5. A uniform solid ring of mass M = 10 kg and radius R = 2 m is initially rotating in mid-air with angular speed 20 rad/s [the Before state in figure (I) below]. It is then placed in contact with a surface (assume that the surface immediately feels the full weight of the ring). (1) M Before Wi (II) wf V Ө ə Ө After Ө wf H ? (a) What kind of frictional force acts on the ring upon contact with the surface? (b) If the ring begins to roll smoothly to the right with speed v = 10 m/s after 5 seconds of contact (the After state), find the coefficient of friction corresponding to the frictional force you mentioned in (a). (c) What is the increase in the thermal energy of the ring/surface system from right before the ring contacts the surface, to when the ring starts to roll smoothly? The ring then rolls smoothly up a ramp of 0 = π/6 rad and H = 5 m [see figure (II)] (d) What is the horizontal distance, from the end of the ramp, at which the ring lands?

The solenoid's inductance and the average emf around the solenoid if the current changes from +3.50 A to −3.50 A in 7.83 ✕ 10−3 s will be 1.01 ×10⁻⁸ H and 9.02 × 10⁻² V.

A coil of wire that carries an electric current is a solenoid. A solenoid is an electromagnet formed by a helical coil of wire.

Which generates a magnetic field when an electric current is passed through the coil.

A solenoid is a form of coil that produces a magnetic field when the electric current is passed through it. A solenoid is created when a conductive wire is used to make a loop.

Given data;

Turns ,N = 465

Length,L= 8.00 cm

Cross-sectional area,A = 3.10 ✕ 10−9 m2.

Solenoid's inductance, L=?

The average emf around the solenoid, E=?

Time of flow,t= 7.83 ✕ 10−3 s

Current changes from +3.50 A to −3.50 A

The inductance of the solenoid is found as;

[tex]\rm L = \frac{\mu_0 AN^2}{L} \\\\ \rm L = \frac{4 \times \pi \times 10^{-7}\times 3.0 \times 10^{-9} \times (465)^2 }{8.00 \times 10^{-2}} \\\\ L= 1.01 \times 10^{-8} \ H[/tex]

The average emf around the solenoid is found as;

[tex]\rm e = L \frac{I_2-I_1}{t} \\\\ \rm e = 1.01 \times 10^{-8} \times \frac{3.50-(-3.50)}{7.83 \times 10^{-3}} \\\\ e =9.02 \times 10^{-12} \ V[/tex]

Hence, the solenoid's inductance will be 1.01× 10⁻⁸ H.

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The velocity of the car at the end of the initial ten-second interval is 15a₂ + 30.

v = u + at

after 10 seconds;

v = 30 + 10a₁  --(1)

after 5 seconds

v₂ = v + 5a₂ ---- (2)

solve (1) and (2) together

v₂ = 30 + 10a₁ + 5a₂

but a₁/a₂=1.5

a₁ = 1.5a₂

v₂ = 30 + 10(1.5a₂) + 5a₂

v₂ = 30 + 15a₂ + 5a₂

v₂ = 30 + 20a₂

from equation (2), v₂ = v + 5a₂

v + 5a₂ = 30 + 20a₂

v = 15a₂ + 30

Thus, the velocity of the car at the end of the initial ten-second interval is 15a₂ + 30.

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Answer:

4.2m/s

Explanation:

0.50x8.4=4.2m/s

The output voltage of a transformer is 40V.

An output voltage is the voltage released by a device, such as a voltage regulator or a generator.

The formula to calculate Output voltage of transformer is:

[tex]\frac{V_{s} }{V_{p} } = \frac{N_{s} }{N_{p} }[/tex]

here,  [tex]V_{s}[/tex] is the secondary voltage, [tex]V_{p}[/tex] is the primary voltage, [tex]N_{s}[/tex] is the number of secondary windings and [tex]N_{p}[/tex] is the number of primary windings.

According to the question.

[tex]N_{p}[/tex] =1200 , [tex]N_{s}[/tex]=200 , [tex]V_{p}[/tex]=240V  ,[tex]V_{s}[/tex]= ?

By using the formula of Output voltage,

[tex]{V_{s} } = \frac{V_{p}N_{s} }{N_{p} }[/tex]

=[tex]\frac{240*200}{1200}[/tex]

[tex]=40V[/tex]

Therefore, 40V is the output voltage of a transformer.

