A machinist turns the power on to a grinding wheel, which is at rest at time t = 0.00 s. The wheel accelerates uniformly for 10 s and reaches the operating angular velocity of The wheel is run at that angular velocity for 37 s and then power is shut off. The wheel decelerates uniformly at until the wheel stops. In this situation, the time interval of angular deceleration (slowing down) is closest to:

Answer:

the question is incomplete, below is the complete question

"An object is undergoing simple harmonic motion along the x-axis. Its position is described as a function of time by x(t) = 5.5 cos(4.4t - 1.8), where x is in meters, the time, t, is in seconds, and the argument of the cosine is in radians.

a.What is the object's velocity, in meters per second, at time t = 2.9?

b.Calculate the object's acceleration, in meters per second squared, at time t = 2.9.  

c. What is the magnitude of the object's maximum acceleration, in meters per second squared?

d.What is the magnitude of the object's maximum velocity, in meters per second?"

a.[tex]v(t)==24.1m/s[/tex]

b.[tex]a(t)=3.79m/s^{2}[/tex]

c.[tex]a_{max}=106.48m/s^{2}[/tex]

d.[tex]v_{max}=24.2m/s[/tex]

Explanation:

the gneral expression for the displacement of object in simple harmonic motion is represented by

[tex]x(t)=Acos(wt- \alpha)\\[/tex]

while the velocity is express as

[tex]v(t)=-Awcos(4.4t-1.8)\\[/tex]

and the acceleration is

[tex]a(t)=-aw^{2}cos(wt- \alpha )\\[/tex]

Note: the angle is in radians

The expression for the displacement from the question is [tex]x(t)=5.5cos(4.4t-1.8)\\[/tex]

comparing, A=5.5, w=4.4,α=1.8

a.To determine the object velocity at t=2.9secs,

we substitute for t in the velocity equation

[tex]v(t)=-5.5*4.4sin(4.4*2.9-1.8)\\v(t)=-24.2sin(10.96)\\[/tex]

[tex]v(t)=-24.2*(-0.9993)\\v(t)==24.1m/s[/tex]

b.To determine the object acceleration at t=2.9secs,

we substitute for t in the acceleration equation

[tex]a(t)=-5.5*4.4^{2} cos(4.4*2.9-1.8)\\a(t)=-106.48cos(10.96)\\[/tex]

[tex]a(t)=-106.48*0.0356\\a(t)=3.79m/s^{2}[/tex]

c. The acceleration is maximum when the displacement equals the amplitude. hence  magnitude of the object acceleration is

[tex]a_{max}=-w^{2}A\\ a_{max}=-4.4^{2}*5.5\\ a_{max}=106.48m/s^{2}[/tex]

d.The maximum velocity is expressed as

[tex]v_{max}=wA\\v_{max}=4.4*5.5\\v_{max}=24.2m/s[/tex]

Answer:

U = 11.85 J

Explanation:

given,

spring is stretched = 23 cm

                             x = 0.23 m

Force experiences = 103 N

we know,

 F = k x

where k is the spring constant

 [tex]k =\dfrac{F}{x}[/tex]

 [tex]k =\dfrac{103}{0.23}[/tex]

       k = 447.83 N/m

energy stored in the spring

 [tex]U =\dfrac{1}{2}kx^2[/tex]

 [tex]U =\dfrac{1}{2}\times 447.83 \times 0.23^2[/tex]

        U = 11.85 J

hence, energy stored in the spring is equal to U = 11.85 J

The elastic potential energy stored in the spring when stretched to a length of 23 cm, given a spring constant of 4 N/cm, is 0.18 J.

The question pertains to how much energy is stored in a spring when it is stretched 23 cm from its equilibrium point, where it experiences a force of 103 N. The potential energy stored in the spring can be calculated using the formula U = 1/2kx². Given that the spring constant is 4 N/cm and the displacement of the spring from its unstretched length is 3 cm (since the unstretched length is 20 cm), the potential energy can be calculated as follows: U = 1/2 * 4 N/cm * (3 cm)² = 0.18 J. Thus, the elastic potential energy contributed by the spring when it is stretched to a length of 23 cm is 0.18 J.

The problem is solved by applying the Ideal Gas Law and Dalton's Law of Partial Pressures. First, we calculate the mass of acetone condensed using these laws and then determine the rate at which acetone is vaporized using the given volumetric flow rate.

The given problem involves a number of gas law principles, but its main focus is on the application of the Ideal Gas Law and Dalton's Law of partial pressures. Initially, we calculate the moles of acetone in the feed using the Ideal Gas Law, and then we find out the moles of acetone in the effluent using Dalton's law. Subtracting gives us the moles of acetone condensed.

(a) Using Ideal Gas Law, we have PV=nRT. Hence, n (acetone, feed) = P (acetone, feed) * V(feed) /RT(feed). To find the moles of acetone in the effluent, we use Dalton's law and the Ideal Gas Law to get n (acetone, effluent) = P(acetone, effluent) * V (effluent) / RT (effluent). Subtracting moles in effluent from moles in feed gives moles condensed. Multiplying by the molar mass of acetone, we get mass of acetone condensed.

(b) The question tells us the volumetric flow rate of the gas leaving the condenser. Therefore, number of moles of acetone vaporized per hour can be calculated using the Ideal Gas Law and then can be converted into mass by using the molar mass of acetone. Hence, rate at which acetone is vaporized = moles (acetone, vaporized per hour) * molar mass (acetone).

