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3 years ago

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Two stars, one classified A4 V and the other F8 V, have the same apparent magnitude. There is no significant amount of absorptio

n of starlight by interstellar material. From this information we know that the A4 V star is ____ the F8 V star.
A) the A4 V star is closer to the Sun than F8 V.
B) the A4 V star is smaller than F8 V.
C) the A4 V star is farther from the Sun than F8 V.
D) both stars are at the same distance from Sun.

1 answer:

vovikov84 [41]3 years ago

4 0

Answer: C

the A4 V star is farther from the sun than F8 V

Explanation:

An A-type main-sequence star (A V) or A dwarf star is a main-sequence (hydrogen- burning) star of spectral type A and luminosity class stars of spectral type, the hot stars, have spectra characterized by very few spectral lines the A4 V star is farther from the sun than F8 V

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2 years ago

dybincka [34]

Answer:

1. 218.55 N

2. $30.96^{o}$

3. $2.1 m/s^{2}$

Explanation:

Part 1;

Net force $F=mg sin \theta$ where m is mass, g is gravitational force and $\theta$ is the angle of inclination

$F= 46*9.8*sin 29^{o}= 218.55N$

Frictional force, $F_{r}$ is given by

$F_{r} = \mu_{s}mg cos \theta$ where $\mu_{s}$ is the coefficient of static friction

$F_{r} = 0.6*46*9.8 cos 29$

$F_{r}=236.57N$

Since $F_{r}>F$ , therefore, the block doesn’t slip and the frictional force acting is mgh=218.55N

Part 2.

Using the relationship that

Frictional force $F_{s} = \mu_{s} mg cos \theta$

$mg sin \theta =\mu_{s} mg cos \theta$

$\mu_{s}= \frac {sin \theta}{cos \theta}$

$\mu_{s}= tan \theta$

The maximum angle of inclination $\theta = tan^{-1} \mu_{s}$

$\theta = tan^{-1} (0.6)$

$\theta= 30.96^{o}$

Part 3:

Net force on the object is given by

$ma = mg sin 38 - \mu_{k} mg cos 38$ where $\mu_{k}$ is the coefficient of kinetic friction

$a = g ( sin 38 - \mu_{k} cos 38 )$

= 9.8 ( sin 38 - (0.51) cos 38 )

= $2.1m/s^{2}$

7 0

3 years ago