Ask question

Ask question

All categories

4 years ago

8

The C-14 decay is used to estimate the age of an old dead tree. The activity of C-14 in the dead tree is determined to have fall

en to 21% of its original value. C-14 has a half-life of 5700 years. Calculate, in years, the age of the dead tree.
I used the relevant formula but somehow my answer does not match with the markscheme. Can you help me?

1 answer:

meriva4 years ago

7 0

From the calculations i did, its 1197. Is that the one you got?

You might be interested in

What is the mass of a man who accelerates 4 m/s2 under the action of a 200 N net force?

Over [174]

Answer:

$\huge \boxed{ \boxed{50 \: kg }}$

Explanation:

The mass of the man can be found by using the formula

$m = \frac{f}{a} \\$

f is the force

a is the acceleration

From the question we have

$m = \frac{200}{4} \\$

We have the final answer as

<h3>50 kg</h3>

Hope this helps you

7 0

3 years ago

What is the momentum of a 5 kg object that has a velocity of 1. 2 m/s? 3. 8 kg â€˘ m/s 4. 2 kg â€˘ m/s 6. 0 kg â€˘ m/s 6. 2 kg â

nevsk [136]

The momentum of a 5kg object that has a velocity of 1.2m/s is 6.0kgm/s.

<h3> MOMENTUM:</h3>

Momentum of a substance is the product of its mass and velocity. That is;

Momentum (p) = mass (m) × velocity (v)

According to this question, an object has a mass of 5kg and velocity of 1.2m/s. The momentum is calculated thus:

Momentum = 5kg × 1.2m/s

Momentum = 6kgm/s.

Therefore, the momentum of a 5kg object that has a velocity of 1.2m/s is 6.0kgm/s.

Learn more about momentum at: brainly.com/question/250648?referrer=searchResults

7 0

3 years ago

Read 2 more answers

A speedboat moving at 29.0 m/s approaches a no-wake buoy marker 100 m ahead. The pilot slows the boat with a constant accelerati

Goshia [24]

Answer:

$t =4.8sec$

Explanation:

From the question we are told that

Velocity of speed boat $V=29.0m/s$

Distance to Marker $d=100$

Acceleration of $a=-3.4m/s^2$

Generally the Newtons 3rd motion equation is given as

$v^2 = u^2 + 2 * a* s$

$v^2 = 29^2 + 2 * -3.4* 100$

$v = \sqrt{161}$

$v=12.68m/s$

Generally the Newton's first equation of motion is given as

$v = u + a*t$

$12.68 = 29 -3.4*t$

$12.68-29 = -3.4t$

$-16.32 = -3.4t$

$t =\frac{-16.32}{-3.4}$

$t =4.8sec$ .

8 0

3 years ago

If the tensile strength of the Kevlar 49 fibers is 0.550 x 106 psi and that of epoxy resin is 11.0 x103 psi, calculate the stren

Elina [12.6K]

Answer:

$\sigma_{tot} =436810 psi$

Explanation:

Notation

$\sigma_{fib} =0.55x10^6 psi$ represent the tensile strength for the Kevlar 49 fibers

$\sigma_{epox} =11x10^3 psi$ represent the tensile strength for the epoxy resin

79% by volume is made of Kevlar 49 fibers

The rest 100-79=21% is made of epoxy resin

Tensile strength is a "measurement of the force required to pull something such as rope, wire, or a structural beam to the point where it breaks". Can be defined as "the maximum amount of tensile stress that it can take before failure".

For this case we need to use a basis of calculus since we don't know the total volume but we can assume a reference value on order to make the calculations.

If we assume a total volume of $V_{tot}=1 in^3$ we can do the follwoing balance:

$\sigma_{tot}V_{tot}=\sigma_{fib}V_{fib} +\sigma_{epox}V_{epox}$

We can replace the values given:

$\sigma_{tot}=(0.55x10^6 psi)(0.79) +(11x10^3 psi)(1-0.79)$

$\sigma_{tot} =436810 psi$

So then the strength of the composite material is : $\sigma_{tot} =436810 psi$

4 0

4 years ago

A 62.0-kg skier is moving at 6.30 m/s on a frictionless, horizontal, snow-covered plateau when she encounters a rough patch 4.90

Klio2033 [76]

Here are the missing questions:
(a) How fast is the skier moving when she gets to the bottom of the hill?
(b) How much internal energy was generated in crossing the rough patch?
Part A
The initial kinetic energy of the skier is:
$E_{k0}=m\frac{v_0^2}{2}$
Part of this energy is then used to do work against the force of friction. Force of friction on the horizontal surface can be calculated using following formula:
$F_f=mg\mu$
The work is simply the force times the length:
$W_f=F_f\cdot L=mg\mu L$
So when the skier passes over the rough patch its energy is:
$E=E_{k0}-W_f$
When the skier is going down the skill gravitational potential energy is transformed into the kinetic energy:
$E_p=E_{k1}\\ mgh=E_{k1}$
So the final energy of the skier is:
$E_f=E_{k0}-W_f+E_{k1}\\ E_f=m\frac{v_0^2}{2}-mg\mu L+mgh=1856.86$J$
This energy is the kinetic energy of the skier:
$E_f=m\frac{v_f^2}{2}\\ v_f=\sqrt{\frac{2E_f}{m}}=7.74\frac{m}{s}$
Part B
We know that skier lost some of its kinetic energy when crossing over the rough patch. This energy is equal to the work done by the skier against the force of friction.
$E_{int}=W_f\\
E_{int}=mg\mu L=894.1$J$

4 0

4 years ago