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Digiron [165]
2 years ago
7

An effort of 200 N is used to lift a load of 800 N by using a lever. If the load is at a distance of 40 cm from the fulcrum then

find the effort distance​
Physics
1 answer:
Alchen [17]2 years ago
7 0

The effort distance​ will be 160 cm.Applying the moment at the center as follows will provide the effort distance:

<h3 /><h3>What is the mechanical advantage?</h3>

Mechanical advantage is a measure of the ratio of output force to input force in a system, it is used to obtain the efficiency of forces in levers and pulleys.

Given data;

Effort,\rm F_e=00 N

Load,\rm P= 400 \ N

Distance from the fulcrum,\rm d=40 \ cm

The effort distance​ is found by applying the moment at the center as;

\rm F_e \times d= P \times  d' \\\\ 200 \  N \times d'=800 \ N \times 40 \ cm \\\\ d'=160 \ cm

Hence, the effort distance​ will be 160 cm

To learn more about the mechanical advantage refer to the link;

brainly.com/question/7638820

#SPJ1

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Resistors 1 and 2− R1 = 50 Ω , R2 = 90 Ω − are connected in series to a 6.0-V battery. Part APart complete What is the potential
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Part A: The voltage across resistor R1 is approximately \rm 2.1 \; V.

Part B: When the value of resistor R1 decreases, the current in this circuit will increase.

Part C: When the value of resistor R1 decreases, the voltage across resistor R1 will decrease.

Explanation:

<h3>Part A</h3>

Resistor R1 and and R2 are connected in series. That's equivalent to a single resistor of R_1 + R_2 = 50 + 90 = 140\; \Omega. The voltage across the two resistor, combined, is equal to \rm 6\; V. Hence by Ohm's Law, the current through the circuit will be equal to \rm \dfrac{6\; V}{140\; \Omega} = \dfrac{3}{70}\; A.

These two resistors are connected in series. The voltage across each of them might differ. However, the current through each of them should both be equal to the current through the circuit. In this case, the current through both R1 and R2 should be equal to \rm \dfrac{3}{70}\; A. Apply Ohm's Law (again) to find the voltage across R1:

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<h3>Part B</h3>

Since the equivalent resistance is equal to R_1 + R_2, when the value of R_1 decreases, the equivalent resistance will also decrease. By Ohm's Law, I = \dfrac{V}{R}. When the value of the denominator ( decreases, the value of the quotient, I the current through the circuit, will increase.

<h3>Part C</h3>

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I(R_1) = I(\text{Circuit}) = I(R_2).

The resistance of R1 decreases, while the current through it increases. Applying Ohm's Law on R1 won't give much useful information. However, since the resistance of R2 stays the same, the voltage across it will increase when its current increases (again by Ohm's Law.)

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V(R_1) + V(R_2) = V(\text{Circuit}) = \rm 6 \; V,

when the voltage across R2 increases, the voltage across R1 will decrease.

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