The rate of rotation of the disk is gradually increased. The coefficient of static friction between the coin and the disk is 0.50. Determine the linear speed of the coin when it just begins to slip.

Answer:

H = 4500 m

Explanation:

        [tex]H = \frac{1}{2}*g*t^{2} = \frac{1}{2} * 10 m/s2*(30s)^{2} = 4500 m[/tex]

The given question is incomplete. The complete question is as follows.

The block has a weight of 75 lb and rests on the floor for which [tex]\mu k[/tex] = 0.4. The motor draws in the cable at a constant rate of 6 ft/sft/s. Neglect the mass of the cable and pulleys.

Determine the output of the motor at the instant [tex]\theta = 30^{o}[/tex].

Explanation:

We will consider that equilibrium condition in vertical direction is as follows.

           [tex]\sum F_{y} = 0[/tex]

         N - W = 0

           N = W

or,      N = 75 lb

Again, equilibrium condition in the vertical direction is  as follows.

        [tex]\sum F_{x} = 0[/tex]

       [tex]T_{2} - F_{k}[/tex] = 0

         [tex]T_{2} = \mu_{k} N[/tex]

                  = [tex]0.4 \times 75 lb[/tex]

                  = 30 lb

Now, the equilibrium equation in the horizontal direction is as follows.

         [tex]\sum F_{x} = 0[/tex]

       [tex]T Cos (30^{o}) + T Cos (30^{o}) = T_{2}[/tex]

          [tex]2T Cos (30^{o}) = T_{2}[/tex]

    or,             T = [tex]\frac{T_{2}}{2 Cos (30^{o})}[/tex]

                        = [tex]\frac{30}{2 Cos (30^{o})}[/tex]

                        = [tex]\frac{30}{1.732}[/tex]

                        = 17.32 lb

Now, we will calculate the output power of the motor as follows.

             P = Tv

                = [tex]17.32 lb \times 6[/tex]

                = [tex]103.92 \times \frac{1}{550} \times \frac{hp}{ft/s}[/tex]

                = 0.189 hp

or,             = 0.2 hp

Thus, we can conclude that output of the given motor is 0.2 hp.

Answer:

The out put power is 0.188 hp.

Explanation:

Given that,

Weight = 75 lb

Coefficient of friction = 0.4

Rate = 6 ft/s

Suppose, Determine the output of the motor at the instant θ = 30°.

For block,

We need to calculate the force in vertical direction

Using balance equilibrium equation in vertical

[tex]\sum{F_{y}}=0[/tex]

[tex]N-W=0[/tex]

[tex]N=W[/tex]

Put the value into the formula

[tex]N=75\ lb[/tex]

Using balance equilibrium equation in horizontal

[tex]\sum{F_{x}}=0[/tex]

[tex]T_{2}-f_{k}=0[/tex]

[tex]T_{2}=\mu_{k}N[/tex]

Put the value into the formula

[tex]T_{2}=0.4\times75[/tex]

[tex]T_{2}=30\ lb[/tex]

For pulley,

We need to calculate the force

Using balance equilibrium equation in horizontal

[tex]\sum{F_{x}}=0[/tex]

[tex]T\cos\theta+T\cos\theta=T_{2}[/tex]

[tex]2T\cos30=T_{2}[/tex]

[tex]T=\dfrac{T_{2}}{2\cos30}[/tex]

Put the value into the formula

[tex]T=\dfrac{30}{2\times\cos30}[/tex]

[tex]T=17.32\ lb[/tex]

We need to calculate the out put power

Using formula of power

[tex]P=Tv[/tex]

Put the value into the formula

[tex]P=17.32\times6[/tex]

[tex]P=103.92\ lb.ft/s[/tex]

[tex]P=0.188\ hp[/tex]

Hence, The out put power is 0.188 hp.

Answer:

#Check explanation below for a detailed description.

Explanation:

A plot of absorbance vs temperature is used to determine  [tex]T_m[/tex] of a duplex DNA.

-The duplex DNA denatures.

-The denaturation creates two single strands which enhances Ultraviolet absorption ( increased temperatures increases absorption rates).

-Higher concentrations of [tex]Sodium \ Chloride[/tex] hampers stability of the DNA.

Answer:

Resistance increases with increase in temperature which depends on power supplied which also depends on voltage.

Thermal expansion will make resistance larger.

Explanation:

Light bulb is a good example of a filament lamp. If we plot the graph of voltage against current we will notice that resistance is constant at constant temperature.

The filament heats up when an electric current passes through it, and produces light as a result.

The resistance of a lamp increases as the temperature of its filament increases. The current flowing through a filament lamp is not directly proportional to the voltage across it.

tensile stress begins to appear in resistor as the temperature rises. Thus, the resistance value increases as the temperature rises. Resistance value can only decrease as the temperature rises in case of thin film resistor with aluminium substrate.

In case of a filament bulb, the resistance will increase as increase in length of the wire. The thermal expansion in this regard is linear expansivity in which resistance is proportional to length of the wire.

Resistance therefore get larger.

I have attached the circuit image missing in the question.

Answer:

A) The range of vo is; -6.6V≤ vo ≤-1V

B) σ = 0.1861

Explanation:

A) First of all, Let VΔ be the voltage from the potentiometer contact to the ground.

