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zaharov [31]
3 years ago
12

Charge flows from low potential to high potential. O True or O False

Physics
1 answer:
inn [45]3 years ago
8 0

Answer:

False

Explanation:

Think of the electric potential in terms of potential energy. If you imagine a place with high elevation (A) and another one at sea level (B), a ball will roll from high potential to low potential (A-->B).

Everything in our universe wants to reach a lower state of energy if no external force is acted upon it. Every object tends to slow down (friction), a radioactive element dissipates energy (an unstable element releases energy to get to a stable state), water in the clouds comes down to the ground (rain experiencing difference in potential energy).

Electric potential is exactly the same, you just can't see it! It flows from higher voltage (which is a synonym for electric potential) to lower voltage.

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Anything that has mass and occupies space is defined as.
forsale [732]

Answer:

matter is the correct answer.

8 0
2 years ago
Which quantity and unit are INCORRECTLY paired?
liq [111]
Force, the unit is Newton, newton is the force to accelerate a mass. so it should be kg m/s^2

joule (J) is equal to Nm not Ns

the unit of work is J and it is correct.
the unit of power is J/s which is equal to W
the unit of of energy is the same with work, which is J which equivalent to kgm2/s2
5 0
3 years ago
A simple model of a hydrogen atom is a positive point charge +e (representing the proton) at the center of a ring of radius a wi
Norma-Jean [14]

Answer:

Now e is due to the ring at a

So

We say

1/4πEo(ea/ a²+a²)^3/2

= 1/4πEo ea/2√2a³

So here E is faced towards the ring

Next is E due to a point at the centre

So

E² = 1/4πEo ( e/a²)

Finally we get the total

Et= E²-E

= e/4πEo(2√2-1/2√2)

So the direction here is away from the ring

8 0
3 years ago
Which of the following graphs represents constant positive acceleration
nika2105 [10]
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5 0
3 years ago
If it is fixed at C and subjected to the horizontal 60-lblb force acting on the handle of the pipe wrench at its end, determine
pickupchik [31]

Answer:

τ = 132.773 lb/in² = 132.773 psi

Explanation:

b = 12 in

F = 60 lb

D = 3.90 in (outer diameter)  ⇒ R = D/2 = 3.90 in/2 = 1.95 in

d = 3.65 in (inner diameter)  ⇒ r = d/2 = 3.65 in/2 = 1.825 in

We can see the pic shown in order to understand the question.

Then we get

Mt = b*F*Sin 30°

⇒ Mt = 12 in*60 lb*(0.5) = 360 lb-in

Now we find ωt as follows

ωt = π*(R⁴ - r⁴)/(2R)

⇒ ωt = π*((1.95 in)⁴ - (1.825 in)⁴)/(2*1.95 in)

⇒ ωt = 2.7114 in³

then the principal stresses in the pipe at point A is

τ = Mt/ωt ⇒ τ = (360 lb-in)/(2.7114 in³)

⇒ τ = 132.773 lb/in² = 132.773 psi

7 0
4 years ago
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