An open-delta three-phase transformer system has one transformer center-tapped to provide a neutral for single-phase voltages. If the voltage from line to center tap is 277 V, what is the high-leg voltage?

Answer:

The initial angular acceleration is [tex]16.8 s^{-2}[/tex]

Explanation:

From the parallel axis theorem the moment of inertia about 25cm mark is

[tex]I = \dfrac{1}{12}ML^2+MD^2[/tex]

since [tex]L = 1m[/tex] & [tex]D = 0.25m[/tex], we have

[tex]I = \dfrac{1}{12}M(1)^2+M(0.25m)^2[/tex]

[tex]I = \dfrac{7}{48} M[/tex]

Now, the gravitational force (equal to the weight of the object) acts as a torque [tex]\tau[/tex] on the center of mass of the rod, which induces angular acceleration [tex]\alpha[/tex]according to

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

since [tex]\tau = Mg D[/tex]

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

[tex]MgD = \dfrac{7}{48} M \alpha[/tex]

solving for [tex]\alpha[/tex] we get

[tex]\boxed{\alpha = \dfrac{48}{7} gD}[/tex]

putting in [tex]g= 9.8m/s^2[/tex] and [tex]D = 0.25m[/tex] we get:

[tex]\boxed{\alpha = 16.8\: s^{-2}.}[/tex]

which is the initial angular acceleration.

The initial angular acceleration of the stick is 16.812 radians per square second.

First, we calculate the resulting moment of inertia by Steiner theorem:

[tex]I_{O} = I_{g} + M\cdot r^{2}[/tex] (1)

Where:

If we know that [tex]I_{g} = \frac{1}{12} \cdot M\cdot L^{2}[/tex], [tex]L = 1\,m[/tex] and [tex]r = 0.25\,m[/tex], then the formula for the moment of inertia is:

[tex]I_{O} = \frac{1}{12}\cdot M + \frac{1}{16}\cdot M[/tex]

[tex]I_{O} = \frac{7}{48}\cdot M[/tex] (2)

We know that initial angular acceleration ([tex]\alpha[/tex]), in radians per square second, is solely due to gravity. By the second Newton's law and the D'Alembert principle, we derive an expression for the initial angular acceleration of the stick:

[tex]\Sigma M = M\cdot g\cdot r = I_{O}\cdot \alpha[/tex] (3)

Where:

By (2) and (3) we have the following simplified expression:

[tex]M\cdot g \cdot r = \frac{7}{48} \cdot M\cdot \alpha[/tex]

[tex]\alpha = \frac{48}{7}\cdot g \cdot r[/tex] (4)

If we know that [tex]g = 9.807\,\frac{m}{s^{2}}[/tex] and [tex]r = 0.25\,m[/tex], then the initial angular acceleration of the stick is:

[tex]\alpha = \frac{48}{7}\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (0.25\,m)[/tex]

[tex]\alpha = 16.812\,\frac{rad}{s^{2}}[/tex]

The initial angular acceleration of the stick is 16.812 radians per square second.

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Answer:

a. yes

Explanation:

The initial speed of the circular saw is:

[tex]\dot n_{o} = \frac{6000}{60} \,\frac{rev}{s}[/tex]

[tex]\dot n_{o} = 100\,\frac{rev}{s}[/tex]

Deceleration rate needed to stop the circular saw is:

[tex]\ddot n = -\frac{100\,\frac{rev}{s} }{2\,s}[/tex]

[tex]\ddot n = - 50\,\frac{rev}{s^{2}}[/tex]

The number of turns associated with such deceleration rate is:

[tex]\Delta n = \frac{\dot n^{2}}{2\cdot \ddot n}[/tex]

[tex]\Delta n = \frac{\left(100\,\frac{rev}{s} \right)^{2}}{2\cdot \left(50\,\frac{rev}{s^{2}} \right)}[/tex]

[tex]\Delta n = 100\,rev[/tex]

Since the measured number of revolutions is lesser than calculated number of revolution, the circular saw meets specifications.

Answer:

3241.35J

Explanation:

No. Of rods = 5

Mass = 2.89kg

Length (L) = 0.827m

W = 507rpm

Kinetic energy of rotation = ½I*ω²

For each rod, the moment of inertia (I) = ML² / 3

I = ML² / 3

I = [2.89*(0.827)²] / 3

I = 1.367 / 3 = 0.46kgm²

ω = 507 rev/min. Convert rev/min to rev/sec.

