8. At temperature 15°C, aluminum rivets have a diameter of 0.501 cm, and holes drilled in a titanium sheet have a diameter of 0.500 cm. If both the aluminum rivets and the titanium sheet are cooled together, at what temperature will the rivets just fit into the appropriate holes in the titanium sheet? Use 25x10-6 (°C)-1 for the coefficient of linear expansion for aluminum, and 8.5x10-6 (°C)-1 for titanium

Answer:

The third image

Explanation:

The one with the thumb pointing to the right

Answer:

3, correct on Edge 2020

Answer:

b) 20 kJ

Explanation:

Efficiency of carnot engine = (T₁ - T₂ ) / T₁  Where T₁ is temperature of hot source  and T₂ is temperature of sink .

T₁ = 270 + 273 = 543K

T₂ = 50 + 273 = 323 K

Putting the given values of temperatures

efficiency = (543 - 323) / 543

= .405

heat input = 50 KJ

efficiency = output work / input heat energy

.405 = output work / 50

output work = 20.25 KJ.

= 20 KJ .

Answer:

a and b are the correct answers

Explanation:

Answer:

A)  physical property; mass.

Explanation:

took the test

Answer:

The angular acceleration of CD player is [tex]13.93\ rad/s^2[/tex].

Explanation:

Initial angular speed of a CD player is 0 and final angular speed is 41.8 rad/s. Time to change the angular speed is 3 s.

It is required to find the angular acceleration. The change in angular speed of the CD player divided by time taken is called its angular acceleration. It can be given by :

[tex]a=\dfrac{\omega_f-\omega_i}{t}\\\\a=\dfrac{41.8-0}{3}\\\\a=13.94\ rad/s^2[/tex]

So, the angular acceleration of CD player is [tex]13.93\ rad/s^2[/tex].

Answer:

0.25916 lb-s/ft^2

Explanation:

Given:-

- The specific gravity of oil, SGo = 0.8

- The specific gravity of sphere, SGo = 8

- Terminal velocity of sphere, v = 2.5 fpm

- The diameter of sphere, D = 0.25 in

Find:-

What is the viscosity of the oil?

Solution:-

- Consider a sphere completely submerged into oil and travelling with terminal velocity ( v ).

- Develop a free body diagram for the sphere. There are forces acting on the sphere.

- The downward acting force is due to the weight of the sphere ( W ):

                         [tex]W = m_s*g[/tex]

Where,

The mass ( m_s ) of the sphere is given as:

                          [tex]m_s = S.G_s*p_w*V_s[/tex]

Where,

              ρ_w : Density of water = 1.940 slugs/ft3

               V_s: The volume of object ( sphere )

- The volume of sphere is expressed as a function of radius:

                        [tex]V_s = \frac{\pi *D^3}{6}[/tex]

Hence,

                        [tex]W = S.G_s*p_w*\frac{\pi*D^3 }{6}* g\\\\W = 8*1.940*\frac{\pi*(0.25/12)^3 }{6}*32\\\\W = 0.00235 lb[/tex]

- One of the upward acting force is the buoyant force ( Fb ) that is proportional to the volume of fluid displaced by the immersed object.

- The buoyant force ( Fb ) is given by:

                    [tex]F_b = S.G_o*p_w*V_s*g[/tex]

- Therefore the buoyant force ( Fb ) becomes:

                    [tex]F_b = 0.8*1.94*\frac{\pi*(0.25/12)^3 }{6} *32\\\\F_b = (4.73451*10^-^6)*(49.664)\\\\F_b = 0.00023 lb[/tex]

- The other upward acting force is the frictional drag ( F_d ) i.e the resistive frictional force acting on the contact points of the sphere and the fluid oil.

- From stokes formulations the drag force acting on a spherical object which is completely immersed in a fluid is given as:

                    [tex]F_d = 3*\pi*D*u*v[/tex]

Where,

                    μ: The viscosity of fluid

                    v : The velocity of object

Therefore,

                       [tex]F_d = 3*\pi*\frac{0.25}{12} *u*0.041666\\\\F_d = 0.00818*u\\[/tex]

- Apply Newton's second law of motion for the sphere travelling in the fluid:

                      [tex]F_n_e_t = m_s*a[/tex]

Where,

                     a: Acceleration of object = 0 ( Terminal velocity condition )

                     [tex]F_n_e_t = 0[/tex]

- Plug in the three forces acting on the metal sphere:

                     [tex]F_d + F_b - W = 0\\\\F_d = W - F_b\\\\0.00818*u = 0.00235 - 0.00023\\\\u = \frac{0.00212}{0.00818} = 0.25916 \frac{lb-s}{ft^2}[/tex]

Answer:

Explanation:

We shall use Ampere's circuital law to find magnetic field at required point.

