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topjm [15]
3 years ago
10

Eric threw a baseball 20 meters in 0.5 seconds. What was the average speed of the baseball to the nearest hundredths of a m/sec.

Physics
2 answers:
anzhelika [568]3 years ago
4 0

Answer:

40m/sec

Explanation:

speed = distance ÷ time

speed = 20 ÷ 0.5

speed = 40 m/sec

UkoKoshka [18]3 years ago
3 0

Answer:

40 m/s

Explanation:

Speed or velocity is distance over time. Therefore,

20 meters/0.5 seconds

= 40 m/s

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It gives our light which we need for probably everything.

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Where does current flows maximum in? series connection or parallel connection?
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Current flow depends on other things in addition to the circuit configuration. If the SAME voltage is applied to some arrangement of the SAME components, the greatest current will occur when they are all in parallel.
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According to the photon energy formula, tripling the frequency of the radiation from a monochromatic source will change the ener
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By a factor of 3

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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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Why are lamps connected in parrells rather than in a series in domestic circuit
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