Answer:
Volume of head = 1000 cm^3
Explanation:
The head of a normal person can be assumed as a sphere with radius 10 cm.
Volume of sphere
, where r is the radius.
We have approximate radius = 10 cm.
Approximate volume of head 
In the given options the closest value to the approximate volume is 1000 cm^3.
So, volume of head = 1000 cm^3
- The two types of grip in table tennis are <u>penhold grip</u> and <u>shakehand grip</u>.
- A <u>serve</u> is a stroke that starts a rally.
- A <u>receive</u> is a stroke to reply to a <u>serve</u>.
- A let is a <u>rally</u> of which the result is <u>not scored</u>.
- A point is a rally of which the result is scored.
<h3>What is table tennis?</h3>
Table tennis can be defined as an indoor sport and recreational activity in which two (2) or four (4) players hit a ping-pong ball back and forth on a table that is divided into halves by a low net, especially through the use of a small-solid bat (racket).
<h3>Types of grip in
table tennis.</h3>
Generally, there are two (2) main types of grip in table tennis and these include:
<h3>The
fundamental skills of table tennis.</h3>
Basically, there are four (4) fundamental skills used in table tennis and these are:
- Forehand drive
- Backhand drive
- Backhand push
- Forehand push.
Read more on table tennis here: brainly.com/question/17358010
Answer:
E = 1,873 10³ N / C
Explanation:
For this exercise we can use Gauss's law
Ф = E. dA =
/ ε₀
Where q_{int} is the charge inside an artificial surface that surrounds the charged body, in this case with the body it has a spherical shape, the Gaussian surface is a wait with radius r = 1.35 m that is greater than the radius of the sphere.
The field lines of the sphere are parallel to the radii of the Gaussian surface so the scald product is reduced to the algebraic product.
The surface of a sphere is
A = 4π r²
E 4π r² = q_{int} /ε₀
The net charge within the Gauussian surface is the charge in the sphere of q1 = + 530 10⁻⁹ C and the point charge in the center q2 = -200 10⁻⁹ C, since all the charge can be considered in the center the net charge is
q_{int} = q₁ + q₂
q_{int} = (530 - 200) 10⁻⁹
q_{int} = 330 10⁻⁹ C
The electric field is
E = 1 / 4πε₀ q_{int} / r²
k = 1 / 4πε₀
E = k q_{int}/ r²
Let's calculate
E = 8.99 10⁹ 330 10⁻⁹/ 1.32²
E = 1,873 10³ N / C