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Answer:

Explanation:

Heres the correct and full question:

For a new TV series "Stupidity Factor contestants are dropped into the ocean (p=1030 kg/m³) along with a Styrofoam p=3oo kg/m³ block that is 2m x 3 m X 20 cm thick. If too many contestants climb aboard the block and it sinks below the water surface, they are declared "stupid" and abandoned at sea. Find the maximum number of 7o kg contestants that the block can hold. (No fractional contestants, please. They hate it when that happens.

answer:

volume of styrofoam block=V=2m x 3m x 0.20m =1.2m³

density of styrofoam=ρs=300kg/m³

mass of styrofoam=ms=v(ρs)=1.2 x 300=360kg

weight of styrofoam = ws=(ms)g=360 x 9.8=3528N

consider number of contestants= n

toatal weight of ccontestants=W=n(70 x 9.8)=n(686N)

since, styrofoam fully submerged into water, then bat force,

B=ρ(vs)g=1030 x 1.2 x 9.8 = 12112.8N

At equilibrium,

B - W - Ws = 0

12112.8 - 686n - 3528 = 0

[tex]n=\frac{12112.8-3528}{686}=12.5[/tex]

n=12 person(contestants)

Answer:negative Z direction

Explanation:

It is given that EM wave is travelling in negative y direction i.e. -[tex]\hat{j}[/tex]

Magnetic Field is in positive x direction i.e. [tex]\hat{i}[/tex]

We know that Electric field and magnetic field are perpendicular to each other and they are mutually perpendicular to direction of wave propagation.

Mathematically if Electric field is in negative z direction it will yield the direction of wave propagation

[tex]=(-\hat{k})\times \hat{i}[/tex]

[tex]=-\hat{j}[/tex]                            

Final answer:

The Paschen and Brackett series of lines in atomic hydrogen can overlap in terms of their wavelength ranges.

Explanation:

The Paschen series of lines in atomic hydrogen occurs when nf = 3, and the Brackett series occurs when nf = 4.

The range of wavelengths in these two series can overlap because the wavelength equation 1/λ = 2π²mk²e⁴/(h³c)(Z²)(1/nf² - 1/ni²) depends on the values of nf and ni, which can be different for each series

For example, if nf = 3 in the Paschen series and nf = 4 in the Brackett series, the wavelengths in these two series can overlap depending on the values of ni.

Answer:

a) [tex]T=1.02 s[/tex]

b) [tex]v=0.28\frac{m}{s}[/tex]

c) [tex]v=0.25\frac{m}{s}[/tex]

Explanation:

a) The period is defined as:

[tex]T=\frac{2\pi}{\omega}[/tex]

[tex]\omega[/tex] is the natural frequency of the system, in this case is given by:

[tex]\omega=\sqrt{\frac{k}{m}}\\\omega=\sqrt{\frac{9.5\frac{N}{m}}{250*10^{-3}kg}}\\\omega=6.16\frac{rad}{s}[/tex]

Now, we calculate the period:

[tex]T=\frac{2\pi}{6.16\frac{rad}{s}}\\T=1.02 s[/tex]

b) According to the law of conservation of energy, we have:

[tex]\frac{kx^2}{2}=\frac{mv^2}{2}\\v=\sqrt{\frac{kx^2}{m}}\\v=\sqrt{\frac{(9.5\frac{N}{m})(4.5*10^{-2}m)^2}{250*10^{-3}kg}}\\v=0.28\frac{m}{s}[/tex]

c) In this case, we have:

[tex]U_i=U_f+K_f\\\frac{kx_i^2}{2}=\frac{kx_f^2}{2}+\frac{mv^2}{2}\\v=\sqrt{\frac{k(x_i^2-x_f^2)}{m}}\\v=\sqrt{\frac{9.5\frac{N}{m}((4.5*10^{-2}m)^2-(2*10^{-2}m)^2)}{250*10^{-3}kg}}\\v=0.25\frac{m}{s}[/tex]

The maximum speed is; 0.28 m/s

The speed when it is located 2.0 cm from its equilibrium position is; 0.25 m/s

A) Formula for period is;

T = √(2π/ω)

where ω which is the natural angular frequency

ω = √(k/m)

we are given;

k = 9.5 N/m

m = 250 g = 0.25 kg

x = 4.5 cm = 0.045 m

Thus;

ω = √(9.5/0.25)

ω = 6.1 rad/s

Thus;

Period is; T = √(2π/6.1)

T = 1.02 s

B) From law of conservation of energy;

maximum speed is;

v = √(kx²/m)

v = √(9.5 * 0.045²/0.25)

v = 0.28 m/s

C) We are now given;

x₁ = 4.5cm = 0.045 m

x₂ = 2 cm = 0.02 m

Thus, speed is gotten from;

v = √((k(x₁ - x₂)/m)

v = √((9.5(0.045 - 0.02)/0.25)

v = 0.25 m/s

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Final answer:

The maximum shear stress developed in the shaft is approximately 7.57 MPa. The angle of twist in the shaft at full power is approximately 3.00°.

