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madam [21]
3 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?
Physics
1 answer:
meriva3 years ago
7 0
From the calculations i did, its 1197. Is that the one you got?
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The car starts from a room with a constant acceleration of 5 ms-2. What path will it pass in 6 seconds and what speed will it re
Kobotan [32]

Answer: The speed will be  30 m/s .

Explanation:

Given: Initial velocity of the car: u = 0 m/s

Constant Acceleration: a = 5 m/s²

Time: t= 6 seconds

To find: Final velocity(v)

Formula:  v = u+at

Substitute values in the formula, we get

v=  0+(5)(6) m/s

⇒ v= 30 m/s

i.e. Final velocity = 30 m/s

Hence, the speed will be 30 m/s .

4 0
2 years ago
Which energy source contributes least to global warming?
Nady [450]
It’s C
solar

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3 0
3 years ago
The center of gravity of a loaded truck depends on how the truck is packed. If it is 4.0 m high and 2.4 m wide, and its CG is 2.
Vitek1552 [10]

The slope of the road can be given as the ratio of the change in vertical

distance per unit change in horizontal distance.

  • The maximum steepness of the slope where the truck can be parked without tipping over is approximately <u>54.55 %</u>.

Reasons:

Width of the truck = 2.4 meters

Height of the truck = 4.0 meters

Height of the center of gravity = 2.2 meters

Required:

The allowable steepness of the slope the truck can be parked without tipping over.

Solution:

Let, <em>C</em> represent the Center of Gravity, CG

At the tipping point, the angle of elevation of the slope = θ

Where;

tan\left(\theta \right) = \dfrac{\overline{AM}}{\overline{CM}}

The steepness of the slope is therefore;

\mathrm{The \ steepness \  of  \ the  \ slope}= \dfrac{\overline{AM}}{\overline{CM}} \times 100

Where;

\overline{AM} = Half the width of the truck = \dfrac{2.4 \, m}{2} = 1.2 m

\overline{CM} = The elevation of the center of gravity above the ground = 2.2 m

\mathrm{The \ steepness \  of  \ the  \ slope}= \dfrac{1.2}{2.2} \times 100 \approx 54.55\%

tan\left(\theta \right) = \mathbf{\dfrac{2.2}{1.2}} = \dfrac{11}{6}

Elevation \ of \ the \ road \ \theta = arctan\left( \dfrac{6}{11} \right)  \approx 28.6 ^{\circ}

The maximum steepness of the slope where the truck can be parked is <u>54.55 %</u>.

Learn more here:

brainly.com/question/20793607

3 0
2 years ago
very fine smoke particles are suspended in air. the translational rms speed of a smoke particle is 2.45 10-3 m/s, and the temper
m_a_m_a [10]

The mass of a particle is 2.2x10⁻¹⁵ kg

Consider smoke particles as an ideal gas

The translational RMS speed of the smoke particles is 2.45x10⁻³ m/s.

<em>v= √3kT/m</em>

<em>where k= 1.38x10⁻²³J/K, T is 288K, and m is the mass of the smoke particle</em>

<em>2.45x10⁻³ = √3x1.38x10⁻²³x288/m</em>

<em>m= 2.2x10⁻¹⁵ kg</em>

Therefore, the mass of a particle is 2.2x10⁻¹⁵ kg.

To learn more about the translational root mean square speed of gases, visit brainly.com/question/6853705

#SPJ4

5 0
1 year ago
Honeybees acquire a charge while flying due to friction with the air. A 100 mg bee with a charge of +23 pC experiences an electr
Sedbober [7]

Answer:

(A) ratio of electric force to weight will be  23.469\times 10^{-10}

(b) Electric field will be E=4.26\times 10^{10}N/C

Explanation:

We have given mass of bee = 100 mg  = m=100\times 10^{-3}=0.1kg

Charge on bee q=23pC=23\times 10^{-12}C

Electric field E = 100 N/C

Weight of the bee W=mg=0.1\times 9.8=0.98N

Electric force on the bee F=qE=23\times 10^{-12}\times 100=23\times 10^{-10}N

So the ratio of electric force on the bee and weight is =\frac{F}{W}=\frac{23\times 10^{-10}}{0.98}=23.469\times 10^{-10}

(B) To hold the bee in air electric force must be equal to weight of bee

So mg=qE

0.1\times 9.8=23\times 10^{-12}E

E=4.26\times 10^{10}N/C

4 0
3 years ago
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