3 years ago

When was the Hubble telescope launched

Physics

2 answers:

SpyIntel [72]3 years ago

5 0

Answer:

April 24, 1990 is the answer

MrRissso [65]3 years ago

4 0

April 24 1990 is the correct answer

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After 24.0 days 2.00 milligrams of an original 128.0. Milligram sample remain what is the half life of the sample

alina1380 [7]

Answer:

4 days

either multiply 128 by .5 until you get to 2 counting each time or use 2 formulas ln(n2/n1)=-k(t2-t1) to get k then input k into ln(2)=k*t1/2

n2 is final amount and n1 is beginning and t is either time elapsed as in the first formula or the actual half life that is t1/2

Explanation:

5 0

3 years ago

At the outer edge of a rotating space habitat, 130 m from the center, the rotational acceleration is g. What is the rotational a

enyata [817]

Answer:

Explanation:

Given:

R1 = 130 m

R2 = 65 m

w^2R = g

Assume, g = 9.81 m/s^2

w^2 = 9.81/130

w = 0.275 rad/s

At R2 = 65 m

g = w^2R

= (0.275^2) × 65

= 4.905 m/s^2

In conclusion,

g × R = k

g1/R1 = g2/R2

g2 = (g1 × 65)/130

= g1 ×1/2

= g1/2

6 0

3 years ago

A balloon filled with helium occupies 20.0 l at 1.50 atm and 25.0◦

bija089 [108]

At stp (standard temperature and pressure), the temperature is T=0 C=273 K and the pressure is p=1.00 atm. So we can use the ideal gas law to find the number of moles of helium:
$pV=nRT$
where p is the pressure (1.00 atm), V the volume (20.0 L), n the number of moles, T the temperature (273 K) and $R=0.082 atm L K^{-1} mol^{-1}$ the gas constant. Using the numbers and re-arranging the formula, we can calculate n:
$n= \frac{pV}{RT}= \frac{(1.00atm)(20.0L)}{(0.082 LatmK^{-1}mol^{-1})(273 K)}=0.89 mol$