A flatbed truck is supported by its four drive wheels, and is moving with an acceleration of 6.3 m/s2. For what value of the coefficient of static friction between the truck bed and a cabinet will the cabinet slip along the bed surface?

Answer:

0.624 kg

Explanation:

We are given that

Mass of one block,[tex]m_1=1.3 kg[/tex]

[tex]v_1=1.3 m/s[/tex]

[tex]v_2=5 m/s[/tex]

[tex]V=2.5 m/s[/tex]

We have to find the mass of second block.

Mass of second block,[tex]m_2=\frac{m_1V-m_1v_1}{v_2-V}[/tex]

Substitute the values

[tex]m_2=\frac{1.3\times 2.5-1.3\times 1.3}{5-2.5}[/tex]

[tex]m_2=0.624 kg[/tex]

Hence, the mass of second block=0.624Kg

Answer:

The mass of the second block is 0.624 kg.              

Explanation:

Given that,

Mass of the block 1, m₁ = 1.3 kg

Speed of block 1, u₁ = 1.3 m/s

Speed of block 2, u₂ = 5 m/s

The final velocity of the combined blocks is 2.5 m/s, V = 2.5 m/s

It is a case of inelastic collision. Using the conservation of linear momentum as :

[tex]m_1u_1+m_2u_2=(m_1+m_2)V\\\\1.3\times 1.3+5m_2=(1.3+m_2)2.5\\\\1.69+5m_2=3.25+2.5m_2\\\\m_2=0.624\ kg[/tex]

So, the mass of the second block is 0.624 kg.              

Answer:

0.685

Explanation:

the background-corrected light measurement at 29mm: 20.7mv- 5.1mv⇒15.6mv

at 32.5mm: 15.8mv- 5.1mv⇒ 10.7mv

So, Normalize the data to the maximum value probably means to set the peak value to unity and scale the remaining data by the same factors.

Therefore, the normalized corrected value at 32.5mm is ⇒ 10.7mv/15.6mv ⇒0.685

Answer:

The amplitude of the motion of the spring is 1 m.

Explanation:

From the conservation of energy;

[tex](PE_{spring})_{max}= PE+KE[/tex]

which is

[tex]\frac{1}{2} kx^2_{max}=\frac{1}{2} kx^2+\frac{1}{2} mv^2[/tex]

Make [tex]x_{max}[/tex] the subject of formula

[tex]x^2_{max}= \frac{kx^2\times mv^2}{k}[/tex]

[tex]x_{max}= \sqrt{\frac{kx^2\times mv^2}{k} }[/tex]

Substitute 320 N/m for k, 0.017 m for x, 3.7 kg for m, and 0.60 m/s for v.

[tex]x_{max}= \sqrt{\frac{(320)(0.017)^2+ (3.7)(0.60)^2}{320} } \\\\=1m[/tex]

Thus, The amplitude of the motion of the spring is 1 m.

Answer: 0.0667m

Explanation:

given

Mass of the object, m = 3.7kg

Force constant of the spring, k = 320 N/m

Speed of the object, v = 0.6 m/s

Distance from equilibrium position, x = 0.017 m

Using the law of conservation of energy. The total energy conserved is

= 1/2kx² + 1/2mv²

= 1/2 * 320 * 0.017² + 1/2 * 3.7 * 0.6²

= 0.046 + 0.666

= 0.712 J

When v = 0, the maximum deflection is Xmas. This Xmas, is also the amplitude. Such that,

Energy = 1/2kx²

0.712 = 1/2 * 320 * x²

1.424 = 320 x²

x² = 1.424 / 320

x² = 0.00445

x = 0.0667

x = 6.67*10^-2 m

Thus, the amplitude is 0.0667 m

Answer:

The distance of the star is [tex]3.40x10^{16}[/tex] meters

Explanation:

It is known that the speed of light has a value of [tex]3x10^{8}m/s[/tex] in vacuum. That is, it travels [tex]3x10^{8]m[/tex] in one second, according with the following equation:    

[tex]v = \frac{x}{t}[/tex]

Where v is the speed, x is the distance and t is the time.

[tex]x = v\cdot t[/tex] (1)

Equation 1 can be used to determine the distance that the light travels in 1 year:  

It is necessary to find how many seconds are in 1 year (365.25 days).

