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2 years ago

A cylinder 0.170 m in diameter rotates in a lathe at 530 rpm . what is the tangential speed of the surface of the cylinder?

1 answer:

8 0

Diameter = 0.170 meter
Circumference = 0.170 π meters

530 rpm = 530 circumferences / minute

   = (530 x 0.170 π meters) / minute

   =    283.06 meter.minute

   =    4.72 meters/second

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what term describes the situation where the frequency of the wind match the frequency of the building? a. velocity b. resonance

saul85 [17]

The answer is B) resonance

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3 years ago

A student is swimming south applying a force of 256 N. The water exerts a westward force of 104 N. If the student has a mass of

grigory [225]

Answer:

$a=2.9\ m/sec^2$

Explanation:

<u>Net Forces and Acceleration</u>

The second Newton's Law relates the net force $F_r$ acting on an object of mass m with the acceleration a it gets. Both the net force and the acceleration are vector and have the same direction because they are proportional to each other.

$\vec F_r=m\vec a$

According to the information given in the question, two forces are acting on the swimming student: One of 256 N pointing to the south and other to the west of 104 N. Since those forces are not aligned, we must add them like vectors as shown in the figure below.

The magnitude of the resulting force $F_r$ is computed as the hypotenuse of a right triangle

$|F_r|=\sqrt{256^2+104^2}$

$|F_r|=276.32\ Nw$

The acceleration can be obtained from the formula

$F_r=ma$

Note we are using only magnitudes here

$\displaystyle a=\frac{F_r}{m}$

$\displaystyle a=\frac{276.32Nw}{95.3Kg}$

$\boxed{a=2.9\ m/sec^2}$

7 0

3 years ago

Oil having a density of 926 kg/m3 floats on water. A rectangular block of wood 3.69 cm high and with a density of 974 kg/m3 floa

zloy xaker [14]

Answer:

the position of the wood below the interface of the two liquids is 2.39 cm.

Explanation:

Given;

density of oil, $\rho _o$ = 926 kg/m³

density of the wood, $\rho _{wood}$ = 974 kg/m³

density of water, $\rho _w$ = 1000 kg/m³

height of the wood, h = 3.69 cm

Based on the density of the wood, it will position across the two liquids.

let the position of the wood below the interface of the two liquids = x

Let the wood be in equilibrium position;

$F_{wood} - F_{oil} - F_{water} = 0\\\\\rho _{wood} .gh - \rho _o .g(h-x) - \rho_w .gx = 0\\\\\rho _{wood} .h - \rho _o (h-x) - \rho_w .x = 0\\\\\rho _{wood} .h -\rho _o h + \rho _o x - \rho_w .x =0\\\\h (\rho _{wood} -\rho _o ) = x( \rho_w - \rho _o)\\\\x =h[\frac{ \rho _{wood} -\rho _o }{\rho_w - \rho _o} ]\\\\x = 3.69\ cm \times [\frac{974 - 926}{1000-926} ]\\\\x = 2.39 \ cm$