507 * 2Πrads/60s = 53.09rad/s

ω = 53.09rad/s

k.e = ½ I * ω²

K.E = ½ * 0.46 * (53.09)²

K.E = 648.27.

But there five (5) rods, so kinetic energy is equal to

K.E = 5 * 648.27 = 3241.35J

The rotational kinetic energy of a propeller with five blades, each modeled as a uniform rod, can be determined using formulas for moment of inertia and rotational kinetic energy.

The kinetic energy of rotating objects depends on their moment of inertia and angular velocity. Given that each propeller blade is modeled as a uniform rod rotating about its end, we can calculate the moment of inertia (I) of all five blades using the formula I=5*(1/3*m*L^2), where m is the mass and L is the length of each rod. After we find the moment of inertia, we can determine the angular velocity (ω) in radian per second, given that the propeller rotates at 507 rpm, by using the conversion factor ω=(507*2*pi)/60. Finally, we calculate the rotational kinetic energy (K) using the formula K=1/2*I*ω^2.

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Answer:

Potential difference across resistor will be 87.66 volt

Explanation:

We have given number of electrons [tex]n=4.1\times 10^{21}[/tex]

Charge on one electron [tex]e=1.6\times 10^{-19}C[/tex]

So total charge [tex]Q=4.1\times 10^{21}\times 1.6\times 10^{-19}=656C[/tex]

Time is given t = 5 min

1 minute = 60 sec

So 5 minute = 5×60 = 300 sec

So current [tex]i=\frac{Q}{t}=\frac{656}{300}=2.1866A[/tex]

Resistance is given R = 40 ohm

Sp from ohm's law potential difference across resistor v = iR = 2.1866×40 = 87.466 volt

The potential difference across the 40 Ω resistor when 4.1 × 1021 electrons have passed through it over a time period of 5 minutes is calculated using Ohm's law, which results in 87.6 Volts.

The subject of this question is based on Ohm's law, an important concept in Physics. Ohm's law states that the current through a conductor between two points is directly proportional to the voltage across the two points.

First, calculate the total charge (Q) passed through the resistor. This can be done by multiplying the number of electrons by the fundamental charge of one electron (Q = electrons x electron charge). Thus, Q = 4.1 × 1021 electrons x 1.602 × 10-19 C = 656.82 Coulombs.

Next, use the formula to obtain the electric current (I) passed through the resistor. Since current (in Amperes) is equal to the total charge (in Coulombs) divided by the time (in seconds) and time here is given in minutes, we first need to convert minutes to seconds. So, 5 min = 5 x 60 = 300 seconds. Hence, I = Q / t = 656.82 C / 300 s = 2.19 Ampere.

Lastly, we can now find the potential difference (V) across the resistor. According to Ohm's law, V = I * R. Thus, V = 2.19 Ampere * 40 Ω = 87.6 Volts.

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Answer:

8 minutes

Explanation:

The time it takes for light to travel from the Sun to Earth can be found using the formula time = distance/speed. Given the speed of light is known to be 300,000 km/s and the distance is 150,000,000 km, it will take light about 500 seconds to travel from the Sun to the Earth.

To calculate the time it takes for light to travel from the Sun to Earth, you need to use the formula for time which is distance/speed. The speed of light is 300,000 km/s and the distance from the Sun to Earth is 150,000,000 km.

So, Time = Distance / Speed = 150,000,000 km / 300,000 km/s = 500 seconds

Therefore, it takes approximately 500 seconds for light to travel from the Sun to Earth.

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Answer: Alloy A is best chosen for the sheet forming operation

Explanation:

taking both alloy A and B into consideration;

For alloy A;

the true strain ration, R₀ = width strain / thickness strain

   R₀ = in (0.65/0.75) / in (0.092/0.1) = 1.716

at 45⁰  and 90⁰, the true strain ratio becomes;

R45⁰ = in (0.63/0.75) / in (0.093/0.1) = 2.4025

R90⁰ = in (0.67/0.75) / in (0.087/0.1) = 0.8099

where R(avg) = (R₀+2Ras+Rao) / A = (1.716+2(2.4025)+0.8099) = 3.0340

R(avg) = 3.0340

For alloy B;

R₀ = in (0.7/0.75) / in (0.082/0.1) = 0.3477

R45⁰ = in (0.69/0.75) / in (0.083/0.1)  = 0.4475

R90⁰ = in (0.71/0.75) / in (0.078/0.1) = 0.2206

R(avg) = (R₀+2Ras+Rao) / A = (0.3477+2(0.4475)+0.2206) = 0.365825

R(avg) = 0.365825

comparing both we have that,

R(avg) for allow A > R(avg) for allow B

∴ Alloy A is the best to be selected for sheet formation operation.

