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Maurinko [17]
3 years ago
8

HELP PLS .. The half-life of potassium-44 is 22 minutes. It decays into calcium-44. After

Physics
2 answers:
anzhelika [568]3 years ago
6 0

Answer:

B. 1/8

Explanation:

did the test !!

Ber [7]3 years ago
3 0

Answer:

1/4

Explanation:

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N76 [4]
Pretty sure it is clockwise if I am not mistaken
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4 years ago
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Calculate the force of gravity between a comet with a mass of 500kg and a small asteroid with a mass of 20kg that is separated b
givi [52]

▪▪▪▪▪▪▪▪▪▪▪▪▪  {\huge\mathfrak{Answer}}▪▪▪▪▪▪▪▪▪▪▪▪▪▪

The equivalent gravitational force is ~

  • F  \approx1.48\times 10 {}^{ - 7}  \: \: N

\large \boxed{ \mathfrak{Step\:\: By\:\:Step\:\:Explanation}}

We know that ~

\huge\boxed{\mathrm{F = \dfrac{ Gm_1m_2}{ r²}}}

where,

  • F = gravitational force

  • m_1 = mass of 1st object = 500 kg

  • m_2 = mass of 2nd object = 20kg

  • G = gravitational constant = 6.674 × {10}^ {-11}

  • r = distance between the objects = 2.12 m

Let's calculate the force ~

  • F = \dfrac{6.674   \times 10 {}^{ - 11} \times 500 \times 20}{(2.12) {}^{2} }

  • F = \dfrac{6.674  \times 10 {}^{ - 11} \times 10 {}^{4} }{4.4944}

  • F =  \dfrac{6.674}{4.4944}  \times 10 {}^{ - 7}

  • F =1.484 \times 10 {}^{ - 7}  \: \: newtons
7 0
2 years ago
why does the velocity in the horizontal direction remain constant while the velocity in the vertical direction changes?
Bond [772]
The horizontal velocity<span> of a projectile is </span>constant<span> (a never </span>changing<span> in value), There is a </span>vertical<span>acceleration caused by gravity; its value is 9.8 m/s/s, down, The </span>vertical velocity<span> of a projectile </span>changes<span> by 9.8 m/s each second, The </span>horizontal<span> motion of a projectile is independent of its </span>vertical<span> motion.</span>
8 0
3 years ago
Assume the following values: d1 = 0.880 m , d2 = 1.11 m , d3 = 0.560 m , d4 = 2.08 m , F1 = 510 N , F2 = 306 N , F3 = 501 N , F4
dsp73

Answer:

= 2630.6 N.m

Explanation:

(FR)x = ΣFx = -F4 = -407 N

(FR)y = ΣFy =-F1-F2 -F3 = -510 - 306 - 501 = -1317 N

(MR)B =ΣM + Σ(±Fd)

= MA + F1(d1 +d2) + F2d2 - F4d3

= 1504 + 510(0.880+1.11) +306(1.11) - 407(0.560)

= 2630.64 N.m (counterclockwise)

6 0
4 years ago
When reaching a boundary between two media (1 and 2), an incident ray is partially reflected and partially refracted. The ray is
podryga [215]

Answer: critical angle, sin^-1 (n2/n1)

Explanation: the angle of incidence at which the retracted ray makes an angle of 90° with the normal is known as the critical angle.

Snell's law defined refraction mathematically as shown below

n1 sin θi = n2 sin θr

n1 = refractive index of the first medium

n2 = refractive index of the second medium

θi = angle of incidence

θr = angle of refraction

When the refrafted ray is perpendicular to the normal, the angle of refraction (θr) is 90° hence making the angle of incidence (θi) the critical angle θc

By substituting these conditions into the Snell's law, we have that

n1 sin θc = n2 sin 90

According to trigonometry, the value of sin 90 is 1, hence we have that

n1 sin θc =n2

sin θc = n2/n1

θc = sin^-1 (n2/n1)

3 0
3 years ago
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