Explanation:

To determine the maximum shear stress developed in the shaft, we can use the formula:

Shear stress (τ) = Torque (T) / Polar Moment of Inertia (J)

To find the torque, we can use the formula:

Torque (T) = Power (P) / Angular Velocity (ω)

We are given that the power is 2500 hp and the angular velocity is 1700 rpm. Converting these values to W and rad/s respectively, we have:

Power (P) = 2500 hp * 745.7 W/hp ≈ 1,864,250 W

Angular Velocity (ω) = 1700 rpm * 2π rad/minute ≈ 17897.37 rad/s

Substituting these values into the torque formula, we have:

Torque (T) = 1,864,250 W / 17897.37 rad/s ≈ 103.98 N*m

Next, we need to find the polar moment of inertia (J) of the shaft. The polar moment of inertia for a solid shaft can be calculated using the formula:

J = (π/2) * (outer diameter^4 - inner diameter^4)

Converting the diameter to meters, we have:

Diameter (d) = 8 in * 0.0254 m/in = 0.2032 m

Substituting this value into the polar moment of inertia formula, we have:

J = (π/2) * (0.2032^4 - (0.2032 - 2 * 3/8)^4)

= (π/2) * (0.2032^4 - 0.1926^4)

0.013743 m^4

Finally, we can calculate the maximum shear stress using the formula:

Shear stress (τ) = 103.98 N*m / 0.013743 m^4 ≈ 7.57 MPa

As for the angle of twist, we can use the formula:

Angle of twist (θ) = T * L / (G * Given that the length of the shaft is 100 ft and Young's modulus (G) for A-36 steel is approximately 200 GPa, we can calculate the angle of twist as follows:

Angle of twist (θ) = 103.98 N*m * 100 ft * 0.3048 m/ft / (200 GPa * 0.013743 m^4)

≈ 0.0524 radians or 3.00°

Answer:

Explanation:

The expression for the calculation of the enthalpy change of a process is shown below as:-

[tex]\Delta H=m\times C\times \Delta T[/tex]

Where,  

[tex]\Delta H[/tex]

is the enthalpy change

m is the mass

C is the specific heat capacity

[tex]\Delta T[/tex]

is the temperature change

Thus, given that:-

Mass of water = 2.4 kg

Specific heat = 4.18 J/g°C

[tex]\Delta T=100-21\ ^0C=79\ ^0C[/tex]

So,  

[tex]\Delta H=2.4\times 4.18\times 79\ J=792.52\ kJ[/tex]

Heat Supplied [tex]Q=VIt[/tex]

where [tex]i=current[/tex]

[tex]V=Voltage[/tex]

[tex]8.5\times 120\times t=2.4\times 4186\times 79[/tex]

[tex]t=778.10 s[/tex]

Final answer:

To determine the time required to boil the water in a coffee maker, calculate the power using P = IV, then the heat needed to raise the water temperature to its boiling point with Q = mcΔT, and finally divide Q by P to find the time.

Explanation:

The question deals with calculating how long it takes an automatic coffee maker to boil 2.4 kg of water using a resistive heating element. The initial temperature of the water is 21 °C, and we are given that the coffee maker is plugged into a 120 V electrical outlet with a current of 8.5 A. The specific heat capacity of water is 4186 J/kg°C. To find the time to boil the water, we'll first calculate the power delivered by the heating element using the relationship P = IV, where P is power, I is current, and V is voltage. Once power is known, we can use the equation Q = mcΔT to find the heat Q required to raise the temperature of water to its boiling point (100 °C), where m is water's mass, c is the specific heat capacity, and ΔT is the change in temperature. Finally, we'll divide quantified heat Q by the power P to find the time t: t = Q/P.

Answer:

V = 0.3724 m³

T = 27.836 N

Explanation:

Given :

m = 3.21 kg  , W= 3.21 * 9.81 m / s² = 31.4901 N

ρ = 8.62 g / cm ³  = 8620 kg / m³

V = m / ρ =  3.21 kg  /  8620 kg / m³

V = 0.3724 m³

when submerged the weight of brass cylinder is equal to the tension in string:

F =  (0.3724m³) * (1000 kg / m³) * (9.81 m/s²²) = 3.653 ≈ 3.65 N

T = 31.4901 N - 3.65 N  

T = 27.836 N

The volume corresponds to the measure of the space occupied by a body. From the given dimensions we can intuit that we are looking to find the Volume of an Cuboid, that is, an orthogonal rectangular prism, whose faces form straight dihedral angles.

Mathematically the volume of this body is given as

[tex]V = lWh[/tex]

Where,

L = Length

W = Width

H = High

[tex]V = (12)(0.65)(13*10^{-2})[/tex]

[tex]V = 1.014m^3[/tex]

Note: The value given for the height was in centimeters, so it was transformed to meters.

Answer:

P₂  = 370 kPa

Explanation:

Boyle's  law: States that the volume of a given mass of gas is inversely proportional to its pressure provided the temperature remains constant. It can be expressed mathematically as,

P₁V₁ = P₂V₂................ Equation 1

Where P₁ = Initial Pressure, V₁ = Initial volume, P₂ = Final Pressure, V₂ = Final Volume.

Making P₂ The subject of the equation above,

P₂ = P₁V₁/V₂..................... equation 2

Substituting these vaues into equation 2,

Where P₁ = 163 kPa = 163000 pa, V₁ =  10 L, V₂ = 4.4 L.

P₂ = 163000(10)/4.4

P₂ = 370454.55 Pa

P₂ = 370000 Pa

P₂  = 370 kPa To 3 significant figure.