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Answer:

47.76°

Explanation:

Magnitude of dipole moment = 0.0243J/T

Magnetic Field = 57.5mT

kinetic energy = 0.458mJ

∇U = -∇K

Uf - Ui = -0.458mJ

Ui - Uf = 0.458mJ

(-μBcosθi) - (-μBcosθf) = 0.458mJ

rearranging the equation,

(μBcosθf) - (μBcosθi) = 0.458mJ

μB * (cosθf - cosθi) = 0.458mJ

θf is at 0° because the dipole moment is aligned with the magnetic field.

μB * (cos 0 - cos θi) = 0.458mJ

but cos 0 = 1

(0.0243 * 0.0575) (1 - cos θi) = 0.458*10⁻³

1 - cos θi = 0.458*10⁻³ / 1.397*10⁻³

1 - cos θi = 0.3278

collect like terms

cosθi = 0.6722

θ = cos⁻ 0.6722

θ = 47.76°

Answer:

The initial angle between the dipole moment and the magnetic field is 47.76⁰

Explanation:

Given;

magnitude of dipole moment, μ = 0.0243 J/T

magnitude of magnetic field, B = 57.5 mT

change in kinetic energy, ΔKE = 0.458 mJ

ΔKE = - ΔU

ΔKE = - (U₂ -U₁)

ΔKE = U₁ - U₂

U₁ -U₂ = 0.458 mJ

[tex](-\mu Bcos \theta_i )- (-\mu Bcos \theta_f) = 0.458 mJ\\\\-\mu Bcos \theta_i + \mu Bcos \theta_f = 0.458 mJ\\\\\mu Bcos \theta_f -\mu Bcos \theta_i = 0.458 mJ\\\\\mu B(cos \theta_f - cos \theta_i ) = 0.458 mJ[/tex]

where;

θ₁ is the initial angle between the dipole moment and the magnetic field

[tex]\theta_f[/tex] is the final angle which is zero (0) since the  dipole moment is aligned with the magnetic field

μB(cos0 - cosθ₁) = 0.458 mJ

Substitute the given values of μ and B

0.0243 x 0.0575 (1 - cosθ₁) = 0.000458

0.00139725 (1 - cosθ₁) = 0.000458

(1 - cosθ₁) = 0.000458 / 0.00139725

(1 - cosθ₁) = 0.327787

cosθ₁ = 1 -  0.327787

cosθ₁ = 0.672213

θ₁ = cos⁻¹ (0.672213)

θ₁ = 47.76⁰

Thus, the initial angle between the dipole moment and the magnetic field is 47.76⁰

Answer:

5.213ft

Explanation:

Z² = x² + y²

x = √(z² - y²)

y = 52ft, dx = 6ft, z = 105ft, dz = ?

d(z² = x² + y²)

2zdz = 2xdx

dz = xdx/z

But x = √(z² - y²)

dz = √(z² - y²)/z * dx

dz = [√(105² - 52²)/105] * 6

dz = √(8521)/ 17.5

dz = 5.213ft

Answer:in a spring tide the sun , moon and the earth are in a straight line which causes regular higher tide and low tide to increase.

Explanation:

In a spring tide, the Moon, Earth, and Sun are all in allignment, which causes the greatest gravitational pull on the water of Earth. Basically, when the Moon and Sun align towards and in the same direction at the Earth, their gravitational pulls join together making the pull stronger. (see image below)

What this very strong pull does is make high tides higher than usual, and low tides lower than usual. This is because the pull on tides is very strong and more water is being pulled to form higher high tides, making there less water for there to be in low tides.

(image is from oceanservice.gov)

Answer:

T1 = 112.07[lb]

T2 = 487.3 [lb]

Explanation:

To solve this problem we must perform a static balance analysis, for this we perform a free body diagram. In this free body diagram we use the angles mentioned in the description of the problem.

Performing a sum of forces on the X-axis equal to zero, we can find an equation that relates the tension of the T1 & T2 cables.

Then we perform a summation of forces on the Y-axis, in which we can find another equation. In this new equation, we replace the previous one and we can find the tension T2.