The point is outside the circumference of two given wires so whole current will be accounted for .

Ampere's circuital law

B = ∫ Bdl = μ₀ I

line integral will be over circular path of radius r = 41 cm .

Total current  I  = 5A -3A = 2A .

∫ Bdl = μ₀ I

2π r B = μ₀ I

2π x .41  B = 4π x 10⁻⁷ x 2

B = 2 x 10⁻⁷ x 2 / .41

= 9.75 x 10⁻⁷ T . It will be along - ve Y - direction.

Answer:

the answer is a

Explanation:

Answer:

A) equal to the battery's terminal voltage.

Explanation:

When the capacitor is fully charged after long hours of charging , its  potential becomes equal to the emf of the battery and its polarity is opposite to that of battery . Hence net emf becomes equal . The capacitor itself becomes a battery which is connected in the circuit with opposite polarity . This results in the net emf and  current becoming zero . There is no charging current when the capacitor is fully charged .

Complete Question

The complete question is shown on the first uploaded image

Answer:

Explanation:

From the question we are told that

     The mass of the bat is [tex]m_b = 0.800 \ kg[/tex]

      The bat length is  [tex]L_b = 0.900 \ m[/tex]

      The distance of the bat's center of mass to the handle end is  [tex]z_c = 0.600 \ m[/tex]

      The moment of inertia of the bat is    [tex]I = 0.0530 \ kg \cdot m^2[/tex]

The objective of the solution is to find  x   which is the distance from the handle of the bat to the point where the baseball hit the bat

Generally the velocity change at the end of the bat is mathematically represented as

         [tex]\Delta v_e = \Delta v_c - \Delta w* z_c[/tex]

         Where  [tex]\Delta v_c[/tex] is the velocity change at the center of the bat  which is mathematically represented as

                [tex]\Delta v_c = \frac{Impulse}{m_b }[/tex]

We are told that the impulse is  J so

              [tex]\Delta v_c = \frac{J}{m_b }[/tex]

And   [tex]\Delta w[/tex] is the change in angular velocity which is mathematically represented as

         [tex]\Delta w = \frac{J (z -z_c)}{I}[/tex]

Now we have that

           [tex]\Delta v_e = \frac{J}{m_b } - [\frac{J (x- z_c)}{I} ] * z_c[/tex]

Before a swing when the bat is at rest the velocity change a the end of the bat handle is zero  and the impulse will be  1

   So  

            [tex]0 = \frac{1}{m_b } - [\frac{J (x- z_c)}{I} ] * z_c[/tex]

=>           [tex]x = \frac{I}{m_b z_c} + m_b[/tex]

substituting values

            [tex]x = \frac{0.530}{0.800 * 0.600} + 0.600[/tex]

           [tex]x = 0.710 \ m[/tex]

Answer:

-7.23 rad/s

Explanation:

Given that

Mass of the cockroach, m = 0.157 kg

Radius of the disk, r = 14.9 cm = 0.149 m

Rotational Inertia, I = 5.92*10^-3 kgm²

Speed of the cockroach, v = 2.92 m/s

Angular velocity of the rim, w = 3.89 rad/s

The initial angular momentum of rim is

Iw = 5.92*10^-3 * 3.89

Iw = 2.3*10^-2 kgm²/s

The initial angular momentum of cockroach about the axle of the disk is

L = -mvr

L = -0.157 * 2.92 * 0.149

L = -0.068 kgm²/s

This means that we can get the initial angular momentum of the system by summing both together

2.3*10^-2 + -0.068

L' = -0.045 kgm²/s

After the cockroach stops, the total inertia of the spinning disk is

I(f) = I + mr²

I(f) = 5.92*10^-3 + 0.157 * 0.149²

I(f) = 5.92*10^-3 + 3.49*10^-3

I(f) = 9.41*10^-3 kgm²

Final angular momentum of the disk is

L'' = I(f).w(f)