Therefore Final pressure = 370 kPa

Answer:

U2 = 47.38m/s = initial velocity of B before impact

Explanation:

An example of the diagram is shown in the attached file because of missing angle of direction in the question

Mass A, B are mass of cars

A = 1965

B =1245

U1 = initial velocity of A = 52km/hr

U2 = initial velocity of B

V = common final velocity of two cars

BU2 = (A + B)*V sin ¤ ...eq1 y plane

AU1 = (A + B) *V cos ¤ ....equ 2plane

From equ 2

V = AU1/(A + B)*cos ¤

Substitute V into equation 1

We have

U2 = (AU1/B)tan ¤ where ¤ = angle of direction which is taken to be 30°

Substitute all parameters to get

U2 = (1965/1245)*52 * tan 30°

U2 = 47.38m/s

To find the corresponding velocity of car B just before impact, we can use the principle of conservation of momentum. By equating the total momentum before the collision to the total momentum after the collision, we can solve for the velocity of car B. Plugging in the given values and calculating, we can find the corresponding velocity of car B just before impact.

To solve this problem, we can use the principle of conservation of momentum. Since the two cars stick together after the collision, the total momentum before the collision is equal to the total momentum after the collision.

Let's define the velocity of car A just before impact as vA and the velocity of car B just before impact as vB.

Using the principle of conservation of momentum, we can write:

(mass of car A * velocity of car A) + (mass of car B * velocity of car B) = (total mass * common velocity)

Plugging in the given values, we have:

(1965 kg * 52 km/h) + (1245 kg * vB) = (3210 kg * common velocity)     (Note: Convert km/h to m/s)

Simplifying the equation, we get:

101820 kg * km/h + 1245 kg * vB = 3210 kg * common velocity

Now, we can solve for the velocity of car B just before impact:

vB = (3210 kg * common velocity - 101820 kg * km/h) / 1245 kg

By substituting the values and calculating, we can find the corresponding velocity of car B just before impact.

Answer:

169.4 m/s

Explanation:

Given that the angle of projectile is θ_1 =450°

The speed of body at maximum height is U cosθ_1 = 150 m/s

The angle in second trail is θ_2 =37°  

From the given data U cosθ_1 = 150 m/s

U = 150 m/s / cosθ_1

=  150m/s / cos45°

=212.13 m/s

The velocity of the projectile at maximum height in second trail= Ucos(θ_2)

=212.13 m/s×cos37°

=169.4 m/s

The velocity of a projectile at its highest point when fired at an angle to the horizontal remains the same if the initial speed is unchanged, regardless of the angle. Hence, even when changing the angle from 45 to 37 degrees, the velocity at the highest point would still be 150 m/s.

The student's question deals with the velocity of a projectile at the highest point in its trajectory. When a projectile is fired upward at an angle, its velocity at the highest point of its trajectory is only composed of the horizontal component because the vertical component of the velocity becomes zero at that point.

In the given scenario, when the projectile is shot at 45 degrees to the horizontal, the speed at the highest point is given as 150 m/s. This speed represents the horizontal component since there's no vertical component at the highest point. When the angle is changed to 37 degrees, the horizontal component of the initial velocity is calculated using the cosine component of the initial speed, hence the velocity at the highest point remains the same as when it was fired at 45 degrees, provided that the initial speed is unchanged.

Therefore, in the second trial, with the angle at 37 degrees, the velocity of the projectile at its highest point would still be 150 m/s, because the horizontal component of the velocity is not affected by the change in angle.

Answer:

Friction force F₂ after doubling the angular speed is same as the friction force at angular speed ω₁

Explanation:

Consider the fig attached below. Forces acted on person are Centripetal force (-mv²/r)  exerted in x direction and reversal normal force N wall exerted on person.

                     [tex]\sum F_{x} =0\\N+ ma_{x}\\-N=m(-\frac{v^{2}}{r})\\-N=m(-r\omega^{2})N=m(r\omega^{2})[/tex] ---(1)

In y direction there is frictional force Fs exerted in upward direction that keeps the person standing without falling which is balanced by weight of person in downward direction.

                     [tex]\sum F_{y} = 0\\F_{s}-mg=0\\F_{s}=mg[/tex]----(2)

from eq 2 it can e seen that frictional force is equal to weight of person exerted in upward direction, it does not depends on angular speed ω₁. So when the angular speed is doubled i.e ω₂ = 2ω₁, frictional force Fs remains same.

Answer:

Part A:

B) mA*vA = mA*v′A*cosθ′A + mB*v′B*cosθ′B

Part B:

C) 0 = mA*v′A*sinθ′A − mB*v′B*sinθ′B

since vAy = 0 m/s

Part C:

θ′B = tan⁻¹(1.0699) = 46.94°

Part D:

v′B = 1.246 m/s

Explanation:

Given:  

mA = 0.117 kg

vA = vAx = 2.80 m/s  

mB = 0.135 kg

vB = 0 m/s

θ′A = 30.0°

v′A = 2.10 m/s

Part A: Taking the x axis to be the original direction of motion of ball A, the correct equation expressing the conservation of momentum for the components in the x direction is

B) mA*vA = mA*v′A*cosθ′A + mB*v′B*cosθ′B

Part B: Taking the x axis to be the original direction of motion of ball A, the correct equation expressing the conservation of momentum for the components in the y direction is