Thus; [(0 - vg)/(2000)] +[(0 - vΔ)/(50,000)] = 0

So, [(- vg)/(2000)] +[(- vΔ)/(50,000)] = 0

Simplifying further; -25 vg - vΔ = 0

From the question, vg = 40mV = 0.04 V

So - 25(0.04) = vΔ

So: vΔ = - 1 V

Now, [vΔ/(σRΔ)] + [(vΔ - 0)/(50,000)] + [(vΔ - vo)/((1 - σ)RΔ))] = 0

So, multiplying each term by RΔ to get; [vΔ/(σ)] + [(vΔ x RΔ)/(50,000)] + [(vΔ - vo)/((1 - σ))] = 0

So RΔ = 100kΩ or 100,000Ω from the question.

So, substituting for RΔ, we get,

[vΔ/(σ)] + [2vΔ] + [(vΔ - vo)/((1 - σ))] = 0

Let's put the value of - 1 for vΔ as gotten before.

So, ( - 1/σ) - 2 + [(-1 - vo)/(1 - σ)] = 0

Now let's make vo the subject of the equation to get;

-1 - vo = (1 - σ)[2 + (1/σ)]

-1 - vo = 2 - 2σ + (1/σ) - 1

-vo = 1 + 2 - 2σ + (1/σ) - 1

-vo = 2 - 2σ + (1/σ)

vo = - 1 (2 - 2σ + (1/σ))

When σ = 0.2; vo = - 1(2 - 0.4 + 5) =

- 1 x 6.6 = - 6.6V

Also when σ = 1;

vo = - 1(2 - 2 + 1) = - 1V

Therefore, the range of vo is;

- 6.6V ≤ vo ≤ - 1V

B) it will saturate at vo = - 7V

So, from;

vo = - 1 (2 - 2σ + (1/σ))

-7 = - 1 (2 - 2σ + (1/σ))

Divide both sides by (-1)

7 = (2 - 2σ + (1/σ))

Now, subtract 2 from both sides to get; 5 = - 2σ + (1/σ)

Multiply each term by α to get;

5σ = - 2σ^(2) + 1

So 2σ^(2) + 5σ - 1 = 0

Solving simultaneously and picking the positive value , we get σ to be approximately 0.1861

Final answer:

The question involves understanding an ideal operational amplifier's output behavior, focusing on its output voltage changes and saturation with respect to variations in σ (sigma). Calculations for the output voltage range given a specific input and understanding the conditions under which the op-amp will saturate are key.

Explanation:

The question revolves around an operational amplifier (op-amp) circuit, exploring its behavior under certain conditions, specifically examining output voltage (vO) variations and saturation point related to the parameter σ (sigma). The op-amp is assumed to be ideal, implying infinite gain, zero input current, and that its input terminals are at the same potential.

a. Range of vO if vI = 40 mV

Given that the op-amp is ideal, the output voltage will depend on the input voltage (vI), the gain settings (σ), and the feedback resistor (R3). In practical scenarios, the gain can be adjusted by changing the value of σ or R3. However, for an ideal op-amp, the input voltage is directly proportional to the output voltage, influenced by σ. With vI = 40 mV and σ ranging between 0.2 and 1.0, the output voltage will vary accordingly, directly proportional to these parameters.

b. Saturation point of the op-amp

Saturation in an op-amp occurs when the output voltage exceeds the power supply limits, meaning the op-amp can no longer amplify the input signal. The specific value of σ at which saturation occurs depends on the supply voltage, the op-amp's maximum output voltage capability, and the configuration of the feedback network. Without specific values for the power supply or feedback network, calculating the exact σ value for saturation is not possible. Yet, in theory, as σ approaches the op-amp's gain limit or if the gain results in an output voltage beyond what the op-amp can deliver based on its power supply, saturation will occur.

Answer:

1)   v = 2.20 10⁵ m / s , 2)  r = 4.86 10⁻³ m,    r = 4.92 10⁻³ m

Explanation:

1) A speed selector is a section where the magnetic and electrical forces have opposite directions, so

           [tex]F_{m} = F_{e}[/tex]

           q v B = q E

           v = E / B

           V = E s

           E = V / s

            v = V / s B

            v = 148 / (0.0016 0.42)

            v = 2.20 10⁵ m / s

2) when the isotopes enter the spectrometer, we can use Newton's second law

             F = m a

Acceleration is centripetal

             a = v² / r

             q v B = m v² / r

              r = m v / qB

The mass of uranium 235 is

           m = 3.903 10⁻²⁵ kg

The radius of this isotope is

             r = 3,903 10⁻²⁵ 2.20 10⁵ / (92  1.6 10⁻¹⁹  1.2)

             r = 4.86 10⁻³ m

The mass of the uranium isotope 238 is

              m = 238 a = 238 1.66 10-27 = 395.08 10-27 kg

The radius is

             r = 395.08 10⁻²⁷ 2.20 10⁵ / (92 1.6 10⁻¹⁹  1.2)

             r = 4.92 10⁻³ m

Final answer:

The speed of the ions entering the mass spectrometer is 352.38 m/s. The difference in the diameters of the orbits of the 238U and 235U ions can be calculated using the formula for the radius of a charged particle's orbit in a magnetic field.

