PHYSICS COLLEGE

A disk of mass 5 kg and radius 1m is rotating about its center. A lump of clay of mass 3kg is dropped onto the disk at a radius of 0.5m , sticking to the disk. If the system is rotating with an angular velocity of 11 rad/s, what is the final angular momentum of the disk h the clay lump?wit? ( Idisk = MR^2/2)

Answers

Answer 1

Answer:

Answer:

$27.5 kgm^2/s$

Explanation:

We can solve for the final angular velocity of the system using the law of momentum conservation

$I_1\omega_1 = I_2\omega_2 = M_2$

Where $I_1 = MR^2/2 = 5*1^2/2 = 2.5 kgm^2$ is the moments of inertia of the disk before. $I_2 = I_1 + mr^2 = 2.5 + 3*0.5^2 = 2.5 + 0.75 = 3.25 kgm^2$ is the moments of inertia of the disk after (if we treat the clay as a point particle). $\omega_1 = 11rad/s$ is the angular speed before.

$2.5*11 = M_2$

$M_2 = 27.5 kgm^2/s$

So the final momentum of the system is 27.5 kgm2/s

Answer 2

Answer:

Answer:

The final angular momentum is 35.75 kg.m²/s

Explanation:

Given;

mass of disk, M = 5 kg

radius of disk, R = 1 m

mass of clay, M = 3 kg

radius of clay, R = 0.5 m

final angular momentum, $\omega _f$ = 11 rad/s

Final angular momentum angular momentum of the disk that the clay lumped with;

$P = I_f\omega_f$

where;

$I_f$ is the final moment of inertia

$I_f = I_(disk) + I _(sand)\n\nI_f = (M_DR^2)/(2) + M_SR^2\n\nI_f = (5*1^2)/(2)+ 3*0.5^2\n\nI_f = 2.5 + 0.75=3.25 \ kg.m^2$

Final angular momentum of the disk;

= $I_f \omega_f$

= 3.25 x 11 = 35.75 kg.m²/s

Therefore, the final angular momentum is 35.75 kg.m²/s

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Write as ordinary number 3 x 10^0

Answers

Answer:

Explanation:

10⁰ = 1 because anything to the power of 0 is 1.

3×1= 3

Box 1 and box 2 are whirling around a shaft with a constant angular velocity of magnitude ω. Box 1 is at a distance d from the central axis, and box 2 is at a distance 2d from the axis. You may ignore the mass of the strings and neglect the effect of gravity. Express your answer in terms of d, ω, m1 and m2, the masses of box 1 and 2. (a) Calculate TB, the tension in string B (the string connecting box 1 and box 2). (b) Calculate TA, the tension in string A (the string connecting box 1 and the shaft).

Answers

Answer:

a) TB = m2 * w^2 * 2*d

b) TA = m1 * w^2 * d + m2 * w^2 * 2*d

Explanation:

The tension on the strings will be equal to the centripetal force acting on the boxes.

The centripetal force is related to the centripetal acceleration:

f = m * a

The centripetal acceleration is related to the radius of rotation and the tangential speed:

a = v^2 / d

f = m * v^2 / d

The tangential speed is:

v = w * d

Then

f = m * w^2 * d

For the string connecting boxes 1 and 2:

TB = m2 * w^2 * 2*d

For the string connecting box 1 to the shaft

TA = m1 * w^2 * d + m2 * w^2 * 2*d

Those who support the alien theory believe that the ancient Egyptians where very primitive. What evidence about the ancient Egyptians suggests that claim is inaccurate?

Answers

Answer:

They had a well-developed agricultural system , they raised domesticated animals , they developed a writing system, and they used early forms of tools.

Final answer:

The claim that the ancient Egyptians were primitive and relied on aliens to build their monuments is inaccurate. Evidence from archaeology and history shows that the ancient Egyptians had advanced knowledge and skills in various fields. Their construction techniques and use of mathematics in building the pyramids are well-documented.

Explanation:

The claim that the ancient Egyptians were primitive and that their accomplishments, such as building the pyramids, were assisted by aliens is inaccurate. There is evidence from archaeology and history that ancient monuments were built by ancient people using their own ingenuity and capabilities. The ancient Egyptians had advanced knowledge in various fields including architecture, engineering, astronomy, and mathematics. For example, their construction techniques and use of mathematics in building the pyramids and other structures are well-documented.

