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satela [25.4K]
3 years ago
5

What was Sylvester James Gates biggest discovery

Physics
2 answers:
Dahasolnce [82]3 years ago
8 0
Sylvester James Gates' biggest discovery was the string theory which involves computer coding. I hope this answer helped.
Mekhanik [1.2K]3 years ago
6 0
Superstring theory and the generalized complex structure
You might be interested in
NEED HELP NOW. I WILL BRAINLIEST
NeTakaya

Answer:

number of Protons=16

number of Neutrons=16

number of electrons=32

Explanation:

Tip:-

<em><u>Always the number of protons = the number of neutrons.</u></em>

<em><u>Add them and you will get the number of electrons.</u></em>

<u><em>Happy to help</em></u>

ADD ME BRAINLIEST AS YOU SAID

YOUR WELCOME

3 0
3 years ago
Which is an example of a solution?
vladimir1956 [14]

Answer:

C

Explanation:

plato

6 0
3 years ago
A coaxial cable has a charged inner conductor (with charge +8.5 µC and radius 1.304 mm) and a surrounding oppositely charged con
Tcecarenko [31]

Complete question:

A 50 m length of coaxial cable has a charged inner conductor (with charge +8.5 µC and radius 1.304 mm) and a surrounding oppositely charged conductor (with charge −8.5 µC and radius 9.249 mm).

Required:

What is the magnitude of the electric field halfway between the two cylindrical conductors? The Coulomb constant is 8.98755 × 10^9 N.m^2 . Assume the region between the conductors is air, and neglect end effects. Answer in units of V/m.

Answer:

The magnitude of the electric field halfway between the two cylindrical conductors is 5.793 x 10⁵ V/m

Explanation:

Given;

charge of the coaxial capable, Q = 8.5 µC = 8.5  x 10⁻⁶ C

length of the conductor, L = 50 m

inner radius, r₁ = 1.304 mm

outer radius, r₂ = 9.249 mm

The magnitude of the electric field halfway between the two cylindrical conductors is given by;

E = \frac{\lambda}{2\pi \epsilon_o r} = \frac{Q}{2\pi \epsilon_o r L}

Where;

λ is linear charge density or charge per unit length

r is the distance halfway between the two cylindrical conductors

r = r_1 + \frac{1}{2}(r_2-r_1) \\\\r = 1.304 \ mm \ + \  \frac{1}{2}(9.249 \ mm-1.304 \ mm)\\\\r = 1.304 \ mm \ + \ 3.9725 \ mm\\\\r = 5.2765 \ mm

The magnitude of the electric field is now given as;

E = \frac{8.5*10^{-6}}{2\pi(8.85*10^{-12})(5.2765*10^{-3})(50)} \\\\E = 5.793*10^5 \ V/m

Therefore, the magnitude of the electric field halfway between the two cylindrical conductors is 5.793 x 10⁵ V/m

5 0
3 years ago
A 0.144-kg baseball is moving toward home plate with a speed of 43 m/s when
algol [13]
I would say 648858. bc yes
4 0
3 years ago
Find the average force exerted by the bat on the ball if the two are in contact for 0.00129 s. Answer in units of N.
Semmy [17]

Answer:

Explanation:

Given

time of contact between bat and ball is t=0.00129 s

suppose u is the incoming velocity and v is the final velocity after collision

m=mass\ of\ ball

Impulse exerted is given by change in momentum of the particle.

Initial momentum P_i=m\times u

Final momentum P_f=m\times v

Change in momentum \Delta P=P_i-P_f

Impulse    J=F_{avg}\cdot t=\Delta P

J=F_{avg}\cdot t=m(u-v)

F_{avg}=\frac{m(v-u)}{t}

F_{avg}=775.2\times m(v-u)\ N                

5 0
3 years ago
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