T1 = 112.07[lb]

T2 = 487.3 [lb]

Answer:

contaminated drinking water

deaths of sea creatures that are used as a food source

limits to potential economic activities such as a fishing

Explanation:

A watershed is a large area that comprises of drainage area of all the surrounding water bodies meeting at a common affluence point before draining into sea or ocean or any other large water body. Pollution in this area can pollute the small water streams flowing through it, thereby polluting the larger water body into which it drains.  

Thus, the water extracted for drinking from such area will be contaminated. Pollution in larger water body can cause death of water creature and hence pose a threat to fishing.  

Answer:

loss of areas for tourism

contaminated drinking water

deaths of sea creatures that are used as a food source

limits to potential economic activities such as fishing

The number of revolutions is 301

Explanation:

Angular acceleration, α = 3.5 rad/s

Time, t = 9s

Number of revolutions, n = ?

We know,

[tex]\alpha = \frac{w}{t}[/tex]

where,

ω = angular velocity

t = time

It can also be written as:

[tex]\alpha = \frac{\pi n}{t X 30}[/tex]

On substituting the value, we get:

[tex]3.5 = \frac{\pi X n}{9 X 30} \\\\n = 300.8[/tex]

Therefore, the number of revolutions is 301

The  answer is 22 complete revolutions.

The number of complete revolutions the disk makes in 9 seconds is given by the formula for angular displacement due to constant angular acceleration:

[tex]\[ \theta = \theta_0 + \omega_0 t + \frac{1}{2} \alpha t^2 \][/tex]

where:

-[tex]\( \theta \)[/tex] is the final angular displacement in radians,

-[tex]\( \theta_0 \)[/tex] is the initial angular displacement, which is zero since the disk starts from rest,

-[tex]\( \omega_0 \)[/tex] is the initial angular velocity, which is also zero since the disk starts from rest,

- [tex]\( \alpha \)[/tex] is the angular acceleration, which is 3.5 [tex]rad/s\( ^2 \)[/tex],

- [tex]\( t \)[/tex] is the time in seconds, which is 9 seconds.

Since the disk starts from rest, [tex]\( \theta_0 = 0 \)[/tex] and [tex]\( \omega_0 = 0 \)[/tex], the equation simplifies to:

[tex]\[ \theta = \frac{1}{2} \alpha t^2 \][/tex]

Substituting the given values:

[tex]\[ \theta = \frac{1}{2} \times 3.5 \, \text{rad/s}^2 \times (9 \, \text{s})^2 \][/tex]

[tex]\[ \theta = \frac{1}{2} \times 3.5 \times 81 \][/tex]

[tex]\[ \theta = 1.75 \times 81 \][/tex]

\[ \theta = 141.75 \, \text{radians} \]

To find the number of complete revolutions, we need to convert the angular displacement from radians to revolutions. Since there are [tex]\( 2\pi \)[/tex] radians in one revolution, we divide the angular displacement by [tex]\( 2\pi \)[/tex]:

[tex]\[ \text{Number of revolutions} = \frac{\theta}{2\pi} \][/tex]

[tex]\[ \text{Number of revolutions} = \frac{141.75}{2\pi} \][/tex]

[tex]\[ \text{Number of revolutions} = \frac{141.75}{6.2831853071796} \][/tex]

[tex]\[ \text{Number of revolutions} \approx 22.57 \][/tex]

Answer:

Check the explanation

Explanation:

Atomization, also called the spraying technique or procedure, refers to a process in which molten metals are broken down into little or tiny drops of liquid through the use of high-speed fluids (liquid as water, gas as air or inert gas) or fluids with centrifugal force, and then solidified into a powdered state.

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Answer:

C. 2.3A

Explanation:

V/Ω=A

9V / 4Ω = 2.25 ≅ 2.3A

The current flowing through the circuit is 2.3 A. So, option C is correct.

Ohm's law states that, the voltage applied across a circuit is directly proportional to the steady current flowing through the circuit and also proportional to the resistance of the circuit.

Here,

Voltage applied in the circuit, V = 9 V

Resistance of the bulb, R = 4 Ω

According to Ohm's law,

V [tex]\alpha[/tex] I

V [tex]\alpha[/tex] R

So, V = IR

Therefore, current flowing through the circuit,

I = V/R

I = 9/4

I = 2.25 A    approximately, 2.3 A

Hence,

The current flowing through the circuit is 2.3 A.

