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UkoKoshka [18]
3 years ago
9

As a train accelerates uniformly, it passes successive 5-kilometer markers while traveling at velocities 10 m/s and 25 m/s. what

is the velocity in km/h when it passes the next marker? round to nearest tenth of a km/h.
Physics
1 answer:
gregori [183]3 years ago
8 0
<span>122.0 km/hr. First let’s make sure all of our units are in the base meter form: i.e. convert 5km to 5000m. (We will convert back to km later). The first thing to do is look at the equation relating velocity, acceleration, and distance: Vf^2 = Vi^2 + 2*a*d, where Vf is final velocity, Vi is initial velocity, a is acceleration, and d is distance. 25^2 = 10^2 + 2*a*5000 =?> 625 = 100 +10000a => a= 0.0525m/s^2. Now that we have acceleration, we can use the same equation again with different numbers.: Vf^2 = Vi^2 + 2*a*d = 25^2 + 2*0. 0525m*5000 = 625 + 525 =1150 => Vf^2 = 1150 => 33.9m/s. Convert to km/hour: 33.9m/s * 1km/1000m *60s/1min * 60min/ 1 hr = 122.0 km/hr.</span>
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<u><em>heyaaaaa</em></u>

<u><em>Momentum before Pb = momentum after Pa</em></u>

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<u><em>Velocity has magnitude (speed) and direction. V = -350/175 = -2m/s </em></u>

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3 years ago
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20 joule is your answer

Answer:

here

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2 years ago
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hello friends,i need your help my home work now in physics,topic:motion.40 marks +brainliest if correct .​
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Answer:

See below

Explanation:

Vertical position is given by

df = do + vo t - 1/2 a t^2      df = final position = 0 (on the ground)

                                           do =original position = 2 m

                                            vo = original <u>VERTICAL</u> velocity = 0

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5 0
1 year ago
The 4kg head of a sledge hammer is moving at 6m/s when it strikes a chisel, driving it into a log. The duration of the impact (o
LUCKY_DIMON [66]

Answer:

The average impact force is 12000 newtons.

Explanation:

By Impact Theorem we know that impact done by the sledge hammer on the chisel is equal to the change in the linear momentum of the former. The mathematical model that represents the situation is now described:

\bar F \cdot \Delta t = m \cdot  (v_{2}-v_{1}) (1)

Where:

\bar F - Average impact force, in newtons.

\Delta t - Duration of the impact, in seconds.

m - Mass of the sledge hammer, in kilograms.

v_{1}, v_{2} - Initial and final velocity, in meters per second.

If we know that \Delta t = 0.0020\,s, m = 4\,kg, v_{1} = -6\,\frac{m}{s} and v_{2} = 0\,\frac{m}{s}, then we estimate the average impact force is:

\bar F = \frac{m\cdot  (v_{2}-v_{1})}{\Delta t}

\bar F = 12000\,N

The average impact force is 12000 newtons.

5 0
3 years ago
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