[tex]365.25 days \cdot \frac{86400s}{1 day}[/tex]  ⇒  [tex]31557600s[/tex]

[tex]x = (3x10^{8}m/s)(31557600s)[/tex]    

[tex]x = 9.46x10^{15}m[/tex]      

Therefore, in 1 year, light travels [tex]9.46x10^{15}[/tex] meters.

If the distance to a star is 3.6 light-years, what is this distance in meters?  

A simple conversion between units can be used to get the distance in meters

[tex]x_{star} = 3.6ly \cdot \frac{9.46x10^{15}m}{1ly}[/tex] ⇒ [tex]3.40x10^{16}m[/tex]

Hence, the distance of the star is [tex]3.40x10^{16}[/tex] meters.              

A light-year, a distance unit used in astronomy, is approximately 9.46 x [tex]10^{15[/tex] meters. Hence, a distance of 3.6 light years would convert to approximately 3.4 x [tex]10^{16[/tex] meters.

A light-year is a unit of distance used in astronomy, which represents the distance light travels in one year. Light travels at a speed of 186,000 miles per second. Over the course of a year, this distance adds up significantly.

To calculate a light-year in meters, we would carry out the following calculation - light travels at 299,792 kilometers per second. This converts to exactly 299,792,000 meters per second. If we multiply this by the number of seconds in a year (60 seconds/minute, 60 minutes/hour, 24 hours/day, 365.25 days/year) we find that one light-year is approximately 9.46 x [tex]10^{15[/tex]  meters.

Therefore, if the distance to a star is 3.6 light-years, then this would convert to about 3.4 x  [tex]10^{16[/tex]  meters.

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Answer:

The mass of the beam is = 29 kg.

Explanation:

A beam with mass 40 kg is shown in figure. Point S is the support point. Point B is the middle point on the beam where mass of the beam acts.

Taking moment about Point S

40 × 21 = [tex]M_{beam}[/tex] × 29

[tex]M_{beam}[/tex] = 29 kg

Therefore the mass of the beam is = 29 kg.

Answer:

0.25

Explanation:

Since there is a vertical displacement of 0.5cm from the crest to the trough, there is half of that displacement to the midline, which is also known as the amplitude. Therefore, the amplitude is 0.5/2=0.25cm. Hope this helps!

Explanation:

The dimension of a single rectangular loop is 0.46 m x 0.68 m.

Magnetic field, B = 3 T

The loop is inclined at an angle of 67 degrees with respect to the normal to the plane of the loop.

It is required to find the magnitude of the average emf induced in the loop if the magnetic field decreases to zero in a time of 0.49 s.

Te induced emf in the loop is given by :

[tex]\epsilon=\dfrac{-d\phi}{dt}\\\\\epsilon=\dfrac{-d(BA)}{dt}\\\\\epsilon=A\dfrac{dB}{dt}\cos\theta\\\\\epsilon=0.46\times 0.68\times \dfrac{3}{0.49}\times \cos(67)\\\\\epsilon=0.74\ V[/tex]

So, the magnitude of the average emf induced in the loop is 0.74 V.

The correct answer for the gravitational field of a point mass is [tex]g = -\dfrac{GM_1M_1}{r^2} * \dfrac{r}{m}[/tex]

The gravitational force between two masses is given by Newton's law of gravitation:

[tex]F = \dfrac{GM_1M_2}{r^2}[/tex]

[tex]F[/tex] is the gravitational force between [tex]M_1[/tex] and [tex]M_2[/tex]

[tex]G[/tex] is the gravitational constant,

[tex]r^2[/tex] is the square of the distance between two masses.

The gravitational force is a vector quantity, and it is directed along the line connecting the two masses.

[tex]g= \dfrac{F}{m}[/tex]

[tex]g[/tex] is the gravitational field.

[tex]F[/tex] is the gravitational force.

Let [tex]m[/tex] be the mass test object.

Substitute the value of gravitational force from Newton's law of gravity:

[tex]g = \dfrac{GM_1M_1}{r^2} * \dfrac{r}{m}[/tex]

[tex]r[/tex]is the unit vector pointing from the mass M to the test mass, which represents the direction of the gravitational field.

