An object moving with uniform acceleration has a velocity of 14.0 cm/s in the positive x-direction when its x-coordinate is 2.94 cm. If its x-coordinate 1.85 s later is −5.00 cm, what is its acceleration?

-19.9m/s²

Explanation:

Given that

vi = 14cm/s

xi = 2.94 cm

xf=-5.00 cm

t= 1.85s

So

Using xf-xi= vi t + 1/2at²

Which is = -5-2.94= 14*1.85+1/2a1.85²

= -7.94= 25.9+1.7a

-33.9= 1.7a

a= - 19.9m/s²

part 1: describe the mass and the surface temperature of dwarf stars.

part 2: explain using complete sentences where on the hr diagram you would expect to find a dwarf star based on its physical properties.

part 1: the mass of a dwarf star is very small and the surface temperature of dwarf stars are high and they have the lowest absolute brightness.

part 2: on the hr diagram you would expect to find a dwarf star based on its physical properties at an absolute brightness of -2 to -4

hope this