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3 years ago

A lever used to lift a heavy box has an input arm of 4 meters and an output arm of 0.8 meters. What is the

Physics

1 answer:

Ede4ka [16]3 years ago

6 0

The mechanical advantage is (4meters) / (0.8meter) = <em><u>5</u></em> .

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A machine does 1200 J of work in 1 min. What is the power developed

densk [106]

Answer:

20 watts

Explanation:

Big brain mode activated:

Power=1200J/60sec

Power=20 watts

8 0

2 years ago

A typical atomic polarizability is 1 × 10-40 (C·m)/(N/C). If the q in p = qs is equal to the proton charge e, what charge separa

Alenkasestr [34]

Answer:

s = 6.25 10⁻²² m

Explanation:

Polarizability is the separation of electric charges in a structure, in the case of the atom it is the result of the separation of positive charges in the nucleus and the electrons in their orbits, macroscopically it is approximated by

p = q s

s = p / q

let's calculate

s = 1 10⁻⁴⁰ / 1.6 10⁻¹⁹

s = 0.625 10⁻²¹ m

s = 6.25 10⁻²² m

We see that the result is much smaller than the size of the atom, therefore this simplistic model cannot be taken to an atomic scale.

6 0

2 years ago

Does this graph represent constant or changing speed? How do you know? Find the average speed

stepan [7]

Constant speed because the time is directly proportional to the speed (2). The average speed is 2 m/s

6 0

2 years ago

What property of equality can be used to solve the equation. n – 14 = 25

Inga [223]

Do 25-14 and you will get your answer

7 0

2 years ago

A certain parallel-plate capacitor is filled with a dielectric for which κ = 5.0. The area of each plate is 0.027 m2, and the pl

alexdok [17]

Answer:

Energy will be $4.776\times 10^{-5}J$

Explanation:

We have given that dielectric constant k = 5

Area of the plate $A=0.027m^2$

Distance between the plates $d=2mm=2\times 10^{-3}m$

Electric field E = 200 kN/C

We know that capacitance is given by $C=\frac{K\epsilon _0A}{d}=\frac{5\times 8.85\times 10^{-12}\times 0.027}{2\times 10^{-3}}=0.597\times 10^{-9}F$

We know that electric field is given by $E=\frac{V}{d}$

So $V=Ed=200\times 10^3\times 2\times 10^{-3}=400volt$

So energy will be $E=\frac{1}{2}CV^2=\frac{1}{2}\times 0.597\times 10^{-9}\times 400^2=4.776\times 10^{-5}J$

6 0

2 years ago