C) 0 = mA*v′A*sinθ′A − mB*v′B*sinθ′B

since vAy = 0 m/s

Part C: Solving these equations for the angle, θ′B , of ball B after the collision and assuming that the collision is not elastic:

mA*vA = mA*v′A*cosθ′A + mB*v′B*cosθ′B

⇒ (0.117)(2.80) = (0.117)(2.10)Cos 30° + (0.135)*v′B*cosθ′B

⇒ v′B*cosθ′B  = 0.8505  ⇒ v′B = 0.8505/cosθ′B

then

0 = mA*v′A*sinθ′A − mB*v′B*sinθ′B

⇒ 0 = (0.117)(2.10)Sin 30° - (0.135)*v′B*sinθ′B

⇒ v′B*sinθ′B  = 0.91   ⇒ v′B = 0.91/sinθ′B

if we apply

0.8505/cosθ′B = 0.91/sinθ′B

⇒ tanθ′B = 0.91/0.8505 = 1.0699

⇒  θ′B = tan⁻¹(1.0699) = 46.94°

Part D: Solving these equations for the speed, v′B , of ball B after the collision and assuming that the collision is not elastic:

if   v′B = 0.91/sinθ′B

⇒ v′B = 0.91/sin 46.94°

⇒ v′B = 1.246 m/s

Conservation of momentum can be used to solve the equations for the angle and the speed of ball B after the collision, using the initial conditions, final conditions, and trigonometric identities. This involves the application of physics concepts, combined with the mathematics of trigonometry.

Part A: The correct equation expressing the conservation of momentum for the components in the x direction would be option B, mAvA=mAv′Acosθ′A+mBv′Bcosθ′B. This equation shows that the initial momentum of ball A (mAvA) equals the sum of the momentum of ball A and ball B after the collision in the x direction.

Part B: For the components in the y direction, the right answer is C, 0=mAv′Asinθ′A−mBv′Bsinθ′B. Since ball B was initially at rest and ball A was moving along the x-axis, there was no momentum in the y direction before the collision. Therefore, the total momentum in the y direction after the collision should also be 0.

Part C and D: To solve these equations for the angle and the speed of ball B after the collision, you need to use these equations in combination with the conservation of kinetic energy formula (1/2*m*v^2) and the trigonometric identities. Detailed solution steps require knowledge of the involved mathematics.

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Answer:

(D) 10 seconds

Explanation:

The Houston Methodist Hospital automated the emergency power supply system (EPSS) testing processes in order to ensure the safety of patients during an outage

There is an emergency power generator that turns on to supply emergency power after normal power shuts down within 10 seconds.

Answer:

D. 10 seconds

Explanation:

The Houston Methodist Hospital has eleven diesen powered generators that turn on at most 10 seconds after the normal power shuts down.

Considering that it is a hospital, this system is really important.

So the correct answer is:

D. 10 seconds

Answer:

The total distance traveled by an object attached to a spring that is pulled to position x=A and then released is 4 A.

Explanation:

We have an small object attached to a relaxed spring whose opposite end is fixed. The spring rests on a frictionless surface. This means the only force acting on the object is the elastic  force of the spring, a conservative force, since the weight and the normal force compensate between them.

The initial position of the object is x=0 then is pulled to position x=A and released. After which it undergoes a simple harmonic motion with an amplitude A. From the position x=A to the equilibrium position in x=0 the object travels a distance A. From the equilibrium position x=0 to maximum negative position in x= -A the object travels again a distance A. Then to return to the original position the object should travel a distance 2 A in reverse direction.

In one full cycle of its motion the object travels a distance 4 A.

To develop this problem it is necessary to apply the concepts related to the conservation of Energy. In this case the definition concerning kinetic energy from the simple harmonic movement.

From the conservation of energy we know that the kinetic energy would be conserved through the work done by the frictional force and the simple harmonic potential energy, in other words:

[tex]KE = PE_s +W_f[/tex]

[tex]KE = \frac{1}{2}kA^2 + W_f[/tex]

Where,

[tex]KE =[/tex] Kinetic Energy

[tex]PE_s =[/tex]Potential Harmonic Simple Energy

[tex]W_f =[/tex] Work made by friction.

Our values are given as,

[tex]m = 10Kg  \rightarrow[/tex] mass

[tex]k = 1400N/m \rightarrow[/tex] Spring constant

[tex]A = 0.1m \rightarrow[/tex]Amplitude

[tex]f_f = 30N \rightarrow[/tex] Frictional Force

Replacing we have,

[tex]KE = \frac{1}{2}kA^2 + W_f[/tex]

[tex]KE =\frac{1}{2} 1400 * 0.1^2 + ( - 30 * 0.1)[/tex]

[tex]KE = 4 J[/tex]

Therefore the Kinetic Energy of the block as it passes through its equlibrium position is 4J.

The kinetic energy of the 10-kg block as it passes through its equilibrium position is 4 Joules. This is calculated by converting the potential energy stored in the spring to kinetic energy and then subtracted the energy lost due to friction.

The first step in solving this problem is to understand the two forces acting on the block in this question: the spring force and the frictional force. The spring potential energy when the block is pulled 10 cm to the right is given by the formula U = 1/2kx^2, where k is the force constant and x is the displacement. Substituting the given values, we have U = 1/2(1400 N/m)(0.1 m)^2 = 7 Joules. This is the initial potential energy stored in the spring. As the block passes through its equilibrium position, this potential energy is fully converted to kinetic energy.

We also need to take into account the work done against frictional force which is equal to the frictional force times the displacement, i.e., W_friction = Friction * displacement = 30N * 0.1m = 3 Joules. This is the energy lost due to friction.

