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3 years ago

8

Why does a compass point toward Earth’s geographic North Pole? Select three options.

1 answer:

Eva8 [605]3 years ago

4 0

Answer:

B,C,E

Explanation:

just figured it out on edge

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A student whirls a rubber stopper attached to a string in a horizontal circle above her head. If the radius of the stopper in ci

algol13

Answer:

$v = 12.57 m/s$

Explanation:

As we know that the radius of the circular motion is given as

$R = 50.00 cm$

time period of the motion is given as

$T = 0.2500 s$

now we know that it is moving with uniform speed

so it is given as

$v = \frac{2\pi R}{T}$

now plug in all data

$v = \frac{2\pi(0.50)}{0.25}$

$v = 12.57 m/s$

7 0

3 years ago

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

4 years ago

A machine part has the shape of a solid uniform sphere of mass 250 g and a diameter of 4.30 cm. It is spinning about a frictionl

zysi [14]

Answer: $\alpha =9.302\ rad/s^2$

Explanation:

Given

mass of sphere $m=250\ gm$

diameter of sphere $d=4.30\ cm$

radius $r=\frac{4.30}{2}\ cm$

$f=0.0200\ N$

friction will provide resisting torque so

$f\times r=I\times \alpha$

where $I=\text{moment of Inertia}$

$f=\text{friction force}$

$\alpha =\text{angular acceleration}$

$I=\frac{2}{5}mr^2$

$0.02\times r=\frac{2}{5}mr^2\times \alpha$

$\alpha =\frac{5}{2r}\times f$

$\alpha =\frac{5}{2}\times \frac{2}{4.3\times 10^{-2}}\times 0.02$

$\alpha =9.302\ rad/s^2$

(b)time taken to decrease its rotational speed by $21\ rad/s$

$t=\dfrac{\Delta \omega }{\alpha }$

$t=\dfrac{21}{9.302}$

$t=2.25\ s$

6 0

3 years ago

In 1687, Isaac Newton published a book explaining his three laws of planetary motion and the law of gravity. The importance of t

alexdok [17]

B. all of nature follows uniform, unchaining laws.

5 0

3 years ago

A typical laboratory centrifuge rotates at 4000 rpm. Testtubes have to be placed into a centrifuge very carefully because ofthe

Bogdan [553]

Answer:

A) a_c = 1.75 10⁴ m / s², B) a = 4.43 10³ m / s²

Explanation:

Part A) The relation of the test tube is centripetal

a_c = v² / r

the angular and linear variables are related

v = w r

we substitute

a_c = w² r

let's reduce the magnitudes to the SI system

w = 4000 rpm (2pi rad / 1 rev) (1 min / 60s) = 418.88 rad / s

r = 1 cm (1 m / 100 cm) = 0.10 m

let's calculate

a_c = 418.88² 0.1

a_c = 1.75 10⁴ m / s²

part B) for this part let's use kinematics relations, let's start looking for the velocity just when we hit the floor

as part of rest the initial velocity is zero and on the floor the height is zero

v² = v₀² - 2g (y- y₀)

v² = 0 - 2 9.8 (0 + 1)

v =√19.6

v = -4.427 m / s

now let's look for the applied steel to stop the test tube

v_f = v + a t

0 = v + at

a = -v / t

a = 4.427 / 0.001

a = 4.43 10³ m / s²

4 0

3 years ago