Explanation:

To find the speed of the ions entering the mass spectrometer, we can use the equation for the force on a charged particle in a magnetic field: F = qvB, where F is the force, q is the charge, v is the velocity, and B is the magnetic field. The force is equal to the electric force, qE, so we can set these two equations equal to each other and solve for v: qvB = qE. Since q is the charge of the ion and E is the electric field strength, we can rearrange this equation to solve for v, giving us v = E/B.

Plugging in the values given in the question, we find that the speed of the ions entering the mass spectrometer is 148 V / 0.42 T = 352.38 m/s.

To find the difference in the diameters of the orbits of the 238U and 235U ions, we can use the formula for the radius of a charged particle's orbit in a magnetic field: r = mv / (qB), where m is the mass of the ion, v is its velocity, q is its charge, and B is the magnetic field. Since the ions are singly ionized, their charges are +e, where e is the charge of an electron. Plugging in the values given in the question, we can calculate the radius of the orbits of the 238U and 235U ions, and then subtract these values to find the difference in their diameters.

Answer:

121550 J

Explanation:

Parameters given:

Mass, m = 0.34kg

Specific heat capacity, c = 14300 J/kgK

Change in temperature, ΔT = 25K

Heat gained/lost by an object is given as:

Q = mcΔT

Since ΔT is positive in this case and also because we're told that heat was transferred to the hydrogen sample, the hydrogen sample gained heat. Therefore, Q:

Q = 0.34 * 14300 * 25

Q = 121550J or 121.55 kJ

Answer:

weight of the bullet is 0.0500 lb

velocity as it travels from block A to block B is 900 ft/s

Explanation:

given data

horizontal velocity = 1500 ft/s

mass block A = 6-lb

mass block B = 4.95 lb

blocks A velocity = 5 ft/s

blocks B velocity = 9 ft/s

solution

we apply here law of conservation of momentum that is

m × v(o) + m1 × (0) + m2 × (0) = m × v(2) + m1 × v1 + m2 × v2     ................1

put here value and we get m

m = [tex]\frac{m1 \times v1 +m2 \times v2}{v(o) - v2}[/tex]   ..............2

m = [tex]\frac{6 \times 5 + 4.95 \times 9}{1500 - 9}[/tex]  

m = 0.0500 lb

and

when here bullet is pass through the A block then moment is conserve that is

m × v(o) + m × v1 + m1 × v1   ............3

v1 = [tex]\frac{m\times v(o)- m1\times v1}{m}[/tex]    

v1 = [tex]\frac{0.0500\times 1500 - 61\times 5}{0.05}[/tex]  

v1 = 900 ft/s

Acceleration of a speck is 0.77 m/s²

Explanation:

Given of the solution-

Diameter (We can represent as d) = 32 cm

radius, (We can represent as r )= 32/2 = 16 cm = 0.16 m

Angular acceleration,(We can represent as  α ) = 46 rpm

[tex]\alpha = 46 * \frac{2\pi }{60}[/tex]

Acceleration, a = ?

We know that the formula for the acceleration is

[tex]a = r * \alpha[/tex]

[tex]a = 0.16 * 46 * \frac{2\pi }{60}\\ \\a = 0.16 * 46 * \frac{2 * 3.14}{60}[/tex]

[tex]a = 0.77m/s^2[/tex]

Therefore, acceleration of a speck is 0.77 m/s²

Answer:

[tex]\dfrac{F}{l}=2\times 10^{-7}\ N/m[/tex]

Explanation:

Given,

current in the wire I₁ = 1 A

                               I₂ = 1 A

distance between them, r = 1 m

using Force per unit length formula

[tex]\dfrac{F}{l}=\dfrac{\mu_0I_1I_2}{2\pi r}[/tex]

[tex]\mu_0 = magnetic\ permeability\ of\ free\ space = 4\pi\times 10^{-7}[/tex]

[tex]\dfrac{F}{l}=\dfrac{4\pi \times10^{-7}\times 1\times 1}{2\pi\times 1}[/tex]

[tex]\dfrac{F}{l}=2\times 10^{-7}\ N/m[/tex]

Hence, the magnitude of force per unit length is equal to [tex]\dfrac{F}{l}=2\times 10^{-7}\ N/m[/tex]

Final answer:

The magnitude of the force per unit length each 1.0 A current-carrying wire experiences when separated by a distance of 1.0 m is 2 x 10⁻⁷ N/m, calculated using the formula for magnetic force between parallel conductors.

Explanation:

When discussing two parallel current-carrying wires, we're addressing the magnetic force that arises between them due to their currents. Specifically, when each wire carries a current of 1.0 A and they are placed 1.0 m apart, the force per unit length that each wire experiences can be calculated using the formula for the magnetic force between two parallel conductors. The equation is FE = (µ0 / 2π) × (I1I2 / r), where µ0 is the magnetic constant (4π x 10-7 T·m/A), I1 and I2 are the currents in the wires, and r is the distance between the wires. Given that each wire carries 1.0 A of current and the separation is 1.0 m, we can plug these values into the formula to find that FE = (4π x 10-7 T·m/A / 2π) × (1.0 A2 / 1.0 m), resulting in a force per unit length of 2 x 10-7 N/m.

