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You have a grindstone (a disk) that is 105.00 kg, has a 0.297-m radius, and is turning at 71.150 rpm, and you press a steel axe against it with a radial force of 46.650 N. Assuming the kinetic coefficient of friction between steel and stone is 0.451. How many turns will the stone make before coming to rest?

Answers

Answer 1

Answer:

Answer:

3.27 turns

Explanation:

To find how many turns (θ) will the stone make before coming to rest we will use the following equation:

$\omega_(f)^(2) = \omega_(0)^(2) + 2\alpha*\theta$

Where:

$\omega_(f)$ : is the final angular velocity = 0

$\omega_(0)$ : is the initial angular velocity = 71.150 rpm

α: is the angular acceleration

First, we need to calculate the angular acceleration (α). To do that, we can use the following equation:

$\alpha = (\tau)/(I)$

Where:

I: is the moment of inertia for the disk

τ: is the torque

The moment of inertia is:

$I = (mr^(2))/(2)$

Where:

m: is the mass of the disk = 105.00 kg

r: is the radius of the disk = 0.297 m

$I = (105.00 kg*(0.297 m)^(2))/(2) = 4.63 kg*m^(2)$

Now, the torque is equal to:

$\tau = -F x r = -\mu*F*r$

Where:

F: is the applied force = 46.650 N

μ: is the kinetic coefficient of friction = 0.451

$\tau = -\mu*F*r = -0.451*46.650 N*0.297 m = -6.25 N*m$

The minus sign is because the friction force is acting opposite to motion of grindstone.

Having the moment of inertia and the torque, we can find the angular acceleration:

$\alpha = (-6.25 N*m)/(4.63 kg*m^(2)) = -1.35 rad/s^(2)$

Finally, we can find the number of turns that the stone will make before coming to rest:

$0 = \omega_(0)^(2) + 2\alpha*\theta$

$\theta = -((\omega_(0))^(2))/(2\alpha) = -((71.150 (rev)/(min))^(2))/(2*(-1.35 (rad)/(s^(2)))*(1 rev)/(2\pi rad)*((60 s)^(2))/((1 min)^(2))) = 3.27 rev = 3.27 turns$

I hope it helps you!

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Two non-conducting slabs of infinite area are given a charge-per-unit area of σ = -16 C/m^2 and σb =+6.0 C/m^2 respectively. A third slab, made of metal, is placed between the first two plates. The charge density σm on the metal slab is 0 (i.e., the slab is uncharged).Required:
Find the magnitude and direction of the electric field.

Answers

Answer:

$E_(C) = -5.65 * 10^(11) \hat{x}$ N/C

$E_(A) = -1.24 * 10^(12) \hat{x}$ N/C

Explanation:

The charge per unit area of the two non-conducting slabs are given by:

$\sigma_(a) = -16 C/m^2$

$\sigma_(b) = 6 C/m^2$

The charge density on the metal $\sigma_(m) = 0$

ε0 = 8.854 x 10-12 C2/N m2

Note that the electric field inside the conductor is zero because it is an equipotential surface.

The diagram attached to this solution typifies the description given in the question:

The electric field in the region C can be calculated by:

$E_(C) = ( |\sigma_(b) |- |\sigma_(a)| )/(2 \epsilon_(0) ) \nE_(C) = (6 - 16 )/(2 * 8.854 * 10^(-12) ) \nE_(C) = -5.65 * 10^(11) \hat{x}$

The electric field in the region A can be calculated by:

$E_(A) = (- |\sigma_(a) |- |\sigma_(b)| )/(2 \epsilon_(0) ) \nE_(A) = (-16 - 6 )/(2 * 8.854 * 10^(-12) ) \nE_(A) = -1.24 * 10^(12) \hat{x}$

Determine the tension in the string that connects M2 and M3.

Answers

thereforemass m1=4.8kg and the tension

in the horizontalspring T2=10N.

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To determine the tension in the string that connects M2 and M3, we can follow these steps:

Step 1: Identify the necessary variables. Given data (for example) could be:
- Mass of M2, which is 5 kg
- Mass of M3, which is 10 kg
- The acceleration due to gravity, which is approximately 9.8 m/s²
- The angle at which the string pulls on M2, which is 30 degrees
- Assume the system is in equilibrium, meaning there is no net acceleration, so the acceleration is 0 m/s²

Step 2: Calculate the weight of M3, which is its mass times the acceleration due to gravity. This is because weight is the force exerted by gravity on an object, which equals the object's mass times the acceleration due to gravity.

