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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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When four people with a combined mass of 310 kg sit down in a 2000-kg car, they find that their weight compresses the springs an

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Answer:

Explanation:

F=kx

x=F/k

F=2000 kg

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effective spring constant=k=F/x

k=2000/9*10^-3=2.2*10^-5

now frequency

f=1/2π√k/m

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f=1/6.28√7.097*10^-8

f=1/6.28*2.7*10^-4

f=0.16*2.7*10^-4

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3 years ago

Suppose an automobile has 2000-joules of kinetic energy. when it moves at twice the speed, what will be its kinetic energy? what

NeX [460]

Answer:

K.Eₓ = 4 K.E

K.Eₓ = 9 K.E

Explanation:

Th formula for the kinetic energy of a body is given as follows:

$K.E = \frac{1}{2}mv^2\\$ ---------------equation (1)

where,

K.E = Kinetic Energy of Automobile

m = mass of automobile

v = speed of automobile

For twice speed:

vₓ = 2v

then,

$K.E_{x} = \frac{1}{2}mv_{x}^2\\K.E_{x} = \frac{1}{2}m(2v)^2\\K.E_{x} = 4\frac{1}{2}mv^2\\$

using equation (1):

For thrice speed:

vₓ = 3v

then,

$K.E_{x} = \frac{1}{2}mv_{x}^2\\K.E_{x} = \frac{1}{2}m(3v)^2\\K.E_{x} = 9\frac{1}{2}mv^2\\$

using equation (1):

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