L''= 9.41*10^-3w(f)

Using the conservation of total angular momentum, we have

-0.068 = 9.41*10^-3w(f) + 0

w(f) = -0.068 / 9.41*10^-3

w(f) = -7.23 rad/s

Therefore, the speed of the lazy Susan after the cockroach stops is -7.23 and is directed in the opposite direction of the initial lazy Susan angular speed

b)

The mechanical energy of the cockroach is not converted as it stops

Answer:

All elements in a row have the same number of electron shells. Each next element in a period has one more proton and is less metallic than its predecessor. Arranged this way, groups of elements in the same column have similar chemical and physical properties, reflecting the periodic law.

Answer:

Option A, Sea Breeze

Explanation:

Ssea breeze is a wind that blows from the ocean or any water body to the nearby land mass. This breeze is cold as compared to the air on land. The water in water bodies has high specific heat capacity and hence takes longer time to cool as compared to the surrounding objects. The warmer air over the land rises upward thereby reducing the pressure on land and hence the sea breeze starts flowing from region of high pressure (i.e above the water body) towards the low pressure region that is the land.

Hence, option A is correct

Answer:

20cal/s

Explanation:

Question:

There are two questions. The first one has been answered:

From the formular, Power = Q/t = (kA∆T)/l

the amount heat depends on the duration of time interval, length of the iron rod, the thermal conductivity of iron and the temperature difference between the ends of the rod.

The amount of heat that flows through the rod by conduction during a given time interval does not depend upon the mass of the iron rod (D).

Second question:

The ends of a cylindrical steel rod are maintained at two different temperatures. The rod conducts heat from one end to the other at a rate of 10 cal/s. At what rate would a steel rod twice as long and twice the diameter conduct heat between the same two temperatures?

Solution:

Power = 10cal/s

Power = energy per unit time = Q/t

Where Q = energy

Power = (kA∆T)/l

k = thermal conductivity of iron

A = area

Area = πr^2

r = radius

Diameter = d = 2r

r = d/2

Area = (πd^2)/4

Length = l

∆T = change in temperature

10 = (kA∆T)/l

For a steel rod with length doubled and diameter doubled:

Let Length (L) = 2l

Diameter (D)= 2d

Area = π [(2d)^2]/4 = (π4d^2)/4

Area = 4(πd^2)/4

Using the formula Power = (kA∆T)/l, insert the new values for A and l

Power = [k × 4(πd^2)/4 × ∆T]/2l

Power = [4k((πd^2)/4) ∆T]/2l

Power = [(4/2)×k((πd^2)/4) ∆T]/l

Power = [2k(A) ×∆T]/l = 2(kA∆T)/l

Power of a steel that has its length doubled and diameter doubled = 2(kA∆T)/l

Recall initial Power = (kA∆T)/l = 10cal/s

And ∆T is the same

2[(kA∆T)/l] = 2 × 10

Power of a steel that has its length doubled and diameter doubled = 20cal/s

Answer:

Check the explanation

Explanation:

Kindly check the attached image below to see the step by step explanation to the question above.

Answer:

the answers the correct one is cη

Explanation:

In this simple pendulum experiment the student observes a significant change in time between each period. This occurs since an approximation used is that the sine of the angle is small, so

              sin θ = θ

with this approach the equation will be surveyed

     d² θ / dt² = - g / L sin θ

It is reduced to

      d² θ / dt² = - g / L θ

in which the time for each oscillation is constant, for this approximation the angle must be less than 10º so that the difference between the sine and the angles is less than 1%

The angle is related to the height of the pendulum

         sin θ = h / L

         h = L sin θ.

Therefore the student must be careful that the height is small.

When reviewing the answers the correct one is cη

Considering the approximation of simple harmonic motion, the correct option is:

(c) Make sure that the difference in height between the pendulum's release position and rest position is not too large.

According to Newton's second law in case of rotational motion, we have;

[tex]\tau = I \alpha[/tex]

Applying this, in the case of a simple pendulum, we get;

[tex]-mg\,sin\,\theta =mL^2 \,\frac{d^2 \theta}{dt^2}[/tex]

On, rearranging the above equation, we get;

[tex]mL^2 \,\frac{d^2 \theta}{dt^2} + mg\,sin\,\theta=0\\\\\implies \frac{d^2 \theta}{dt^2} +\frac{g}{L} sin \,\theta=0[/tex]

Now, if angular displacement is very small, i.e.; the bob of the pendulum is only raised slightly.

