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KatRina [158]
3 years ago
13

90cm uniform lever has a load 30N suspended at 15cm from one of it's end. If the fulcrum is at the center of gravity. The force

that must be applied at it's other hand to keep it in horizontal equilibrium is.​
Physics
1 answer:
solniwko [45]3 years ago
3 0

Answer: F = 20 N

Explanation:

I will ASSUME that the fulcrum is at the center of gravity of the lever arm, This means that the lever arm itself creates no moment about the fulcrum because there is no moment arm for that particular force.

To solve, we sum moments about any convenient point to zero (zero because there is no acceleration in the F = ma equation)

The easiest convenient point is the fulcrum

30((90/2) - 15) - F(90/2) = 0

           30(30) = F(45)

                    F = 900/45 = 20 N

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A student lifts 100 N of books 1 m to his shoulder. He walks forward
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3 years ago
Two charged point-like objects are located on the x-axis. The point-like object with charge q1 = 4.60 µC is located at x1 = 1.25
mylen [45]

Answer:

a) the total electric potential is 2282000 V

b) the total electric potential (in V) at the point with coordinates (0, 1.50 cm) is 1330769.23 V

Explanation:

Given the data in the question and as illustrated in the image below;

a) Determine the total electric potential (in V) at the origin.

We know that; electric potential due to multiple charges is equal to sum of electric potentials due to individual charges

so

Electric potential at p in the diagram 1 below is;

Vp = V1 + V2

Vp = kq1/r1 + kq2/r2

we know that; Coulomb constant, k = 9 × 10⁹ C

q1 = 4.60 uC = 4.60 × 10⁻⁶ C

r1 = 1.25 cm = 0.0125 m

q2 = -2.06 uC = -2.06 × 10⁻⁶ C

location x2 = −1.80 cm; so r2 = 1.80 cm = 0.018 m

so we substitute

Vp = ( 9 × 10⁹ × 4.60 × 10⁻⁶/ 0.0125 ) + ( 9 × 10⁹ × -2.06 × 10⁻⁶ / 0.018 )

Vp = (3312000) + ( -1030000 )

Vp = 3312000 -1030000

Vp = 2282000 V

Therefore, the total electric potential is 2282000 V

b)

the total electric potential (in V) at the point with coordinates (0, 1.50 cm).

As illustrated in the second image;

r1² = 0.015² + 0.0125²

r1 = √[ 0.015² + 0.0125² ]

r1 = √0.00038125

r1 = 0.0195

Also

r2² = 0.015² + 0.018²

r2 = √[ 0.015² + 0.018² ]

r2 = √0.000549

r2 = 0.0234

Now, Electric Potential at P in the second image below will be;

Vp = V1 + V2

Vp = kq1/r1 + kq2/r2

we substitute

Vp = ( 9 × 10⁹ × 4.60 × 10⁻⁶/ 0.0195 ) + ( 9 × 10⁹ × -2.06 × 10⁻⁶ / 0.0234 )

Vp = 2123076.923 + ( -762962.962 )

Vp = 2123076.923 -792307.692

Vp =  1330769.23 V

Therefore, the total electric potential (in V) at the point with coordinates (0, 1.50 cm) is 1330769.23 V

4 0
3 years ago
Two small plastic spheres are given positive electrical charges. When they are a distance of 14.8cm apart, the repulsive force b
sdas [7]

Answer:

Explanation:

Case I: They have same charge.

Charge on each sphere = q

Distance between them, d = 14.8 cm = 0.148 m

Repulsive force, F = 0.235 N

Use Coulomb's law in electrostatics

F=\frac{Kq_{1}q_{2}}{d^{2}}

By substituting the values

0.235=\frac{9\times10^{9}q^{2}}{0.148^{2}}

q=7.56\times10^{-7}C

Thus, the charge on each sphere is q=7.56\times10^{-7}C.

Case II:

Charge on first sphere = 4q

Charge on second sphere = q

distance between them, d = 0.148 m

Force between them, F = 0.235 N

Use Coulomb's law in electrostatics

F=\frac{Kq_{1}q_{2}}{d^{2}}

By substituting the values

0.235=\frac{9\times 10^{9}\times 4q^{2}}{0.148^{2}}

q=3.78\times10^{-7}C

Thus, the charge on second sphere is q=3.78\times10^{-7}C and the charge on first sphere is 4q = 4\times 3.78\times 10^{-7}=1.51 \times 10^{-6} C.

6 0
3 years ago
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