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Ivahew [28]
3 years ago
5

The electrons of an atom store nuclear energy.

Physics
2 answers:
Makovka662 [10]3 years ago
5 0

Answer:

False

Explanation:

This statement is False.

The atom is made up of nucleus in which there are neutrons and protons and electrons which are revolving around the atom's nucleus.

nuclear energy is basically energy the source of which is nucleus of the atom. there are two methods by which nuclear energy is produces.

1) Nuclear fission

2) nuclear fusion

In nuclear fission, the neutron is bombarded on the nucleus of an atom of an element. It splits and produces a daughter nucleus, neutrons and huge amount of energy

In nuclear fusion, two nuclei of the atoms fuse and produce huge amount of energy.

In both of above mentioned cases, there is neither any role of electrons to produce energy nor in storing nuclear energy.

So, the given statement is wrong

Key words: Nucleus, Nuclear energy, electron, Nuclear fission, Nuclear fusion, neutrons, atom

Sladkaya [172]3 years ago
4 0
False. The nuclear energy is found within the nucleus. Electrons are located outside the nucleus.
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Lostsunrise [7]
B.) Speeding up......
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3 years ago
You have a set of calipers that can measure thicknesses of a few inches with an uncertainty of ± 0.005 inches. You measure the t
Bond [772]

Answer:

a) x = (0.0114 ± 0.0001) in , b) the number of decks is 5

Explanation:

a) The thickness of the deck of cards (d) is measured and the thickness of a card (x) is calculated

        x = d / 52

        x = 0.590 / 52

        x = 0.011346 in

Let's look for uncertainty

       Δx = dx /dd Δd

       Δx = 1/52 Δd

       Δx = 1/52  0.005

       Δx = 0.0001 in

The result of the calculation is

        x = (0.0114 ± 0.0001) in

b) You want to reduce the error to Δx = 0.00002, the number of cards to be measured is

           #_cards = n 52

The formula for thickness is

           x = d / n 52

Uncertainty

          Δx = 1 / n 52  Δd

         n = 1/52 Δd / Δx

         n = 1/52 0.005 / 0.00002

         n = 4.8

Since the number of decks must be an integer the number of decks is 5

3 0
3 years ago
4. O/N 15/P11/Q11
anygoal [31]

Answer:

B) 4500 Pa

Explanation:

As pressure is force per unit area,

P = F/A

It stands to reason that the smallest pressure for a given force is when it is shared by the largest area.

The possible areas are

0.30(0.40) = 0.12 m²

0.30(0.50) = 0.15 m²

0.40(0.50) = 0.20 m²

The pressure when the face with the largest area (0.20 m²) is down is

P = 900 / 0.20 = 4500 N/m² or 4500 Pa

the other possible pressures would be

900/0.15 = 6000 Pa

900/0.12 = 7500 Pa

which are both larger than our solution.

5 0
3 years ago
Using 6400 km as the radius of Earth, calculate how high above Earth’s surface you would have to be in order to weigh 1/16th of
FrozenT [24]
Gravity obeys the inverse square law.  At 6400 km above the center of the Earth (Earth's surface) you weigh x.  Twice that reduces your weight to 1/4th.  Four times that height reduces your weight to 1/16th.  4 times 6400 km is 25,600 km.  But that is above the center of the earth, and the question requests the height above the surface, so we deduct 6400 km to arrive at our final answer:  19,200 km.

Incidentally, it doesn't exactly work the opposite way.  At the center of the Earth the mass would be equally distributed around you, and you would therefore be weightless.
6 0
3 years ago
What is the moment of inertia of a 2.0 kg, 20-cm-diameter disk for rotation about an axis (a) through the center, and (b) throug
FinnZ [79.3K]

Answer:

(a) I=0.01 kg.m²

(b) I=0.03 kg.m²

Explanation:

Given data

Mass of disk M=2.0 kg

Diameter of disk d=20 cm=0.20 m

To Find

(a) Moment of inertia through the center of disk

(b) Moment of inertia through the edge of disk

Solution

For (a) Moment of inertia through the center of disk

Using the equation of moment  of Inertia

I=\frac{1}{2}MR^{2}\\  I=\frac{1}{2}(2.0kg)(0.20m/2)^{2}\\  I=0.01 kg m^{2}

For (b) Moment of inertia through the edge of disk

We can apply parallel axis theorem for calculating moment of inertia

I=(1/2)MR^{2}+MD\\ Here\\D=R\\I=(1/2)(2.0kg)(0.20m/2)^{2}+(2.0kg)(0.20m/2)^{2}\\  I=0.03kgm^{2}

8 0
3 years ago
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