Then, [tex]sin\, \theta \approx \theta[/tex]

[tex]\implies \frac{d^2 \theta}{dt^2} +\frac{g}{L} \,\theta=0[/tex]

This is now in the form of the equation of a simple harmonic motion.

[tex]\frac{d^2 \theta}{dt^2} +\omega^2 \,\theta=0[/tex]

Comparing both these equations, we can say that;

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

[tex]T=2\pi\sqrt{\frac{L}{g}}[/tex]

This relation for the time period can only be obtained if the angular displacement is very less.

So, the correct option is;

Option (c): Make sure that the difference in height between the pendulum's release position and rest position is not too large.

Learn more about simple harmonic motion here:

https://brainly.com/question/26114128

Answer:

Explanation:

The problem is based on Newton's law of cooling .

According to Newton's law

dQ / dt = k ( T - T₀ ) ,

dT / dt = k' ( T - T₀ )          ; dT / dt is rate of fall of temperature.

T is average  temperature of hot body , T₀ is temperature of surrounding .

In the first case rate of fall of temperature = (210 - 191) / 2.5

= 7.6 degree / s

average temperature T = (210 + 191) /2

= 200.5  

Putting in the equation

7.6 = k' ( 200.5  - 64 )

k' = 7.6 / 136.5

= .055677

In the second case :---

In the second case, rate of fall of temperature = (191 - 156) / t  

= 35 / t   , t is time required.

average temperature T = (156 + 191) /2

= 173.5  

Putting in the equation

35 / t = .05567 ( 173.5 - 64 )

t = 5.74 minute .

Answer:

L= 276.4 mm

Explanation:

Given that

E= 180 GPa

d= 3.7 mm

F= 1890 N

ΔL= 0.45 mm

We know that ,elongation due to load F in a cylindrical bar is given as follows

[tex]\Delta L =\dfrac{FL}{AE}[/tex]

[tex]L=\dfrac{\Delta L\times AE}{F}[/tex]

Now by putting the values in the above equation we get

[tex]L=\dfrac{0.45\times 10^{-3}\times \dfrac{\pi}{4}\times (3.7\times 10^{-3})^2\times 108\times 10^9}{1890}\ m[/tex]

L=0.2764 m

L= 276.4 mm

Therefore the length of the specimen will be 276.4 mm

Answer: (a) and (b) => check attached file.

(c). Picture (a) and (b) will both remain the same.

Explanation:

IMPORTANT: The solution to the question (a) and (b) that is  (a) Draw a neat snapshot mode labeled vector picture of the wave. (b) Draw a neat movie mode labeled vector picture of the wave is there in the ATTACHED FILE/PICTURE.

It is also worthy of note to know that in anything Electromagnetic wave, the magnetic field, the Electric Field and their direction of propagation are perpendicular to each other.

Therefore, knowing the fact above we can say that in yellow light, the magnetic field is in the y-direction and the Electric Field is in the z-direction.

Hence, the solution to option C is given below;

(C).If the wave were to represent blue light instead of yellow light, picture (a) will remain the same because both light are Electromagnetic wave, although the wavelength will have to change. Picture (b) will also remain the same because they are both Electromagnetic waves and possess similar properties.

Answer:

The center of mass move with the velocity of -3.487 m/s.

Explanation:

Given values of block A.

Mass of block A, (M1) = 4 kg

Speed of block A, (V1) = 2 m/s

Given values of block B.

 Mass of block B, (M2) = 8.4 kg

Speed of block B, (V2) = -6.1 m/s

Below is the formula to find the velocity of center of mass.