Answer:

Option as B is correct At the same speed as before

Explanation:

As we know the relation between speed of the wave and tension in string

The speed of wave in stretched string

ν = [tex]\sqrt{\frac{T}{\mu} }[/tex]  

speed of wave is the directly proportional to the square root of tension as mentioned in question tension of string is unaffected when in linear mass density is constant,  so we can say that the  speed of wave will  be the same  

Option as B is correct At the same speed as before  

If you suddenly begin to wiggle more rapidly without appreciably affecting the tension, you will cause the waves to move down the string at the same speed as before (Option b).

A wave can be defined as a type of disturbance that contains energy independently of particle motion.

In conclusion, if you suddenly begin to wiggle more rapidly without appreciably affecting the tension, you will cause the waves to move down the string at the same speed as before (Option b).

Learn more in:

https://brainly.com/question/12215474

Answer:

Explanation:

capacitance of parallel plate capacitor

c = ε A / d , d is distance between plates , A is surface area , ε is constant

As d becomes two times , Capacitance c = 1/ 2 times ie c / 2

potential V = Q / C

Q is constant , potential

v = Q / c /2

= 2 . Q / C

= 2 V

So potential difference becomes 2 times.

NEW P D = 398 X 2

= 796 V.

Answer:

|Ax| =1.03 m  (directed towards negative x-axis)

|Ay|= 5.30 (directed towards negative y-axis)

Explanation:

Let A is a vector = 5.4 m

θ = 259°

to Find Ax, Ay

Sol:

according the condition it lies in 3rd quadrant

we know that Horizontal Component Ax = A cos θ

Ax = 5.4 Cos 259°

Ax = - 1.03 m

|Ax| =1.03 m  (directed towards negative x-axis)

Now Ay = A sin θ

Ay = 5.4 Sin 259°

Ay = -5.30

|Ay|= 5.30 (directed towards negative y-axis)

(a) -1.030m

(b) -5.301m

Given a vector F in the xy plane, of magnitude F and in a direction θ counterclockwise from the positive direction of the x-axis;

The x-component ([tex]F_{X}[/tex]) of vector F is given by;

[tex]F_{X}[/tex] = F cos θ    ---------------------(i)

And;

The y-component ([tex]F_{Y}[/tex]) of vector F is given by;

[tex]F_{Y}[/tex] = F sin θ            -----------------------(ii)

Now to the question;

Let the vector be A

Therefore;

The magnitude of vector A is A = 5.4m

The direction θ of A counterclockwise from the positive direction of the x-axis = 259°

(a) The x-component ([tex]A_{X}[/tex]) of the vector A is therefore given by;

[tex]A_{X}[/tex] = A cos θ       ------------------------(iii)

Substitute the values of θ and A into equation (iii) as follows;

[tex]A_{X}[/tex] = 5.4 cos 259°

[tex]A_{X}[/tex] = 5.4 x (-0.1908)

[tex]A_{X}[/tex] = -1.030

Therefore, the x-component of the vector is -1.030m

(b) The y-component ([tex]A_{Y}[/tex]) of the vector A is therefore given by;

[tex]A_{Y}[/tex] = A sin θ       ------------------------(iv)

Substitute the values of θ and A into equation (iv) as follows;

[tex]A_{Y}[/tex] = 5.4 sin 259°

[tex]A_{Y}[/tex] = 5.4 x (-0.9816)

[tex]A_{Y}[/tex] = -5.301m

Therefore, the y-component of the vector is -5.301m

Explanation:

A trapeze is rotating with 1 rotation per second .

Thus its angular velocity ω = 2π n

here n is the number of rotations per second

Thus ω = 2π b because n = 1 in this case

Suppose the moment of inertia of his is = I

Then angular momentum L₁ = I  ω = 2 I π

In the second case , the moment of inertia becomes = 0.4 I

Let his angular velocity is  ω₀

Thus angular momentum L₂ = 0.4 I  ω₀

Because no external torque is applied , therefore angular momentum will remain constant .

Thus L₁ = L₂

Therefore  2 I π = 0.4 I x 2 n₀ π

here n₀ is the number of rotations per second

n₀ = [tex]\frac{1}{0.4}[/tex] = [tex]\frac{5}{2}[/tex] = 2.5

Answer:

2.2 m/s

Explanation:

solution:

To calculate change in stored energy at desired extension

ΔU = 1/2*k*(δx)^2

     = 1/2*3700*(0.37^2-0.180^2)

     = 201 N.m

use work energy theorem

ΔU = ΔK = 1/2*m*v^2 = 201

              = 2.2 m/s

note:

calculation maybe wrong but method is correct.

Explanation:

Below is an attachment containing the solution.

Answer:

vi = 4.77 ft/s

Explanation:

Given:

- The radius of the surface R = 1.45 ft

- The Angle at which the the sphere leaves

- Initial velocity vi

- Final velocity vf

Find:

Determine the sphere's initial speed.