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Answer:

The separation distance between the two charges must be 82704.2925 m

Explanation:

Given:

Two negative charges that are both q = -3.8 C

Force of 19 N

Question: How far apart are the two charges, s = ?

First, you need to get the electrostatic force of this two negative charges:

[tex]F=\frac{kq}{s^{2} } \\s=\sqrt{\frac{kq}{F} }[/tex]

Here

k = electric constant of the medium = 9x10⁹N m²/C²

Substituting values:

[tex]s=\sqrt{\frac{9x10^{9}*(-3.8)^{2} }{19} } =82704.2925m[/tex]

Answer:

Explanation:

Young modulus ε = 1.4 × 10¹⁰ Pa

ΔL = 1% of the original length = 0.01 x where x is the original length

cross sectional area = 3.0 cm² =( 3 .0 / 10000) m²= 0.0003 m²

ε = Stress / strain

stress = ε × strain

stress = F /A

F force = ε × A × ( ΔL / L) =  1.4 × 10¹⁰ Pa × 0.0003 m² × 0.01 = 4.2 × 10⁴ N

b) F net = F max - mg ( weight) = 84000 - ( 70 × 9.8 m/s² ) ( F is double since the stress on the two leg is equally distributed)

f net = ma = 84000 - ( 70 × 9.8 m/s² )

a =  (84000 - ( 70 × 9.8 m/s² )) = 1190.2 m/s²

v = u + at

where final velocity equal zero

- u = -at since it coming downwards

u = at = 1190.2 m/s² × 0.03s = 35.706 m/s

using conservation of energy

1/2 mv² = mgh

1/2v²/ g = h

h = 0.5 × (35.706 m/s )² / 9.8 = 65.04 m

Answer:

The answer is -x direction.

Explanation:

According to the right hand rule for electromagnetism, when the thumb points up, index finger points forward and the middle finger is perpendicular to the index finger, the thumb points in the direction of the magnetic force, the index finger in the direction of the charge movement and the middle finger in the direction of the electromagnetic lines. If the electric field is pointing in the -z direction which is into the screen and the magnetic field is in the +y direction which is upwards, then the magnetic lines are in the -x direction.

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Answer:

The temperature change in Celsius is 46°C.

The temperature change in Fahrenheit is 82.8°F.

Explanation:

A degree of Celsius scale is equal to that of kelvin scale; therefore,

[tex]\Delta C = \Delta K = 283K-237K\\\\ \boxed{\Delta C = 46^oC}[/tex]

A degree in Fahrenheit is 1.8 times the Celsius degree; therefore

[tex]\Delta F = 1.8(46^o)[/tex]

[tex]\boxed{\Delta F = 82.8^oF}[/tex]

Hence, the temperature change in Celsius is 46°C, and the temperature change in Fahrenheit is 82.8°F.

Final answer:

The temperature change of 46 degrees from Kelvin to Celsius in Granville, North Dakota, is equivalent to a change of 82.8 degrees Fahrenheit.

Explanation:

On January 21, 1918, the temperature in Granville, North Dakota, changed from 237 K to 283 K within 12 hours. To convert the change in temperature to the Celsius scale, we first recognize that 0 degrees Celsius is equivalent to 273.15 K. Therefore, the initial temperature in Celsius would have been 237 K - 273.15 K = -36.15  extdegree C, and the final temperature would have been 283 K - 273.15 K = 9.85  extdegree C. The change in the Celsius scale is then 9.85  extdegree C - (-36.15  extdegree C) = 46  extdegree C

To convert the change in temperature to the Fahrenheit scale, we start with the change in Celsius and use the conversion formula (F = C  imes \frac{9}{5} + 32). Since we are looking for the change, we only need to convert the difference in Celsius to Fahrenheit. Therefore, the change in Fahrenheit is 46  extdegree C  imes \frac{9}{5} = 82.8  extdegree F.

Answer:

  h ’= 3.75 inch

Explanation:

For this exercise we use the definition of magnification which is the height of the image over the height of the object

                 m = h ’/ h

                  h’= m h

Let's calculate

               h ’= 5  3/4

              h ’= 3.75 inch

Velocity changes with time and this is called acceleration.

The question is incomplete but I will try to explain the meaning of acceleration to you. The term acceleration refers to the change of velocity with time.

a = Δv/t

Hence, velocity changes with time and this is called acceleration.

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The increase of the pressure of the system would lead to an increase of acceleration.

