According to the question,

Mass,

Final velocity,

Initial velocity,

By using Newton's 2nd law of motion, we get

→ Impulse, [tex]I = m(v-u)[/tex]

By substituting the values, we get

                     [tex]= 2.4[2.5-(1.5)][/tex]

                     [tex]= 2.4(2.5+1.5)[/tex]

                     [tex]= 2.4\times 4[/tex]

                     [tex]= 9.6 \ Ns[/tex]

Thus the above answer is right.    

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Answer:

6

Explanation:

We are given that

[tex]\theta=2.12^{\circ}[/tex]

Slid width,a=0.110 mm=[tex]0.11\times 10^{-3} m[/tex]

[tex]1mm=10^{-3} m[/tex]

Wavelength,[tex]\lambda=582 nm=582\times 10^{-9}[/tex] m

[tex]1nm=10^{-9} m[/tex]

We have to find the number of diffraction maxima are contained in a region of the Fraunhofer single-slit pattern.

[tex]asin\theta=\frac{2N+1}{2}\lambda[/tex]

Using the formula

[tex]0.11\times 10^{-3}sin(2.12)=\frac{2N+1}{2}(582\times 10^{-9})[/tex]

[tex]2N+1=\frac{0.11\times 10^{-3}sin(2.12)\times 2}{582\times 10^{-9}}[/tex]

[tex]2N+1=13.98[/tex]

[tex]2N=13.98-1=12.98[/tex]

[tex]N=\frac{12.98}{2}\approx 6[/tex]

Hence, 6 diffraction maxima are contained in a region of the Fraunhofer single-slit pattern

Answer:

A) M

Explanation:

The three blocks are set in series on a horizontal frictionless surface, whose mutual contact accelerates all system to the same value due to internal forces as response to external force exerted on the box of mass M (Newton's Third Law). Let be F the external force, and F' and F'' the internal forces between boxes of masses M and 2M, as well as between boxes of masses 2M and 3M. The equations of equilibrium of each box are described below:

Box with mass M

[tex]\Sigma F = F - F' = M\cdot a[/tex]

Box with mass 2M

[tex]\Sigma F = F' - F'' = 2\cdot M \cdot a[/tex]

Box with mass 3M

[tex]\Sigma F = F'' = 3\cdot M \cdot a[/tex]

On the third equation, acceleration can be modelled in terms of F'':

[tex]a = \frac{F''}{3\cdot M}[/tex]

An expression for F' can be deducted from the second equation by replacing F'' and clearing the respective variable.

[tex]F' = 2\cdot M \cdot a + F''[/tex]

[tex]F' = 2\cdot M \cdot \left(\frac{F''}{3\cdot M} \right) + F''[/tex]

[tex]F' = \frac{5}{3}\cdot F''[/tex]

Finally, F'' can be calculated in terms of the external force by replacing F' on the first equation:

[tex]F - \frac{5}{3}\cdot F'' = M \cdot \left(\frac{F''}{3\cdot M} \right)[/tex]

[tex]F = \frac{5}{3} \cdot F'' + \frac{1}{3}\cdot F''[/tex]

[tex]F = 2\cdot F''[/tex]

[tex]F'' = \frac{1}{2}\cdot F[/tex]

Afterwards, F' as function of the external force can be obtained by direct substitution:

[tex]F' = \frac{5}{6}\cdot F[/tex]

The net forces of each block are now calculated:

Box with mass M

[tex]M\cdot a = F - \frac{5}{6}\cdot F[/tex]

[tex]M\cdot a = \frac{1}{6}\cdot F[/tex]

Box with mass 2M

[tex]2\cdot M\cdot a = \frac{5}{6}\cdot F - \frac{1}{2}\cdot F[/tex]

[tex]2\cdot M \cdot a = \frac{1}{3}\cdot F[/tex]

Box with mass 3M

[tex]3\cdot M \cdot a = \frac{1}{2}\cdot F[/tex]

As a conclusion, the box with mass M experiments the smallest net force acting on it, which corresponds with answer A.

