Ball B is hanging from an inextensible cord. An identical ball A is released from rest when it is just touching the cord and drops through the vertical distance hA = 8 in. before striking ball B. Assuming e = 0.9 and no friction, determine the resulting maximum vertical displacement hB of ball B.

Answer:

434°C=813.2 °F

434°C=707.15 K

Explanation:

Fahrenheit is a temperature scale that sets the freezing point of water to 32 degrees Fahrenheit (° F) and the boiling point to 212 ° F (at normal atmospheric pressure). The conversion from degrees celsius to gradis fahrenheit is done by: ℉ = ℃ * 1.80+ 32.00

So, being 434°C: ℉ = 434℃ * 1.80+ 32.00= 813.2

Then: 434°C=813.2 °F

Kelvin is the temperature unit of the International System. It is one of the seven basic temperature units. Its symbol in the international system of units is K. Zero on the Celsius scale or degrees Celsius (0 ° C) is defined as the equivalent of 273.15 K. Then the conversion of Census degrees to Kelvin degrees is done by: K = ℃ + 273.15

Being 434 °C: K=434 °C+273.15=707.15

Then: 434°C=707.15 K

Answer:

U = 12,205.5 J

Explanation:

In order to calculate the internal energy of an ideal gas, you take into account the following formula:

[tex]U=\frac{3}{2}nRT[/tex]        (1)

U: internal energy

R: ideal gas constant = 8.135 J(mol.K)

n: number of moles = 10 mol

T: temperature of the gas = 100K

You replace the values of the parameters in the equation (1):

[tex]U=\frac{3}{2}(10mol)(8.135\frac{J}{mol.K})(100K)=12,205.5J[/tex]

The total internal energy of 10 mol of Oxygen at 100K is 12,205.5 J

Answer:

a, 71.8° C, 51° C

b, 191.8° C

Explanation:

Given that

D(i) = 200 mm

D(o) = 400 mm

q' = 24000 W/m³

k(r) = 0.5 W/m.K

k(s) = 4 W/m.K

k(h) = 25 W/m².K

The expression for heat generation is given by

q = πr²Lq'

q = π . 0.1² . L . 24000

q = 754L W/m

Thermal conduction resistance, R(cond) = 0.0276/L

Thermal conduction resistance, R(conv) = 0.0318/L

Using energy balance equation,

Energy going in = Energy coming out

Which is = q, which is 754L

From the attachment, we deduce that the temperature between the rod and the sleeve is 71.8° C

At the same time, we find out that the temperature on the outer surface is 51° C

Also, from the second attachment, the temperature at the center of the rod was calculated to be, 191.8° C

Answer:

0.9432 m/s

Explanation:

We are given;

Mass of swimmer;m_s = 64.38 kg

Mass of log; m_l = 237 kg

Velocity of swimmer; v_s = 3.472 m/s

Now, if we consider the first log and the swimmer as our system, then the force between the swimmer and the log and the log and the swimmer are internal forces. Thus, there are no external forces and therefore momentum must be conserved.

So;

Initial momentum = final momentum

m_l × v_l = m_s × v_s

Where v_l is speed of the log relative to water

Making v_l the subject, we have;

v_l = (m_s × v_s)/m_l

Plugging in the relevant values, we have;

v_l = (64.38 × 3.472)/237

v_l = 0.9432 m/s

Answer:

Following are the answer to this question:

Explanation:

In option (a):

In option (b):

Answer:

The deceleration of the car is [tex]\approx -0.065m/s^{2}[/tex]

Explanation:

to solve this, we will have to apply the knowledge that will be got from the equations of motion.

There are several equations of motion, and depending on the parameters given in the problem, we can choose the perfect equation that can best be used to solve the problem.

In this case, since we are given the velocity and time, and we are solving for the acceleration, we will use this formula

[tex]v = u +at[/tex]

where v= final velocity = 0

u = initial velocity = 135Km/h [tex]\approx 0.278 m/s[/tex]

t= time = 4.29 seconds.

