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ss7ja [257]
2 years ago
8

The two types of grip in table tennis are 1. ___________ and 2. _____________.

Physics
1 answer:
pshichka [43]2 years ago
8 0
  1. The two types of grip in table tennis are <u>penhold grip</u> and <u>shakehand grip</u>.
  2. A <u>serve</u> is a stroke that starts a rally.
  3. A <u>receive</u> is a stroke to reply to a <u>serve</u>.
  4. A let is a <u>rally</u> of which the result is <u>not scored</u>.
  5. 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:

  • Shakehand grip
  • Penhold grip

<h3>The fundamental skills of table tennis.</h3>

Basically, there are four (4) fundamental skills used in table tennis and these are:

  1. Forehand drive
  2. Backhand drive
  3. Backhand push
  4. Forehand push.

Read more on table tennis here: brainly.com/question/17358010

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Nataliya [291]
I think is between A&B
I think I would answer B
6 0
3 years ago
An electric heater rated 500 W operates 24 hours/ day. what is the cost of energy to operate it for 20 days at Rs. 2.00 per kWh
Rina8888 [55]

Answer:

Rs. 480.00

Explanation:

1kW = 1000W

therefore 500W = 0.5kW

20 × 24hrs = 480hrs in total.

0.5kW × 480hrs = 240kWh

if rs. 2 for 1kWh

then, 240kWh × 2 = Rs. 480.

4 0
2 years ago
Read 2 more answers
The notes produced by a tuba range in frequency from approximately 45 Hz to 375 Hz. Find the possible range of wavelengths in ai
taurus [48]

Answer:

The possible range of wavelengths in air produced by the instrument is 7.62 m and 0.914 m respectively.

Explanation:

Given that,

The notes produced by a tuba range in frequency from approximately 45 Hz to 375 Hz.

The speed of sound in air is 343 m/s.

To find,

The wavelength range for the corresponding frequency.

Solution,

The speed of sound is given by the following relation as :

v=f_1\lambda_1

Wavelength for f = 45 Hz is,

\lambda_1=\dfrac{v}{f_1}

\lambda_1=\dfrac{343}{45}=7.62\ m

Wavelength for f = 375 Hz is,

\lambda_2=\dfrac{v}{f_2}

\lambda_2=\dfrac{343}{375}=0.914\ m/s

So, the possible range of wavelengths in air produced by the instrument is 7.62 m and 0.914 m respectively.

6 0
3 years ago
4. A ball is thrown vertically upward from the ground with a velocity of 30m/s. (a) how long will it take to rise to the highest
yarga [219]

All the answers are:

a) The time that will it take to rise to the highest point is 3.06 seconds.

b) The ball will rise to a height of 45.87 meters.

c) The time at which the ball will have a velocity of 10 m/s upward is 2.04 seconds.

The time when the ball has 10 m/s downward is 1.02 seconds.

d) The displacement of the ball will be zero at 6.12 seconds.

e) The time when the magnitude of the ball's velocity is equal to half its velocity of projection is 1.53 seconds.

f) The ball's displacement is equal to half the maximum height to which it rises after 0.90 seconds.

g) In each moment (upward and downward) the magnitude of the acceleration is the value of g (9.81 m/s²) and is a vector in the negative y-direction.

Let's calculate the values for each case.

a) At the highest point, the final velocity is 0, so we can use the following equation.  

v_{f}=v_{i}-gt (1)

Where:

  • v(i) is the initial velocity
  • v(f) is the final velocity
  • g is the acceleration due to gravity (9.81 m/s²)

We know that v(i) = 30 m/s.

0=30-9.81t

Solve it for t:

t=3.06\: s

Hence, the time is 3.06 s.

b) At the highest point, the final velocity is 0, so we can use the following equation.  

v_{f}^{2}=v_{i}^{2}-2gh (2)

0=v_{i}^{2}-2gh

We know that the initial velocity is 30 m/s.

0=30^{2}-2gh

Solving it for h we have:  

h=\frac{30^{2}}{2*9.81}

h=45.87 \: m

Then, the height is 45.87 m.

c) Using equation (1) we can find the time (t).

10=30-(9.81t)

So, the time elapsed to get 10 m/s is:

t_{upward}=2.04\: s

We know the upward time is equal to the downward time. So the time from v=10 m/s to v=0 m/s will be.

t_{upward}=2.04+t  

t=1.02\: s

This is the time when the ball has 10 m/s downward.          

Therefore, the time upward is 2.04 s, and the time downward is 1.02 s.

d) It will be when the ball returns to the ground.

t=2t_{upward}

t=2*3.06      

t=6.12\: s

The displacement will be zero after 6.12 s.  

e) Here we need to find the time when v(f) is 15 m/s

15=30-gt

t=\frac{15}{9.81}  

t=1.53\: s

The time when the v(f) is 15 m/s is 1.53 s.

f) Here, we need to find t when h = 45.87/2 m = 22.94 m

We can use the next equation:

[tex]h=v_{i}t-0.5gt^{2}/tex]

[tex]22.94=30t-0.5*9.81*t^{2}/tex]

Solving this quadratic equation, t will be:

[tex]t=0.90\: s/tex]

Hence, the ball's displacement is equal to half the maximum h, at 0.90 s.

g) In each moment the magnitude of the acceleration is the value of g (9.81 m/s²) and is a vector in the negative y-direction.

Learn more about vertical motion here:

brainly.com/question/13966860

I hope it helps you!

3 0
3 years ago
A circular coil consists of N = 410 closely winded turns of wire and has a radius R = 0.75 m. A counterclockwise current I = 2.4
Svetllana [295]

The complete question is;

A circular coil consists of N = 410 closely winded turns of wire and has a radius R = 0.75 m. A counterclockwise current I = 2.4 A is in the coil. The coil is set in a magnetic field of magnitude B = 1.1 T.

a. Express the magnetic dipole moment μ in terms of the number of the turns N, the current I, and radius

R.

b. Which direction does μ go?

Answer:

A) μ = 1738.87 A.m²

B) The direction of the magnetic moment will be in upward direction.

Explanation:

We are given;

The number of circular coils;

N = 410

The radius of the coil;R = 0.75m

The current in the coils; I = 2.4 A

The strength of magnetic field;

B =1.1T

The formula for magnetic dipole moment is given as;

μ = NIA

Where;

N is number of turns

I is current

A is area

Now, area; A = πr²

So, A = π(0.75)²

Thus,plugging in relevant values, the magnetic dipole moment is;

μ = 410 * 2.4 * π(0.75)²

μ = 1738.87 A.m²

B) According to Fleming's right hand rule, the direction of the magnetic moment comes out to be in upward direction.

7 0
3 years ago
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