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Inessa [10]
3 years ago
8

A baseball is hit so that it travels straight upward after being struck by the bat. A fan observes that it takes 3.00 s for the

ball to reach its maximum height. Find (a) the ball’s initial velocity and (b) the height it reaches.
Physics
1 answer:
Kay [80]3 years ago
3 0
To answer this problem first is to analyze and understand the problem first and then consider the given and also the gravity that is exert in the ball. Base on my calculation and the use of the formulas the answer are the following:
A. The initial velocity is <span>29.43 m/s
b. The height it reaches is 132.435m</span>
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The music is in ___meter? <br>a.free time<br>b.duple<br>c.triple<br>d.quadruple<br>​
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D.quadruple is the answer
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2 years ago
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According to the Big Bang Theory, the density and temperature of the Universe is ___________ now than in the past.
34kurt

Answer:

According to the Big Bang Theory, the density and temperature of the Universe is <u>lower</u> now than in the past.

<em>Hope</em><em> this</em><em> answer</em><em> correct</em><em> </em><em>:</em><em>)</em>

5 0
3 years ago
A piano wire with mass 2.95 g and length 79.0 cm is stretched with a tension of 29.0 N . A wave with frequency 105 Hz and amplit
Romashka-Z-Leto [24]

The concept needed to solve this problem is average power dissipated by a wave on a string. This expression ca be defined as

P = \frac{1}{2} \mu \omega^2 A^2 v

Here,

\mu = Linear mass density of the string

\omega =  Angular frequency of the wave on the string

A = Amplitude of the wave

v = Speed of the wave

At the same time each of this terms have its own definition, i.e,

v = \sqrt{\frac{T}{\mu}} \rightarrow Here T is the Period

For the linear mass density we have that

\mu = \frac{m}{l}

And the angular frequency can be written as

\omega = 2\pi f

Replacing this terms and the first equation we have that

P = \frac{1}{2} (\frac{m}{l})(2\pi f)^2 A^2(\sqrt{\frac{T}{\mu}})

P = \frac{1}{2} (\frac{m}{l})(2\pi f)^2 A^2 (\sqrt{\frac{T}{m/l}})

P = 2\pi^2 f^2A^2(\sqrt{T(m/l)})

PART A ) Replacing our values here we have that

P = 2\pi^2 (105)^2(1.8*10^{-3})^2(\sqrt{(29.0)(2.95*10^{-3}/0.79)})

P = 0.2320W

PART B) The new amplitude A' that is half ot the wavelength of the wave is

A' = \frac{1.8*10^{-3}}{2}

A' = 0.9*10^{-3}

Replacing at the equation of power we have that

P = 2\pi^2 (105)^2(0.9*10^{-3})^2(\sqrt{(29.0)(2.95*10^{-3}/0.79)})

P = 0.058W

8 0
2 years ago
If an electric wire is allowed to produce a magnetic field no larger than that of the Earth (0.503 X 104 T) at a distance of 15
dem82 [27]

Answer:

37.725 A

Explanation:

B = magnitude of the magnetic field produced by the electric wire = 0.503 x 10⁻⁴ T

r = distance from the wire where the magnetic field is noted = 15 cm = 0.15 m

i = magnitude of current flowing through the wire = ?

Magnetic field by a long wire is given as

B = \frac{\mu _{o}}{4\pi }\frac{2i}{r}

Inserting the values

0.503\times 10^{-4} = (10^{-7})\frac{2i}{0.15}

i = 37.725 A

6 0
2 years ago
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