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Fudgin [204]
4 years ago
14

One knife is longer from the cutting edge to the back of the blade. It's easier to push the longer knife through a pear and more

surface is in contact with the pear. Why is this an example of mechanical advantage?
a. the juices of the pear help the knife slide through


b. the longer the knife, the stronger the output force


c. the knife changes the downward force into a sideways force


d. not enough information to tell


WILL MARK BRAINLYEST PLS HELPPP
Physics
1 answer:
AlladinOne [14]4 years ago
6 0

By the formula of torque balance we know that

\tau = r\times F

here we know that

r = distance of force

F = applied force

so here we know that as we will take large edge knife then the distance of force will increase

due to this the torque will also increase in it so it is easy to pear it

So here we can say

F_{in}r_1 = F_{out}r_2

F_{out} = \frac{r_1}{r_2} F_{in}

so output force will increase in this case

So here the correct answer would be

<em>b. the longer the knife, the stronger the output force</em>

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Answer:

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3 years ago
Two 10-cm-diameter charged rings face each other, 25.0cm apart. Both rings are charged to +20.0nC. What is the electric field st
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Answer:

Part A:

E_{midpoint}=0

Part B:

E_{center}=2711.7558 N/C

Explanation:

Part A:

Formula of Electric Field Strength:

E=\frac{1}{4\pi\epsilon}\frac{xQ}{(x^2+R^2)^{3/2}}

Where:

x is the distance from the ring

R is the radius of the ring

\epsilon is constant permittivity of free space=8.854*10^-12 farads/meter

Q is the charge

For right Ring E at the midpoint can be calculated as:

x for right plate=25/2=12.5 cm=0.125 m

Radius=R=10/2=5 cm=0.05 m

E_{right}=\frac{1}{4\pi8.854*10^{-12}}\frac{(0.125)*(20*10^{-19})}{((0.125)^2+(0.05)^2)^{3/2}}\\E_{right}=9208.1758 N/C

For Left Ring E at the midpoint can be calculated as:

Since charge on both plates is +ve and same in magnitude, the electric field will be same for both plates.

E_{left}=\frac{1}{4\pi8.854*10^{-12}}\frac{(0.125)*(20*10^{-19})}{((0.125)^2+(0.05)^2)^{3/2}}\\E_{left}=9208.1758 N/C

Electric Field at midpoint:

Both rings have same magnitude but the direction of fields will be opposite as they have same charge on them.

E_{midpoint}=E_{left}-E_{right}\\E_{midpoint}=9208.1758-9208.1758\\E_{midpoint}=0

Part B:

At center of left ring:

Due to left ring Electric field at center is zero because x=0.

E_{left}=\frac{1}{4\pi8.854*10^{-12}}\frac{(0)*(20*10^{-19})}{((0)^2+(0.05)^2)^{3/2}}\\E_{left}=0 N/C

Due to right ring Electric field at center of left ring:

Now: x=25 cm= o.25 m (To the center of left ring)

E_{right}=\frac{1}{4\pi8.854*10^{-12}}\frac{(0.25)*(20*10^{-19})}{((0.25)^2+(0.05)^2)^{3/2}}\\E_{right}=2711.7558 N/C

Electric Field Strength at center of left ring is same as that of right ring.

E_{center}=2711.7558 N/C

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-19.5m

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