Answer:

(a) B = 2.85 × [tex]10^{-6}[/tex] Tesla

(b) I =  I = 0.285 A

Explanation:

a. The strength of magnetic field, B, in a solenoid is determined by;

r = [tex]\frac{mv}{qB}[/tex]

⇒ B = [tex]\frac{mv}{qr}[/tex]

Where: r is the radius, m is the mass of the electron, v is its velocity, q is the charge on the electron and B is the magnetic field

B = [tex]\frac{9.11*10^{-31*1.0*10^{4} } }{1.6*10^{-19}*0.02 }[/tex]

  = [tex]\frac{9.11*10^{-27} }{3.2*10^{-21} }[/tex]

B = 2.85 × [tex]10^{-6}[/tex] Tesla

b. Given that; N/L = 25 turns per centimetre, then the current, I, can be determined by;

B = μ I N/L

⇒    I = B ÷ μN/L

where B is the magnetic field,  μ is the permeability of free space = 4.0 ×[tex]10^{-7}[/tex]Tm/A, N/L is the number of turns per length.

I = B ÷ μN/L

 = [tex]\frac{2.85*10^{-6} }{4*10^{-7} *25}[/tex]

I = 0.285 A

Answer:

θ = 12.60°

Explanation:

In order to calculate the angle below the horizontal for the velocity of the hockey puck, you need to calculate both x and y component of the velocity of the puck, and also you need to use the following formula:

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

θ: angle below he horizontal

vy: y component of the velocity just after the puck hits the ground

vx: x component of the velocity

The x component of the velocity is constant in the complete trajectory and is calculated by using the following formula:

[tex]v_x=v_o[/tex]

vo: initial velocity = 28.0 m/s

The y component is calculated with the following equation:

[tex]v_y^2=v_{oy}^2+2gy[/tex]         (2)

voy: vertical component of the initial velocity = 0m/s

g: gravitational acceleration = 9.8 m/s^2

y: height

You solve the equation (2) for vy and replace the values of the parameters:

[tex]v_y=\sqrt{2gy}=\sqrt{2(9.8m/s^2)(2.00m)}=6.26\frac{m}{s}[/tex]

Finally, you use the equation (1) to find the angle:

[tex]\theta=tan^{-1}(\frac{6.26m/s}{28.0m/s})=12.60\°[/tex]

The angle below the horizontal is 12.60°

The angle below the horizontal of the velocity of the puck just before it hits the ground is 12.60°.

Given the following data:

To find the angle below the horizontal of the velocity of the puck just before it hits the ground:

First of all, we would determine the horizontal and vertical components of the hockey puck.

For horizontal component:

[tex]V_y^2 = U_y^2 + 2aS\\\\V_y^2 = 0^2 + 2(9.81)(2)\\\\V_y^2 = 39.24\\\\V_y = \sqrt{39.24} \\\\V_y = 6.26 \; m/s[/tex]

For vertical component:

[tex]V_x = U_x\\\\V_x = 28.0 \;m/s[/tex]

Now, we can find the angle by using the formula:

[tex]\Theta = tan^{-1} (\frac{V_y}{V_x} )[/tex]

Substituting the values, we have:

[tex]\Theta = tan^{-1} (\frac{6.26}{28.0} )\\\\\Theta = tan^{-1} (0.2236)\\\\\Theta = 12.60[/tex]

Angle = 12.60 degrees.

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Answer:

3.08*10^-6 m

Explanation:

Given that

Total shearing force, F = 640 N

Shear modulus, S = 1*10^9 N/m²

Height of the cylinder, L = 0.7 cm

Diameter of the cylinder, d = 4.3 cm

The solution is attached below.

We have our shear deformation to be 3.08*10^-6 m

Answer:

The beam used is a negatively charged electron beam with a velocity of

v = E / B

Explanation:

After reading this long statement we can extract the data to work on the problem.

* They indicate that when the beam passes through the plates it deviates towards the positive plate, so the beam must be negative electrons.

* Now indicates that the electric field and the magnetic field are contracted and that the beam passes without deviating, so the electric and magnetic forces must be balanced

           [tex]F_{e} = F_{m}[/tex]

           q E = qv B

           v = E / B

this configuration is called speed selector

They ask us what type of beam was used.

The beam used is a negatively charged electron beam with a velocity of v = E / B

Answer:

a) A positive current will be induced in the coil

Explanation:

Electromagnetic induction is the induction of an electric field on a conductor due to a changing magnetic field flux. The change in the flux can be by moving the magnet relative to the conductor, or by changing the intensity of the magnetic field of the magnet. In the case of this electromagnets, the gradual increase in the the electromagnet's field strength will cause a flux change, which will in turn induce an electric current on the coil.