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Evidence might be used to demonstrate their falsity. must apply to all potential solutions to a scientific topic.Option A is correct.

A scientific hypothesis is the main element in the scientific method and experiment.

It is said that a hypothesis is a scientific hypothesis. The majority must agree on it, or it must be done after thorough experimentation.

Only scientific inquiries can have their answers provided by scientific endeavor. Scientific concerns have an impact on the research's design as well.

Evidence might be used to demonstrate their falsity. must apply to all potential solutions to a scientific question.

Hence option A is correct.

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1. The total mass of the rod is 61.48 kg.

2. The x-coordinate of the center of the mass for this rod is  172.6021 m.

The density is the ratio of the mass and the volume of the object. It is denoted by ρ.

ρ = mass/Volume = m/V

Given a long thin solid rod lies along the positive x-axis. One end is at x = 1.80 m and the other at x = 3.40 m. The linear mass density is

λ = ax³ + bx,

where λ is measured in kg/m, and the constants have the following values: a = 1.70 kg/m⁴ and b = 2.20 kg/m²

So, λ = 1.70x³ + 2.20x

1. The total mass of the rod is

m =₁.₈³⁴ ∫ [1.70x³ + 2.20x] dx

m = [ 1.70x⁴/4 + 2.20x² /2 ]₁.₈³⁴

Putting the limits, we have

m = 69.51 - 8.02548

m = 61.48 kg

Thus, the total mass of the rod is 61.48 kg

2. The x-coordinate of the center of the mass for this rod

Xcm =∫ xλdx/m

Xcm =∫[ x ( 1.70x³ + 2.20x) /m ]dx

Xcm =∫[( 1.70x⁴ + 2.20x²) /m ]dx

On integrating and putting the limits  x = 1.80 m to x = 3.40 m

Xcm = = [ {1.70(3.4)⁵/5 + 2.20(3.4)³ /3} - {1.70(1.8)⁵/5 + 2.20(1.8)³ /3} ]

Xcm = 183.3034 - 10.70133

Xcm = 172.6021 m

Thus,  the x-coordinate of the center of the mass for this rod is  172.6021 m.

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(a)The value of W for the given condition will be 237.3 N.

(b)The distance must the fulcrum be moved to restore balance will be 0.070 m.

A location is established in relation to an object or set of objects in the center of mass. It is the system's average position across all of its components.

Given data;

Weight is applied to the right end of the rod, W₁= 225 N

Rod's weight,W₂ = 255

W₃ is the Weight at "x" distance from the left end.

x1  is the location of W₁ with respect to the left end, where L = 1.80 m

m is the location of the rod's center of mass from the left end.

The position of W from the left end; x₃ = 50 cm = 0.50 m

Xcm is the position of the center of gravity of the rod at the fulcrum from the left end;

Given that the fulcrum is balanced in this situation, the rod's center of gravity will be there.

Xcm= L - 75 cm

Xcm=1.80 m - 0.75 m

Xcm= 1.05 m

The value of x₂

x₂= 1.80 m/2 = 0.90 m

Currently, the horizontal center of gravity is determined by

[tex]\rm X_{cm} =\frac{ (W_1x_1 + W_2x_2 + W_3*x_3)}{(W_1 + W_2 + W_3)}[/tex]

Substitute the values we get;

[tex]\rm 1.05 = \fraca{ (225 \times 1.80 + 255\times 0.90 + W\times 0.50)}{(225 + 255 + W)}\\\\ W = \frac{ (225\times 1.80 + 255\times 0.90 - 480 \times 1.05)}{(1.05 - 0.50)}\\\\ W = 237.3 \ N[/tex]

After moving the fulcrum 20.0 cm to the right,

The location of the W from the left end;

x₃ = .50 cm + 20 cm

x₃  = 70 cm

x₃  = 0.70 m

W₁, W₂, W, x₁, and x₂ have fixed values. Therefore, based on these values, the new center of gravity will be:

[tex]\rm X_{cm} =\frac{ (W_1x_1 + W_2x_2 + W_3*x_3)}{(W_1 + W_2 + W_3)}[/tex]

[tex]\rm X_{cm} = \frac{(225\times 1.80 + 255\times 0.90 + 237.3\times 0.70)}{(225 + 255 + 237.3)}\\\\ X_{cm} = 1.12 m[/tex]

Fulcrum must be shifted by;

Y = (1.12 - 1.05)

Y= 0.070 meters.

