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sasho [114]
2 years ago
12

Hi,please i need help with this one in physics.hurry and correct i need it.Topic:Equilibrium.​

Physics
1 answer:
Rudik [331]2 years ago
5 0

The half-meter rule (easy math) is 0.5 meters or 50 centimeters since a meter is 1 meters long, which is equivalent to 100 centimeters. Therefore, we shall apply the 50 cm rule.

A 50 cm rule's center of mass is now 25 cm away.

Additionally, according to the data, the object is pivoted at 15 cm, while the 40 g object is hung at 2 cm from the rule's beginning. Using a straightforward formula, we can compare the two situations: the distance from the pivot to the center of the mass times the mass of the 40 g object divided by 2 cm must equal the distance from the pivot to the center of the mass times mass of the 10 x g object

The result of the straightforward computation must be 52g.

Most simplified version:

the center of mass of the rule is at the 25 cm mark

⇒ 40 g * (15 cm - 2 cm)

⇒ = M * (25 cm - 15 cm)

#SPJ2

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7 0
3 years ago
A single-turn circular loop of wire of radius 5.0 cm lies in a plane perpendicular to a spatially uniform magnetic field. During
Artemon [7]

Answer:

Induced EMF,\epsilon=0.0143\ volts

Explanation:

Given that,

Radius of the circular loop, r = 5 cm = 0.05 m

Time, t = 0.0548 s

Initial magnetic field, B_i=200\ mT=0.2\ T

Final magnetic field, B_f=300\ mT=0.3\ T

The expression for the induced emf is given by :

\epsilon=\dfrac{d\phi}{dt}

\phi = magnetic flux

\epsilon=\dfrac{d(BA)}{dt}

\epsilon=A\dfrac{d(B)}{dt}

\epsilon=A\dfrac{B_f-B_i}{t}

\epsilon=\pi (0.05)^2\times \dfrac{0.3-0.2}{0.0548}

\epsilon=0.0143\ volts

So, the induced emf in the loop is 0.0143 volts. Hence, this is the required solution.

5 0
3 years ago
A parallel-plate air capacitor is made from two plates 0.190 m square, spaced 0.770 cm apart. It is connected to a 120 V battery
Naddik [55]

Answer:

aaksj

Explanation:

a) the capacitance is given of a plate capacitor is given by:

C = \epsilon_0*(A/d)

Where \epsilon_0 is a constant that represents the insulator between the plates (in this case air, \epsilon_0 = 8.84*10^(-12) F/m), A is the plate's area and d is the distance between the plates. So we have:

The plates are squares so their area is given by:

A = L^2 = 0.19^2 = 0.0361 m^2

C = 8.84*10^(-12)*(0.0361/0.0077) = 8.84*10^(-12) * 4.6883 = 41.444*10^(-12) F

b) The charge on the plates is given by the product of the capacitance by the voltage applied to it:

Q = C*V = 41.444*10^(-12)*120 = 4973.361 * 10^(-12) C = 4.973 * 10^(-9) C

c) The electric field on a capacitor is given by:

E = Q/(A*\epsilon_0) = [4.973*10^(-9)]/[0.0361*8.84*10^(-12)]

E = [4.973*10^(-9)]/[0.3191*10^(-12)] = 15.58*10^(3) V/m

d) The energy stored on the capacitor is given by:

W = 0.5*(C*V^2) = 0.5*[41.444*10^(-12) * (120)^2] = 298396.8*10^(-12) = 0.298 * 10 ^6 J

6 0
3 years ago
Read 2 more answers
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