[tex]g = -\dfrac{GM_1M_1}{r^2} * \dfrac{r}{m}[/tex]

The negative sign indicates the direction of the gravitational force is attractive.

The gravitational field is [tex]g = -\dfrac{GM_1M_1}{r^2} * \dfrac{r}{m}[/tex]

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The gravitational field of a point mass is given by -GM/(r^2)*r^, where the negative sign indicates the attractive nature (opposite direction to r^) of gravity.

The gravitational field g of a point mass using the same reasoning as for the electric field would be analogous but with some slight modifications to account for the attractive nature of gravity, unlike the repulsive nature of electric charges of the same sign. The formula is given by -GM/(r^2)*r^, where G is the gravitational constant, M is the mass of the object, and r is the distance to the object. The negative sign indicates that the force is attractive and acts in the direction opposite to r (the vector pointing directly away from the mass).

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Answer: 1.91*10^8 N/m²

Explanation:

Given

Radius of the steel, R = 10 mm = 0.01 m

Length of the steel, L = 80 cm = 0.8 m

Force applied on the steel, F = 60 kN

Stress on the rod, = ?

Area of the rod, A = πr²

A = 3.142 * 0.01²

A = 0.0003142

Stress = Force applied on the steel/Area of the steel

Stress = F/A

Stress = 60*10^3 / 0.0003142

Stress = 1.91*10^8 N/m²

From the calculations above, we can therefore say, the stress on the rod is 1.91*10^8 N/m²

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

The chart shows the voltage of four electric currents. A two-column table with 2 row titled Voltage of Currents. The first column labeled Current has entries W, X, Y, Z. The second column labeled Volts (volts) has entries 9.0, 1.5, 3.0, 4.5.

Which is best supported by the data in the chart?

A. Current W flows at a higher rate than Current Z.

B. Current Y flows at a lower rate than Current X.

C. Current X has a lower potential difference than Current Y.

D. Current Z has a greater potential difference than Current W.

Answer: C. Current X has a lower potential difference than Current Y.

Explanation: The table described is show below:

Current                        Volts (volts)

   W                                      9.0

   X                                       1.5

   Y                                       3.0

   Z                                       4.5

Current is defined by the flow of charge that passes through a material, called conductor in an amount of time:

I = [tex]\frac{Q}{t}[/tex]

where:

Q is charge in Coulomb

t is time in seconds

I is the current in Ampere: A = C/s

According to the question, the currents are not defined by charge or numbers, so it's difficult to determine its value. So, alternatives A and B aren't correct.

Voltage is the difference in electrical potential between two points. It is because of the voltage that the charges flow.

From the table, Current W has the biggest voltage of all the others. So, alternative D is incorrect.

Analysing the voltages of Current X and Y, we observe that they are 1.5 and 3.0, respectively. So, volts or the potential diference of current X is lower than Y, making alternative C the correct one.

Answer:

Current X has a lower potential difference than Current Y.

Explanation:

Answer:

Light travels fast in air but i dont know about water

Answer:

light slower in water than in Air

Explanation:

Air's refractive index is about 1.0003, while water's is about 1.3. This means that light is “slower” in water than air. This is because it's more likely to hit a molecule and then get re-emitted, lengthening the amount of time the light takes to get through a certain distance of the medium.

Answer:

Explanation:

Given that,

One fragment is 7 times heavier than the other

Let one fragment mass be M

Let this has a velocity v

And the other 7M

And this a velocity V

Initially the fragment is at rest u = 0

Applying conservation of momentum

Momentum is given as p=mv

Initial momentum = final momentum

Po = Pf

(M+7M) × 0 = 7M •V − Mv

0 = 7M•V - Mv

Divide both sides by M

0 = 7V -v

v = 7V

Since friction decelerates the masses to zero speed, we can calculate the NET work on the individual blocks and relate this quantity to the change in kinetic energy of each block

The workdone by the 7M mass is

Distance moved by 7M mass is 6.8m, Then, d =6.8m

W = fr × d

Where fr = µkN

When N=W =mg, where m=7M

N= 7Mg

fr = −µk × 7mg

Then, W(7m) = −7µk•Mg×d

W(7m) = −7µk•Mg×6.8

W(7m) = −47.6 µk•Mg

Then, same procedure,

Let distance move by the small mass be m

Work done by M mass

W(m) = −µk•Mg×d'

Since it is a wordone by friction, that is why we have a negative sign.

