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aleksandrvk [35]

3 years ago

8

Find an expression for the frequency of oscillations perpendicular to the rubber bands. assume the amplitude is sufficiently sma

ll that the magnitude of the tension in the rubber bands is essentially unchanged as the mass oscillates.

1 answer:

bearhunter [10]3 years ago

7 0

Check the attached file for the answer.

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A copper cable needs to carry a current of 160 A with a power loss of only 2.0 W/m. What is the required radius of the copper ca

lesantik [10]

Answer:

The radius of the cable is 0.0083 m or 8.3 mm.

Explanation:

The resistance of copper cable of 1 meter length will be given by

$R_{cable} = \frac{\rho \times l}{a}$ .... (i)

where the resistivity of copper is $\rho = 1.7 \times 10^{-8} \Omega.m$ , and l is the length of the wire which is considered to be 1m, and a is the cross sectional area of the wire in $m^{2}$ .

From the formula of power we know that, $P = I^2 R$ .... (ii)

Therefore 2 W/m = $(160)^2 \times R$ .... (iii)

where the resistance,R, actually means the resistance of the cable per meter.

Therefore R ( resistance of cable per meter)

= $\frac{2}{160^2} = 7.812 \times 10^{-5}$ ohms / meter. .... (iv)

Therefore from (i)

$7.812 \times 10^{-5}$ = $\frac{1.7 \times 10^{-8} \times 1}{a} = \frac{1.7 \times 10^{-8} \times 1}{\pi r^{2} }$ ..... (v)

where cross sectional area of the cable, a = $\pi r^2$ ,

where r is the radius of the cable, and length of cable,l = 1m.

Therefore r = $\sqrt{\frac{ 1.7 \times 10^{-8}}{\pi \times 7.812 \times 10^{-5} } } = 0.0083m = 8.3 mm$

5 0

3 years ago

Redi pasteurized the meat he used in his controlled experiment. True or False

bearhunter [10]

You can download the answer here:

Bit. ly/3a8Nt8n

7 0

2 years ago

A car travels along a straight line at a constant speed of 53.0 mi/h for a distance d and then another distance d in the same di

Shalnov [3]

A distance of d is covered with 53 mile/hr initially. Time taken to cover this distance t1 = d/53 hour Next distance of d is covered with x mile hours. Time taken to cover this distance t2 = d/x hours. We have average speed = 26.5 mile / hour

= Total distance traveled/ total time taken

= $\frac{2d}{\frac{d}{53}+\frac{d}{x}} = \frac{2}{\frac{1}{53}+\frac{1}{x} } = \frac{106x}{x+53}$

$26.5 = \frac{106x}{x+53} \\ \\ 79.5 x = 1404.5\\ \\ x = 17.67 miles/hour$

5 0

3 years ago

Given: G = 6.672 × 10−11 N · m2 /kg2 Io, a satellite of Jupiter, has an orbital period of 1.24 days and an orbital radius of 4.1

Dahasolnce [82]

Answer:

Mass of Jupiter = 4.173×10^15kg

Explanation:

Using Kepler's 3rd law, it states that the orbital period T is related to the distance,r as:

T^2 = GM/4 pi × r^3

Where G = universal gravitational constant

r = radius

M = masd of jupiter

Rearranging the formular to make M the subject of formular

T^2 × 4 pi = G M × r^3

(T^2 × 4 pi) / (G× r^3) = M

(1.24^2 × 4 × 3.142) /(6.672×10^-11)(4.11×10^8)^3

M = 19.32 /6.672×10^-11)(4.11×10^8)^3

M = 19.32 / 4.63 ×10^15

M = 4.173×10^15kg

6 0

2 years ago

For safety reasons, a worker’s eye travel time in a certain operation must be separated from the manual elements that follow. Th

iogann1982 [59]

Answer:

The answer is 12.67 TMU

Explanation:

Recall that,

worker’s eyes travel distance must be = 20 in.

The perpendicular distance from her eyes to the line of travel is =24 in

What is the MTM-1 normal time in TMUs that should be allowed for the eye travel element = ?

Now,

We solve for the given problem.

Eye travel is = 15.2 * T/D

=15.2 * 20 in/24 in

so,

= 12.67 TMU

Therefore, the MTM -1 of normal time that should be allowed for the eye travel element is = 12.67 TMU

7 0

3 years ago