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Marina86 [1]
2 years ago
10

The table lists several radioisotopes that have technological uses. Which radioisotope would be used by both archaeologists and

botanists (scientists who study plants)? uranium-238 carbon-14 iodine-131 technetium-99
Physics
2 answers:
ElenaW [278]2 years ago
8 0

Answer:

B. carbon-14

Explanation:

Vinil7 [7]2 years ago
5 0
Carbon-14
Since carbon can be found in plants and items that are studied by archaeologist 
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( 8c5p79) A certain force gives mass m1 an acceleration of 13.5 m/s2 and mass m2 an acceleration of 3.5 m/s2. What acceleration
Talja [164]

Answer:

4.725 m/s^{2}

Explanation:

We know that from Newton's second law of motion, F=ma hence making acceleration the subject then a=\frac {F}{m}  where a is acceleration, F is force and m is mass

Also making mass the subject of the formula m=\frac {F}{a}

For m1= \frac {F}{13.5} and m2=\frac {F}{3.5} hence F=(m2-m1)a= (\frac {F}{3.5}-\frac {F}{13.5})a=0.2116402116\\\frac {1}{a}=0.2116402116\\a=4.725 m/s^{2}

4 0
3 years ago
What is the potential energy of two charges of +4.6 μC and +1.0 μC that are separated by a distance of 10.0 cm?
Artist 52 [7]

Answer:

U = 0.413 J

Explanation:

the potential energy between two charges q1 and q2 is given by the following formula:

U=k\frac{q_1q_2}{r}    (1)

k: Coulomb's constant = 8.98*10^9 NM^2/C^2

q1: first charge = 4.6 μC = 4.6*10^-6 C

q2: second charge = 1.0 μC*10^-6 C

r: distance between charges = 10.0 cm = 0.10 m

You replace the values of all variables in the equation (1):

U=(8.98*10^9Nm^2/C^2)\frac{(4.6*10^{-6}C)(1.0*10^{-6}C)}{0.10m}=0.413\ J

Hence, the energy between charges is 0.413 J

3 0
3 years ago
The energy processed and used by living beings is <br>​
Molodets [167]

Answer:

Nutrition , heat , wind

Explanation:

6 0
2 years ago
your friend rides her bicycle across town at a constant speed. Describe how you could determine her speed
Alchen [17]
You measure the total distance the bike travels, and the total time it takes to travel that distance. You then divide the distance by the time
6 0
3 years ago
Calculate the magnitude of the linear momentum for the following cases. (a) a proton with mass 1.67 10-27 kg, moving with a spee
abruzzese [7]

Answer:

<h2>8.0995×10^-21 kgms^-1</h2>

Explanation:

Mass of proton :

m_P=1.67\times 10^-^2^7\:kg\\

Speed of Proton:

v_P=4.85\times 10^6

Linear Momentum of a particle having mass (m) and velocity (v) :

-> p =m->v\:\:\: (1)

Magnitude of momentum :

p=mv\:\:\: (2)

Frome equation (2), magnitude of linear momentum of the proton :

p_P=m_P\:v_P\\\\p_P=1.67\times 10^-^2^7 \:kg\times4.85\times 10^6\:ms^-^1\\\\p_P= 8.0995\times 10^-^2^1\:kgms^-^1

7 0
2 years ago
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