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3 years ago

Carbon-14 has a half-life of 5,700 years. How long will it take for 6.25% of the Carbon-14 to be remaining?

Physics

1 answer:

IRISSAK [1]3 years ago

4 0

Answer:

22,800 years

Explanation:

Half life equation:

A = A₀ (½)^(t / T)

where A is the final amount,

A₀ is the initial amount,

t is time,

and T is the half life.

0.0625 = (½)^(t / 5700)

log 0.0625 = (t / 5700) log 0.5

4 = t / 5700

t = 22,800

It takes 22,800 years.

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An airplane flies 12 m/s due north with a velocity of 35.11 m/s. how far east does it fly?

goblinko [34]

The plane's velocity of 35.11 m/s is actually due in a north-eastward direction. The 12 m/s velocity is the vertical component of the plane's velocity, hence it is pointing northwards. We will use the formula:

Vy = Vsin∅

To determine the angle ∅ at which the plane is flying. This is:

12 = 35.11 * sin∅
∅ = 20.0 degrees

The eastward velocity is:

Vx = Vcos∅
Vx = 35.11 * cos(20)
Vx = 33.0 m/s

The plane's eastward velocity is 33.0 m/s

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3 years ago

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An astronaut travels to a star system 3.9 lyly away at a speed of 0.90 cc . Assume that the time needed to accelerate and decele

kakasveta [241]

Total time elapsed is =8.2y

The starting event is the astronaut leaving Earth. The finishing event is the astronaut arriving at the star system. The time between these events on Earth is:

Δt=3.9ly/0.9c

Δt=4.3y

For the astronaut, two events occur at the same position and can be measured with just one clock. Hence,

Δτ

$= \sqrt{1 - \frac{v {}^{2} }{c {}^{2} } } \times Δt$

Δτ

$= \sqrt{1 - ( \frac{0.9c}{c} ) {}^{2} } \times (4.3y)$

Δτ=1.8ly

The total elapsed time is:

T elapsed=Δt+3.9

T elapsed=4.3+3.9

T elapsed=8.2y

learn more about time from here: brainly.com/question/28208983

#SPJ4

4 0

2 years ago

(Please help asap)

siniylev [52]

The answer would be B

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4 years ago

A student standing on the ground throws a ball straight up. The ball leaves the student's hand with a speed of 19.0 m/s when the

german

Answer: 4 s

Explanation:

Given

The ball leaves the hand of student with a speed of $u=19\ m/s$

When the hand is $h=2.5\ m$ above the ground

Using the equation of motion we can write

$h=ut+\dfrac{1}{2}at^2$

Substitute the values

$\Rightarrow 2.5=-19t+0.5\times 9.8t^2\\\Rightarrow 4.9t^2-19t-2.5=0\\\\\Rightarrow t=\dfrac{19\pm \sqrt{(-19)^2-4\times 4.9\times (-2.5)}}{2\times 19}\\\Rightarrow t=4.0049\quad [\text{Neglecting the negative value of }t]$

Thus, the ball will take 4 s to hit the ground.

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3 years ago

If an object has a mass of 10 kilograms, how much does it weigh in newtons?

schepotkina [342]

10 kilograms of mass weighs 98.1 newtons on Earth,
16.2 newtons on the Moon, 37.1 newtons on Mars,
and other weights in other places.

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3 years ago