Solution:

- Newton's second law of motion in centripetal direction is given as:

                         m*g*cos(θ) - N = m*v^2 / R

Where, m: mass of sphere

             g: Gravitational Acceleration

             θ: Angle with the vertical

             N: Normal contact force.

- The sphere leaves surface at θ = 34°. The Normal contact is N = 0. Then we have:

                         m*g*cos(θ) - 0 = m*vf^2 / R

                         g*cos(θ) = vf^2 / R    

                         vf^2 = R*g*cos(θ)

                         vf^2 = 1.45*32.2*cos(34)

                        vf^2 = 38.708 ft/s

- Using conservation of energy for initial release point and point where sphere leaves cylinder:

                          ΔK.E = ΔP.E

                          0.5*m* ( vf^2 - vi^2 ) = m*g*(R - R*cos(θ))

                          ( vf^2 - vi^2 ) = 2*g*R*( 1 - cos(θ))

                          vi^2 =  vf^2 - 2*g*R*( 1 - cos(θ))

                          vi^2 = 38.708 - 2*32.2*1.45*(1-cos(34))

                          vi^2 = 22.744

                           vi = 4.77 ft/s

a) -10,800 N/C

b) [tex]4.79\cdot 10^{10}m/s^2[/tex]

c) [tex]4.88\cdot 10^9[/tex] times g

Explanation:

a)

The magnitude of the electric field produced by a single-point charge is given by

[tex]E=\frac{kQ}{r^2}[/tex]

where

k is the Coulomb's constant

Q is the charge producing the  field

r is the distance at which the field is calculated

In this problem:

[tex]Q=-3.0\cdot 10^{-5}C[/tex] is the c harge producing the field

[tex]r=5.0 m[/tex] is the distance at which we want to calculate the field

Substituting,

[tex]E=\frac{(9.0\cdot 10^9)(-3.0\cdot 10^{-5})}{(5.0)^2}=-10,800 N/C[/tex]

where the negative sign indicates that the direction of the field is towards the charge producing the field (for a negative charge, the electric field is inward, towards the charge)

b)

The force experienced by a charged particle in an electric field is given by

[tex]F=qE[/tex]

where

q is the magnitude of the charge

E is the electric field

Moreover, the force on an object can be written as:

[tex]F=ma[/tex]

where

m is the mass

a is the acceleration

Combining the two equations,

[tex]ma=qE\\a=\frac{qE}{m}[/tex]

In this problem:

[tex]q=1.6\cdot 10^{-19}C[/tex] is the charge of the proton

[tex]E=0.50 kN/C=500 N/C[/tex] is the strength of the electric field

[tex]m=1.67\cdot 10^{-27} kg[/tex] is the mass of the proton

Substituting, we find the acceleration of the proton:

[tex]a=\frac{(1.6\cdot 10^{-19})(500)}{(1.67\cdot 10^{-27})}=4.79\cdot 10^{10}m/s^2[/tex]

c)

The acceleration due to gravity is the acceleration at which every object near the Earth's surface falls down, in absence of air resistance, and it is given by

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

On the other hand, the acceleration of the proton in this problem is:

[tex]a=4.79\cdot 10^{10} m/s^2[/tex]

To find how many times greater is this acceleration than that due to gravity, we can divide the acceleration of the proton by the value of g. Doing so, we find:

[tex]\frac{a}{g}=\frac{4.79\cdot 10^{10}}{9.81}=4.88\cdot 10^9[/tex]

So, it is [tex]4.88\cdot 10^9[/tex] times greater than g.

Final answer:

The electric field at the given point is -1.08×10⁳ N/C, which points towards the origin. The acceleration of a proton in a 0.50 kN/C electric field is 4.79×10ⁱ³ m/s². This acceleration is approximately 4.88×10ⁱ¹ times greater than the acceleration due to gravity.

Explanation:

To calculate the electric field at a point due to a point charge, we use Coulomb's law and the expression for the electric field, E = kQ/r², where k is Coulomb's constant (8.99×10⁹ N⋅m²/C²), Q is the charge, and r is the distance from the charge. Placing a charge of -3.0×10⁻⁵ C at the origin, the electric field at the point x= 5.0 m on the x-axis is calculated as follows:

E =  kQ/r²
=  (8.99×10⁹)(-3.0×10⁻⁵) / (5.0)²
=  -1.08×10⁳ N/C

The electric field points towards the negative charge, which is towards the origin.

To determine the acceleration of a proton in an electric field, we use the formula F = qE, where F is the force, q is the charge of the proton (1.60×10⁻⁵ C), and E is the electric field strength. The acceleration a is then found using Newton's second law, F = ma:

F = qE
= (1.60×10⁻⁵)(0.50×10⁳)
= 8.00×10⁻⁴ N

a = F/m
= 8.00×10⁻⁴ / 1.67×10⁻⁲⁷
=  4.79×10ⁱ³ m/s²

To find how many times greater this is compared to the acceleration due to gravity, we simply divide the proton's acceleration by the acceleration due to gravity (9.81 m/s²).

Number of times greater = a / g
=  4.79×10ⁱ³ / 9.81
=
≈ 4.88×10ⁱ¹ times

In the scenario described, increasing the height of a mass M in a vertical spring-mass system will result in both gravitational and spring potential energy. Thus, the potential energy in Case 2 is the sum of gravitational potential energy (m*g*D) and the elastic potential energy of the spring (1/2*k*D^2).