Using conservation of energy

Work done by 7M mass is equal to work done by M mass

W(7m) = W(m)

−47.6 µk•Mg = −µk•Mg×d

Then, M, g and µk cancels out

We are left with

-46.7 = -d

Then, d = 46.7m

Answer:

Distance the lighter fragment slides before stopping= 333.2 m

Explanation:

Gravitational acceleration = g

Mass of the lighter fragment = m

Mass of the heavier fragment = 7m

Velocity of the lighter fragment after the explosion = V1

Velocity of the heavier fragment after the explosion = V2

The object is at rest before the explosion hence the total momentum of the system is zero.

mV1 + 7mV2 = 0

V1 = -7V2

Coefficient of friction = [tex]\mu[/tex]

Friction force on the lighter fragment = f1 =[tex]\mu mg[/tex]

Friction force on the heavier fragment = f2 =[tex]7\mu mg[/tex]

Deceleration of the lighter fragment due to friction = a1

ma1 = -f1

ma1 = [tex]-\mu mg[/tex]

a1 =[tex]-\mu g[/tex]

Deceleration of the heavier fragment due to friction = a2

7ma2 = -f2

7ma2 = -7[tex]\mu mg[/tex]

a2 = [tex]-\mu g[/tex]

Final velocity of the lighter fragment = V3 = 0 m/s

Final velocity of the heavier fragment = V4 = 0 m/s

Distance traveled by the lighter fragment before coming to rest = d1

Distance traveled by the heavier fragment before coming to rest= d2 =6.8 m

V4² = V2² + 2a₂d₂

[tex],[/tex]

[tex](0)^2 = V_2^2 + (2)(-\mu g)d_2[/tex]

[tex]d_2 = \frac{V_2^2}{2 \mu g}[/tex]

V₃² = V₁² + 2a₁d₁

[tex](0) = (-7V_2)^2 + 2(-\mu g)d_1[/tex]

[tex]d_1 = \frac{49V_2^2}{2 \mu g}[/tex]

d₁ = 49d₂

d₁ = (49)(6.8)

d₁ = 333.2 m

Distance the lighter fragment slides before stopping = 333.2 m

To calculate the electric-field magnitude at points inside and outside a uniformly charged insulating sphere, we apply Gauss's Law, considering the proportion of enclosed charge within a Gaussian surface for points inside, and treating the sphere as a point charge for points outside.

We apply Gauss's Law to understand the electric-field magnitude around a uniformly charged insulating sphere, which states that the electric flux through a closed surface is proportional to the enclosed charge. We'll calculate the electric field (E) at two distances from the centre of the sphere: 4.00 cm, which is inside the sphere, and 6.00 cm, which is outside the sphere.

For the inside point (r = 4.00 cm), we only consider the charge enclosed by a Gaussian surface of radius 4.00 cm. Using the fact that charge is distributed uniformly throughout the volume of the sphere, we calculate the enclosed charge (q_enclosed) by using the ratio of volumes: q_enclosed = (Q * (4/3 * π * r^3)) / (4/3 * π * R^3), where Q is the total charge of the sphere. R is the radius of the sphere.

The electric field inside the sphere at 4.00 cm is then E = (1 / (4 * π * ε_0)) * (q_enclosed / r^2). We substitute the values using the given radius and charge and solve for E.

For the outside point (r = 6.00 cm), we use the formula for the electric field outside a spherically symmetric charge distribution, which is E = (1 / (4 * π * ε_0)) * (Q / r^2), and solve for E using the sphere's total charge and the distance of 6.00 cm.

