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zysi [14]
3 years ago
10

The speed of a car is decreased uniformly from 30. meters per second to 10. meters per second in 4.0 seconds. What was the car's

acceleration ?
Physics
2 answers:
Westkost [7]3 years ago
8 0

Hello!

The speed of a car is decreased uniformly from 30. meters per second to 10. meters per second in 4.0 seconds. What was the car's acceleration ?

We have the following data:

V (final velocity) = 10 m/s

Vo (initial velocity) = 30 m/s

ΔV  (speed interval)  = V - Vo → ΔV  = 10 - 30 → ΔV  = - 20 m/s

ΔT (time interval) = 4.0 seconds

a (average acceleration) = ? (in m/s²)

Formula:

\boxed{a = \dfrac{\Delta{V}}{\Delta{T}}}

Solving:

a = \dfrac{\Delta{V}}{\Delta{T}}

a = \dfrac{-20\:\dfrac{m}{s}}{4.0\:s}

\boxed{\boxed{a = - 5\:\frac{m}{s^2}}}\:\:\:\:\:\:\bf\green{\checkmark}

________________________________

\bf\red{I\:Hope\:this\:helps,\:greetings ...\:Dexteright02!}

masha68 [24]3 years ago
3 0
Acceleration (magnitude anyway) = (change in speed) / (time for the change) .

Change in speed = (10 - 30) = -20 m/s

Time for the change = 4.0sec

Magnitude of acceleration = -20/4  =  <em>-5 m/s² </em>
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At a certain location, Earth has a magnetic field of 0.60 ✕ 10−4 T, pointing 75° below the horizontal in a north-south plane. A
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Answer with Explanation:

We are given that

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A thin stream of water flows smoothly from a faucet and falls straight down. at one point the water is flowing at a speed of v1
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<span>The formulas are, v1d1² = v2d2² ........ (1) h = (v2²-v1²)/2g ...... (2) Given that, v1 = 1.71 m/s we assume that the stream has decreased by a factor d2 =0.805d1 then, v1d1² = v2 (0.805d1)² cancelled both side d1² then we get, v1 = v2 (0.805)² v1 = v2 (0.648025) Sub v1 = 1.71, 1.71 = v2 (0.648025) v2 = 1.71/0.648025 v2 = 2.638787083831642 v2 = 2.64 m/s The vertical distance formula, h = (v2²-v1²)/2g We know that value of gravity constant is 9.8 m/s² h = {(2.64)² - (1.71)²)/2(9.8) h = {(6.9696) - (2.9241)}/19.6 h = (4.0455)/19.6 h = 0.2064030612244898 h = 0.21 cm Therefore, the vertical distance h = 0.21 cm.</span>
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What is the wavelength and frequency of a photon emitted by transition of an electron from a n- orbit to a n-1 orbit'?
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Answer:

\lambda=9.12\times 10^{-8}}\times \frac {{{{(n-1)}^2}\times n^2}}{1-2n}\ m

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Explanation:

E_n=-2.179\times 10^{-18}\times \frac{1}{n^2}\ Joules

For transitions:

Energy\ Difference,\ \Delta E= E_f-E_i =-2.179\times 10^{-18}(\frac{1}{n_f^2}-\frac{1}{n_i^2})\ J=2.179\times 10^{-18}(\frac{1}{n_i^2} - \dfrac{1}{n_f^2})\ J

n_i=n\ and\ n_f=n-1

Thus solving it, we get:

\Delta E=2.179\times 10^{-18}(\frac{1}{n^2} - \dfrac{1}{{(n-1)}^2})\ J

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Where,  

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c is the speed of light having value 3\times 10^8\ m/s

So,

\frac {h\times c}{\lambda}=2.179\times 10^{-18}(\frac{1-2n}{{{(n-1)}^2}\times n^2}})\ J

\lambda=\frac {6.626\times 10^{-34}\times 3\times 10^8}{2.179\times 10^{-18}}\times \frac {{{{(n-1)}^2}\times n^2}}{{1-2n}}\ m

So,

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So,

h\times \nu=2.179\times 10^{-18}\frac{1-2n}{{{(n-1)}^2}\times n^2}}

\nu=\frac {2.179\times 10^{-18}}{6.626\times 10^{-34}}\frac{1-2n}{{{(n-1)}^2}\times n^2}}\ s^{-1}

\nu=3.29\times 10^{15}\frac{1-2n}{{{(n-1)}^2}\times n^2}}\ s^{-1}

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Pablo preto lift a barbell. During which stage of the lift does Pablo do work on the barbell
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While he is lifting the dumbell as the definition of work done = moving a mass through a distance = F x d
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