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3 years ago

14

Which of the following methods is most likely to be used to determine the absolute age of animal bones that are less than 45,000

years old? A. Radiocarbon B. Potassium- argon dating C. Uranium- lead dating D. Relative dating

1 answer:

Leni [432]3 years ago

6 0

Radiocarbono la opcion a

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You pull a solid nickel ball with a density of 8.91 g/cm3 and a radius of 1.40 cm upward through a fluid at a constant speed of

Sunny_sXe [5.5K]

Answer:

$P = 1.090\,N$

Explanation:

The constant speed means that ball is not experimenting acceleration. This elements is modelled by using the following equation of equilibrium:

$\Sigma F = P - W + F_{D}$

$\Sigma F = P - \rho \cdot V \cdot g + c\cdot v = 0$

Now, the exerted force is:

$P = \rho \cdot V \cdot g - c\cdot v$

The volume of a sphere is:

$V = \frac{4\cdot \pi}{3}\cdot R^{3}$

$V = \frac{4\cdot \pi}{3}\cdot (0.014\,m)^{3}$

$V = 1.149\times 10^{-5}\,m^{3}$

Lastly, the force is calculated:

$P = (8910\,\frac{kg}{m^{3}} )\cdot (1.149\cdot 10^{-5}\,m^{3})\cdot (9.81\,\frac{m}{s^{2}} )+(0.950\,\frac{kg}{s})\cdot (0.09\,\frac{m}{s} )$

$P = 1.090\,N$

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3 years ago

Two wires with the same resistance have the same diameter but different lengths. If wire 1 has length L 1 and wire 2 has length

SVEN [57.7K]

Answer with Explanation:

We are given that

Length of wire 1= $L_1$

Length of wire 2= $L_2$

Resistivity of copper wire= $\rho_1=1.7\times 10^{-5}\Omega-m$

Resistivity of aluminum wire= $\rho_2=2.82\times 10^{-5}\Omega-m$

Wire 1=Copper wire

Wire 2=Aluminum wire

Diameter of both wires are same and resistance of both wires are also same.

We know that

Resistance = $\frac{\rho l}{A}$

When diameter of wires are same then their cross section area are also same .

$l=\frac{RA}{\rho}$

When resistance and area are same then the length of wire depend upon the resistivity of wire .

The length of wire is inversely proportional to resistivity.

When resistivity is greater then the length of wire will be short and when the resistivity is small then the length of wire will be large.

$\rho_1$

Therefore, $L_1>L_2$

Hence, the length of wire 1 (copper wire) is greater than the length of wire 2 (aluminum).

$\frac{L_1}{L_2}=\frac{\frac{RA}{1.7\times 10^{-5}}}{\frac{RA}{2.82\times 10^{-5}}}=1.66$

$L_1=1.66L_2$

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3 years ago

BRAINLIST!!!!! USE 3-7 SENTENCES!!!!

Savatey [412]

Answer: The author uses metaphors to explain how childeren taking on certain traits by (put what story is about here). The effect is that (put what happened at the end of story here).

Explanation:

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3 years ago

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An ac series circuit contains a resistor of 15 ohms, an inductor of 11.5 mH and a variable capacitor. If the frequency of the ap

SpyIntel [72]

Answer:

C) 20.23 μF

Explanation:

R = resistance of the resistor = 15 ohm

L = inductance of the inductor = 11.5 mH = 0.0115 H

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C = Capacitance of the capacitor

For resonance to be possible

$2\pi fL = \frac{1}{2\pi fC}$

$2(3.14) (330) (0.0115) = \frac{1}{2 (3.14) (330) C}$

C = 20.23 x 10⁻⁶ F

C = 20.23 μF

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Radio waves are a type of electromagnetic wave that can travel when there is no medium, whereas sound waves are a type of mechanical wave that cannot travel if there is no medium

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2 years ago