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3 years ago

How can radioactive dating and relative dating be used together to date the age of a rock strata or an organism?

Physics

1 answer:

Otrada [13]3 years ago

7 0

Answer:

It compares the the difference between a radioactive element remaining in specimen to the amount of the radioactive element that would have been originally trapped in the specimen. This is done by comparing the ratio of the relative abundance of this radioactive element to its non radioactive isotope in nature to their ratio remaining in the specimen and comparing it to the half-life of the radioactive isotope.

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Find the slope of the line containing the two points (1,−1) and (−5,−3). (1 point)

grandymaker [24]

Answer:

(-4,-2)

Explanation:

4 0

2 years ago

Please select the word from the list that best fits the definition

lina2011 [118]

Answer:

A subculture.

Explanation:

Aspects of a group that is based on the shared values and belief systems and the norms and behaviors aspects that are only a part of the subsection of society are called as subcultures. Like the goths, hippies, etc as its nor accepted by the entire population.

3 0

3 years ago

Read 2 more answers

what is the power of an electrical device which operates with a current of 12.4 A and a potential difference of 12 V

jek_recluse [69]

148.8 Watts

Explanation:

P = VI

= (12 V)(12.4 A)

= 148.8 Watts

3 0

3 years ago

A transformer connected to a 120-v(rms) at line is to supply 13,000V(rms) for a neon sign.

lyudmila [28]

Answer:

(a) 108

(b) 110.500 kW

Solution:

As per the question:

Voltage at primary, $V_{p} = 120\ V$ (rms voltage)

Voltage at secondary, $V_{s} = 13000\ V$ (rms voltage)

Current in the secondary, $I_{s} = 8.50\ mA$

Now,

(a) The ratio of secondary to primary turns is given by the relation:

$\frac{N_{s}}{N_{p}} = \frac{V_{s}}{V_{s}}$

where

$N_{p}$ = No. of turns in primary

$N_{s}$ = No. of turns in secondary

$\frac{N_{s}}{N_{p}} = \frac{13000}{120}$ ≈ 108

(b) The power supplied to the line is given by:

Power, P = $V_{s}I_{s} = 13000\times 8.50 = 110.500\ kW$

$\frac{V_{s}}{V_{p}} = \frac{I_{p}}{I_{s}}$

$\frac{13000}{120} = \frac{I_{p}}{8.50}$

$I_{p} = \frac{13000}{120}\times 8.50 = 920.84\ A$

6 0

3 years ago

In an internal combustion engine, air at atmostpheric

notsponge [240]

Answer:

1302 K or 1029 C

Explanation:

Air at atmospheric pressure has pressure of 1 atm

20 C = 20 + 273 = 293 K

Assume ideal gas, according to the ideal gas law:

$\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}$

Where P1, V1 and T1 are the pressure, volume and temperature of the gas before the compression and P2, V2 and T2 are the pressure, volume and temperature of the gas after the compression

$T_2 = T_1\frac{P_2V_2}{P_1V_1}$

Since the gas is compressed to 1/9 of its original volume, V2/V1 = 1/9:

$T_2 = 293\frac{40}{9} = 1302 K$ or 1029 C

8 0

3 years ago