W has been moved in the appropriate direction, thus the fulcrum should likewise be moved in the appropriate direction.

Hence, the value of W for the given condition will be 237.3 N. and the distance must the fulcrum be moved to restore balance will be 0.070 m.

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In infrared satellite images, land appears black and clouds appear white. C

Infrared satellite images are images produced as a result of electromagnetic radiations reflected or emitted from a target surface in the infrared position of the electromagnetic spectrum.

On infrared images, the clouds emit very bright white color because of the low temperature and the land appears to be black because of the hot temperature.

Therefore, in infrared satellite images, land appears black and clouds appear white. C

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Answer:

  19.6 m/s

Explanation:

The velocity is the product of acceleration and time.

__

Here, the acceleration is due to gravity.

  v = at

  v = (9.8 m/s²)(2 s) = 19.6 m/s

The stone hits the ground with a speed of 19.6 m/s.

A stone is dropped from the top of a tower which touches the ground after 2 seconds.

Hence, The stone hit the ground with the velocity speed of 19m/s².

- When you're solving velocity, you're determining how fast an object moves from its original position, with respect to a frame of reference, and a function of time.

- That means an object's velocity will be equal to the object's speed and direction of motion.

"Problem has been solve"

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Answer:

liquid turns into vapor when the liquid molecules achieves the energy to break down the form or when gets highly excited. The temp required for this is called latent heat of vaporization. As the liquid molecules attain this energy through heat it suddenly changes form into vapor , and when the heat is continued , molecule after molecule receives this energy and changes to vapor .So if you continue heating the liquid all the liquid will change form to vapor state leaving residue , ie the dissolved particles .

liquid turns into vapor when the liquid molecules achieves the energy to break down the form or when gets highly excited. The temp required for this is called latent heat of vaporization. As the liquid molecules attain this energy through heat it suddenly changes form into vapor , and when the heat is continued , molecule after molecule receives this energy and changes to vapor .So if you continue heating the liquid all the liquid will change form to vapor state leaving residue , ie the dissolved particles .I’m no expert but this is what i think happens …

The angle at which it should be banked for a traffic speed of 100 km/h is

14.57°.

To find the angle of banking following formula can be used,

[tex]tan \alpha = v^{2} /rg[/tex]

Given v = 100km/h or

          = 100  x 1000 / 3600

          = 27.7 m/s

Hence,

[tex]tan \alpha = (27.7)^{2} / 300 * 9.81[/tex]

[tex]tan\alpha = 0.26[/tex]

[tex]\alpha = tan^{-1} 0.26[/tex]

[tex]\alpha = 14.57[/tex]°

Hence, the angle of banking of the road will be 14.57°

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The total time the ball is in the air is 1.47 s and the maximum horizontal distance covered is 58.7 m.

The time of motion of the ball is calculated as follows;

h = vt + ¹/₂gt²

55 = (50sin37)t + ¹/₂(9.8)t²

55 = 30.1t + 4.9t²

4.9t² + 30.1t - 55 = 0

Solve using formula method;

t = 1.47 s

X = vxt

where;

X = (50 x cos37) x 1.47

X = 58.7 m

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Answer:

kg

Explanation:

the highr and jebad to kilogram mizans

Answer:

w=m×g

Explanation:

w is given and take g as 9.8m/s^2

Answer:

cost=$15

Explanation:

given;

P= 2100 W

t= 12'= 720 sec=0.2h

V= 120 V

required;

Energy?

cost?

solution;

Energy(kWh) = power*time(in hour)

E=2100*0.2= 420 kWh

cost= 420 kWh* $0.25/kWh

cost= $15

since blow dryer use 120-V

Magnetic stripes provide evidence of seafloor spreading due to their occurrence in the region of sea floor.

When the Earth's magnetic field reverses, a new stripe starts to form. Such magnetic patterns enable the scientist to recognize the occurrence of sea-floor spreading and provides strongest evidence for the theory of plate tectonics.

So we can conclude that magnetic stripes provide evidence of seafloor spreading due to their occurrence in the region of sea floor.