As pressure equals to the force upon action. Here F is the force and the A the area. As Newton's second law       F = ma  We substitute      P A = m a

Hence the pressure is directly proportional to acceleration.

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Answer:

[tex]v_{f} \approx 6.647\,\frac{m}{s}[/tex]

Explanation:

The final speed of the cyclist is determined by applying the Principle of Energy Conservation:

[tex]\frac{1}{2}\cdot m\cdot v_{o}^{2} + m\cdot g\cdot h_{o} = \frac{1}{2}\cdot m\cdot v_{f}^{2} + m\cdot g\cdot h_{f}[/tex]

[tex]\frac{1}{2} \cdot v_{o}^{2} + g\cdot (h_{o}-h_{f}) = \frac{1}{2}\cdot v_{f}^{2}[/tex]

[tex]v_{f}^{2}=v_{o}^{2} + 2\cdot g \cdot (h_{o}-h_{f})[/tex]

[tex]v_{f} = \sqrt{v_{o}^{2}+2\cdot g \cdot (h_{o}-h_{f})}[/tex]

[tex]v_{f} = \sqrt{v_{o}^{2}-2\cdot g \cdot \Delta s \cdot \sin \theta}[/tex]

[tex]v_{f} = \sqrt{(9\,\frac{m}{s} )^{2}-2\cdot (9.807\,\frac{m}{s} )\cdot (12\,m)\cdot \sin 9^{\textdegree}}[/tex]

[tex]v_{f} \approx 6.647\,\frac{m}{s}[/tex]

Answer:

the final speed of the 10.85 m/s.

Explanation:

Given that,

Slope with respect to horizontal, [tex]\theta=9^{\circ}[/tex]

Distance travelled, d = 12 m

Initial speed of the cyclist, u = 9 m/s

We need to find the final speed of the cyclist. Let h is the height of the sloping surface such that,

[tex]h=d\times sin\thetah\\\\=12\times sin(9) \\\\h = 1.877 m[/tex]

v is the final speed of the cyclist. It can be calculated using work energy theorem as

[tex]\dfrac{1}{2}m(v^2-u^2)\\\\=mgh\dfrac{1}{2}(v^2-u^2)\\\\=gh\dfrac{1}{2}\times (v^2-(9.0)^2)\\\\=9.8\times 1.87\\\\v = 10.85 m/s[/tex]

Thus,the final speed of the 10.85 m/s.

Answer:

[tex]W=173.48J[/tex]

Explanation:

information we know:

Total force: [tex]F=45N[/tex]

Weight: [tex]w=100N[/tex]

distance: [tex]4m[/tex]

vertical component of the force: [tex]F_{y}=12N[/tex]

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

In this case we need the formulas to calculate the components of the force (because to calculate the work we need the horizontal component of the force).

horizontal component: [tex]F_{x}=Fcos\theta[/tex]

vertical component: [tex]F_{y}=Fsen\theta[/tex]

but from the given information we know that [tex]F_{y}=12N[/tex]

so, equation these two [tex]F_{y}=Fsen\theta[/tex] and [tex]F_{y}=12N[/tex]

[tex]Fsen\theta =12N[/tex]

and we know the force [tex]F=45N[/tex], thus:

[tex]45sen\theta=12[/tex]

now we clear for [tex]\theta[/tex]

[tex]sen\theta =12/45\\\theta=sin^{-1}(12/45)\\\theta =15.466[/tex]

the angle to the horizontal is 15.466°, with this information we can calculate the horizontal component of the force:

[tex]F_{x}=Fcos\theta[/tex]

[tex]F_{x}=45cos(15.466)\\F_{x}=43.37N[/tex]

whith this horizontal component we calculate the work to move the crate a distance of 4 m:

[tex]W=F_{x}*D\\W=(43.37N)(4m)\\W=173.48J[/tex]

the work done is W=173.48J

To calculate the work done by the force in moving the crate, we use the formula Work = Force x Distance x cos(theta). We are given the force, distance, and the vertical component of the force, so we can find the horizontal component using trigonometry. Substituting the values into the formula gives us the work done as 174 J.

To calculate the work done by a force, we use the formula:

Work = Force x Distance x cos(theta)

In this case, the force is 45 N, the distance is 4 m, and the vertical component of the force is 12 N. We need to find the horizontal component of the force, which can be calculated using the given information.