[tex]a = \frac{v - u}{t}[/tex]

[tex]a =\frac{0-0.278}{4.29} \approx 0.065m/s^{2}[/tex]

Hence, the deceleration of the car is [tex]\approx -0.065m/s^{2}[/tex]

Answer:

the positive terminal has higher potential(voltage) than the negative. Any terminal at the positive terminal has higher potential

Explanation:

ΔV =Vtop - Vbottom

Answer:

The magnetic field at a distance of 19.8 cm from the wire is 1.591 mT

Explanation:

Given;

first magnetic field at first distance, B₁ = 2.50 mT

first distance, r₁ = 12.6 cm = 0.126 m

Second magnetic field at Second distance, B₂ = ?

Second distance, r₂ = ?

Magnetic field for a straight wire is given as;

[tex]B = \frac{\mu I}{2 \pi r}[/tex]

Where:

μ is permeability

B is magnetic field

I is current flowing in the wire

r distance to the wire

[tex]Let \ \frac{\mu I}{2\pi} \ be \ constant; = K\\\\B = \frac{K}{r} \\\\K = Br\\\\B_1r_1 = B_2r_2\\\\B_2 =\frac{B_1r_1}{r_2} \\\\B_2 = \frac{2.5*10^{-3} *0.126}{0.198} \\\\B_2 = 1.591 *10^{-3}\ T\\\\B_2 = 1.591 \ mT[/tex]

Therefore, the magnetic field at a distance of 19.8 cm from the wire is 1.591 mT

Answer: the angular frequency is 2.31 rad/s

Explanation:

The data we have is:

Radial acceleration A = 27.9 m/s^2

Beam length r = 5.21m

The radial acceleration is equal to the velocity square divided the radius of the circle (the lenght of the beam in this case)

And we can write the velocity as:

v = w*r where r is the radius of the circle, and w is the angular frequency.

w = 2pi*f

where f is the "normal" frequency.

So we have:

A = (v^2)/r = (r*w)^2/r = r*w^2

We can replace the values and find w.

27.9m/s^2 = 5.21m*w^2

√(27.9/5.21) = w = 2.31 rad/s

Answer:

about 1 meter

Explanation:

The distance past the edge that the man will get before the plank starts to tip is; 0.4285 m

We are given;

Mass of plank; m = 30 kg

Length of plank; L = 6m

Mass of man; M = 70 kg

Since the plank has 2 supports which are the deck of the ship, then it means that, we can take moments about the right support before the 2m stick out of the plank.

Thus;

Moment of weight of plank about the right support;

τ_p = mg((L/2) - 2)

τ_p = 30 × 9.8((6/2) - 2)

τ_p = 294 N.m

Moment of weight of man about the right support;

τ_m = Mgx

where x is the distance past the edge the man will get before the plank starts to tip.

τ_m = 70 × 9.8x

τ_m = 686x

Now, moment of the board is counterclockwise while that of the man is clockwise. Thus;

τ_m = τ_p

686x = 294

x = 294/686

x = 0.4285 m

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Answer:

F = 103.54N

Explanation:

In order to calculate the magnitude of the applied force, you take into account that the forces on the box are the applied force F and the weight of the box W.

The box moves with a constant velocity. By the Newton second law you have that the sum of forces must be equal to zero.

Furthermore, you have that the sum of forces are given by:

[tex]F-Wsin\theta=0[/tex]                (1)

F: applied force = ?

W: weight of the box = Mg = (25kg)(9.8m/s^2) = 245N

θ: degree of the incline = 25°

You solve the equation (1) for F:

[tex]F=Wsin\theta=(245N)sin(25\°)=103.54N[/tex]          (2)

The applied force on the box is 103.54N

Answer:

6.3 rev/s

Explanation:

The new rotation rate of the satellite can be found by conservation of the angular momentum (L):

[tex] L_{i} = L_{f} [/tex]

[tex] I_{i}*\omega_{i} = I_{f}*\omega_{f} [/tex]

The initial moment of inertia of the satellite (a solid sphere) is given by:

[tex] I_{i} = \frac{2}{5}m_{s}r^{2} [/tex]