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Answer:

M = [tex]\frac{qrB^{2} }{E}[/tex]

Explanation:

considering the magnetic force in the second region derive a mathematical expression that equates the mass of the particle to other variables

In a magnetic field

q = charge, M = mass of particle, E = electric field,B= magnetic field

qvb = [tex]\frac{mv^{2} }{r}[/tex] = therefore  m = [tex]\frac{qrb}{v}[/tex]  (equation 1)

note : the particle passes through the region undeflected

therefore : qvb = qE  therefore (E = VB)  

 hence  v = [tex]\frac{E}{B}[/tex] ( equation 2 )

insert equation 2 into equation 1

m = [tex]\frac{qrB^{2} }{E}[/tex]

Answer:

a.)r=4RE/4

b.)r=2RE

c.)ZERO

Explanation:

a.) given v= 0.462 which is V= 0.466Ve

Since the projectile is shot directly away from earth surface the speed it escape with;

v=√2GM/RE......................eqn(1)

M is the Earth's Mass

RE is the Radius

From the Law of conservation of Energy

K+U=0...............................eqn(2)

K₁+U₁=K₂+U₂....................eqn(3)

where K₁ and U₁ is initial kinetic and potential energy

K₂ and U₂ are final kinetic and potential energy

Kinetic Energy (K.E) decrease with time as the projectile moves up and there is decrease in Potential Energy (P.E) , it will let to a point where K.E will turn to zero i.e K₂=0

U₂=K₁U₂ ..........................eqn(3)

From Gravitational Law

U₁= -GMm/RE ..................(5)

U₂= -GMm/r .....................(6)

Where "r" is the distance

v= 0.462√2GM/RE

v= √GM/2RE

GM/4RE - GM/RE = -GM/r

r= 4RE/3

b.) "r" is calculated by this equation;

K₁=0.466

K₁= 1/4MVe².............................eqn(9)

substitute eqn(1) into eqn (9) then

1/4m2GM/RE=0.466GM/2Re

GM/2RE - GM/RE =-GM/r

r=2RE

c.)The potential energy and kinetic energy is the same in terms of their size both in different directions, while the potential energy face outward, the kinetic energy face inward therefore the least initial mechanical energy

required at launch if the projectle is to escape is ZERO

Answer:

The number of revolutions of the rotor required to turn the probe is 118 revolutions

Explanation:

Given;

rotational inertia of the electric motor,  Im = 4.36 x 10⁻³ kg·m²

rotational inertia of the probe, Ip = 6.07 kg·m²

the angular position of the probe, θ = 30.6°

From the principle of conservation of angular momentum;

[tex]I_m \omega _m = I_p \omega _p \\\\Also;\\\\I_m \theta _m = I_p \theta _p[/tex]

where;

[tex]\omega _m[/tex] is the angular velocity of the electric motor

[tex]\omega _p[/tex] is the angular velocity of the probe

[tex]\theta _m[/tex] is the angular position of the electric motor

[tex]\theta _p[/tex] is the angular position of the probe

[tex]\theta _m = \frac{I_p \theta_p}{I_m} \\\\\theta _m = \frac{6.07* 30.6^o}{4.36*10^{-3}} = 42601.4^o[/tex]

360° = One revolution

42601.4° = ?

Divide 42601.4°  by 360°

= 118 revolutions

Therefore, the number of revolutions of the rotor required to turn the probe is 118 revolutions

Final answer:

The maximum rate at which the magnetic field strength is changing in a closed loop conductor can be determined using Faraday's law. The rate is equal to the maximum induced emf divided by the area of the loop.

Explanation:

The maximum rate at which the magnetic field strength is changing can be determined using Faraday's law of electromagnetic induction. According to Faraday's law, the induced emf (voltage) in a closed loop conductor is equal to the negative rate of change of magnetic flux through the loop. In this case, the magnetic field is perpendicular to the plane of the loop, so the magnetic flux is given by the product of the magnetic field strength and the area of the loop. Therefore, the maximum rate of change of magnetic field strength is equal to the maximum induced emf divided by the area of the loop.

So, the maximum rate at which the magnetic field strength is changing is given by: (d(B)/dt) = (Emax / A)

where (d(B)/dt) is the rate of change of magnetic field strength, Emax is the maximum induced emf, and A is the area of the loop.

Answer:

A

Explanation:

Peaks are in opposite direction because change in magnetic field at one end of the coil is opposite to the change in magnetic field at other end of the coil. Faraday's law predict that the direction of induced voltage is dependent on the nature of change in magnetic field

Answer:

Explanation:

An equilibrium is a state in which opposing forces or influences are banned.

An example of equilibrium is in economics when supply and demand are equal. An example of equilibrium is when you are calm and steady. An example of equilibrium is when hot air and cold air are entering the room at the same time so that the overall temperature of the room does not change at all.

