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4 years ago

Fossils are dated using radioactive... a. hydrogen b. helium c. carbon

Physics

2 answers:

OlgaM077 [116]4 years ago

7 0

Answer:

carbon

Explanation:

Setler79 [48]4 years ago

5 0

Answer:

fossils are dated using radioactive C. carbon

Explanation:

hope this helps!

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An electron (charge −e=−1.60×10−19C−e=−1.60×10−19C) is at rest at a distance 1.00×10−10 m 1.00×10−10m from the center of a nucle

shtirl [24]

Answer:

$-9.45\times10^{-16} \text{ J}$

Explanation:

According to Coulomb's law, the force of attraction between two point charges, $q_1$ and $q_2$ , separated by a distance $d$ is given by

$F = k\dfrac{q_1q_2}{d^2}$

$k$ is a constant with a value of $9\times10^9\text{ F/m}$ .

When we substitute the values from the question,

$F = (9\times10^9\text{ F/m})\dfrac{(-1.60\times10^{-19} \text{ C})\times(+1.31\times10^{-17}\text{ C})}{(1.00\times10^{-10}\text{ m})^2} = -1.89\times10^{-6} \text{ N}$

This value is negative because it is in a direction towards the positive charge.

The work done in moving the electron from the nucleus is

$W = F\times r$

$W = (-1.89\times 10^{-6} \text{ N})\times(5.00\times10^{-10}\text{ m}) = -9.45\times10^{-16} \text{ J}$

This is negative because work is done on the electron, not by it.

3 0

3 years ago

A current-carrying wire oriented north-south and laid over a compass deflects the compass 8° to the east. What is the magnitude

zhenek [66]

$2.8 \times 10^{-6}\ T$ is the magnitude of the magnetic field made by the current

<u>Explanation:</u>

Given data:

$\theta=8^{\circ}$

Magnetic field of earth, $B_{\text {earth}}=2 \times 10^{-5}\ \text {tesla}$

We need to find the magnetic field of wire, $B_{\text {wire }}$

The compass needle moves toward a direction of magnetic field. The current in wire makes a magnetic field in available space where the compass is on the ground. The vector sum of the Earth's magnetic field and the wire's magnetic field represents the net magnetic field, as shown in the attached drawing, expressing the angle:

$\tan \theta=\frac{B_{\text {wire}}}{B_{\text {earth}}}$

By substituting the given values, we get

$B_{\text {wire}}=\tan \theta \times B_{\text {earth}}=\tan 8^{\circ} \times 2 \times 10^{-5}=0.1405 \times 2 \times 10^{-5}$

$B_{\text {wire}}=0.28 \times 10^{-5}=2.8 \times 10^{-6}\ T$

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3 years ago

A 0.0140 kg bullet traveling at 205 m/s east hits a motionless 1.80 kg block and bounces off it, retracing its original path wit

makvit [3.9K]

Answer:

Final velocity of the block = 2.40 m/s east.

Explanation:

Here momentum is conserved.

Initial momentum = Final momentum

Mass of bullet = 0.0140 kg

Consider east as positive.

Initial velocity of bullet = 205 m/s

Mass of Block = 1.8 kg

Initial velocity of block = 0 m/s

Initial momentum = 0.014 x 205 + 1.8 x 0 = 2.87 kg m/s

Final velocity of bullet = -103 m/s

We need to find final velocity of the block( u )

Final momentum = 0.014 x -103+ 1.8 x u = -1.442 + 1.8 u

We have

2.87 = -1.442 + 1.8 u

u = 2.40 m/s

Final velocity of the block = 2.40 m/s east.

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4 years ago

There was a major collision of an asteroid with the Moon in medieval times. It was described by monks at Canterbury Cathedral in

LekaFEV [45]

Answer:

c = 3.00E108 m/s = 3.00E5 km/s

t = S / v = 3.84E5 / 3.00E5 = 1.28 sec

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3 years ago

Electron dot diagrams for four elements are shown. Which element would be<br> found in Group 14?

Aleonysh [2.5K]

Answer:

A (2)

Explanation:

Ap3x

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3 years ago