For M3, this calculation would be M3 * g = 10 kg * 9.8 m/s² = 98 N (Newtons).

Step 3: Determine the force exerted by M2 that acts along the line of the string. This won't be the full weight of M2, because the string pulls at an angle. This component of the force can be calculated using the sine of the angle, because sine gives us the ratio of the side opposite the angle (here, the force along the string) to the hypotenuse (here, the full weight of M2) in a right triangle.

The horizontal component of the force of M2 is then M2 * g * sin(30deg) = 5 kg * 9.8 m/s² * sin(30deg) = 24.5 N.

Step 4: The tension in the string is the force M3 exerts on it, which is its weight, minus the component of M2's weight that acts along the string. This is because M2 and M3 are pulling in opposite directions, so they subtract from each other.

The tension in the string is then the weight of M3, 98 N, minus the horizontal (along the string) component of M2's weight, 24.5 N.

So, the tension in the string is 98 N - 24.5 N = 73.5 N.

This is the force that the string needs to exert in order to keep M2 and M3 connected and in equilibrium.

Learn more about Tension in a string here:

brainly.com/question/34111688

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The planet uranus is tilted nearly on its side so that its axis or rotation is only 8 degress abway from its orbit plane. if you lived at latitude 45 degrees on uranus for what fraction of the uranian year would answer?

Answers

The rotation of Uranus, like that of Venus, is retrograde and its axis of rotation is inclined almost ninety degrees above the plane of its orbit. During its orbital period of 84 years one of the poles is permanently illuminated by the Sun while the other remains in the shade. Exactly its rotation period is equivalent to 17 hours and 14 Earth minutes and its translation period is equivalent to 84 years, 7 days and 9 Earth hours.

Only a narrow band around the equator experiences a rapid cycle of day and night, but with the Sun very low on the horizon as in the polar regions of the Earth. On the other side of the orbit of Uranus, the orientation of the poles in the direction of the Sun is inverse. Each pole receives about 42 years of uninterrupted sunlight, followed by 42 years of darkness. Therefore an observer at latitude of 45 degrees in Uranus will probably experience a long winter night that is equivalent to one third of the year uranium.

A ball having a mass of 500 grams is dropped from a height of 9.00 meters what is its kinetic energy when it hits the ground

Answers

Answer:

where m = mass, g = acceleration due to gravity (9.8 m/s^2), h = height

Given m = 500g = 0.5 kg, h = 9 meters

0.5*9*9.8 = 44.1 joules

Explanation:

Answer:44.1

Explanation:

What is the density of the paint if the mass of a tin containing 5000 cm3 paint is 7 kg. If the mass of the empty tin, including the lid is 0.5 kg.

Answers

We are given:

Mass of the Paint bucket (with paint) = 7000 grams

Mass of the paint bucket (without paint) = 500 grams

Volume of Paint in the Bucket = 5000 cm³

Mass of Paint in the Bucket:

To get the mass of the paint in the bucket, we will subtract the mass of the bucket from the mass of the paint bucket (with paint)

Mass of Paint = Mass of Paint bucket (with paint) - Mass of the paint Bucket (without paint)

Mass of Paint = 7000 - 500

Mass of Paint = 6500 grams

Density of the Paint:

We know that density = Mass / Volume

Density of Paint = Mass of Paint / Volume occupied by Paint

Density of Paint = 6500/5000

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A plank 2.00 cm thick and 13.0 cm wide is firmly attached to the railing of a ship by clamps so that the rest of the board extends 2.00 m horizontally over the sea below. A man of mass 68.4 kg is forced to stand on the very end. If the end of the board drops by 5.20 cm because of the man's weight, find the shear modulus of the wood.

Answers

Answer:

9.93 MPa

Explanation:

Given:

- mass of the man = 68.4 kg

- Deflection dx = 5.2 cm

- thickness of plank t = 2.0 cm

- width of plank w = 13.0 cm

- Length subtended L = 2.0 m

Find:

Shear Modulus of Elasticity S :

S = shear stress / shear strain

Shear stress = F / A

Shear stress = 68.4*9.81 / 0.02*0.13

Shear stress = 258078.4615 Pa

Shear strain = dx / L

Shear Strain = 0.052 / 2

Shear Strain = 0.026

Hence,

S = 258078.4615 / 0.026

S = 9.93 MPa

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