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nlexa [21]
2 years ago
8

Learning Task 1: Define or describe the following words relat are not familiar with the word, you may ask assistance from y your

answer in your notebook. 1. table 2. tennis C 3. ping-pong 4. net 5. ball 6. racket​
Physics
1 answer:
Tems11 [23]2 years ago
3 0

Answer:

1.) TABLE = The table is 2.74 m (9.0 ft) long, 1.525 m (5.0 ft) wide, and 76 cm (2.5 ft) high with any continuous material so long as the table yields a uniform bounce of about 23 cm (9.1 in) when a standard ball is dropped onto it from a height of 30 cm (11.8 in), or about 77%.

2.) TENNIS = Tennis is a racket sport that can be played individually against a single opponent (singles) or between two teams of two players each (doubles). Each player uses a tennis racket that is strung with cord to strike a hollow rubber ball covered with felt over or around a net and into the opponent's court.

3.) PING-PONG = Table tennis, also known as ping-pong and whiff-whaff, is a sport in which two or four players hit a lightweight ball, also known as the ping-pong ball, back and forth across a table using small rackets. ... Spinning the ball alters its trajectory and limits an opponent's options, giving the hitter a great advantage.

4.) NET = This is stretched across the centter of the table by a cord attached to a post at either end. It measures 6ft long and the ball must pass over it for a rally to continue.

5.) BALL = The ball, which is spherical and hollow, was once made of white celluloid. Since 1969 a plastic similar to celluloid has been used. The ball, which may be coloured white, yellow, or orange, weighs about 0.09 ounce (2.7 grams) and has a diameter of about 1.6 inches (4 cm).

6.) RACKET = A table tennis racket is made up of two distinct parts - a wooden blade which incorporates the handle together with table tennis rubbers affixed to each side of the blade using water-based glue.

Explanation:

All my answers are

about the tools of the

game of tennis

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A particle with charge − 2.74 × 10 − 6 C −2.74×10−6 C is released at rest in a region of constant, uniform electric field. Assum
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Answer:

241.7 s

Explanation:

We are given that

Charge of particle=q=-2.74\times 10^{-6} C

Kinetic energy of particle=K_E=6.65\times 10^{-10} J

Initial time=t_1=6.36 s

Final potential difference=V_2=0.351 V

We have to find the time t after that the particle is released and traveled through a potential difference 0.351 V.

We know that

qV=K.E

Using the formula

2.74\times 10^{-6}V_1=6.65\times 10^{-10} J

V_1=\frac{6.65\times 10^{-10}}{2.74\times 10^{-6}}=2.43\times 10^{-4} V

Initial voltage=V_1=2.43\times 10^{-4} V

\frac{\initial\;voltage}{final\;voltage}=(\frac{initial\;time}{final\;time})^2

Using the formula

\frac{V_1}{V_2}=(\frac{6.36}{t})^2

\frac{2.43\times 10^{-4}}{0.351}=\frac{(6.36)^2}{t^2}

t^2=\frac{(6.36)^2\times 0.351}{2.43\times 10^{-4}}

t=\sqrt{\frac{(6.36)^2\times 0.351}{2.43\times 10^{-4}}}

t=241.7 s

Hence, after 241.7 s the particle is released has it traveled through a potential difference of 0.351 V.

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