According to Lenz law, the induced current acts in such a way as to negate the motion or action that is producing it. A positive current will be induced on the coil so as to repel any form of attraction between the north pole of the electromagnet and the coil. This law obeys the law of conservation of energy, since work has to be done to move the move them closer to themselves.

Answer: Part 1: Propellant Fraction (MR) = 8.76

Part 2: Propellant Fraction (MR) = 1.63

Explanation: The Ideal Rocket Equation is given by:

Δv = [tex]v_{ex}.ln(\frac{m_{f}}{m_{e}} )[/tex]

Where:

[tex]v_{ex}[/tex] is relationship between exhaust velocity and specific impulse

[tex]\frac{m_{f}}{m_{e}}[/tex] is the porpellant fraction, also written as MR.

The relationship [tex]v_{ex}[/tex] is: [tex]v_{ex} = g_{0}.Isp[/tex]

To determine the fraction:

Δv = [tex]v_{ex}.ln(\frac{m_{f}}{m_{e}} )[/tex]

[tex]ln(MR) = \frac{v}{v_{ex}}[/tex]

Knowing that change in velocity is Δv = 9.6km/s and [tex]g_{0}[/tex] = 9.81m/s²

Note: Velocity and gravity have different measures, so to cancel them out, transform km in m by multiplying velocity by 10³.

Part 1: Isp = 450s

[tex]ln(MR) = \frac{v}{v_{ex}}[/tex]

ln(MR) = [tex]\frac{9.6.10^{3}}{9.81.450}[/tex]

ln (MR) = 2.17

MR = [tex]e^{2.17}[/tex]

MR = 8.76

Part 2: Isp = 2000s

[tex]ln(MR) = \frac{v}{v_{ex}}[/tex]

ln (MR) = [tex]\frac{9.6.10^{3}}{9.81.2.10^{3}}[/tex]

ln (MR) = 0.49

MR = [tex]e^{0.49}[/tex]

MR = 1.63

Answer:

λ = 509 nm

Explanation:

In order to calculate the wavelength of the light you use the following formula:

[tex]y=m\frac{\lambda D}{d}[/tex]   (1)

where

y: distance of the mth fringe to the central peak = 3.30 mm = 3.30*10^-3 m

m: order of the bright fringe = 1

D: distance from the slits to the screen = 3.24 m

d: distance between slits = 0.500 mm = 0.500*10^-3 m

You first solve the equation (1) for λ, and then you replace the values of the other parameters:

[tex]\lambda=\frac{dy}{mD}\\\\\lambda=\frac{(0.500*10^{-3}m)(3.30*10^{-3}m)}{(1)(3.24m)}=5.09*10^7m\\\\\lambda=509*10^{-9}m=509nm[/tex]

The wavelength of the light is 509 nm

Answer:

μ_k = 0.1773

Explanation:

We are given;

Initial velocity;u = 20 m/s

Final velocity;v = 0 m/s (since it comes to rest)

Distance before coming to rest;s = 115 m

Let's find the acceleration using Newton's second law of motion;

v² = u² + 2as

Making a the subject, we have;

a = (v² - u²)/2s

Plugging relevant values;

a = (0² - 20²)/(2 × 115)

a = -400/230

a = -1.739 m/s²

From the question, the only force acting on the puck in the x direction is the force of friction. Since friction always opposes motion, we see that:

F_k = −ma - - - (1)

We also know that F_k is defined by;

F_k = μ_k•N

Where;

μ_k is coefficient of kinetic friction

N is normal force which is (mg)

Since gravity acts in the negative direction, the normal force will be positive.

Thus;

F_k = μ_k•mg - - - (2)

where g is acceleration due to gravity.

Thus,equating equation 1 and 2,we have;

−ma = μ_k•mg

m will cancel out to give;

-a = μ_k•g

μ_k = -a/g

g has a constant value of 9.81 m/s², so;

μ_k = - (-1.739/9.81)

μ_k = 0.1773

The coefficient of kinetic friction between the hockey puck and ice is equal to 0.178

Given the following data:

Scientific data:

To determine the coefficient of kinetic friction between the hockey puck and ice:

First of all, we would calculate the acceleration of the hockey puck by using the third equation of motion.

[tex]V^2 = U^2 + 2aS\\\\0^2 =20^2 + 2a(115)\\\\-400=230a\\\\a=\frac{-400}{230}[/tex]

Acceleration, a = -1.74 [tex]m/s^2[/tex]

Note: The negative signs indicates that the hockey puck is slowing down or decelerating.