Answer:

a state in which opposing forces or influences are balanced.

Explanation:

In a chemical reaction, chemical equilibrium is the state in which both reactants and products are present in concentrations which have no further tendency to change with time, so that there is no observable change in the properties of the system.

Answer:

The resultant vector A + B is given by 7.2 m at an angle of 26° east of north,

C

Explanation:

Resolving the vectors to vertical and horizontal component;

Vertical;

Vector A = 6sin30

Vector B = 4sin60

Resultant vertical = 6sin30 + 4sin60 = 6.464m

Horizontal;

Vector A = 6cos30

Vector B = -4cos60

Resultant horizontal = 6cos30 - 4cos60 = 3.196m

Resultant R = √(6.464^2 + 3.196^2) = 7.2m

Tanθ = 6.464/3.196

θ = taninverse (6.464/3.196) BN

θ = 64° north of East.

Or

26° east of north

The resultant vector A + B is given by 7.2 m at an angle of 26° east of north,

The resultant vector A + B is given by 7.2 m at an angle of 26° east of north. Hence, option (c) is correct.

Given data:

The magnitude of vector A is, A = 6.0 m.

The direction of vector A is, 30° north of east.

The magnitude of vector B is, B = 4.0 m.

The direction of vector B is, 30° west of north.

The quantity having both the magnitude as well as the magnitude are known as vector quantities.

Resolving the vectors to vertical and horizontal component;

Along the Vertical direction;

Vector A = 6sin30

Vector B = 4sin60

Resultant vertical vector = 6sin30 + 4sin60 = 6.464 m

Along the Horizontal direction;

Vector A = 6cos30

Vector B = -4cos60

Resultant horizontal vector = 6cos30 - 4cos60 = 3.196 m

Now, the resultant vector is calculated as,

[tex]R=\sqrt{6.464^{2}+3.196^{2}}\\\\R = 7.2 \;\rm m[/tex]

And the resultant direction of vectors is,

[tex]tan \theta = 6.464/3.196\\\\\theta = tan^{-1} (6.464/3.196) \\\theta =64 ^{\circ}[/tex]( North of East)

θ = 64° north of East.

Or

26° east of north

Thus, we can conclude that the  resultant vector A + B is given by 7.2 m at an angle of 26° east of north. Hence, option (c) is correct.

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Answer:

b)  the result we got can be termed approximation because we are neglecting the shear stress acting on the two ends of the cylinder. Here we have considered only the share stress acting on the curved surface area only.

Explanation:

check attachment for solution to A

The torque required to rotate the inner cylinder of a rotating viscometer with a fluid of viscosity μ can be calculated using fluid mechanics. However, this solution is approximated and based on a linear velocity profile in the gap, which can change with a significant increase in the gap.

In a rotating viscometer with a fluid of viscosity μ between two concentric cylinders, the torque required to rotate the inner cylinder at a constant speed can be computed using fluid mechanics principles. The shear stress on the fluid due to the rotating cylinder is the product of the viscosity and the velocity gradient across the gap, given by (ωi*ri)/ (ro - ri). This shear stress relates to the force acting on a unit area of the fluid as F/A, and the torque T is the product of the force and the radius of the inner cylinder. Following the equations, we'll obtain T = 2π * L * μ * ωi * (((ro^2 - ri^2) / ln (ro/ri)).

This solution is an approximation as it's based on the assumption that the velocity profile in the gap between the cylinders is linear, which holds true when the gap is thin. If the gap increases disproportionately (i.e., the outer radius R0 were to increase), the linear approximation of the velocity profile may no longer hold. In this scenario, edge effects and non-linear velocity profiles have to be considered, potentially modifying the expected torque value.

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Answer:

-2.2394181 J amount of kinetic energy is lost.