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Answer:

337k

Explanation:

First, let us find the difference between the given two temperatures.

Difference = 85°C - 21°C

                  = 64°C

And now we have to write the temperature in kelvins.

To convert Celcius to Kelvins you can add 273 to the temperature in Celcius.

Let us find it now.

64°C + 273 = 337k

Therefore,

64°C ⇒ 337k

Answer:

it alters the rate of chemical reaction

hope it is useful

The velocity of the third piece is 124.02 m/s at 1.05⁰ below the horizontal.

The velocity of the third piece is calculated from the principle of conservation of linear momentum.

mu = m₁u₁ + m₂u₂ + m₃u₃

where;

6(22)sin(0) = 2(10)sin32 - 1(8)sin(28) + 3u₃y

0 = 6.84 + 3u₃y

u₃y = -6.84/3

u₃y = -2.28 m/s

ΔP = Pf - Pi = J

where;

2640(0.1) = 2(10)cos32 +  1(8)cos(28) + 3u₃x - 6(22)

264 = -108 + 3u₃x

3u₃x = 372

u₃x = 372/3

u₃x = 124 m/s

u₃ = √(124² + 2.28²)

u₃ = 124.02 m/s

tanθ = u₃y/u₃x

tanθ = 2.28/124

tanθ = 0.018

θ = 1.05⁰ (below the horizontal)

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The work done by the force on the box 588 Joule.

It is the energy that is stored in an object due to its position relative to some zero position.

According to the question,

Given mass of the box (m)  = 6.0 kg

Given displacement of the box =  10.00 m

Hence , the height of the box  (h) = 10.00 m

Acceleration due to gravity (g)  =  9.8 m/s²

Since ,

The upward force was applied to the box, the box moves up to the height.

Therefore, the work done by the ball is the potential energy gained by the ball.

By using the formula of potential energy

P.E. = mgh

= 6 × 9.8 × 10 Joule  = 588 Joule

As we know .

Work done (W) = Potential energy (P.E.)

Work done (W) = 588 Joule

Thus,  588 Joule is the work done by the force on the box.

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The rod is stretched due to the force of gravitational attraction.

The magnitude of the deformation force on the rod is  -GMm×((1/r² + 1/(L+r)²)

The force of attraction felt by a person which is directed at the center of a planet or Earth is called as the gravity.

The force of attraction is directly proportional to the product of masses of the object and inversely proportional to the square of distance between them.

F = GMm/R²

Given are two small (green) objects, each of mass m, are separated by a solid, massless rod of length L. They are located so that one of the objects is located at a distance of r away from the center of a uniform spherical planet with mass M.  Assume that m is very small so that you can ignore the gravitational force between the two small (green) objects.

Let the mass closer to the planet is A and the other is B

FA = Force on planet A = -GMm/r²

FB = Force on planet B = -GMm/(L+r)²

Net Force on the rod is given by the addition of the individual forces.

Fnet= FA + FB = -GMm/r² -GMm/(L+r)²

Fnet = -GMm×((1/r² + 1/(L+r)²)

Thus,  the magnitude of the deformation force on the rod is derived above.

A experience more force than B, so A will stretch out from B. Hence the force is stretching the rod.

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Hello!

Recall the period of an orbit is how long it takes the satellite to make a complete orbit around the earth. Essentially, this is the same as 'time' in the distance = speed * time equation. For an orbit, we can define these quantities:

[tex]d = 2\pi r[/tex] ← The circumference of the orbit

speed = orbital speed, we will solve for this later

time = period

Therefore:

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

Where 'r' is the orbital radius of the satellite.

First, let's solve for 'v' assuming a uniform orbit using the equation:
[tex]v = \sqrt{\frac{Gm}{r}}[/tex]

G = Gravitational Constant (6.67 × 10⁻¹¹ Nm²/kg²)

m = mass of the earth (5.98 × 10²⁴ kg)

r = radius of orbit (1.276 × 10⁷ m)

Plug in the givens:
[tex]v = \sqrt{\frac{(6.67*10^{-11})(5.98*10^{24})}{(1.276*10^7)}} = 5590.983 m/s[/tex]

Now, we can solve for the period:

[tex]T = \frac{2\pi (1.276*10^7)}{5590.983} =\boxed{ 14339.776 s}[/tex]

The displacement is determined to be 5.11 m after every direction is plotted on the page.

