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slavikrds [6]
3 years ago
11

The half-life of carbon-14 is 5730 years. how long will it take for 7/8 of a sample of carbon-14 to decay

Physics
2 answers:
wlad13 [49]3 years ago
5 0
The half life of Carbon-14 is 5730 years, how many years would it take for 7/8 of the original amount to decay? 

<span>Can somebody please help with this problem. I *think* I understand the basics of what a half life is. If I learned correctly, its the amount it takes for half of a sample to decay. It should also happen exponentially, 1/2 remaining after one half life, 1/4 after the second, 1/16 after the third etc. I'm still a little shaky though. Could somebody please clarify what exactly a half life is and how it can be determined (i.e. how to find the time it would take for 7/8 to decay) </span>


andrezito [222]3 years ago
3 0

Answer:

B) 17,190 years

Explanation:

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A camcorder has a power rating of 15 watts. If the output voltage from its battery is 5 volts, what current does it use?
MrMuchimi

Formula

W = E * I

Givens

E = 5 volts

W = 15 watts

I = ?

Solution

W = E * I

15 = 5 * I

15/5 = I

I = 3 amps.   Answer

5 0
2 years ago
*please refer to photo* An electric field of magnitude 5.25 ✕ 10^5N/C points due south at a certain location. Find the magnitude
kvv77 [185]

Answer:

Approximately 3.86\; {\rm N} (given that the magnitude of this charge is -7.35\; {\rm \mu C}.)

Explanation:

If a charge of magnitude q is placed in an electric field of magnitude E, the magnitude of the electrostatic force on that charge would be F = E\, q.

The magnitude of this charge is q = 7.35\; {\rm \mu C}. Apply the unit conversion 1\; {\rm \mu C} = 10^{-6}\; {\rm C}:

\begin{aligned} q &= 7.35\; {\mu C} \times \frac{10^{-6}\; {\rm C}}{1\; {\mu C}} = 7.35\times 10^{-6}\; {\rm C}\end{aligned}.

An electric field of magnitude E = 5.25\times 10^{5}\; {\rm N \cdot C^{-1}} would exert on this charge a force with a magnitude of:

\begin{aligned}F &= E\, q \\ &= 5.25 \times 10^{5}\; {\rm N \cdot C^{-1}} \times (-7.35\times 10^{-6}\; {\rm C}) \\ &\approx 3.86\; {\rm N}\end{aligned}.

Note that the electric charge in this question is negative. Hence, electrostatic force on this charge would be opposite in direction to the the electric field. Since the electric field points due south, the electrostatic force on this charge would point due north.

4 0
2 years ago
If a 500-pound object is moved 200 feet how much work is being done?
Paha777 [63]

Answer:

D

Explanation:

Work = Distance x Mass

4 0
2 years ago
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4 0
3 years ago
Read 2 more answers
The knee extensors insert on the tibia at an angle of 30 degrees (from the longitudinal axis of the tibia), at a distance of 3 c
Reika [66]

Answer:

the knee extensors must exert 15.87 N

Explanation:

Given the data in the question;

mass m = 4.5 kg

radius of gyration k = 23 cm = 0.23 m

angle ∅ = 30°

∝ = 1 rad/s²

distance of 3 cm from the axis of rotation at the knee r = 3 cm = 0.03 m

using the expression;

ζ = I∝

ζ = mk²∝

we substitute

ζ = 4.5 × (0.23)² × 1

ζ  = 0.23805 N-m

so

from; ζ = rFsin∅

F = ζ / rsin∅

we substitute

F = 0.23805 / (0.03 × sin( 30 ° )

F = 0.23805 / (0.03 × 0.5)

F F = 0.23805 / 0.015

F = 15.87 N

Therefore, the knee extensors must exert 15.87 N

7 0
2 years ago
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