Average velocity of Mike Phelps swimming 100 m race in the 50 m long pool (2 laps) is approximately equal to *

Why are chemical processes unable to produce the same amount of energy flowing out of the sun as nuclear fusion?

julsineya [31]

Answer:

Explanation:

One of the major differences between nuclear reactions and chemical reactions is that nuclear reactions involve larger amount of energy than chemical energy. <u>This is because the force between the protons and neutrons in the nucleus of an atom is much higher than the force of attraction between electrons and the positively charged nucleus, hence nuclear reactions involves/requires a larger amount of energy (because it's reactions involve the nucleus) than chemical reactions (because it's reactions involve the electrons)</u>.

Thus, during nuclear fusion, two light nuclei are bombarded against one another to produce a larger/heavier nuclei with the release of large amount of energy (because the forces between the protons and neutrons are much higher) unlike when two atoms/molecules are chemically combined together to form a new molecule with the rearrangement of electrons in the valence shells of the participating molecules.

3 0

4 years ago

Dolphins and other sea creatures can leap to great heights by swimming straight up and exiting the water at a high speed. A 200

scoundrel [369]

Answer:

i don't know sorry

Explanation:

so if a pie is baked

5 0

3 years ago

Please answer someone!!

Naily [24]

Answer:

10m

Explanation:

3 0

3 years ago

a length of wire is cut into five equal pieces. the five pieces are then connected in parallel, with the resulting resistance be

Natali [406]

The equivalent resistance of n resistors connected in parallel is given by
$\frac{1}{R_{eq}} = \frac{1}{R_1}+ \frac{1}{R_2}+...+ \frac{1}{R_n}$ (1)

In our problem, the resulting resistance of the 5 pieces connected in parallel is $R_{Eq}=2.00 \Omega$ , and since the 5 pieces are identical, their resistance R is identical, so we can rewrite (1) as
$\frac{1}{R_{Eq} }= \frac{1}{2 \Omega}= \frac{1}{R}+ \frac{1}{R} + \frac{1}{R} + \frac{1}{R} + \frac{1}{R} = \frac{5}{R}$
From which we find $R= 5 \cdot 2 \Omega = 10 \Omega$ .

So, each piece of wire has a resistance of $10 \Omega$ . Before the wire was cut, the five pieces were connected as they were in series. The equivalent resistance of a series of n resistors is given by
$R_{Eq}=R_1 + R_2 + ...+R_n$
So if we apply it at our case, we have
$R_{eq}=R+R+R+R+R=5 R= 5\cdot 10 \Omega= 50 \Omega$

therefore, the resistance of the original wire was $50 \Omega$ .

5 0

3 years ago

A block of mass 10.1kg rests on a slope an angle of 43.9∘ relative to the horizontal. What is the size of the force parallel to

Julli [10]

Cncnfnjfjfjfjjffjjfjffjfjfjfjfjf

8 0

3 years ago