What is the half of 3.24<br> Plzzz help me

ASK YOUR TEACHER A meter stick is found to balance at the 49.7-cm mark when placed on a fulcrum. When a 51.5-gram mass is attach

Simora [160]

Answer:

0.114 kg or 114 g

Explanation:

From the diagram attaches,

Taking the moment about the fulcrum,

sum of clockwise moment = sum of anticlockwise moment.

Wd = W'd'

Where W = weight of the mass, W' = weight of the meter rule, d = distance of the mass from the fulcrum, d' = distance of the meter rule.

make W' the subject of the equation

W' = Wd/d'................ Equation 1

Given: W = mg = 0.0515(9.8) = 0.5047 N, d = (39.2-16) = 23.2 cm, d' = (49.7-39.2) = 10.5 cm

Substitute these values into equation 1

W' = 0.5047(23.2)/10.5

W' = 1.115 N.

But,

m' = W'/g

m' = 1.115/9.8

m' = 0.114 kg

m' = 114 g

6 0

3 years ago

Two large, parallel, nonconducting sheets of positive charge face each other. What is at points (a) to the left of the sheets, (

alexira [117]

Answer:

a)The electric Field will be zero at the point between the sheets

b) $E_1=\dfrac{\sigma}{\epsilon_0}$

c) $E_2=\dfrac{\sigma}{\epsilon_0}$

Explanation:

Let $\sigma$ be the surface charge density of the of the non conducting parallel sheet.Let consider a Gaussian surface in the form of of cylinder such that its cross-sectional is A . Then there will be flux only due to cross sectional area as the curved sectional is perpendicular to the the electric field so the Electric Flux due to it is zero.