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alex41 [277]
3 years ago
6

Which statements about radiocarbon dating are true?

Physics
2 answers:
seropon [69]3 years ago
8 0
Radiocarbon saying assume the half life for radioistopes change over time. The half life of C -14 is 5730 years
Reika [66]3 years ago
8 0

both are the correct answers:

B. It's useful for dating some organic (once-living) things.

D. It's useful for dating some materials that are thousands of years old.

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An object of mass kg is released from rest m above the ground and allowed to fall under the influence of gravity. Assuming the f
IgorLugansk [536]

Answer:

Explanation:

From, the given information: we are not given any value for the mass, the proportionality constant and the distance

Assuming that:

the mass = 5 kg and the proportionality constant = 50 kg

the distance of the mass above the ground x(t) = 1000 m

Let's recall that:

v(t) = \dfrac{mg}{b}+ (v_o - \dfrac{mg}{b})^e^{-bt/m}

Similarly, The equation of mption:

x(t) = \dfrac{mg}{b}t+\dfrac{m}{b} (v_o - \dfrac{mg}{b}) (1-e^{-bt/m})

replacing our assumed values:

where v_=0 \ and \ g= 9.81

x(t) = \dfrac{5 \times 9.81}{50}t+\dfrac{5}{50} (0 - \dfrac{(5)(9.81)}{50}) (1-e^{-(50)t/5})

x(t) = 0.981t+0.1 (0 - 0.981) (1-e^{-(10)t}) \ m

\mathbf{x(t) = 0.981t-0.981(1-e^{-(10)t}) \ m}

So, when the object hits the ground when x(t) = 1000

Then from above derived equation:

\mathbf{x(t) = 0.981t-0.981(1-e^{-(10)t}) \ m}

1000= 0.981t-0.981(1-e^{-(10)t}) \ m

By diregarding e^{-(10)t} \ m

1000= 0.981t-0.981

1000 + 0.981 = 0.981 t

1000.981 = 0.981 t

t = 1000.981/0.981

t = 1020.36 sec

7 0
3 years ago
nder process​ costing, costs assigned to goods sold are transferred to the Cost of Goods Sold account from the​ ________. A perp
Reika [66]

Answer:

B

Explanation:

Under process​ costing, costs assigned to goods sold are transferred to the Cost of Goods Sold account from the​  Finished goods inventory account

The simple reason can be that Earlier than they are transferred to cost of goods sold account and being sold, they are finished goods right before it. So, they are actually transferred from finished goods inventory account only.

6 0
3 years ago
Take the regular compass and hold it so the case is vertical. Now use it to investigate the direction of the coil’s magnetic fie
Sliva [168]

Answer:

Please find the answer in the explanation

Explanation:

Take the regular compass and hold it so the case is vertical. Now use it to investigate the direction of the coil’s magnetic field at locations other than the central axis.

What happens as you move away from the center axis toward the coil? The direction of the magnetic compass needle will move in an opposite direction since the direction of the induced voltage is reversed.

What happens above the coil?

the needle on the magnetic compass will be deflected. Since compasses work by pointing along magnetic field lines

Outside the coil? The magnetic compass needle will experience no deflection. Since there is no induced voltage or current.

Below the coil?

The needle will move in an opposite direction.

3 0
3 years ago
If the mass of each ball were the same, but the velocity of ball A were twice as much as ball B, what do you think would happen
Margaret [11]
Given momentum is conserved and is mv

if
mass is the same and va=2vb
then momentum is m(2vb + vb)

if collision is elastic (bounces off)
then m(2vb + vb) = m(vfa + vfb)
3vb = vfa + vfb
meaning final velocities would be different

if collision is inelastic (sticks together)
then m(2vb + vb) = (m+m)vf
meaning final velocity would be same

is this what you wanted?
5 0
3 years ago
If you know that the period of a pendulum is 1.87 seconds, what is the length of that pendulum? (assume that we are on earth and
laiz [17]

The length of the pendulum is 0.087 m. Option d is correct.

<h3>What is Simple harmonic motion?</h3>

Simple harmonic motion is periodic motion caused by a restoring force that is proportionate to the deviation from equilibrium.

Simple harmonic motion is periodic motion but many other conditions are dependent.

The time period of the pendulum is found as;

\rm T= 2 \pi \sqrt{\frac{L}{g} } \\\\ \rm 1.87 \ sec= 2 \times 3.14  \sqrt{\frac{L}{9.81 m/s^2} } \\\\ L=0.087 \ m

The length of the pendulum is 0.087 m

Hence, option d is correct.

To learn more about the simple harmonic motion refer to the link;

brainly.com/question/17315536

#SPJ1

4 0
3 years ago
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