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2 years ago

5

The Olympic record for running the 200 m dash is 19.3 seconds. What is the average speed for this record?

2 answers:

Hoochie [10]2 years ago

8 0

Answer:

A

Explanation:

v = change of X / change of T

v = 200/19.3

g100num [7]2 years ago

3 0

Answer:

A. 10.36 m/s

Explanation:

The Olympic record for running the 200 m dash is 19.3 seconds. What is the average speed for this record?

A. 10.36 m/s

B. 3960 m/s

C. 219.3 m

D. 0.0965 m/s

speed is the change in distance per time

speed is scalar quantity and hence as no direction but only magnitude. time and distance are also scalar quantity

200/19.3

speed=10.36m/s

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1) A rock is dropped from a cliff with a height of 200 m. With what velocity will the rock hit the ground

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A rock is dropped from a 200 m high cliff. How long does it take to fall (a) the first 100 m and (b) the last 50 m?

The basic equation you want is:

s=at22

Solving for t:

t=2sa−−−√

We’ll assume a=9.8 , so 2a−−√=14.9−−−√≈0.4518

So, for (a)s=100 , so t=0.4518100−−−√=4.518

The total time is 0.4518200−−−√≈6.389

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So the time to fall the last 50 m is 6.389 - 5.533 = 0.856 seconds

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1 year ago

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A bug crawls 3.0 mm east, 4.0mm north, and then 5.0 mm at 45 north of east. Draw a diagram showing its displacements and determi

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2 years ago

How much heat does it take to raise the temperature of 10.0 kg of water by 1.0 C?

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Answer:The specific heat capacity of water is 4,200 joules per kilogram per degree Celsius (J/kg°C). This means that it takes 4,200 J to raise the temperature of 1 kg of water by 1°C.

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2 years ago

A guitar player tunes the fundamental frequency of a guitar string to 560 Hz. (a) What will be the fundamental frequency if she

lawyer [7]

Answer:

(a) if she increases the tension in the string is increased by 15%, the fundamental frequency will be increased to 740.6 Hz

(b) If she decrease the length of the the string by one-third the fundamental frequency will be increased to 840 Hz

Explanation:

(a) The fundamental, f₁, frequency is given as follows;

$f_1 = \dfrac{\sqrt{\dfrac{T}{\mu}} }{2 \cdot L}$

Where;

T = The tension in the string

μ = The linear density of the string

L = The length of the string

f₁ = The fundamental frequency = 560 Hz

If the tension in the string is increased by 15%, we will have;

$f_{(1 \, new)} = \dfrac{\sqrt{\dfrac{T\times 1.15}{\mu}} }{2 \cdot L} = 1.3225 \times \dfrac{\sqrt{\dfrac{T}{\mu}} }{2 \cdot L} = 1.3225 \times f_1$

$f_{(1 \, new)} = 1.3225 \times f_1 = (1 + 0.3225) \times f_1$

$f_{(1 \, new)} = 1.3225 \times f_1 =\dfrac{132.25}{100} \times 560 \ Hz = 740.6 \ Hz$

Therefore, if the tension in the string is increased by 15%, the fundamental frequency will be increased by a fraction of 0.3225 or 32.25% to 740.6 Hz

(b) When the string length is decreased by one-third, we have;

The new length of the string, $L_{new}$ = 2/3·L

The value of the fundamental frequency will then be given as follows;

$f_{(1 \, new)} = \dfrac{\sqrt{\dfrac{T}{\mu}} }{2 \times \dfrac{2 \times L}{3} } =\dfrac{3}{2} \times \dfrac{\sqrt{\dfrac{T}{\mu}} }{2 \cdot L} = \dfrac{3}{2} \times 560 \ Hz = 840 \ Hz$

When the string length is reduced by one-third, the fundamental frequency increases to one-half or 50% to 840 Hz.

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2 years ago