Learn more about Ancient Egyptians

saad has mass 80kg when resting on the ground at the equator what will be the centripetal acceleration on saad if the radius of earth is 6.4×10^6 meter

Answers

I did try to solve. I hope it is correct, below is the solution:

put everything in s.i units
then the answer what u wrote is acceleration to get is divide by mass(80)G=6.011*10^-11
M=6*10^24
R=6.4*10^6
m=80

Hope it helps.

A 500 W heating coil designed to operate from 110 V is made of Nichrome 0.500 mm in diametera.Assuming the resistivity of the nichrome remains constant at is 20.0 degrees C value find the length of wire used.b. Now consider the variation of resistivity with temperature. What power is delivered to the coil of part (a) when it is warmed to 1200 degrees C.?

Answers

(a) Length of the wire is 3.162 m

(b)Power delivered to the coil is 339.7 W

Electrical Power:

The electrical power is given by

P = V² / R

R = V² / P

Resistance of the heating coil, R

R = (110² / 500)

R = 12100 / 500

R = 24.2 Ω

Now the resistivity of a wire is given by

ρ= RA/L

here ρ = 1.50×10⁻⁶ Ωm

so after rearranging we get:

L = RA / ρ

Now, the radius of wirer = 0.5 / 2 mm = 0.25 mm = 2.5×10⁻⁴ m

So the cross sectional area can be calculated as follows

$A = \pi r^2\n\nA = \pi * (2.5*10^(-4))^2\n\nA = 1.96*10^(-7) m^2$

hence,

$L = (24.2 *1.96*10^(-7) / 1.50*10^(-6)) \n\nL = 3.162\; m$

(b)The dependency of resistance with temperature is as follows:

R = R₀[1 + αΔT]

α = $4*10^(-4)^\;oC^(-1)$ for Nichrome

$R' = R [1 + \alpha (1200 - 20) ]\n\nR' = R[1 + \alpha (1180) ]\n\nR' = 24.2[ 1 + 4*10^(-4) * 1180 ]\n\nR' = 24.2[1 + 0.472]\n\nR' = 24.2 * 1.472\n\nR' = 35.62 \;\Omega$

So the power generated is :

P = V² / R

P = (110² / 35.62)

P = 12100/ 35.62

P = 339.70 watts

Learn more about electrical power:

brainly.com/question/26174188

Answer:

a) 3.162 m

b) 339.7 W

Explanation:

Assume ρ = 1.50*10^-6 Ωm, and

α = 4.000 10-4(°C)−1 for Nichrome

To solve this, we would use the formula

P = V² / R

So when we rearrange and make R subject of formula, we have

R = V² / P

Resistance of the heating coil, R

R = (110² / 500)

R = 12100 / 500

R = 24.2 ohms

Recall the formula for resistivity of a wire

R = ρ.L/A

Again, in rearranging and making L subject of formula, we have

L = R.A / ρ

To make it uniform, we convert our radius from mm to m.

Diameter, D = 0.5 mm

Radius of wire = 0.5 / 2 mm = 0.25 mm = 0.00025 m

We then use this radius to find our area

A = πr²

A = π * 0.00025²

A = 1.96*10^-7 m²

And finally, we solve for L

L = (24.2 * 1.96*10^-7 / 1.50*10^-6) =

L = 3.162 m

(b)

Temperature coefficient of resistance.

R₁₂₀₀ = R₂₀[1 + α(1200 - 20.0) ]

R₁₂₀₀ = R₂₀[1 + α(1180) ]

R₁₂₀₀ = 24.2[ 1 + 4.*10^-4 * 1180 ]

R₁₂₀₀ = 24.2[1 + 0.472]

R₁₂₀₀ = 24.2 * 1.472

R₁₂₀₀ = 35.62 ohms

Putting this value of R in the first formula from part a, we have

P = V² / R

P = (110² / 35.62)

P = 12100/ 35.62

P = 339.70 watts

a trcuk weighs four times as much as a stationary car. if teh truck coasts into the car at 12 km/s and they stick toegther, what is their final velocity