Where [tex]m_{s}[/tex]: is the satellite mass and r: is the satellite's radium

[tex] I_{i} = \frac{2}{5}m_{s}r^{2} = \frac{2}{5}1900 kg*(4.6 m)^{2} = 1.61 \cdot 10^{4} kg*m^{2} [/tex]

Now, the final moment of inertia is given by the satellite and the antennas (rod):

[tex] I_{f} = I_{i} + 2*I_{a} = 1.61 \cdot 10^{4} kg*m^{2} + 2*\frac{1}{3}m_{a}l^{2} [/tex]

Where [tex]m_{a}[/tex]: is the antenna's mass and l: is the lenght of the antenna

[tex] I_{f} = 1.61 \cdot 10^{4} kg*m^{2} + 2*\frac{1}{3}150.0 kg*(6.6 m)^{2} = 2.05 \cdot 10^{4} kg*m^{2} [/tex]

So, the new rotation rate of the satellite is:

[tex] I_{i}*\omega_{i} = I_{f}*\omega_{f} [/tex]

[tex]\omega_{f} = \frac{I_{i}*\omega_{i}}{I_{f}} = \frac{1.61 \cdot 10^{4} kg*m^{2}*8.0 \frac{rev}{s}}{2.05 \cdot 10^{4} kg*m^{2}} = 6.3 rev/s[/tex]  

Therefore, the new rotation rate of the satellite is 6.3 rev/s.

I hope it helps you!  

Explanation:

In the given question, the two metal spheres were hanged with the nylon thread.

When these two spheres were brought close together, they attracted each other. The attraction between these spheres is the result of the opposite charges between them.

The possible ways by which these two metal spheres can be charged are by induction that is touching the metal or by rubbing them.

During induction, the same charges are transferred to each sphere. In this case, either both the spheres will be negatively charged or positively charged.

It is not possible that after the sphere touch each other they will cling together because the same charge repels each other and during touching, if one sphere is neutral, then the charged one will transfer the same charge. And as we know that same charge repel each other therefore they will repel each other.

Answer:

The temperature is  [tex]T = 168.44 \ K[/tex]

Explanation:

From the question ewe are told that

   The rate of heat transferred is    [tex]P = 13.1 \ W[/tex]

     The surface area is  [tex]A = 1.55 \ m^2[/tex]

      The emissivity of its surface is  [tex]e = 0.287[/tex]

Generally, the rate of heat transfer is mathematically represented as

           [tex]H = A e \sigma T^{4}[/tex]

=>         [tex]T = \sqrt[4]{\frac{P}{e* \sigma } }[/tex]

where  [tex]\sigma[/tex] is the Boltzmann constant with value  [tex]\sigma = 5.67*10^{-8} \ W\cdot m^{-2} \cdot K^{-4}.[/tex]

substituting value  

             [tex]T = \sqrt[4]{\frac{13.1}{ 0.287* 5.67 *10^{-8} } }[/tex]

            [tex]T = 168.44 \ K[/tex]

Answer is a. Doubles

when the loops are increased in the coil then the magnetic field created doubles

Answer:

Vi = 4.8 m/s

Explanation:

First we need to find the magnitude of constant tangential acceleration. For that purpose we use the following formula between points A and C:

a = (Vf - Vi)/t

where,

a = constant tangential acceleration from A to C = ?

Vf = Final Velocity at C = 5 m/s

Vi = Initial Velocity at A = 4.5 m/s

t = time taken to move from A to C = 4 s

Therefore,

a = (5 m/s - 4.5 m/s)/4 s

a = 0.125 m/s²

Now, applying the same equation between points B and C:

a = (Vf - Vi)/t

where,

a = constant tangential acceleration from A to B = 0.125 m/s²

Vf = Final Velocity at C = 5 m/s

Vi = Initial Velocity at B = ?

t = time taken to move from B to C = 1.6 s

Therefore,

0.125 m/s² = (5 m/s - Vi)/1.6 s

Vi = 5 m/s - (0.125 m/s²)(1.6 s)

Vi = 4.8 m/s

Answer:

