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MAVERICK [17]
3 years ago
8

car was moving in a straight road of length 320 km it covered 240 km with an average velocity 75 km/hr then it ran out of fuel a

nd stopped for 0.6 hr until it was refueled and completed its journey with a velocity 100 km/hr so the average velocity during the whole journey
Physics
1 answer:
Stella [2.4K]3 years ago
6 0

The average velocity of the car for the whole journey is 69.57 km/h.

The given parameters:

  • <em>Length of the road, L = 320 km</em>
  • <em>Distance covered = 240 km at 75 km/h</em>
  • <em>time spent refueling, t₂ = 0.6 hr</em>
  • <em>Final velocity, = 100 km/hr</em>

The time spent by the before refueling is calculated as follows;

t = \frac{d}{v} \\\\t_1 = \frac{240}{75} \\\\t_1 = 3.2 \ hours

The time spent by the car for the remaining journey;

t_3 = \frac{320 - 240}{100} \\\\t_3 = 0.8 \ hr

The total time of the journey is calculated as follows;

t = t_1 + t_2 + t_3\\\\t = 3.2 \ hr \ + \ 0.6 \ hr \ + \ 0.8 \ hr\\\\t = 4.6 \ hours

The average velocity of the car for the whole journey is calculated as follows;

v = \frac{total \ distance }{total \ time} \\\\v = \frac{320}{4.6} \\\\v = 69.57 \ km/h

Learn more about average velocity here: brainly.com/question/6504879

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Explanation:

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Accuracy has two definitions:

More commonly, it is a description of systematic errors, a measure of statistical bias; low accuracy causes a difference between a result and a "true" value. ISO calls this trueness.

Alternatively, ISO defines[1] accuracy as describing a combination of both types of observational error above (random and systematic), so high accuracy requires both high precision and high trueness.

Precision is a description of random errors, a measure of statistical variability.

In simpler terms, given a set of data points from repeated measurements of the same quantity, the set can be said to be accurate if their average is close to the true value of the quantity being measured, while the set can be said to be precise if the values are close to each other. In the first, more common definition of "accuracy" above, the two concepts are independent of each other, so a particular set of data can be said to be either accurate, or precise, or both, or neither.

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Using the conversion factor from eV (electron volts) to joules, determine which energy line for an electron dropping from one en
Tom [10]

Explanation:

Wavelength in an emission spectrum,  \lambda=435\ nm=435\times 10^{-9}\ m  

The energy of an electron is given by :

E=\dfrac{hc}{\lambda}

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c is the speed of light

For 435 nm, the energy of the electron will be :

E=\dfrac{6.63\times 10^{-34}\times 3\times 10^8\ m/s}{435\times 10^{-9}}

E=4.57\times 10^{-19}\ J

We know that 1\ eV=1.6\times 10^{-19}\ J

So, E=\dfrac{4.57\times 10^{-19}}{1.6\times 10^{-19}}

So, E = 2.86 eV

The energy of the electron dropping from one energy level is 2.86 eV. We know that,

\dfrac{hc}{\lambda}=E_{n_2}-E_{n_1}

From the given energy levels :

E_5-E_2=-0.544-(-3.403)

E_5-E_2=2.859\ eV

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5 0
4 years ago
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A spring with a spring constant of 400 n / M has a mass hung on it so it stretches 8 cm. calculate how much mass the spring is s
I am Lyosha [343]

Answer:

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Explanation:

Given parameters:

Spring constant = 400N/m

Extension  = 8cm  = 0.08m

Unknown:

Amount of mass the spring is supporting  = ?

Solution:

To solve this problem:

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k is the spring constant

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  So;

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Mass;

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          32  = mass x 9.8

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So yeah, the student is right. Galileo gave us this theory long ago.

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