[tex]Velocity = \frac{M1V1 + M2V2}{M1 + M2} \\[/tex]

[tex]= \frac{4 \times 2 + 8.4 \times (-6.1) }{4 + 8.4} \\[/tex]

[tex]= \frac{- 43.24}{12.4}\\[/tex]

[tex]= - 3.487 m/s[/tex]

Answer:147 joules

Explanation:

Mass=m=5kg

Acceleration due to gravity=g=9.8m/s^2

Height=h=3 meter

Potential energy=m x g x h

Potential energy=5 x 9.8 x 3

Potential energy=147 joules

Answer: R.A.M

Explanation:

Answer:

D. the linear velocity of the point of contact (relative to the inclined surface) is zero

Explanation:

The force of friction emerges only when there is relative velocity between two objects . In case of perfect rolling , there is no sliding so relative velocity between the surface and the point of contact is zero . In other words the velocity of point of contact becomes zero , even though , the whole body is in linear motion . It happens due  point of contact having two velocities which are equal and opposite . One of the velocity is in forward direction and the other velocity which is due to rotation is in backward direction . So net velocity of point of contact becomes zero . Due to absence of sliding , displacement due to friction  becomes zero . Hence work done by friction becomes zero.

Answer:

There will be no change in the angular momentum of the system.

Explanation:

Total angular momentum of the system  will remain unchanged . We can apply law of conservation of momentum because no external torque is acting on the system . There is increase in the momentum of inertia due to dropping of ball of putty . In order to conserve angular momentum , the system decreases its angular velocity . Hence the final angular momentum remains unchanged .  

Answer:

 6 m/s

Explanation:

Given that :

mass of the block   m =  200.0 g  = 200 × 10⁻³ kg

the horizontal spring constant   k  =  4500.0 N/m

position of the block (distance x) = 4.00 cm  = 0.04 m

To determine the speed the block will be traveling when it leaves the spring; we applying the  work done on the spring as it is stretched (or compressed) with the kinetic energy.

i.e [tex]\frac{1}{2} kx^2 = \frac{1}{2} mv^2[/tex]

[tex]kx^2 = mv^2[/tex]

[tex]4500* 0.04^2 = 200*10^{-3} *v^2[/tex]

[tex]7.2 =200*10^{-3}*v^{2}[/tex]

[tex]v^{2} =\frac{7.2}{200*10^{-3}}[/tex]

[tex]v =\sqrt{\frac{7.2}{200*10^{-3}}}[/tex]

v = 6 m/s

Hence,the speed the block will be traveling when it leaves the spring is  6 m/s

Answer:

In electromagnetic waves, energy is transferred through vibrations of electric and magnetic fields. In sound waves, energy is transferred through vibration of air particles or particles of a solid through which the sound travels. In water waves, energy is transferred through the vibration of the water particles.

Answer:

a) 8.67m

b) 1000N

Explanation:

(a) To find the distance you use the second Newton Law for both car and trailer, in order to calculate the dis-acceleration of the system:

[tex]F=ma\\\\a_=\frac{F}{m}=\frac{5000N+4000N}{1400kg+1200kg}=3.46\frac{m}{s^2}[/tex]

once you have this value, you use the the following kinematic equation to calculate the distance traveled by both car and trailer:

[tex]v^2=v_o^2-2ax\\\\x=\frac{-v^2+v_o^2}{2a}[/tex]

v: final velocity=0

vo: initial velocity = 72km/h = 60 m/s

by replacing the values of these parameters you obtain for x:

[tex]x=\frac{-0m/s+60m/s}{2(3.46m/s^2)}\\\\x=8.67m[/tex]

(b) The horizontal component of the force exerted by the trailer hitch is given by:

[tex]F_T=5000N-4000N=1000N[/tex]

Answer:

Time of flight=3.5 seconds

Speed at maximum height is 0

Explanation:

Φ=60°

initial velocity=u=20m/s

Acceleration due to gravity=g=9.8 m/s^2

Total time of flight=T

Final speed=v

question 1:

T=(2 x u x sinΦ)/g

T=(2 x 20 x sin60)/9.8

T=(2 x 20 x 0.8660)/9.8

T=34.64/9.8

T=3.5 seconds

Question 2

Speed at maximum height is 0

Answer:

Question 3: 2.25 watts

Question 4: 405 joules

Explanation:

question 3:

Current =0.5 amps

Voltage =4.5 volts

Power= current x voltage

Power=0.5 x 4.5

power=2.25 watts

Question 4

Current =0.5 amps

Voltage =4.5v

Time=3 minutes

Time =3x60

Time =180 seconds

Energy=current x voltage x time

Energy =0.5 x 4.5 x 180

Energy =405 joules