Given that the vertical component of the force is 12 N, we can use the trigonometric relationship between the horizontal and vertical components of a force to find the horizontal component as follows:

sin(theta) = vertical component / total force

sin(theta) = 12 N / 45 N

sin(theta) = 0.267

theta = sin^(-1)(0.267)

theta = 15.1 degrees

Once we have the horizontal component of the force, we can calculate the work done as follows:

Work = Force x Distance x cos(theta)

Work = 45 N x 4 m x cos(15.1 degrees)

Work = 180 J x cos(15.1 degrees)

Work = 180 J x 0.966

Work = 174 J

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Question:

a.  It would decrease by a factor of 14 .

b. It would increase by a factor of 4.

c. It would remain unchanged.

d. It would increase by a factor of 2.

e. It would decrease by a factor of 12 .

Answer:

The correct option is;

d. It would increase by a factor of 2

Explanation:

Here we have

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

The formula for maximum mass is given by the following relation;

[tex]v_{max1} =A\sqrt{\frac{k}{m} }[/tex]

Where:

[tex]v_{max}[/tex] = Maximum speed of mass on spring

k = Spring constant

m = Mass attached to the spring

A = Amplitude of the oscillation

Therefore, when k is increased by a factor of 4 we have;

[tex]v_{max2} =A\sqrt{\frac{4 \times k}{m} } = 2 \times A\sqrt{\frac{ k}{m} }[/tex]

Therefore, [tex]v_{max2} = 2 \times v_{max1}[/tex], that is the velocity will increase by a factor of 2.

Answer:

a) P ( X ≤ 49 ) = 0.9149

b) P ( X ≥ 48 ) = 0.2119

c) 0.9876

Explanation:

Solution:-

- Lets define a random variable X: the maximum speed of a moped.

- The random variable follows normal distribution with the following parameters:

                    X ~ Norm ( u , σ^2 )

Where,

              u = Mean = 46.6 km/h

              σ = Standard deviation 1.75 km/h

Hence,

                    X ~ Norm ( 46.6 , 1.75^2 ).

a) The probability that the maximum speed of mopeds is atmost 49 km/h?

- To evaluate the probability of P ( X ≤ 49 ). We will find the standard Z-score value using the following formula:

                       [tex]Z-score = \frac{x-u}{s} \\\\ Z-score = \frac{49-46.6}{1.75} \\\\Z-score = 1.37142[/tex]

- Now use the standard normal tables to determine the required probability of:

                     P ( Z < 1.37142 ) = 0.9149

Hence,

                     P ( X ≤ 49 ) = 0.9149  

b) The probability that the maximum speed of mopeds is at-least 48 km/h?

- To evaluate the probability of P ( X ≥ 48 ). We will find the standard Z-score value using the following formula:

                       [tex]Z-score = \frac{x-u}{s} \\\\ Z-score = \frac{48-46.6}{1.75} \\\\Z-score = 0.8[/tex]

- Now use the standard normal tables to determine the required probability of:

                     P ( Z ≥ 0.8 ) = 0.2119

Hence,

                     P ( X ≥ 48 ) = 0.2119                          

c) The probability that the maximum speed of mopeds differs from mean by at-most 2.5 standard deviation.

- The required probability is the standard error from the mean value "u" of 2.5 standard deviation.

- We don't need to evaluate the test statistics as we are already given the standard error about mean.

- Using, the standard normal Z-score: The required probability is:

                P ( -2.5 < Z < 2.5 ) = 2*P ( Z < 2.5 ) - 1

                                              = 2*0.9938 - 1

                                              = 0.9876

Answer: 0.88 m/s

Explanation:

given

Force constant of the spring, k = 965 N/m

Compression of the spring, x = 3.7 cm = 0.037m

Mass of the block, m = 1.7 kg

To solve this, we would be using law of conservation of energy, where initial energy is all kinetic and elastic energy is elastic but potential.

E(f) = E(i)

1/2kx² = 1/2mv²

kx² = mv²

965 * 0.037² = 1.7 * v²

1.321 = 1.7 v²

v² = 1.321 / 1.7

v² = 0.777

v = 0.88 m/s

Thus, the initial speed of the block is v = 0.88 m/s

Answer: you'll see cyan color on the screen

Explanation:

Saturating the red cone causes them to stop functioning, hence you can't perceive the red part of white light. White light is made up of three main colors which are blue, red and green. When one can no longer perceive the red part of light, one is left with the grean and blue part. The green and blue part of light will superimpose to give a cyan color.