Explanation:

Let Vf be the final speed

(48.7) (9.8) = (48.7 + 1100) Vf

477.26 = 1148.7 Vf

=>   Vf = 0.4155 m/s

KE initiali = 1/2 (0.0487) (9.8)²

KE initial = 1/2 (0.0487) (96.04)

KE initial = 1/2 (4.677148)

KE initial = 2.338574 J

KE final = 1/2 (1.1 + 0.0487) (0.4155)²

KE final = 1/2 (1.1487‬) (0.17264025)

KE final = 1/2 (0.198311855175)

KE final = 0.0991559 J

KE (lost) = KEf - KEi

             = 0.0991559 J - 2.338574 J

             = -2.2394181 J

Answer:

The value of b 0.7351 kg/m

Explanation:

Given that;

Mass of sky diver = 82 kg

Velocity = 33.4 m/s

f = −bv²

It is a resistance force, therefore the negative sign is ignored.

since; f = mg

∴ mg = bv²

b = mg / v ²  ........ (1)

At terminal velocity a = 0

Put parameters in (1)

b = (82 × 10) / (33.4)²

b = 820 / 1,115.56

b = 0.7351

Answer:

The same in both the regions of constructive interference and the regions of destructive interference.

Explanation:

Interference is a phenomenon which occurs when two waves meet while moving along the same medium . The amplitude formed as a result of the interference could be greater, lower, or the same amplitude.

Constructive and destructive interference result from the interaction of waves that are correlated or coherent with each other. This is because arose from the same source or they have the same or nearly the same frequency.

The waves being coherent, arising from the same source and having the same frequency explains why it’s the same in both the regions of constructive interference and the regions of destructive interference.

Answer:

Explanation:

Constructive interference occurs when the maxima of two waves add together (the two waves are in phase), so that the amplitude of the resulting wave is equal to the sum of the individual amplitudes. Whereas in destructive interference opposite is true.

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Answer:

Answer is in the following attachment

Explanation:

Answer:

The answer is 12.67 TMU

Explanation:

Recall that,

worker’s eyes travel  distance must be = 20 in.

The perpendicular distance from her eyes to the line of travel is =24 in

What is the MTM-1 normal time in TMUs that should be allowed for the eye travel element = ?

Now,

We solve for the given problem.

Eye travel is = 15.2 * T/D

=15.2 * 20 in/24 in

so,

= 12.67 TMU

Therefore, the MTM -1 of normal time that should be allowed for the eye  travel element is = 12.67 TMU

Answer:

a. The energy transformed from mechanical to internal in the particle—hoop is 12.21 Joules

b. The total number of revolutions the particle makes before stopping is 1.43 revolutions

Explanation:

a.

Given

m = mass of particle = 0.350-kg

u = initial speed of 10.00 m/s

v = final speed = 5.50 m/s

r = radius = 0.600 m

We assume that the floor is horizontal;

This means that F = mg.

We also assume the rotational kinetic energy to be negligable.

Having listed the assumptions, we proceed as follows;

Let ∆E represent The energy transformed from mechanical to internal in the particle hoop.

This is given by

∆E = KE1 - KE2

Where KE1 = ½mu²

KE2 = ½mv²

So, ∆E = KE1 - KE2 becomes

∆E = ½mu² - ½mv²

∆E = ½m(u² - v²)

∆E = ½ * 0.350 * (10² - 5.5²)

∆E = 12.20625

∆E = 12.21J (Approximated)

Hence, the energy transformed from mechanical to internal in the particle—hoop is 12.21 Joules

b.

Let n = number of revolutions

The relationship between n and the energy is

1/n = (KE1 - KE2)/KE1

Make n the subject of formula

n = KE1 / (KE1 - KE2)

n = ½mu² / (½mu² - ½mv²) --- Simplify

n = ½mu² / (½m(u² - v²)) ---- Divide through by ½m

n = u² / (u² - v²)

n = 10² / (10² - 5.5²)

n = 1.433691756272401

n = 1.43 rev

Hence, the total number of revolutions the particle makes before stopping is 1.43 revolutions

Final answer:

The energy lost to friction in one revolution is 11.9875 J, and the particle makes a total of 2 revolutions before coming to a stop.

Explanation:

Energy Loss Due to Friction

The energy transformed from mechanical to internal in the particle—hoop—floor system as a result of friction in one revolution (a) is determined by the change in kinetic energy of the particle. The initial kinetic energy is given by ½ mv², where ‘m’ is the mass of the particle and ‘v’ is the initial velocity. The final kinetic energy is likewise given by ½ mv² using the final velocity. The difference in these energies gives the energy lost to friction.