The situation described in the question deals with principles of Physics, specifically potential energy and its application to the spring-mass system. Consider both cases with the mass M hanging from a spring with a constant k in a vertical orientation. In the first case, the system is at equilibrium and thus the potential energy is defined as zero.

In the second case, the mass M is lifted upward a vertical distance D from the equilibrium position, it now possesses gravitational potential energy in addition to the elastic potential energy of the spring. This total energy is preserved as long as no external force behaves on the system. The gravitational potential energy gained by the mass is equal to m*g*D where m is the mass, g is the acceleration due to gravity and D is the vertical distance lifted.

The potential energy stored in the spring when it is stretched or compressed by a distance 'x' is given by the formula U = 1/2*k*x^2, where 'k' is the spring constant and 'x' is the displacement from the equilibrium position (in this case is D).

So, combining both energies, the total potential energy in Case 2, when the mass has been lifted a vertical distance D is m*g*D + 1/2*k*D^2.

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Final answer:

The skier is moving at 11.1 m/s when she gets to the bottom of the hill. This solution is derived using the principles of energy conservation and work done by friction.

Explanation:

A 62.0 kg skier is moving at 6.90 m/s on a frictionless, horizontal, snow-covered plateau when she encounters a rough patch 4.50 m long. The coefficient of kinetic friction between this patch and her skis is 0.300. After crossing the rough patch and returning to friction-free snow, she skis down an icy, frictionless hill 2.50 m high. To determine how fast the skier is moving when she gets to the bottom of the hill, we analyze the problem using principles of energy conservation and work done by friction.

We start with the equation that equates the final kinetic and potential energies to the initial energies along with the work done by friction. This equation is 0.5 mv² + 0 = 0.5 mu² + μmgl + mgh, where v is the final velocity, μ is the coefficient of friction, g is the acceleration due to gravity (9.8 m/s²), l is the length of the rough patch, and h is the height of the hill.

Plugging in the given values,

we have: 0.5 v² = 0.5 x 6.9 x 6.9 + 0.3 x 9.8 x 4.5 + 9.8 x 2.50. S

implifying, we get v² = 123.07, which leads to v = 11.1 m/s.

Thus, the skier's speed at the bottom of the hill is 11.1 m/s.

Answer:

(a). The maximum height by first ball is [tex]1.8545\times10^{4}\ m[/tex]

The maximum height by second ball is [tex]5.1020\times10^{4}\ m[/tex]

(b). The total mechanical energy of the ball–Earth system at the maximum height for each ball is [tex]1.0\times10^{7}\ J[/tex]

Explanation:

Given that,

Mass of cannonball = 20.0 kg

Speed = 1000 m/s

Angle with horizontal= 37.08

Fired angle = 90.08

We need to calculate the speed of the ball

Using formula of speed

[tex]v_{y}=v\sin\theta_{H}[/tex]

Put the value into the formula

[tex]v_{y}=1000\times\sin37.08[/tex]

[tex]v_{y}=602.9\ m/s[/tex]

(a). We need to calculate the maximum height by first ball

Using conservation of energy

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

[tex]h= \dfrac{v^2}{2g}[/tex]

Put the value into the formula

[tex]h=\dfrac{(602.9)^2}{2\times9.8}[/tex]

[tex]h=1.8545\times10^{4}\ m[/tex]

We need to calculate the maximum height by second ball

Using conservation of energy

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

[tex]h= \dfrac{v^2}{2g}[/tex]

Put the value into the formula

[tex]h=\dfrac{(1000)^2}{2\times9.8}[/tex]

[tex]h=5.1020\times10^{4}\ m[/tex]

(b). We need to calculate the total mechanical energy of the ball–Earth system at the maximum height for each ball

Using formula of energy

[tex]E=\dfrac{1}{2}mv^2[/tex]

[tex]E=\dfrac{1}{2}\times20\times1000^2[/tex]

[tex]E=1.0\times10^{7}\ J[/tex]

Hence, (a). The maximum height by first ball is [tex]1.8545\times10^{4}\ m[/tex]

The maximum height by second ball is [tex]5.1020\times10^{4}\ m[/tex]

(b). The total mechanical energy of the ball–Earth system at the maximum height for each ball is [tex]1.0\times10^{7}\ J[/tex]

Answer:

W / n = - 9133 J / mol, W / n = 3653 J / mol , e = 0.600

Explanation:

The Carnot cycle is described by

      [tex]e= 1 - Q_{c} / Q_{H} = 1 - T_{c} / T_{H}[/tex]

In this case they indicate that the final volume is

         V = 3V₀

In the part of the heat absorption cycle from the source is an isothermal expansion

         W = n RT ln (V₀ / V)

         W / n = 8.314 1000 ln (1/3)

          W / n = - 9133 J / mol

During the part of the isothermal compression in contact with the cold focus, as in a machine the relation of volumes is maintained in this part is compressed three times

            W / n = 8.314 400 (3)

           W / n = 3653 J / mol

The efficiency of the cycle is

            e = 1- 400/1000

            e = 0.600

The velocity of A is 5.16m/s²

Explanation:

Given-

mass of cylinder A, mₐ = 2kg

mass of cylinder B, mb = 10kg

Distance, s = 2m

Velocity of A, v = ?