From Newton's Second Law of Motion, we have:

[tex]\sum F_x = F_k + F_n =0\\\\F_k =- F_n\\\\\mu mg =-ma\\\\\mu = \frac{-a}{g}\\\\\mu = \frac{-(-1.74)}{9.8}\\\\\mu = \frac{1.74}{9.8}[/tex]

Coefficient of kinetic friction = 0.178

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Complete question:

The exit nozzle in a jet engine receives air at 1200 K, 150 kPa with negligible kinetic energy. The exit pressure is 80 kPa, and the process is reversible and adiabatic. Use constant specific heat at 300 K to find the exit velocity.

Answer:

The exit velocity is 629.41 m/s

Explanation:

Given;

initial temperature, T₁ = 1200K

initial pressure, P₁ = 150 kPa

final pressure, P₂ = 80 kPa

specific heat at 300 K, Cp = 1004 J/kgK

k = 1.4

Calculate final temperature;

[tex]T_2 = T_1(\frac{P_2}{P_1})^{\frac{k-1 }{k}[/tex]

k = 1.4

[tex]T_2 = T_1(\frac{P_2}{P_1})^{\frac{k-1 }{k}}\\\\T_2 = 1200(\frac{80}{150})^{\frac{1.4-1 }{1.4}}\\\\T_2 = 1002.714K[/tex]

Work done is given as;

[tex]W = \frac{1}{2} *m*(v_i^2 - v_e^2)[/tex]

inlet velocity is negligible;

[tex]v_e = \sqrt{\frac{2W}{m} } = \sqrt{2*C_p(T_1-T_2)} \\\\v_e = \sqrt{2*1004(1200-1002.714)}\\\\v_e = \sqrt{396150.288} \\\\v_e = 629.41 \ m/s[/tex]

Therefore, the exit velocity is 629.41 m/s

Answer:Work is the energy required to move an object from one point to another. while power is the energy transferred per unit time.

Answer:

-67 m/s

Explanation:

We are given that

Mass of ball,m=0.4 kg

Initial speed,u=14 m/s

Impulse,I=-32.4 N-s

Time,t=27 ms=[tex]27\times 10^{-3} s[/tex]

We have to find the mass's velocity in the x- direction.

We know that

[tex]Impulse=mv-mu[/tex]

Substitute the values

[tex]-32.4=0.4v-0.4(14)[/tex]

[tex]-32.4+0.4(14)=0.4 v[/tex]

[tex]-26.8=0.4v[/tex]

[tex]v=\frac{-26.8}{0.4}=-67m/s[/tex]

Answer:

  r₂ = 1,586 m

Explanation:

For this problem we are going to solve it by parts, let's start by finding the sound intensity when we are 25 m

         β = 10 log (I / I₀)

where Io is the sensitivity threshold 10⁻¹² W / m²

          I₁ / I₀ = [tex]e^{\beta/10}[/tex]

          I₁ = I₀  e^{\beta/10}

let's calculate

          I₁ = 10⁻¹² e^{25/10}

          I₁ = 1.20 10⁻¹¹ W / m²

the other intensity in exercise is

          I₂ = 10⁻¹² e^{80/10}

          I₂ = 2.98 10⁻⁹ W / m²

now we use the definition of sound intensity

          I = P / A

where P is the emitted power that is a constant and A the area of ​​the sphere where the sound is distributed

         P = I A

the area a sphere is

         A = 4π r²

we can write this equation for two points of the found intensities

          I₁ A₁ = I₂ A₂

where index 1 corresponds to 25m and index 2 to the other distance

          I₁ 4π r₁² = I₂ 4π r₂²

          I₁ r₁² = I₂ r₂²

           r₂ = √ (I₁ / I₂) r₁

let's calculate

           r₂ = √ (1.20 10⁻¹¹ / 2.98 10⁻⁹) 25

           r₂ = √ (0.40268 10⁻²) 25

           r₂ = 1,586 m

Answer:

5.95m/s to 2 decimal places

Explanation:

In physics speed is measured in metres per second so convert 8mins to seconds

8x60=420 seconds

The formula needed:

Speed (m/s)= Distance (m)/Time (s)

2500/420=5.95m/s

"1 watt" means 1 joule of energy per second.

75 W means 75 joules/sec .

Energy = (75 Joule/sec) x (12 min) x (60 sec/min)

Energy = (75 x 12 x 60) (Joule-min-sec / sec-min)

Energy = 54,000 Joules

Answer:

2 Al(s) + 3Cl₂(g) → 2AlCl₃(s)

This is a combination reaction.