The 40g mass will be attached at 69 cm

Explanation:

First, make a sketch of the meterstick with the masses placed on it;

--------------------------------------------------------------------------

               ↓                    Δ                      ↓

             20 g.................50 cm.................40g

                         38 cm                  y cm  

Apply principle of moment;

sum of clockwise moment = sum of anticlockwise moment

40y = 20 (38)

40y = 760

y = 760 / 40

y = 19 cm

Therefore, the 40g mass will be attached at 50cm + 19cm = 69 cm

              12cm             50 cm              69cm

--------------------------------------------------------------------------

               ↓                    Δ                      ↓

             20 g.................50 cm.................40g

                         38 cm                 19 cm                                              

Answer:

1.1ohms

Explanation:

According to ohms law E = IR

If potential difference of a battery is 2.2 V when it is connected across a resistance of 5 ohm and if suddenly the voltage Falls to 1.8V then the current in the 5ohms resistor I = V/R = 1.8/5

I = 0.36A (This will be the load current).

Before we can calculate the value of the internal resistance, we need to know the voltage drop across the internal resistance.

Voltage drop = 2.2V - 1.8V = 0.4V

Then we calculate the internal resistance using ohms law.

According to the law, V = Ir

V= voltage drop

I is the load current

r = internal resistance

0.4 = 0.36r

r = 0.4/0.36

r = 1.1 ohms

Answer:

1.7 m/s

Explanation:

Relevant Data provided as per the question below:-

Radius = 15.0 cm

Time = 0.56 s.

Based on the above information

The computation of the speed of the object is shown below:-

[tex]Velocity = \frac{2\times \pi \times Radius}{Time}[/tex]

[tex]Velocity = \frac{2\times \frac{22}{7} \times 0.15}{0.56}[/tex]

[tex]= \frac{0.942857}{0.56}[/tex]

= 1.683 m/s

or

= 1.7 m/s

Therefore for computing the speed of the object or velocity we simply applied the above formula by considering the pi and all other given data

Answer:

Specific gravity of other fluid = .854 (Approx)

Explanation:

Given:

Mass of water = 35 g

Mass of filled bottle with water = 98.44 g

Mass of filled bottle with fluid = 89.22 g

Computation:

Mass of water = 98.44g - 35g = 63.44g

Density of water = 1000 g/L

Volume of bottle = 63.44/1000 = 0.06344L

Mass of other liquid = 89.22g - 35g = 54.22g

Density of other liquid = 54.22g/0.06344L = 854.665826 g/L

Water has a specific gravity = 1

So ,  specific gravity of other fluid

1000 / 854.665826 = 1 / specific gravity of other fluid

Specific gravity of other fluid = .854 (Approx)

Answer:

The total mechanical energy does not change if the value of the mass is changed. That is, remain the same

Explanation:

The total mechanical energy of a spring-mass system is equal to the elastic potential energy where the object is at the amplitude of the motion. That is:

[tex]E=U=\frac{1}{2}kA^2[/tex]       (1)

k: spring constant

A: amplitude of the motion = 2.0cm

As you can notice in the equation (1), the total mechanical energy of the system does not depend of the mass of the object. It only depends of the amplitude A and the spring constant.

Hence, if you use a mass of 0.40kg the total mechanical energy is the same as the obtained with a mas 0.20kg

Remain the same

After using a mass of 0.40 kg the total mechanical energy is the same as the obtained with a mass of 0.20 kg

 If the object is at the amplitude of the motion, the total mechanical energy of a spring-mass system is equal to the elastic potential energy.