Kinetic Energy Initial = ⅓ (0.350 kg)(10.00 m/s)² = 17.5 J
Kinetic Energy Final = ⅓ (0.350 kg)(5.50 m/s)² = 5.5125 J
Energy Transformed = Kinetic Energy Initial - Kinetic Energy Final = 17.5 J - 5.5125 J = 11.9875 J

Total Number of Revolutions

For part (b), to find the total number of revolutions the particle makes before stopping, assuming a constant frictional force, we can use a deceleration approach or an energy approach. The work done against friction for each subsequent revolution will be the same; thus each revolution results in the same amount of energy loss until the particle comes to a stop. By dividing the total initial kinetic energy by the energy lost per revolution, we find the number of revolutions.

Revolutions = Total Initial Kinetic Energy / Energy Transform per Revolution = 17.5 J / 11.9875 J ≈ 1.46 revolutions
Since it cannot complete a partial revolution after losing all kinetic energy, we round down to get 1 full revolution. Therefore, the particle makes a total of 1 + 1 = 2 revolutions before stopping.

Final answer:

A leading voltage across a device at 500 Hz compared to a resistor suggests inductive elements within the device. Voltage leads current in inductors, which can be determined by comparing phase differences using a voltmeter parallel to the device.

Explanation:

When measuring the voltage across a device and comparing it to the voltage across a resistor at a specific frequency, such as 500 Hz, the phase difference provides information about the type of elements in the device. If the voltage across the device leads the voltage across the resistor, this indicates the presence of inductive reactance, suggesting that there are inductive elements within the unknown circuitry.

Reactance and impedance play key roles in AC circuits. The voltage drop across a resistor is in phase with the current, meaning it neither leads nor lags the current. However, in an inductor, the voltage leads the current, and in a capacitor, the voltage lags the current. Therefore, measuring a leading voltage implies inductive characteristics. To evaluate this, a voltmeter would be placed in parallel with the device, ensuring minimal disturbance to the circuit as high resistance in the voltmeter minimizes current flow through it.

For components like inductors and capacitors, the relationship between voltage and current is not linear; thus, they are not ohmic devices like resistors. This is evident in an RLC series circuit, where the impedance at a given frequency depends on the resistance (R), inductance (L), and capacitance (C) of the circuit.

The voltage leading the resistor's voltage at 500 Hz suggests the device contains capacitive elements. A phase difference measurement can confirm this. Capacitors cause the voltage to lead the current in AC circuits.

If you observe that the voltage across a device leads the voltage across a resistor at 500 Hz, it is likely that the device includes a capacitive element. In AC circuits, capacitors cause the voltage to lead the current, which means that the voltage across the capacitor will peak before the voltage across a purely resistive element.

To confirm the presence of a capacitive component, you can perform a more detailed analysis:

This phase difference can be analyzed using an oscilloscope to visualize the waveform and confirm the capacitive nature of the device.

Answer:

Explanation:

The guy wire is making a right angled triangle  with the ground and stop sign . It makes an angle of 51 degree with the ground. In this triangle stop sign is the perpendicular and distance from the base on the ground forms the base of the triangle . Wire forms the hypotenuse.

base / hypotenuse = cos51

base = hypotenuse x cos51

= 8 x cos51

= 5.03 ft .

The distance of the stake with which guy wire was attached from the foot of the stop sign is 5.03 ft .

Given that,

Magnetic field strength is

B = 1.8T

The wedding ring has a diameter of

d = 2.23 cm = 0.023m

Time take t = 0.32 secs

A. Current induced

From ohms law

V= iR, given that R = 0.01Ω

So, we need to get the induced emf

Using

ε = -NdΦ / dt

Where Φ = BA

ε = -A ∆B / ∆t

ε = -¼πd²(B2-B1) / (t2-t1)

ε = -¼ × π × 0.023² × -1.8 / 0.32

ε = 0.0023371 V

Then

I = ε / R

I = 0.002337 / 0.01

I =0.2337 A

B. Power discippated?

Power is given as

P = iV

P = 0.2337 × 0.002337

P = 0.0005462 W

P = 0.56 mW

C. The magnetic field at the centre of the ring.

The electric field at the centre of the ring is zero because each part of the ring will cause a symmetrical opposite magnitude at every point,

Then, B = 0 T