Ca(OH)₂(aq) + H₂SO₄(aq) → CaSO₄(aq) + 2H₂O(l)

This is a replacement reaction.

Explanation:

A combination reaction is a reaction in which two reagents are combined into one product. The reaction has the following general form:

A + B → AB

where A and B represent any two chemical substances.

2 Al(s) + 3Cl₂(g) → 2AlCl₃(s)

This is a combination reaction because a single compound forms from two or more reacting species.

Double Substitution, Double Displacement or Metastasis Reactions are those in which two elements found in different compounds exchange their positions forming two new compounds. These chemical reactions do not present changes in the number of oxidation or relative load of the elements. So they are not considered redox reactions.

The solvent of the double displacement reactions usually is water and the reagents and products are usually ionic compounds (cations or anions are exchanged), although they can also be acids or bases.

In general, this type of reaction can be expressed as:

AB + CD ⇒ AD + CD

In the reaction:

Ca(OH)₂(aq) + H₂SO₄(aq) → CaSO₄(aq) + 2H₂O(l)

This is a replacement reaction because it is a double replacement reaction in which the ions are exchanged to form new compounds.

Answer:

   θ = 4.78º

with respect to the vertical or 4.78 to the east - north

Explanation:

This is a velocity compound exercise since it is a vector quantity.

The plane takes a direction, the air blows to the west and the result must be to the north, let's use the Pythagorean theorem to find the speed

                  v_fly² = v_nort² + v_air²

                  v_nort² = v_fly² + - v_air²

Let's use trigonometry to find the direction of the plane

        sin θ = v_air / v_fly

        θ = sin⁻¹ (v_air / v_fly)

let's calculate

        θ = sin⁻¹ (10/120)

         θ = 4.78º

with respect to the vertical or 4.78 to the north-east

Answer:

  c.  the net work a spring does on a body is zero when the body returns to its initial position

Explanation:

A force is conservative when the net work done over any path that returns to the initial position is zero. Choice C matches that definition.

An ideal spring of the kind used in physics problems has the characteristic that it applies the same force at the same distance always. So any work required to extend or compress the spring is reversed when the reverse motion takes place.

Answer:

The velocity of the bird is [tex]v_f = 4.87 \ m/s[/tex]

Explanation:

From the question we are told that  

    The mass of the bird  is [tex]m_1 = 300 \ g = 0.3 \ kg[/tex]

    The initial speed of the bird is  [tex]u_1 = 6.2 \ m/s[/tex]

     The mass of the insect is [tex]m_2 = 10 \ g = 0.01 \ kg[/tex]

       The speed of the insect is [tex]u_ 2 =-35 \ m/s[/tex]

The negative sign is because it is moving in opposite direction  to the bird

According to the principle of linear momentum conservation

       [tex]m_1 u_1 + m_2 u_2 = (m_1 + m_2 )v_f[/tex]

substituting values

        [tex](0.3 * 6.2 ) + (0.01 * (-35)) = (0.3 + 0.01 )v_f[/tex]

    [tex]1.51 = 0.31 v_f[/tex]

     [tex]v_f = 4.87 \ m/s[/tex]

The Final velocity of Bird =  4.87 m/s

Mass of the bird = 300 g = 0.3 kg

Velocity of bird = 6.2 m/s

Momentum of Bird = Mass of bird [tex]\times[/tex] Velocity of Bird = 0.3 [tex]\times[/tex] 6.2 =  1.86 kgm/s

Mass of the insect = 10 g = 0.01 kg

Velocity of insect =   - 35 m/s

Momentum of the Insect = Mass of Insect [tex]\times[/tex] Velocity of Insect = - 0.35  kgm/s

According to the law of conservation of momentum We can write that

In the absence of external forces on the system , the momentum of system remains conserved in that particular direction.

The bird opens the mouth and enjoys the free lunch  hence

Let the final velocity of bird is [tex]v_f[/tex]

Initial momentum of the system = Final momentum of the system

1.86 -0.35 = [tex]v_f[/tex] ( 0.01 + 0.3 )

1.51 =  [tex]v_f[/tex] 0.31

[tex]v_f[/tex] = 4.87 m/s

The Final velocity of Bird =  4.87 m/s

For more information please refer to the link below

https://brainly.com/question/18066930