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motikmotik
3 years ago
5

Starting with 100 grams of uranium-238, after one half life has gone by, how many grams of uranium-238 will remain?

Physics
1 answer:
KatRina [158]3 years ago
7 0

Answer:

50

Explanation:

100 (1/2) = 50

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Tres personas, A, B, C, jalan una caja con ayuda de cuerdas cuya masa es despreciable. Si la persona A aplica −3 en dirección ho
Yuri [45]

Answer:

El valor de la fuerza que debe ejercer la persona C debe ser de -2 para que la caja esté en equilibrio físico.

Explanation:

Si la caja debe hallarse en equilibrio físico, entonces se debe satisfacer la siguiente ecuación:

F_{A} + F_{B} + F_{C} = 0 (1)

Si sabemos que F_{A} = -3 y F_{B} = 5, entonces el valor de la fuerza que debe ejercer la persona C debe ser:

F_{C} = -F_{A}-F_{B}

F_{C} = -(-3)-5

F_{C} = -2

El valor de la fuerza que debe ejercer la persona C debe ser de -2 para que la caja esté en equilibrio físico.

3 0
3 years ago
A 0.20 kg mass is oscillating at a small angle from a light string with a period of 0.78 s.
Scorpion4ik [409]

Answer:

L = 15 cm

Explanation:

T = 2π√(L/g)

L = g(T/2π)²

L = 9.8(0.78/2π)²

L = 0.151027... m

L = 15 cm

7 0
3 years ago
Read 2 more answers
In the 1920s what did Edmund hubble notice about the galaxies
zlopas [31]
Hubble noticed that the galaxies were moving away from us, which meant the universe was expanding.

This is why constellations change over time. In some years, the Big Dipper won't actually look like a dipper anymore.
5 0
3 years ago
If a charge at 60c flow in a conductor for 30 second then the current that flow in a conductor is​
saw5 [17]

Explanation:

<h3>Given</h3>

- Charge = 60c

time = 30 sec

<h3>To find -</h3>

current

<h3>Solution </h3>

Current = Charge/time

I = V/T

I = 60/30

I = 2 ampere

More to know -

I = Current

V = Charge

T = Time

3 0
3 years ago
Damping is negligible for a 0.135-kg object hanging from a light, 6.30-N/m spring. A sinusoidal force with an amplitude of 1.70
IceJOKER [234]

Answer:

0.72 Hz minimum frequency

Explanation:

When the damping is negligible,Amplitude is given as

A = (F/m)/[\sqrt{(\omega ^{^{2}}-\omega _{o}^{2}})^2

here \omega _{o}^{2}= k/m = (6.30)/(0.135) = 46.67 N/m kg

F / mA = 1.70/(0.135)(0.480) = 26.2 N/m kg

From the above equation , rearranging for ω,

\omega ^{2}= \omega _{o}^{2}\pm F/m

⇒ ω² =46.67 ± 26.2 = 72.87 or 20.47

⇒ ω = 8.53 or 4.52 rad/s

Frequency = f

ω=2 π f

⇒ f = ω / 2π =  8.53 /6.28  or 4.52 / 6.28 = 1.36 Hz or 0.72 Hz

The lower frequency is 0.72 Hz and higher is 1.36 Hz

8 0
3 years ago
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