[tex]\bold {E = U =\dfrac 1{2} kA^2}[/tex]

Where,

k: spring constant  

A: amplitude of motion = 2.0 cm

In the equation, the total mechanical energy only depends of the amplitude and the spring constant. It does not depend on the mass of the object. It  

Therefore, after using a mass of 0.40 kg the total mechanical energy is the same as the obtained with a mass of 0.20 kg

To know more about mechanical energy,

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Complete Question

The complete question is shown on the first uploaded image

Answer:

the change in energy of the block due to friction is  [tex]E = -126 \ J[/tex]

Explanation:

From the question we are told that

    The  frictional force is  [tex]F_f = 21 \ N[/tex]

    The mass of the block is  [tex]m_b = 8 \ kg[/tex]

    The length of the ramp is  [tex]l = 6 \ m[/tex]

    The height of the block is  [tex]h = 2 \ m[/tex]

The change in energy of the block due to friction is mathematically represented as

     [tex]\Delta E = - F_s * l[/tex]

The negative sign is to show that the frictional force is acting against the direction of the block movement

  Now substituting values

      [tex]\Delta E = -(21)* 6[/tex]

      [tex]\Delta E = -126 \ J[/tex]

Answer:

Explanation:

From the given information:

(a)

the average magnetic flux through each turn of the inner solenoid can be calculated by the formula:

[tex]\phi _ 1 = B_1 A[/tex]

[tex]\phi _ 1 = ( \mu_o \dfrac{N_i}{l} i_1)(\pi ( \dfrac{d}{2})^2)[/tex]

[tex]\phi _ 1 = ( 4 \pi *10^{-7} \ T. m/A ) ( \dfrac{310}{20*10^{-2} \ m }) (0.130 \ A) ( \pi ( \dfrac{2.20*10^{-2} \ m }{2})^ 2[/tex]

[tex]\phi_1 = 9.625 * 10^{-8} \ Wb[/tex]

(b)

The mutual inductance of the two solenoids is calculated by the formula:

[tex]M = 23 *\dfrac{9.625*10^{-8} \ Wb}{0.130 \ A}[/tex]

M = [tex]1.703 *10^{-5}[/tex] H

(c)

the emf induced in the outer solenoid by the changing current in the inner solenoid can be calculate by using the formula:

[tex]\varepsilon = -N_o \dfrac{d \phi_1}{dt}[/tex]

[tex]\varepsilon = -M \dfrac{d i_1}{dt}[/tex]

[tex]\varepsilon = -(1.703*10^{-5} \ H) * (1800 \ A/s)[/tex]

[tex]\varepsilon = -0.030654 \ V[/tex]

[tex]\varepsilon = -30.65 \ V[/tex]

Answer:

The magnitude of the electric force on the particle in terms of the variables given is, F = qE

The magnitude of the magnetic force on the particle in terms of the variables given is, F = q (v x B)

Explanation:

Given;

a charged particle, q

magnitude of electric field, E

magnitude of magnetic field, B

The magnitude of the electric force on the particle in terms of the variables given;

F = qE

The magnitude of the magnetic force on the particle in terms of the variables given;

F = q (v x B)

where;

v is the constant velocity of the charged particle

Answer:

The magnitude of the electric force acting on a charged particle moving through an electric field = |qE|

The magnitude of the magnetic force of a charged particle moving at a particular velocity through a magnetic field = |qv × B|

Explanation:

The electric force acting on a charged particle, q, moving through an electric field, E, is given as a product of the charge on the particle (a scalar quantity) and the electric field (a vector quantity).

Electric force = qE

The magnitude of the electric force = |qE|

That is, magnitude of the product of the charge and the electric field vector.

The magnetic force acting on a charged particle, q, moving with a velocity, v, through a magnetic field, B is a vector product of qv [a product of the charge of the particle (a scalar quantity) and the velocity of the particle (a vector quantity)] and B (a vector quantity).

It is given mathematically as (qv × B)

The magnitude of the magnetic force is the magnitude of the vector product obtained.

Magnitude of the magnetic force = |qv × B|

Hope this Helps!!!

