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tatiyna
3 years ago
5

If object A has more mass than object B, what will object A need to accelerate at the same rate as object B?

Physics
1 answer:
Leni [432]3 years ago
3 0

Answer:

More force

Explanation:

Object A has more mass than object B

  For object A to accelerate at the same rate as object B, it will need more force.

According to Newton's second law of motion "the net force on a body is the product of its mass and acceleration".

  Net force  = mass x acceleration

Now, if a body has more mass and needs to accelerate at the same rate as another one with a lower mass, the force on it must be increased.

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The Nucleus of the Atom is in the center of the Atom, not in the outer rings & orbitals.
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Answer:

true

Explanation:

this the nucleus is located at the centre and contains protons and neutrons

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3 years ago
When working with two-dimensional motion, choose the direction of _________________ to be a positive direction.
emmainna [20.7K]

Answer:

two dimensional to the motion means motion that takes place into different directions or coordinators at the same time the simplest motion would be an object moving liner in one dimension and example of liner movement would be car moving along a state rod and ball thrown straight from the ground

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Which of these is an example of energy transfer by radiation?
andrezito [222]

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3 years ago
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Each plate of a parallel‑plate capacitor is a square of side 3.63 cm, and the plates are separated by 0.473 mm. The capacitor is
NARA [144]

Answer:

E= 55.53 x 10³ V/m

Explanation:

Given that

a=  3.63 cm

Area ,A= a²

distance ,d= 0.473 mm

Stored energy ,U = 8.49 nJ

Value of capacitor given as

C=\dfrac{\varepsilon _oA}{d}

By putting the values

C=\dfrac{8.85\times 10^{-12}\times 3.63^2\times 10^{-4}}{0.473\times 10^{-3}}

C=2.46 x 10⁻¹¹ F

U=\dfrac{1}{2}CV^2

V=Voltage difference

V=\sqrt{\dfrac{2U}{C}}

V=\sqrt{\dfrac{2\times 8.49\times 10^{-9}}{2.46\times 10^{-11}}}

V=26.27 V

V= E d

E=Electric filed

26.27 = E x 0.473 x 10⁻³

E= 55.53 x 10³ V/m

7 0
4 years ago
Of all the hydrogen in the oceans, 0.0300 % of the mass is deuterium. The oceans have a volume of 317 million mi³.(a) If nuclear
jenyasd209 [6]

Of all the hydrogen in the oceans, 0.0300 % of the mass is deuterium. The oceans have a volume of 319 million mi³. If nuclear fusion were controlled and all the deuterium in the oceans were fused to ⁴₂He, the joules of energy  released is E = 6.912×10^-^1^2J

<h3>How is the energy in joules calculated for a controlled nuclear fusion with all the deuterium in oceans fused to ⁴₂He?</h3>

Given the percentage of hydrogen in the ocean = 0.300%

Volume of the ocean = 319 million mi³

Total volume of the ocean, V = 319 × 10^6 mi³

V = 1.330×10^1^8 m^3

Density of water = 1000 kg/m^3

Avogadro's number =6.022 × 10^2^3

There are two hydrogen atoms in a water molecule

Molar mass of Hydrogen =2.016×10^-^3 kg/mole

Molar mass of deuterium molecule = 4.028204 × 10^-^3kg/mole

Total mass of sea = Volume × Density

= 1.330 × 10^2^1 kg

Total number of water molecules available in the given mass of water =

[Mass of sea water / Molar mass of water] × Avogadro's number

= 4.446 × 10^4^6

Total mass of hydrogen in the given mass of water =

[Number of water molecules × Molar mass of hydrogen molecule] /  Avogadro's number

= 1.488 × 10^2^0 kg

Total mass of deuterium in the calculated mass of hydrogen =

Mass of hydrogen × [ density / 100] =

4.464 × 10^1^6 kg

Total number of deuterium atoms available=

[Total mass of deuterium / Molar mass of deuterium molecule] × Avogadro's number × 2 = 1.335 × 10^4^6

The combined process consumes 6 deuterium atoms and produces two helium atoms and a total of

E = 43.2×10^6 eV of energy.

Therefore the energy release in the consumption of 6 atoms

E=43.2×10^6×1.6×10^1^910^-^1^9

E=6.912×10^-^1^2J

To learn more about nuclear fusion refer

brainly.com/question/23765509

#SPJ4

4 0
2 years ago
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