Read the list of substances

1 answer:

8 0

D. Copper and Gold
Copper and gold are used a lot in electrical applications.
Air and wood are insulators and glass is also kinda bad...

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Which of these properties is the best measure of a star's brightness? (1 point)

storchak [24]

Answer:

ABSOLUTE MAGNITUDE, C

Explanation:

Absolute magnitude is how bright the star is from a distance of 10 parsecs. Apparent magnitude is how bright the star appears from Earth, which can vary, because some stars may be farther or closer away. Age and size are not very accurate either. For example, at the end of their lifetimes, very big start may explode in a violent and bright supernova, and outshine the stars near them. As for size, as many stars several times smaller than the sun shine much brighter than it, making it inaccurate.

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2 years ago

The accepted standard measurement for mass is the A) liter. B) pound. C) Newton. D) kilogram.

andreev551 [17]

D) Kilograms is the standard unit of mass

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3 years ago

Geology belongs to which main category of science?

sergey [27]

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3 years ago

An automobile whose speed is increasing at a rate of 0.800 m/s2 travels along a circular road of radius 10.0 m.

Ket [755]

(a) $a_t = 0.800 m/s^2$

The tangential acceleration component of the car is simply equal to the change of the tangential speed divided by the time taken:

$a_t = \frac{\Delta v}{\Delta t}$

This rate of change is already given by the problem, 0.800 m/s^2, so the tangential acceleration of the car is

$a_t = 0.800 m/s^2$

(b) $a_c = 0.9 m/s^2$

The centripetal acceleration component is given by

$a_c = \frac{v^2}{r}$

where

v is the tangential speed

r is the radius of the trajectory

When the speed is v = 3.00 m/s, the centripetal acceleration is (the radius is r = 10.0 m):

$a_c = \frac{(3.00 m/s)^2}{10.0 m}=0.9 m/s^2$

(c) $1.2 m/s^2, 48.4^{\circ}$

The centripetal acceleration and the tangential acceleration are perpendicular to each other, so the magnitude of the total acceleration can be found by using Pythagorean's theorem:

$a=\sqrt{a_t^2+a_c^2}=\sqrt{(0.8 m/s^2)^2+(0.9 m/s^2)^2}=1.2 m/s^2$