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3 years ago

What evidence did Alfred Wagner have to support his idea of continental drift

1 answer:

8 0

Alfred Wagner's evidence began at the simplest form that the continents appeared to<em> fit</em> together like pieces to a puzzle. It wasn't only that though. The evidence showed that the same rocks and fossils existed on separate continents - the same continents that if pieced together would be connected.

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Mike mixes two chemicals in a container. The container quickly becomes very hot.

Kipish [7]

The chemicals are reactive with one another

7 0

2 years ago

Read 2 more answers

Which has the reverse chemical reaction as

Mazyrski [523]

Answer:

please further explain your question

4 0

2 years ago

1. What is the difference between accidental discoveries and scientific investigations?

Alenkasestr [34]

Answer:

the discovery will always need to be tested more where if you're investigating you should already be in a more controlled environment

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2 years ago

Which of the following conditions would speed up the rate of decomposition? (Select

Anna11 [10]

Answer:

E and B

Explanation:

the warmer the weather the faster it will decompose, this why is why sometimes bodies last longer frozen, or in cool temperatures.

that statement takes a out.

An illness shouldn't effect a decomposing body

Weight also shouldn't effect a decomposing body

Being buried can make bodies decompose fast because of all the bugs and other animals that will get down there and decompose it faster

Answer: B,E

4 0

3 years ago

We can also perform a similar calculation for the mass defect and binding energy for nuclear reactions using the masses of the a

sukhopar [10]

Answer:

See Explanation

Explanation:

$\frac{235}{92} U + \frac{1}{0} n ---->\frac{137}{52} Te + \frac{97}{40} Zr +2\frac{1}{0} n$

Hence the mass defect is;

[235.04393 + 1.00867] - [ 136.92532 + 96.91095 + 2(1.00867)]

= 236.0526 - 235.85361

= 0.19899 amu

Since 1 amu = 1.66 * 10^-27 Kg

0.19899 amu = 0.19899 * 1.66 * 10^-27 = 3.3 * 10^-28 Kg

Binding energy = Δmc^2

Binding energy = 3.3 * 10^-28 Kg * (3 * 10^8)^2 = 2.97 * 10^-11 J

ii) $\frac{10}{5}B + \frac{1}{0}n-----> \frac{7}{3} Li + \frac{4}{2} He + Energy$

Hence the mass defect is;

[10.01294 + 1.00867] - [7.01600 + 4.00260]

= 11.02161 - 11.0186

= 0.00301 amu

Since 1 amu = 1.66 * 10^-27 Kg

0.00301 amu = 0.00301 * 1.66 * 10^-27 = 4.997 * 10^-30 Kg

Binding energy = Δmc^2

Binding energy = 4.997 * 10^-30 Kg * (3 * 10^8)^2 = 4.5 * 10^-13 J

7 0

2 years ago