Answer:

[tex]v_B=3.78\times 10^5\ m/s[/tex]

Explanation:

It is given that,

Charge on helium nucleus is 2e and its mass is [tex]6.63\times 10^{-27}\ kg[/tex]

Speed of nucleus at A is [tex]v_A=6.2\times 10^5\ m/s[/tex]

Potential at point A, [tex]V_A=1.5\times 10^3\ V[/tex]

Potential at point B, [tex]V_B=4\times 10^3\ V[/tex]

We need to find the speed at point B on the circle. It is based on the concept of conservation of energy such that :

increase in kinetic energy = increase in potential×charge

[tex]\dfrac{1}{2}m(v_A^2-v_B^2)=(V_B-V_A)q\\\\\dfrac{1}{2}m(v_A^2-v_B^2)={(4\times 10^3-1.5\times 10^3)}\times 2\times 1.6\times 10^{-19}=8\times 10^{-16}\\\\v_A^2-v_B^2=\dfrac{2\times 8\times 10^{-16}}{6.63\times 10^{-27}}\\\\v_A^2-v_B^2=2.41\times 10^{11}\\\\v_B^2=(6.2\times 10^5)^2-2.41\times 10^{11}\\\\v_B=3.78\times 10^5\ m/s[/tex]

So, the speed at point B is [tex]3.78\times 10^5\ m/s[/tex].

Answer:

88.29 N

Explanation:

mass of the ball = 4.5 kg

weight of the ball will be = mass x acceleration due to gravity(9.81 m/s^2)

weight W = 4.5 x 9.81 = 44.145 N

centrifugal forces Tc act on the ball as it swings.

At the top point of the vertical swing,

Tension on the rope = Tc - W.

At the bottom point of the vertical swing,

Tension on the rope = Tc + W

therefore,

difference in tension between these two points will be;

Net tension = tension at bottom minus tension at the top

= Tc + W - (Tc - W) = Tc + W -Tc + W

= 2W

imputing the value of the weight W, we have

2W = 2 x 44.145 = 88.29 N

Answer:

The work done by the centripetal force is always, zero.

Explanation:

The formula for the work done by a force on an object is given as follows:

W = F d Cos θ

where,

W = Magnitude of the Work Done

F = Force applied to the body

θ = Angle between the direction of force and direction of motion of the object

In case of the circular motion, the force is the centripetal force. The centripetal force is always directed towards the center of the circle. While, the object moves in a direction, which is tangential to the circle. Hence, the angle between them is always 90°. Therefore,

W = F d Cos 90°

W = F d (0)  

W = 0 J

Hence, the work done by the centripetal force is always, zero.

Answer:

[tex]\large \boxed{\text{C. 2.3 m/s}}[/tex]

Explanation:

Data:

[tex]m_{\text{A} } = \text{0.20 kg};\,v_{\text{Ai}} = \text{3.0 m/s}\\m_{\text{B} } = \text{0.40 kg};\,v_{\text{Bi}} = \text{2.0 m/s}\\[/tex]

Calculation:

This is a perfectly inelastic collision.  The two carts stick together after the collision and move with a common final velocity.

The conservation of momentum equation is

[tex]\begin{array}{rcl}m_{\text{A}}v_{\text{Ai}} +m_{\text{B}} v_{\text{Bi}}&=&(m_{\text{A}} + m_{\text{B}})v_{\text{f}}\\0.20\times 3.0 + 0.40\times 2.0 & = & (0.20 + 0.40)v_{\text{f}}\\0.60 + 0.80 & = & 0.60v_{\text{f}}\\1.40 & = & 0.60v_{\text{f}}\\v_{\text{f}}&=& \dfrac{1.40}{0.60}\\\\& = & \textbf{2.3 m/s}\\\end{array}\\\text{The centre of mass has a velocity of $\large \boxed{\textbf{2.3 m/s}}$}[/tex]

Answer:

a. ac = 1844.66 m/s²

b. Fc = 265.63 N

Explanation:

a.

The centripetal acceleration of the ball is given as follows:

ac = v²/r

where,

ac = centripetal acceleration = ?

v = speed of ball = (87 mph)(1 h/ 3600 s)(1609.34 m / 1 mile) = 38.9 m/s

r = radius of path = 82 cm = 0.82 m

Therefore,

ac = (38.9 m/s)²/0.82 m  

ac = 1844.66 m/s²

b.

The centripetal force is given as:

Fc = (m)(ac)

Fc = (0.144 kg)(1844.66 m/s²)

Fc = 265.63 N