Finally, we subtract the work done by the frictional force from the potential energy to achieve the kinetic energy. Therefore, the kinetic energy of the block as it passes through its equilibrium position is K = U - W_friction = 7J - 3J = 4 Joules.

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To solve this problem we will use the concepts related to Torque as a function of the Force in proportion to the radius to which it is applied. In turn, we will use the concepts of energy expressed as Work, and which is described as the Torque's rate of change in proportion to angular displacement:

[tex]\tau = Fr[/tex]

Where,

F = Force

r = Radius

Replacing we have that,

[tex]\tau = Fr[/tex]

[tex]\tau = 21cm (\frac{1m}{100cm})* 550N[/tex]

[tex]\tau = 11.55Nm[/tex]

The moment of inertia is given by 2.5kg of the weight in hand by the distance squared to the joint of the body of 24 cm, therefore

[tex]I = 0.25Kg\cdot m^2 +(2.5kg)(0.24m)^2[/tex]

[tex]I = 0.394kg\cdot m^2[/tex]

Finally, angular acceleration is a result of the expression of torque by inertia, therefore

[tex]\tau = I\alpha \rightarrow \alpha = \frac{\tau}{I}[/tex]

[tex]\alpha = \frac{11.55}{0.394}[/tex]

[tex]\alpha = 29.3 rad/s^2[/tex]

PART B)

The work done is equivalent to the torque applied by the distance traveled by 60 °° in radians [tex](\pi / 3)[/tex], therefore

[tex]W = \tau \theta[/tex]

[tex]W = 11.5* \frac{\pi}{3}[/tex]

[tex]W = 12.09J[/tex]

Answer:

A. 36,000 W

Explanation:

[tex]m[/tex] = mass of the car = 900 kg

[tex]v_{o}[/tex] = Initial speed of the car = 20 m/s

[tex]v_{f}[/tex] = Final speed of the car = 0 m/s

[tex]W[/tex] = Work done by the brakes on the car

Magnitude of work done on the car by the brakes is same as the change in kinetic energy of the car.hence

[tex]W = (0.5) m (v_{o}^{2} - v_{f}^{2})\\W = (0.5) (900) ((20)^{2} - (0)^{2})\\W = 180000 J[/tex]

[tex]t[/tex] = time taken by the car to come to stop = 5 s

[tex]P[/tex] = Average power produced by the car

Average power produced by the car is given as

[tex]P = \frac{W}{t} =\frac{180000}{5} \\P = 36000 W[/tex]

Answer:

[tex]5.2728\times 10^{-25}\ kgm/s[/tex]

Explanation:

h = Planck's constant = [tex]6.626\times 10^{-34}\ m^2kg/s[/tex]

[tex]\Delta x[/tex] = Uncertainity in position = [tex]10^{-10}\ m[/tex]

[tex]\Delta p[/tex] = Uncertainty in momentum

According to the Heisenberg uncertainity principle we have

[tex]\Delta x\Delta p=\dfrac{h}{4\pi}\\\Rightarrow \Delta p=\dfrac{h}{4\pi\Delta x}\\\Rightarrow \Delta p=\dfrac{6.626\times 10^{-34}}{4\pi\times 10^{-10}}\\\Rightarrow \Delta p=5.2728\times 10^{-25}\ kgm/s[/tex]

The minimum uncertainty in its momentum is [tex]5.2728\times 10^{-25}\ kgm/s[/tex]

Answer:

(a) The length of the pendulum on Earth is 36.8cm

(b) The length of the pendulum on Mars is 13.5cm

(c) Mass suspended from the spring on Earth is 0.37kg

(d) Mass suspended from the spring on Mars is 0.36kg

Explanation:

Period = 1.2s, free fall acceleration on Earth = 9.8m/s^2, free fall acceleration on Mars = 3.7m/s^2

( a) Length of pendulum on Earth = [( period ÷ 2π)^2] × acceleration = (1.2 ÷ 2×3.142)^2 × 9.8 = 0.0365×9.8 = 0.358m = 35.8cm

(b) Length of the pendulum on Mars = (1.2÷2×3.142)^2 × 3.7 = 0.0365×3.7 = 0.135cm = 13.5m

(c) Mass suspended from the spring on Earth = (force constant×length in meter) ÷ acceleration = (10×0.358) ÷ 9.8 = 0.37kg

(d) Mass suspended from the spring on Mars = (10×0.135)÷3.7 = 0.36kg

The length of a pendulum with a period of 1.2 s on Earth is approximately 36.95 cm, while on Mars it is around 16.99 cm. The mass suspended from a spring that would result in a period of 1.2 s on Earth is approximately 0.722 kg, and on Mars it is approximately 0.329 kg.

(a) What length of pendulum has a period of 1.2 s on Earth? cm

Using the equation for the period of a pendulum, T = 2π √(L/g), where T is the period, L is the length of the pendulum, and g is the acceleration due to gravity, we can solve for L. Rearranging the equation, we have L = (T/2π)² * g.

Given that the free-fall acceleration on Earth is approximately 9.8 m/s², substituting the values into the equation, we have:

L = (1.2/2π)² * 9.8 = 0.3695 m = 36.95 cm

(b) What length of pendulum would have a 1.2-s period on Mars? cm

Using the same equation, L = (T/2π)² * g, we can substitute the values for the period and acceleration due to gravity on Mars:

L = (1.2/2π)² * 3.7 = 0.1699 m = 16.99 cm.