Answer:

Cyan color

Explanation:

When an individual stares at a red object and immediately look at a white area afterward, there will be an image formation almost immediately that is the same size and shape as the previous one. The only difference is there will be a color change to blue-green, or cyan.

This is due to the eyes generally using the red, blue and green cone cells to perceive white light. Since the red cone cells are fatigued due to long view of it and cone cells being saturated , the blue and green color which are the remaining color cone cells use to perceive white light is viewed. This gives rise to the blue green or cyan color.

Complete Question

The complete question is shown on the first uploaded image

Answer:

The net force is  [tex]F_{net}= 6.44 *10^{-4} N[/tex]

Explanation:

Generally the net force is a force that come up due to the unequal centripetal force(A difference in centripetal force ) and it is mathematically represented  as

                     [tex]F_{net} = \Delta F_{cen}[/tex]

and the difference in centripetal force  [tex]\Delta F_{cen}[/tex] is mathematically represented as

               [tex]\Delta F_{cen} = \Delta m* rw^2[/tex]

Which the difference in mass multiplied by the centripetal acceleration

  Substituting 10 mg = [tex]10 *10^{-3}g[/tex] for [tex]\Delta m[/tex] , 12 cm = [tex]\frac{12}{100} = 0.12m[/tex] for radius

  and  70,000 rpm = [tex]70,000 *[\frac{2 \pi rad}{1 rev}][\frac{1 min}{60s} ] = 7326.7 rad/s[/tex]

              [tex]F_{net} = \Delta F_{cen} = 10*10^{-3} * 0.12 * 7326.7[/tex]

                     [tex]F_{net}= 6.44 *10^{-4} N[/tex]

Answer:

Note that the emf induced is

emf = B d v cos (A)

---> v = emf / [B d cos (A)]

where

B = magnetic field

d = distance of two rails

v = constant speed

A = angle of rails with respect to the horizontal

Also, note that

I = emf/R

where R = resistance of the bar

Thus,

I = B d v cos (A) / R

Thus, the bar experiences a magnetic force of

F(B) = B I d = B^2 d^2 v cos (A) / R, horizontally, up the incline.

Thus, the component of this parallel to the incline is

F(B //) = F(B) cos(A) = B I d = B^2 d^2 v cos^2 (A) / R

As this is equal to the component of the weight parallel to the incline,

B^2 d^2 v cos^2 (A) / R = m g sin (A)

where m = the mass of the bar.

Solving for v,

v = [R m g sin (A) / B^2 d^2 cos^2 (A)]   [ANSWER, the constant speed, PART A]

******************************

v = [R m g sin (A) / B^2 d^2 cos^2 (A)]

Plugging in the units,

m/s = [ [ohm * kg * m/s^2] / [T^2 m^2] ]

Note that T = kg / (s * C), and ohm = J * s/C^2

Thus,

m/s = [ [J * s/C^2 * kg * m/s^2] / [(kg / (s * C))^2 m^2] ]

= [ [J * s/C^2 * kg * m/s^2] / [(kg^2 m^2) / (s^2 C^2)]

As J = kg*m^2/s^2, cancelling C^2,,

= [ [kg*m^2/s^2 * s * kg * m/s^2] / [(kg^2 m^2) / (s^2)]

Cancelling kg^2,

= [ [m^2/s^2 * s * m/s^2] / [(m^2) / (s^2)]

Cancelling m^2/s^2,

= [s * m/s^2]

Cancelling s,

=m/s   [DONE! WE SHOWED THE UNITS ARE CORRECT! ]

Explanation:

Given that,

Mass of the truck, m = 2268 kg

Speed of the truck, v = 30 m/s

The friction force has an average magnitude of 700 N, f = -700 N

(a) The initial kinetic energy of the truck is given by :

[tex]K=\dfrac{1}{2}mv^2\\\\K=\dfrac{1}{2}\times 2268\times (30)^2\\\\K=1.02\times 10^6\ J[/tex]

(b) Finally, the stops due to an effective friction force that the road, final speed is 0. Let d is the stopping distance of the truck. using third equation of motion to find it as :

[tex]v^2-u^2=2ad\\\\d=\dfrac{-u^2}{2a}[/tex]

Since, f = ma

[tex]d=\dfrac{-u^2m}{2f}\\\\d=\dfrac{-(30)^2\times 2268}{-2\times 700}\\\\d=1458\ m[/tex]

So, the stopping distance of the truck is 1458 meters.