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-BARSIC- [3]
2 years ago
14

Explain how sharpening a knife changes is MA (mechanical advantage).

Physics
1 answer:
stepladder [879]2 years ago
7 0

Answer:

Sharpening a knife reduces it's height as a wedge thus giving it more efficiency.

Explanation:

Mechanical advantage refers to the degree or quantity to which a force is increased or amplified by using a tool or a machine.

Mechanical Advantage can be calculated as follows:

Lenght of Incline (LI) / Height of Incline (HI)

Assume I have constructed a hedge with an inclination that run 10 meters from point x to point y (which is the back of a truck. Point y to the floor is 5 meters.

If   LI = 10m and

HI= 5m

The mechanical advantage is given as 10/5 = 2

The angle of incline is also called the effort distance. From our calculations above, by sacrificing horizontal movement, I reduced the amount of effort it would have cost to lift up that object from ground up -  a distance of 5 meters.

Think of a knife as a form of a similar wedge.

The more a knife is sharpening of a knife further reduces its angle of elevation. Thats why it becomes more efficient to use, hence our mechanical advantage.

Cheers

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a certain projetor uses a concave mirror for projecting an object's image on a screen .it produces on image that is 4 times bigg
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Answer:

f = 1 m

Explanation:

The magnification of the lens is given by the formula:

M = \frac{q}{p}

where,

M = Magnification = 4

q = image distance = 5 m

p = object distance = ?

Therefore,

4 = \frac{5\ m}{p}\\\\p = \frac{5\ m}{4}\\\\p = 1.25\ m

Now using thin lens formula:

\frac{1}{f}=\frac{1}{p}+\frac{1}{q}\\\\\frac{1}{f} = \frac{1}{1.25\ m}+\frac{1}{5\ m}\\\\\frac{1}{f} = 1\ m^{-1}\\\\

<u>f = 1 m</u>

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2 years ago
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What is something the screw can do that the other simple machines can't
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Olivia places her pet frog on a line to observe the frog’s motion. The line is divided into sections that measure 1 centimeter e
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Answer:

Explanation:

Let the forward displacement is taken is positive, and the backward displacement is taken is negative.

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4 0
3 years ago
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A series RLC circuit with R = 15.0 ohms, C = 4.70 uF, and L = 25.0mH is connected to an AC voltage source with V(t) = 75.0 Vrms
Brilliant_brown [7]

(a) The rms current in the circuit is 2.58 A.

(b) The rms voltage of Vab is 38.7 V, Vbc is 158.83 V, Vcd is 222.93 V, Vbd is 64.11 V, and Vad is 75 V.

(c) The average rate at which energy is dissipated in each of the 3 circuit elements is 193.23 W.

<h3>Capacitive reactance of the circuit</h3>

The capacitive reactance of the circuit is calculated as follows;

X_c = \frac{1}{\omega C} = \frac{1}{2\pi fC} = \frac{1}{2\pi \times 550 \times 4.7 \times 10^{-6}} \\\\X_c = 61.56 \ ohms

<h3>Inductive reactance of the circuit</h3>

Xl = ωL

Xl = 2πfL

Xl = 2π x 550 x 25 x 10⁻³

Xl = 86.41 ohms

<h3>Impedance of the circuit</h3>

Z = \sqrt{R^2 + (X_L - X_C)^2} \\\\Z = \sqrt{15^2 + (86.41 -61.56)^2 } \\\\Z = 29.03 \ ohms

<h3>rms current in the circuits</h3>

I_{rms} = \frac{V_{rms}}{Z} \\\\I_{rms} = \frac{75}{29.03} \\\\I_{rms} = 2.58 \ A

<h3>rms voltage in resistor (Vab)</h3>

V_{ab} = I_{rms} R\\\\V_{ab} = 2.58 \times 15\\\\V_{ab} = 38.7 \ V

<h3>rms voltage in capacitor (Vbc)</h3>

V_{bc} = I_{rms} X_c\\\\V_{bc} = 2.58 \times 61.56\\\\V_{bc} = 158.83 \ V

<h3>rms voltage in inductor (Vcd)</h3>

V_{cd} = I_{rms} X_l\\\\V_{cd} = 2.58 \times 86.41\\\\V_{cd} = 222.93\ V

<h3>rms voltage in capacitor and inductor (Vbd)</h3>

V_{bd} = I_{rms} \times (X_l - X_c)\\\\V_{bd} = 2.58 \times (86.41 - 61.56)\\\\V_{bd} = 64.11 \ V

<h3>rms voltage in resistor, capacitor and inductor (Vad)</h3>

V_{ad} = I_{rms} \times \sqrt{R^2 + (X_l- X_c)^2} \\\\V_{ad} = 2.58 \times Z \\\\V_{ad} = 2.58 \times 29.03\\\\V_{ad} = 75 \ V

<h3>Average rate of energy dissipation in the 3 circuit element</h3>

P = I_{rms}^2 Z\\\\P = (2.58)^2 \times 29.03\\\\P = 193.23 \ W

Learn more about RLC series circuit here: brainly.com/question/15595203

6 0
2 years ago
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