(c) Find the mass suspended from this spring that would result in a period of 1.2 s on Earth. kg

For a spring-mass system, the period is given by T = 2π √(m/k), where T is the period, m is the mass, and k is the spring constant. Rearranging the equation, we have m = (T/2π)² * k.

Given that the spring constant is 10 N/m, substituting the values into the equation, we have:

m = (1.2/2π)² * 10 = 0.722 kg.

(d) Find the mass suspended from this spring that would result in a period of 1.2 s on Mars. kg

Using the same equation, m = (T/2π)² * k, we can substitute the values for the period and spring constant:

m = (1.2/2π)² * 10 = 0.329 kg.

Answer

given,

diameter = 6 m

depth = h = 1.5 m

Atmospheric pressure = P₀ = 10⁵ Pa

a) absolute pressure

P = P₀ + ρ g h

P = 10⁵ + 1000 x 10 x 1.5

P = 1.15 x 10⁵ Pa

b) When two person enters into the pool

mass of the two person = 150 Kg

    weight of water level displaced is equal to the weight of person

    ρ V g = 2 m g

     [tex]V = \dfrac{2 m}{\rho}[/tex]

     [tex]V = \dfrac{2\times 150}{1000}[/tex]

            V = 0.3 m³

Area of pool = [tex]\dfrac{\pi}{4}d^2[/tex]

                    =[tex]\dfrac{\pi}{4}\times 6^2[/tex]

                    = 28.27 m²

height of the water rise

        [tex]h = \dfrac{V}{A}[/tex]

        [tex]h = \dfrac{0.3}{28.27}[/tex]

               h = 0.0106 m

pressure increased

           P = ρ g h

           P = 1000 x 10 x 0.0106

           P = 106.103 Pa

a. the absolute pressure at the bottom of the pool is approximately 116.02 kPa.

b. the pressure increase at the bottom of the pool after the persons enter and float is approximately 0.147 kPa.

To find the absolute pressure at the bottom of the pool, we can use the hydrostatic pressure formula:

(a) Absolute pressure at a depth in a fluid:

P=P_0 +ρ⋅g⋅h

Where:

P is the absolute pressure at the given depth.

P_0 is the atmospheric pressure (which we'll assume to be 1 atm, or approximately 101.3 kPa).

ρ is the density of the fluid (water), which is about 1000 kg/m³.

g is the acceleration due to gravity, approximately 9.81 m/s².

h is the depth below the surface.

Given that the diameter of the pool is 6.00 m, the radius (r) is 3.00 m, and the depth (h) is 1.50 m, we can find the pressure at the bottom of the pool:

P=101.3kPa+(1000kg/m³)⋅(9.81m/s²)⋅(1.50m)

Solve for P:

P≈101.3kPa+14,715Pa≈116,015Pa≈116.02kPa

So, the absolute pressure at the bottom of the pool is approximately 116.02 kPa.

(b) To find the pressure increase at the bottom of the pool after two persons with a combined mass of 150 kg enter and float, we need to consider the additional weight of the water displaced by the persons. This additional weight will create an increase in pressure at the bottom of the pool.

The buoyant force (Fb) on an object submerged in a fluid is given by Archimedes' principle:

Fb =ρ⋅V⋅g

Where:

Fb is the buoyant force.

ρ is the density of the fluid (water), which is 1000 kg/m³.

V is the volume of water displaced, which is equal to the volume of the submerged part of the persons.

g is the acceleration due to gravity (9.81 m/s²).

The volume of water displaced is equal to the volume of the persons submerged, which is given by the cross-sectional area (A) of the pool base times the depth (h) to which they are submerged:

V=A⋅h

A=π⋅r^2

Given that the radius (r) is 3.00 m and the depth (h) is 1.50 m, we can calculate the area and then the volume:

A=π⋅(3.00m)^2 ≈28.27m²

V=28.27m²⋅1.50m≈42.41m³

Now, we can calculate the buoyant force:

Fb=(1000kg/m³)⋅(42.41m³)⋅(9.81m/s²)≈416,068N

Since pressure is force per unit area, we need to calculate the additional pressure (ΔP) at the bottom of the pool:

ΔP= Fb/A

​A=π⋅(3.00m)^2 ≈28.27m²

ΔP= 416,068N / 28.27m² ≈14,699Pa≈0.147kPa

So, the pressure increase at the bottom of the pool after the persons enter and float is approximately 0.147 kPa.

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Answer:

1002.2688 m/s

Explanation:

g = Acceleration due to gravity = 9.81 m/s²

h = The length of a string = 2 m

m = Mass of block = 1.6 kg

[tex]m_2[/tex] = Mass of bullet = 0.01 kg

Here, the potential energy of the fall will balance the kinetic energy of the bullet

[tex]mgh=\dfrac{1}{2}mv^2\\\Rightarrow v=\sqrt{2gh}\\\Rightarrow v=\sqrt{2\times 9.81\times 2}\\\Rightarrow v=6.26418\ m/s[/tex]

Velocity of block is 6.26418 m/s

As the momentum of system is conserved we have

[tex]mv=m_2u\\\Rightarrow u=\dfrac{mv}{m_2}\\\Rightarrow u=\dfrac{1.6\times 6.26418}{0.01}\\\Rightarrow u=1002.2688\ m/s[/tex]

The magnitude of velocity just before hitting the block is 1002.2688 m/s

Answer:

10.791 m/s

5.93505 m

Explanation:

m = Mass of ball

[tex]v_f[/tex] = Final velocity

[tex]v_i[/tex] = Initial velocity

[tex]t_f[/tex] = Final time

[tex]t_i[/tex] = Initial time

g = Acceleration due to gravity = 9.81 m/s²

From the momentum principle we have

[tex]\Delta P=F\Delta t[/tex]

Force

[tex]F=mg[/tex]

So,

[tex]m(v_f-v_i)=mg(t_f-t_i)\\\Rightarrow v_i=v_f-g(t_f-t_i)\\\Rightarrow v_i=0-(-9.81)(1.1-0)\\\Rightarrow v_i=10.791\ m/s[/tex]

The speed that the ball had just after it left the hand is 10.791 m/s

As the energy of the system is conserved

[tex]K_i=U\\\Rightarrow \dfrac{1}{2}mv_i^2=mgh\\\Rightarrow h=\dfrac{v_i^2}{2g}\\\Rightarrow h=\dfrac{10.791^2}{2\times 9.81}\\\Rightarrow h=5.93505\ m[/tex]

The maximum height above your hand reached by the ball is 5.93505 m

The speed at which the ball had just after it left hand is 10.791 m/s and maximum height above your hand reached by the ball is 5.94 m.

When the two objects collides, then the initial collision of the two body is equal to the final collision of two bodies by the principal of momentum.

The net force using the momentum principle can be given as,

[tex]F_{net}=\dfrac{\Delta P}{\Delta t}[/tex]

Momentum of an object is the force of speed of it in motion. Momentum of a moving body is the product of mass times velocity. Therefore, the above equation for initial and final velocity and time can be written as,

[tex]F_{net}=\dfrac{m(v_f-v_i)}{(t_f-t_i)}[/tex]

The mass of the ball is 1 kg, and it takes 2.2 s to go up and down. The net force acting on the body is mass times gravitational force.

Therefore, the above equation can be written as,

[tex]mg=\dfrac{m(v_f-v_i)}{(t_f-t_i)}\\g=\dfrac{(v_f-v_i)}{(t_f-t_i)}\\[/tex]

As the final velocity is zero and initial time is also zero. Therefore,

[tex](-9.81)=\dfrac{(0-v_i)}{(1.1-0)}\\v_i=10.791 \rm m/s[/tex]

Using the energy principle, we can equate the kinetic energy of the system to the potential energy of the system as,

[tex]\dfrac{1}{2}mv_i^2=mgh\\\dfrac{1}{2}(10.791)^2=(9.81)h\\h=5.94\rm m[/tex]

Thus, the speed at which the ball had just after it left hand is 10.791 m/s and maximum height above your hand reached by the ball is 5.94 m.

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Answer:

Explanation:

Answer:

She can swing 1.0 m high.

Explanation:

Hi there!

The mechanical energy of Jane (ME) can be calculated by adding her gravitational potential (PE) plus her kinetic energy (KE).

The kinetic energy is calculated as follows:

KE = 1/2 · m · v²

And the potential energy:

PE = m · g · h

Where:

m = mass of Jane.

v = velocity.

g = acceleration due to gravity (9.8 m/s²).

h = height.

Then:

ME = KE + PE

Initially, Jane is running on the surface on which we assume that the gravitational potential energy of Jane is zero (the height is zero). Then:

ME = KE + PE      (PE = 0)

ME = KE

ME = 1/2 · m · (4.5 m/s)²

ME = m · 10.125 m²/s²

When Jane reaches the maximum height, its velocity is zero (all the kinetic energy was converted into potential energy). Then, the mechanical energy will be:

ME = KE + PE      (KE = 0)

ME = PE

ME = m · 9.8 m/s² · h

Then, equallizing both expressions of ME and solving for h:

m · 10.125 m²/s² =  m · 9.8 m/s² · h

10.125 m²/s² / 9.8 m/s²  = h

h = 1.0 m

She can swing 1.0 m high (if we neglect dissipative forces such as air resistance).

Final answer:

The work done on the refrigerator during its trip down the ramp is 1350 J.

Explanation:

To calculate the work done while unloading the refrigerator, we need to determine the displacement of the refrigerator and the force applied.

The displacement is given by the length of the ramp, which is 5.0 m.

The force applied is given as 270 N.

To calculate the work done, we use the formula:

Work = Force x Displacement x Cosine(angle)

In this case, the angle is 0 degrees since the force is applied horizontally.

Therefore, the work done on the refrigerator is:

Work = 270 N x 5.0 m x Cos(0°) = 1350 J.

Answer:

w = 7.89 10⁻² rad/s

Explanation:

We will solve this exercise with the conservation of the annular moment, let's write it in two moments

Initial. With the insect in the center

      L₀ = I w₀

End with the bug on the edge

     [tex]L_{f}[/tex]= I w +  [tex]I_{bug}[/tex] w

The moments of inertia are

For a rod

       I = 1/3 M L²

For the insect, taken as a particle

       I = m L²

The system is formed by the rod and the insect, this is isolated, therefore the external torque is zero and the angular momentum is conserved

      L₀ =  [tex]L_{f}[/tex]

      I w₀ = I w + [tex]I_{bug}[/tex] w

      w = I / (I +  [tex]I_{bug}[/tex]) w₀

      w = I / (I + m L²) w₀

Let's calculate

      w = 1.43 10⁻³ / (1.43 10⁻³ + 5 10⁻³ 0.620²)²   0.185

      w = 1.43 10⁻³ / 3.352 10³ 0.185

      w = 7.89 10⁻² rad/s