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valkas [14]
3 years ago
15

State clearly how the pith ball electroscope may be used to test for the kind of charge on a body

Physics
1 answer:
PilotLPTM [1.2K]3 years ago
5 0
An object with a known charge is first brought close to the ball so as to induce the opposite charge. Lets say we brought a positive object, so the ball will be negatively charged. Afterwards, the object with the unknown charge is brought closer. If the ball is attracted, then the object is positively charged and if it is repelled then the object is negatively charged.
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In a tug-of-war game on one campus, 15 students pull on a rope at both ends in an effort to displace the central knot to one sid
mojhsa [17]

Answer:

Net pull = 110 N to the left

Explanation:

Group the different pulls according to the direction (right or left)

2 pull 196 N each to the right

4 pull 98 N each to the left

5 pull 62 N each to the left

3 pull 150 N each to the right

1 pull 250 N to the left

Since positive direction is to the right, the pulls to the left will have a minus (-)

Net Force= 2(196)+4(-98)+5(-62)+3(150)+1(-250) \\Net Force = -110

The resulting force is negative, meaning the direction is to the left

6 0
3 years ago
An open pipe, 0.29 m long, vibrates in the second overtone with a frequency of 1,227 Hz. In this situation, the fundamental freq
Andru [333]

Answer:

f = 409 Hz

Explanation:

We have,

Length of the open organ pipe, l = 0.29 m

Frequency of vibration of second overtone, f_2 = 1227 Hz

It is required to find the fundamental frequency of the pipe. For the open organ pipe, the frequency of second overtone is given by :

f_2=\dfrac{3v}{2l}

v is speed of sound

Let f is the fundamental frequency. It is given by :

f=\dfrac{v}{2l}

The relation between f and f₂ can be written as :

f_2=3f\\\\f=\dfrac{f_2}{3}\\\\f=\dfrac{1227}{3}\\\\f=409\ Hz

So, the fundamental frequency of the pipe is 409 Hz.              

6 0
3 years ago
A 55-kg woman is wearing high heels.
Grace [21]

Answer:

Pressure, P=1.90\times 10^7\ Pa        

Explanation:

It is given that,

Mass of the woman, m = 55 kg

Diameter of the circular cross section, d = 6 mm

Radius, r = 3 mm = 0.003 m

Let P is the pressure exerted on the floor. It is equal to the force acting on woman per unit area. It is given by :

P=\dfrac{F}{A}

P=\dfrac{mg}{\pi r^2}

P=\dfrac{55\times 9.8}{\pi (0.003)^2}

P=1.90\times 10^7\ Pa

So, the pressure exerted on the floor is 1.90\times 10^7\ Pa. Hence, this is the required solution.

7 0
3 years ago
Select the correct answer. Which of the following is true for balancing equations? A. There must be an equal number of atoms of
Sliva [168]
I think it’s B I could be wrong but I tried lol
8 0
3 years ago
How is a light element like helium is formed in stars and how is a heavier element like gold formed by stars?
aliina [53]

Answer:

C

Explanation:

Because *censored*. Then the element *censored* your answer

7 0
3 years ago
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