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svetoff [14.1K]
3 years ago
5

Describe the relationship between amperage, voltage, and, as well as resstance.

Physics
1 answer:
Nikolay [14]3 years ago
5 0

Answer:

I=\frac{V}{R}

Explanation:

Ohms law says that the current in the conductor between the two points is directly related or directly proportional to the voltage by keeping a proportionality of constant which is called resistance.

Mathematically Ohms law can be written as,

I=\frac{V}{R}

Here, I is the current measure in ampere or amperage, V is voltage measure in voltage and R is resistance measure in Ohm.

Now the relation between the amperage, voltage and resistance will be,

I=\frac{V}{R}

Amperage=Voltage/Resistance

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An object has the acceleration graph shown in (Figure 1). Its velocity at t=0s is vx=2.0m/s. Draw the object's velocity graph fo
timama [110]

Answer:

Explanation:

We may notice that change in velocity can be obtained by calculating areas between acceleration lines and horizontal axis ("Time"). Mathematically, we know that:

v_{b}-v_{a} = \int\limits^{t_{b}}_{t_{a}} {a(t)} \, dt

v_{b} = v_{a}+ \int\limits^{t_{b}}_{t_{a}} {a(t)} \, dt

Where:

v_{a}, v_{b} - Initial and final velocities, measured in meters per second.

t_{a}, t_{b} - Initial and final times, measured in seconds.

a(t) - Acceleration, measured in meters per square second.

Acceleration is the slope of velocity, as we know that each line is an horizontal one, then, velocity curves are lines with slopes different of zero. There are three region where velocities should be found:

Region I (t = 0 s to t = 4 s)

v_{4} = 2\,\frac{m}{s}  +\int\limits^{4\,s}_{0\,s} {\left(-2\,\frac{m}{s^{2}} \right)} \, dt

v_{4} = 2\,\frac{m}{s}+\left(-2\,\frac{m}{s^{2}} \right) \cdot (4\,s-0\,s)

v_{4} = -6\,\frac{m}{s}

Region II (t = 4 s to t = 6 s)

v_{6} = -6\,\frac{m}{s}  +\int\limits^{6\,s}_{4\,s} {\left(1\,\frac{m}{s^{2}} \right)} \, dt

v_{6} = -6\,\frac{m}{s}+\left(1\,\frac{m}{s^{2}} \right) \cdot (6\,s-4\,s)

v_{6} = -4\,\frac{m}{s}

Region III (t = 6 s to t = 10 s)

v_{10} = -4\,\frac{m}{s}  +\int\limits^{10\,s}_{6\,s} {\left(2\,\frac{m}{s^{2}} \right)} \, dt

v_{10} = -4\,\frac{m}{s}+\left(2\,\frac{m}{s^{2}} \right) \cdot (10\,s-6\,s)

v_{10} = 4\,\frac{m}{s}

Finally, we draw the object's velocity graph as follows. Graphic is attached below.

3 0
4 years ago
A ship's propeller of diameter 3 m makes 10.6 revolutions in 30s. What is the angular velocity of the propeller?
ycow [4]

Answer:

The angular velocity of the propeller is 2.22 rad/s.

Explanation:

The angular velocity (ω) of the propeller is:  

\omega = \frac{\Delta \theta}{\Delta t}                              

Where:

θ: is the angular displacement = 10.6 revolutions

t: is the time = 30 s

\omega = \frac{\Delta \theta}{\Delta t} = \frac{10.6 rev*\frac{2\pi rad}{1 rev}}{30 s} = 2.22 rad/s

Therefore, the angular velocity of the propeller is 2.22 rad/s.

I hope it helps you!

5 0
3 years ago
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