Let acceleration due to gravity, g = 10m/s²

We know,

[tex]a = \frac{mb * g - ma * g}{ma + mb} \\\\a = \frac{10 * 10 - 2 * 10}{ 2 + 10} \\\\a = \frac{80}{12} \\\\a = 6.67m/s^2[/tex]

We know,

[tex]v = \sqrt{2as}[/tex]

[tex]v = \sqrt{2 X 6.67 X 2} \\\\v = \sqrt{26.68} \\\\v = 5.16m/s^2[/tex]

Therefore, the velocity of A is 5.16m/s²

Answer:

Explanation:

in this case, flux and area vectors are parallel . therefore, flux is just the dot product of electric field and area vector.

(flux=EAcos0 =EA)

flux through the face x face

Physics homework question answer, step 1, image 1

the flux through -x face

Physics homework question answer, step 1, image1

Step 2

the flux through y face,

Physics homework question answer, step 2, image 2

the flux through -y face,

Physics homework question answer, step 2, image 2

the flux through the z faces are zero,

therefore, the net flux is,

Physics homework question answer, step 2, image 3

...

Final answer:

The total electric flux IS ΦE = aL2 + (a + bL)L2 + cL3 net charge is q = ε0 ( aL2 + (a + bL)L2 + cL3 )

Explanation:

The question asks to find the total electric flux ΦE through the surface of a cube with an electric field given by E = (a + bx)x + cy, and to determine the net charge q inside the cube.

Part A: Total Electric Flux through the Cube's Surface

To determine the electric flux through the cube, we need to use Gauss's Law, which relates the electric flux through a closed surface to the charge enclosed by the surface.

The electric flux ΦE through a surface is defined by the integral of the electric field E dotted with the differential area dA over the entire surface. Since the sides of the cube are parallel to the coordinate planes, the electric field components perpendicular to the x, y, and z faces will contribute to the flux.

In this case, only the x-component of the electric field (a + bx)x contributes to the flux through the surfaces perpendicular to the x-axis, which are at x = 0 and x = L.

The y-component cy of the electric field contributes to the flux through the surfaces perpendicular to the y-axis, which are at y = 0 and y = L. There is no z-component to the electric field, so the surfaces perpendicular to the z-axis will have no flux.

The total electric flux through the cube is thus:

ΦE = ∫( (a + bx) x + cy ) · dA

For the two faces perpendicular to the x-axis:

ΦE, x=0 = ∫( a x ) · dA = aL2 (at x = 0)

ΦE, x=L = ∫( (a + bL)x ) · dA = (a + bL)L2 (at x = L)

For the two faces perpendicular to the y-axis:

ΦE, y=0 = 0 (since cy is 0 at y = 0)

ΦE, y=L = ∫( cy ) · dA = cL3 (at y = L)

Summing these up:

ΦE = aL2 + (a + bL)L2 + cL3

Part C: Net Charge Inside the Cube

The net charge q inside the cube can be found using Gauss's Law, which states "the total electric flux out of a closed surface is equal to the charge enclosed divided by the permittivity of free space (epsilon_0)."

q = ΦE ε0

Substituting the expression for ΦE we have:

q = ε0 ( aL2 + (a + bL)L2 + cL3 )

Final answer:

The equilibrium temperature of a mixture of two amounts of water will be closer to the initially cooler water due to the specific heat capacity of water.

Explanation:

The equilibrium temperature of a mixture of two amounts of water will be closer to the initially cooler water.

When different temperatures of water are mixed, heat is transferred between them until they reach a common equilibrium temperature. The amount of heat transferred is determined by the specific heat capacity of water, which is greater than most common substances. As a result, water undergoes a smaller temperature change for a given heat transfer. Therefore, the equilibrium temperature will be closer to the initially cooler water.

Answer:

The pebble reaches a height of 11.716 m above the starting point.

Explanation:

Given:

Mass of pebble (m) = 50.3 g = 0.0503 kg [1 g = 0.001 kg]

Spring constant (k) = 320.0 N/m

Elongation of catapult (x) = 0.190 m

Height of fly (h) = ?

Air resistance is ignored. So, conservation of energy holds true.

The energy stored in the catapult on stretching it is elastic potential energy. This elastic potential energy is transferred to the pebble in the form of gravitational potential energy. Therefore,

Elastic potential energy = Gravitational potential energy

[tex]\frac{1}{2}kx^2=mgh\\\\h=\frac{kx^2}{2mg}[/tex]

Plug in the given values and solve for 'h'. This gives,

[tex]h=\frac{(320.0\ N/m)(0.190\ m)^2}{2(0.0503\ kg)(9.8\ m/s^2)}\\\\h=\frac{11.552\ Nm}{0.986\ N}\\\\h=11.716\ m[/tex]

Therefore, the pebble reaches a height of 11.716 m above the starting point.

