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3 years ago

In which year was the first Badminton game played in the Olympics

Physics

2 answers:

Viefleur [7K]3 years ago

6 0

It was actually 1992 when it was included

Kisachek [45]3 years ago

4 0

Year 1972
if I'm not wrong :)

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An air-track glider attached to a spring oscillates between the 10 cm mark and the 60 cm mark on the track. The glider completes

Assoli18 [71]

Answer:

a) Time period is 3.3 seconds

b) The frequency is 0.3030 Hz

c) amplitude is 0.25 m

d) maximum speed is 0.476 m/s

Explanation:

Given the data in the question;

a) Period

Time Period T = Time taken for one oscillation

T = 33s / 10 = 3.3 seconds

Therefore, Time period is 3.3 seconds

b) Frequency

we know that frequency is the inverse of time period

so;

Frequency f = 1/T = 1 / 3.3 s

Frequency f = 0.3030 Hz

Therefore, The frequency is 0.3030 Hz

c) amplitude

amplitude A = $\frac{1}{2}$ ( 60 cm - 10 cm )

A = $\frac{1}{2}$ × 50 cm

A = 25 cm

A = 0.25 m

Therefore, amplitude is 0.25 m

d) maximum speed of the glider

maximum speed $V_{max}$ = ωA

and ω = 2π/T

so maximum speed $V_{max}$ = $\frac{2\pi }{T}$ A

so we substitute

so maximum speed $V_{max}$ = $\frac{2\pi }{3.3}$ × 0.25 m

so maximum speed $V_{max}$ = 0.476 m/s

Therefore, maximum speed is 0.476 m/s

5 0

2 years ago

A car with an initial velocity of O m/s and a mass of 1500 kg reaches a velocity of 15 m/s in 5

Flura [38]

Answer:the answer can be find out by the following steps;

Explanation:

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3 years ago

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The tires of a car make 75 revolutions as the car reduces its speed uniformly from 95 km/hr to 55 km.hr. the tires have a diamet

sveta [45]

<span>95 km/h = 26.39 m/s (95000m/3600 secs) 55 km/h = 15.28 m/s (55000m/3600 secs) 75 revolutions = 75 x 2pi = 471.23 radians radius = 0.80/2 = 0.40m v/r = omega (rad/s) 26.39/0.40 = 65.97 rad/s 15.28/0.40 = 38.20 rad/s s/((vi + vf)/2) = t 471.23 /((65.97 + 38.20)/2) = 9.04 secs (vf - vi)/t = a (38.20 - 65.97)/9.04 = -3.0719 The angular acceleration of the tires = -3.0719 rad/s^2 Time is required for it to stop (0 - 38.20)/ -3.0719 = 12.43 secs How far does it go? 65.97 - 38.20 = 27.77 M</span>

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3 years ago

What is the freezing point of radiator fluid that is 50% antifreeze by mass? k f for water is 1.86 ∘ c/m?

densk [106]

Ethylene glycol is termed as the primary ingredients in antifreeze.
The ethylene glycol molecular formula is C₂H₆O₂.
Molar mass of C₂H₆O₂ is = (2×12) +(6×1) + (216) = 62g/mol
Now that antifreeze by mass is 50%, then there is 1kg of ethylene glycol which is present in 1kg of water.
ΔTf = Kf×m
ΔTf = depression in the freezing point.
= freezing point of water freezing point of the solution
= O°c - Tf
= -Tf
Kf = depression in freezing constant of water = 1.86°C/m
M is the molarity of the solution.
=(mass/molar mass) mass of solvent in kg
=1000g/62 (g/mol) /1kg
=16.13m
If we plug the value we get
-Tf = 1.86 × 16.13 = 30
Tf = -30°c

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3 years ago

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Quarks are thought to be the basic component of what?

vladimir2022 [97]

<span>Quarks are thought to be the basic component of protons and newtons.</span>

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2 years ago

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