Displacement is defined as the shortest distance between the two points.

The steps in which Jerry walks;

5m to East

6m to N 60' E,

16m to N30'W,

Turn

3√3 m to the south.

Each and every direction is plotted on the page and obtained that the displacement is found as 5.11 m.

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[tex]\large \bold {ANSWER}[/tex]

Considering that the pulley is fixed, the force applied should be equal to the weight of the object - of 400N.

[tex]\large \bold {EXPLANATION}[/tex]

Pulleys or pulleys are mechanical tools used to assist in the movement of objects and bodies. There are two types of pulleys: fixed and movable. While the fixed pulley changes the direction of force, the moving pulley helps to decrease the force needed to move the object or body in question.

As the statement only tells us a pulley, we must consider that it is fixed, because generally when it is mobile, this information is highlighted in the question.

In this way, a fixed pulley only changes the direction of the applied force. Thus, the force must have the same magnitude as the weight of the object to be moved. If the bucket weighs 400N, the force applied to the pulley must be 400N.

Therefore, having a fixed pulley, the force applied must be equal to the weight of the object, and will be 400N.

The initial speed of the object with mass "m" is 4.79 m/s and the initial speed of the object with mass "2m" is 1.81 m/s.

The initial speed of the two masses is calculated by applying the principle of conservation of linear momentum as follows;

m₁u₁ + m₂u₂ = v(m₁ + m₂)

[tex]m_1u_1sin(\theta_1) + m_2u_2sin(\theta_2) = (m_1 + m_2)vsin(\theta_3)\\\\mu_1sin(0) + 2mu_2sin(90) = (m + 2m)(2)sin(37)\\\\2mu_2 = 3.61m\\\\u_2 = \frac{3.61 m}{2m} \\\\u_2 = 1.81 \ m/s[/tex]

[tex]m_1u_1sin(\theta_1) + m_2u_2sin(\theta_2) = (m_1 + m_2)vsin(\theta_3)\\\\mu_1cos(0) + 2mu_2cos(90) = (m + 2m)(2)cos(37)\\\\mu_1 = 4.79 m\\\\u_1 = 4.79 \ m/s[/tex]

Thus, the initial speed of the object with mass "m" is 4.79 m/s and the initial speed of the object with mass "2m" is 1.81 m/s.

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The drag coefficient (C) is determined from terminal velocity, density of the medium, projected area and mass of the object.

Terminal velocity is the constant speed of a freely falling object when the resistance of the medium through which it is falling prevents further acceleration.

The magnitude of terminal velocity is calculated as;

[tex]V_T = \sqrt{\frac{2mg}{\rho A C} }[/tex]

where;

Thus, the drag coefficient (C) is determined from terminal velocity, density of the medium, projected area and mass of the object.

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The forces a piano tuner applies to stretch the steel piano wire willl be 1.4 × 10¹¹ N.

Force is defined as the push or pulls applied to the body. Sometimes it is used to change the shape, size, and direction of the body. Force is defined as the product of mass and acceleration. Its unit is Newton.

The application of a force may be used to describe the steel as an elastic element within a certain range of applied force.

Given data;

Young modulus, E=2.10 × 10¹¹ N/m²

Cross-sectional area,A

Final length,[tex]\rm L_f = 1.45 m = 1450 \ mm[/tex]

Initial length,[tex]\rm L_i = 7.5 mm[/tex]

[tex]\rm F = \frac{(L_f-L_i)(E)(A)}{L_1} \\\\ \rm F = \frac{(1450 - 7.5)(2.0 \times 0^{11})(0.746)}{1450} \\\\ F = 1.4 \times 10^{11} \ N[/tex]

Hence, the force a piano tuner applies to stretch the steel piano wire willl be 1.4 × 10¹¹ N.

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Answer:

Case 1: Balancing length when cell is shunted l1 = 150 cm = 1.5 m, Value of shunt = R = 5 ohm. Case 2: Balancing length when cell is shunted l1 = 175 cm = 1.75 m, Value of shunt = R = 10 ohm. Example 06: A cell balances against a length of 250 cm on potentiometer wire when it is shunted by a resistance of 10 ohms