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Masja [62]
3 years ago
13

The maximum speed limit on interstate 10 is 75 miles per hour. how many meters per second is this

Physics
1 answer:
Dvinal [7]3 years ago
8 0

Answer:

<h2>33.53m/s</h2>

Explanation:

Given the maximum speed limit on interstate 10 as 75 miles per hour, to get the speed in meter per seconds, we need to convert the given speed to meter per seconds.

Using the conversion 1 mile = 1609.34m and 1 hour = 3600 seconds

75 miles perhour = 75miles/1 hour

75miles/1 hour (in m/s) = 75miles*1609.34m* 1 hour/1mile * 1 hour * 3600s *

= 75 *1609.34m* 1 /1 * 1 * 3600s

= 120,700.5m/3600s

= 33.53m/s

<em>Hence the maximum speed limit on interstate 10 in metre per seconds is 33.53m/s</em>

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a ball is thrown straight up into the air with a speed of 13 m/s. if the ball has a mass of 0.25 kg, how high does the ball go?
evablogger [386]
<h2>Hello!</h2>

The answer is: 8.62m

<h2>Why?</h2>

There are involved two types of mechanical energy: kinetic energy and potential energy, in two different moments.

<h2>First moment:</h2>

Before the ball is thrown, where the potential energy is 0.

<h2>Second moment: </h2>

After the ball is thrown, at its maximum height, the Kinetic Energy turns to 0 (since at maximum height,the speed is equal to 0) and the PE turns to its max value.

Therefore,

E=PE+KE

Where:

PE=m.g.h

KE=\frac{1*m*v^{2}}{2}

<em>E</em> is the total energy

<em>PE</em> is the potential energy

<em>KE</em> is the kinetic energy

<em>m</em> is the mass of the object

<em>g</em> is the gravitational acceleration

<em>h </em>is the reached height of the object

<em>v</em> is the velocity of the object

Since the total energy is always constant, according to the Law of Conservation of Energy, we can write the following equation:

KE_{1}+PE_{1}=KE_{2}+PE_{2}

Remember, at the first moment the PE is equal to 0 since there is not height, and at the second moment, the KE is equal to 0 since the velocity at maximum height is 0.

\frac{1*m*v^{2}}{2}+m.g.(0)=\frac{1*m*0^{2}}{2}+m.g.h\\\frac{1*m*v_{1} ^{2}}{2}=m*g*h_{2}

So,

h_{2}=\frac{1*m*v_{1} ^{2}}{2*m*g}\\h_{2}=\frac{1*v_{1} ^{2}}{2g}=\frac{(\frac{13m}{s})^{2} }{2*\frac{9.8m}{s^{2}}}\\h_{2}=8.62m}

Hence,

The height at the second moment (maximum height) is 8.62m

Have a nice day!

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3 years ago
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Answer:

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LAW 2: For a given metal, there exists a certain frequency below which the photoelectric emission does not take place. This frequency is called threshold frequency.

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LAW 3: For a frequency greater than the threshold frequency, the kinetic energy of photoelectrons is dependent upon frequency or wavelength but not on the intensity of light.

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LAW 4: Photoelectric emission is an instantaneous process. The time lag between incidence of radiations and emission of electron is 10^-9 seconds.

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I attempted to answer and got 0m, please explain how to get to the answer.
mafiozo [28]

The maximum height to which the ball attain before falling back down is 1147.96 m

<h3>Data obtained from the question</h3>

The following data were obtained from the question:

  • Initial velocity (u) = 150 m/s
  • Final velocity (v) = 0 m/s (at maximum height)
  • Acceleration due to gravity (g) = 9.8 m/s²
  • Maximum height (h) =?

<h3>How to determine the maximum height </h3>

The maximum height reached by the ball can be obtained as illustrated below:

v² = u² – 2gh (since the ball is going against gravity)

0² = 150² – (2 × 9.8 × h)

0 = 22500 – 19.6h

Collect like terms

0 – 22500 = –19.6h

–22500 = –19.6h

Divide both side by –19.6

h = –22500 / –19.6

h = 1147.96 m

Thus, the maximum height reached by the ball is 1147.96 m

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