Answer:

The horizontal force that must be applied to the box to cause it to start sliding along the surface is 162N

Explanation:

To start sliding the box on the surface it must overcome its static frictional force  under equilibrium condition

The net force on the box is

F - fs = 0

F = fs = us N = us mg

Force = ( 0.3) x ( 55 kg) x ( 9.8 m/s^2) = 161.7 approximately 162 N

Horizontal force is defined as the forces that equal and opposite in the direction. The horizontal resultant force is always zero.

Given that:

Mass of box = 55 Kg

Coeeficient of Static friction = 0.30

Coefficient of kinetic friction = 0.20

The box if have to slide, then it would have to overcome the static force, such that:

F - Fs = 0

F = Fs

The force will be:

F = [tex]\rm \mu \times mass \times g[/tex]

F = [tex]0.3 \times 55 \times 9.8[/tex]

F = 161.7 Newton

Thus, the force required to slide the box is 161.7 Newton.

To know more about force, refer to the following link:

https://brainly.com/question/4456579

Answer:

When light goes from one material into another material having a HIGHER index of refraction, its speed and wavelength decrease, but its frequency stays the same.

The correct option is E

Explanation:

Refraction is the bending of light ray as it crosses the boundary between the two media of different densities, thus causing a change in it's direction.

When light goes from one material of lower refractive to another material of higher refractive index, the speed of light reduces, because from law of refraction

aNg= (speed of light in air)/(speed of light in glass)

Snell's Law

So generally,

n1/n2 = v2/v1

Let n1 be incident index

Let n2 be refracted index

v2 is refracted speed

v1 is incident speed

Since given that, n1<n2

Then, the right side of the equation will always be less than 1 i.e n1/n2

Then, v2=v1• ( n1/n2)

Since n1/n2 is less that one this show that the speed v2 will reduce.

Since we know that the speed decrease,

We also know that speed, wavelength and speed is related by

v=fλ

Since, the speed is directly proportional to wavelength this shows that as the speed reduces the wavelength also reduces where frequency is the constant of proportionality

v∝λ. .

then f is constant of proportional

v=fλ

So as speed reduces, the wavelength reduces.

frequency of a light wave does not depend on the medium, while wavelength and speed do.

its speed and wavelength decrease, but its frequency stays the same.

Then the answer is E

Final answer:

The back emf of the motor decreases due to reduced speed caused by dirt in the shaft, which, in turn, causes the current drawn from the wall socket to increase.

Explanation:

When dirt prevents the shaft of an electric motor from turning rapidly, the back emf produced by the motor decreases. Since the back emf is proportional to the motor's angular velocity, any decrease in speed results in a decrease in back emf. Because the back emf opposes the applied voltage from the wall socket, a decrease in back emf implies that the voltage across the motor's coil will increase, causing the motor to draw a larger current from the socket. Therefore, the current that the motor draws from the wall socket will increase. The correct answer to the question is (c) The back emf decreases, and the current drawn from the socket increases.

Answer:

a. 79.1 N

b. 344 J

c. 344 J

d. 0 J

e. 0 J

Explanation:

a. Since the crate has a constant velocity, its net force must be 0 according to Newton's 1st law. The push force [tex]F_p[/tex] by the worker must be equal to the friction force [tex]F_f[/tex] on the crate, which is the product of friction coefficient μ and normal force N:

Let g = 9.81 m/s2

[tex]F_p = F_f = \mu N = \mu mg = 0.26 * 31 * 9.81 = 79.1 N[/tex]

b. The work is done on the crate by this force is the product of its force [tex]F_p[/tex] and the distance traveled s = 4.35

[tex]W_p = F_ps = 79.1*4.35 = 344 J[/tex]

c. The work is done on the crate by friction force is also the product of friction force and the distance traveled s = 4.35

[tex]W_f = F_fs = -79.1*4.35 = -344 J[/tex]

This work is negative because the friction vector is in the opposite direction with the distance vector

d. As both the normal force and gravity are perpendicular to the distance vector, the work done by those forces is 0. In other words, these forces do not make any work.

e. The total work done on the crate would be sum of the work done by the pushing force and the work done by friction

[tex]W_p + W_f = 344 - 344 = 0 J[/tex]

Answer:

(A) 79N

(B) W = 344J

(C) Wf= -344J

(D) W = 0J

(E) W = 0J

Explanation:

Please see attachment below.

Option d is correct; relativity formulas are not needed because the Lorentz factor (γ-1) is much less than 1, indicating that the electron's velocity is non-relativistic after being accelerated through a potential difference of 0.200 kV.

We need to determine whether we should use relativity formulas for the electrons accelerated through a potential difference. When electrons gain kinetic energy (K) by acceleration through a potential difference (V), the energy they gain can be expressed as Ve=K=(γ-1)mc2. Here, γ stands for Lorentz factor, m is the mass of electron, and c is the speed of light. Given that the potential difference ('V') is 0.200 kV, and the rest mass energy ('Eo') of an electron is 0.511 MeV, we calculate:

γ-1= (0.200 keV) / (0.511 MeV) ≈ 0.3916 × 10-3, which implies that γ is approximately 1. So, γ-1 << 1, which in turn means that the electron's velocity is much less than the speed of light, and the relativistic effect can be neglected. Thus, we do not need to use special relativity.

The correct answer is d.