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Novay_Z [31]
3 years ago
14

Describe Ohm's law and its limitations;

Physics
1 answer:
siniylev [52]3 years ago
5 0
<span>The Ohm's Law is a powerful and simple tool to analyze electric circuits. It shows the relationship between voltage (V) and current (I) in a conductor. But, in order to properly apply to circuits (real circuits), its limitations must be understood.   Most conductors, resistance is greatly unaffected by current or voltage (resistance is stable property).
Limitation: this law does not apply to rectifiers or semiconductors or to the conduction of electricity through gases.</span>
Formula: current (I) is equal to voltage (V) over resistance (R)


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If a motorcycle accelerates uniformly from rest at 5 m / s2, how long does it take for it to reach a speed of 20 m / s.
irina1246 [14]

Answer:

4 seconds

Explanation:

Given:

v₀ = 0 m/s

v = 20 m/s

a = 5 m/s²

Find: t

v = at + v₀

20 m/s = (5 m/s²) t + 0 m/s

t = 4 s

6 0
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The carbon cycle relies on producers and consumers to survive. What is the role of each in the carbon cycle?
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8 0
3 years ago
An object with little mass won't require a lot of force to move.
Anna35 [415]

Answer:

True

Explanation:

Heavier objects (objects with more mass) are more difficult to move and stop. Heavier objects (greater mass) resist change more than lighter objects. Example: Pushing a bicycle or a Cadillac, or stopping them once moving. The more massive the object (more inertia) the harder it is to start or stop.

7 0
3 years ago
A stone of mass 150g is rotated in a horizontal circle at 10m/s which is attached to the end of a 1m long. what will be the acce
erastova [34]

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8 0
2 years ago
Read 2 more answers
A large, cylindrical water tank with diameter 3.60 m is on a platform 2.00 m above the ground. The vertical tank is open to the
zysi [14]

To solve this problem it is necessary to apply the concepts related to the geometry of a cylindrical tank and its respective definition.

The volume of a tank is given by

V = \frac{\pi d^2}{4}h

Where

d = Diameter

h = Height

Considering that there are two stages, let's define the initial and final volume as,

V_0 = \frac{\pi d^2}{4}H

V_f = \frac{\pi d^2}{4}h

We know as well by definition that

1gal = 3.785*10^{-3}m^3

Then we have for the statement that

V_f = V_0 -1gal

V_f = V_0 - 3.785*10^{-3}

Replacing the previous data

\frac{\pi d^2}{4}h = \frac{\pi d^2}{4}H- 3.785*10^{-3}

\frac{\pi (3.6)^2}{4}h = \frac{\pi (3.6)^2}{4}(2)- 3.785*10^{-3}

Solving to get h,

h = 1.99963m

Therefore the change is

\Delta h = H-h

\Delta h = 2- 1.99963

\Delta h = 3.7*10^{-4}m=0.37mm

Therefore te change in the height of the water in the tank is 0.37mm

4 0
3 years ago
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