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Sergeeva-Olga [200]
3 years ago
11

Suppose a wheel with a tire mounted on it is rotating at the constant rate of 2.59 times a second. A tack is stuck in the tire a

t a distance of 0.329 m from the rotation axis. Noting that for every rotation the tack travels one circumference, find the tack's tangential speed.__________m/s
what is the tacks radial acceleration?
___________m/s^2
Physics
1 answer:
Vesna [10]3 years ago
6 0

Answer:

Tangential speed=5.4 m/s

Radial acceleration=88.6m/s^2

Explanation:

We are given that

Angular speed=2.59 rev/s

We know that

1 revolution=2\pi rad

2.59 rev=2\pi\times 2.59=5.18\pi=5.18\times 3.14=16.27 rad/s

By using \pi=3.14

Angular velocity=\omega=16.27rad/s

Distance from axis=r=0.329 m

Tangential speed=r\omega=16.27\times 0.329=5.4m/s

Radial acceleration=\frac{v^2}{r}

Radial acceleration=\frac{(5.4)^2}{0.329}=88.6m/s^2

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The formulas used to analyze the horizontal and vertical motion of projectiles launched at an angle involve the use of tangent, cosine and sine.

<h3>What is vertical motion of a projecile?</h3>

The vertical motion of a projectile is affected by gravity and the velocity of vertical motion given by the following formula;

Vy = Vsinθ

<h3>What is horizontal motion of a projecile?</h3>

The horizontal velocity of a projectile is given by the following formula;

Vx = Vcosθ

<h3>Direction of the motion</h3>

The direction of the motion is calculated as follows;

tanθ  = Vy/Vx

Thus, the formulas used to analyze the horizontal and vertical motion of projectiles launched at an angle involve the use of tangent, cosine and sine.

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4 0
2 years ago
How to read a micrometer on a clark cm-100 vickers hardness tester
allsm [11]

Answer:

Explanation:

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covered to the extent of warranty provided by the original manufacturer and this warranty does not cover any

equipment, new or used, purchased from anyone other than the LECO Corporation. All replacement parts shall

be covered under warranty for a period of thirt days from date of purchase. LECO MAKES NO OTHER

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THE GOODS SOLD HEREUNDER, WHETHER AS TO MERCHANTABILITY, FITNESS FOR PURPOSE, OR

OTHERWISE.

Expendable items such as crucibles, combustion tubes, chemicals and items of like nature are not covered by

this warranty.

LECO's sole obligation under this warranty shall be to repair or replace any part or parts which, to our

satisfaction, prove to be defective upon return prepaid to LECO Corporation, St. Joseph, Michigan. This

obligation does not include labor to install replacement parts, nor does it cover any failure due to accident, abuse,

neglect, or use in disregard of instructions furnished by LECO. In no event shall damages for defective goods

exceed the purchase price of the goods, and LECO SHALL NOT BE LIABLE FOR INCIDENTAL OR

CONSEQUENTIAL DAMAGES WHATSOEVER.

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facts upon which the claim is based. Authorization must be obtained from LECO prior to returning any other

parts. This warranty is voided by failure to comply with these notice requirements.

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The warranty on LECO equipment remains valid only when genuine LECO replacernent parts are employed.

Since LECO has no control over the quality or purity of consumable products not manufactured by LECO, the

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5 0
2 years ago
An archer shoots an arrow with a mass of 45.0 grams from bow pulled
Sladkaya [172]

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The force the archer need to pull in order to achieve the height is approximately 101.8 N

Explanation:

By energy conservation principle, puling an elastic bow with a force, for a given distance, performs work which is converted to the potential energy of the arrow at height

The given parameters are;

The mass of the arrow, m = 45.0 grams = 0.045 kg

The distance the elastic bow is pulled, d = 65.0 cm = 0.65 m

The height at which the arrow is reaches, h = 150.0 meters

Let 'F', represent the force the archer need to pull in order to achieve the height

Work done, W = Force × Distance moved in the direction of the force

Therefore;

The work done in pulling the arrow, W = F × d

By energy conservation, we have;

The work done in pulling the arrow, W = The potential energy gained by the arrow, P.E.

W = P.E.

The potential energy gained by the arrow, P.E. = m·g·h

Where;

m = The mass of the arrow

g = The acceleration due to gravity = 9.8 m/s²

h = The height the arrow reaches

∴ by plugging in the values, P.E. = 0.045 kg ×9.8 m/s² × 150 m = 66.15 J

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Also, W = P.E. = 66.15 J

∴ W = F × 0.065 m = 66.15 J

F × 0.065 m = 66.15 J

F = 66.15 J/(0.65 m) = 1323/13 N ≈ 101.8 N

The force the archer need to pull in order to achieve the height, F ≈ 101.8 N.

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Anastaziya [24]

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phases of moon

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