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astra-53 [7]
2 years ago
9

A 1430 kg is moving at 25.6 m/s when a force is applied, in the direction of the cars motion. The car speeds up to 31.3 m/s. If

the force is applied for 5.4 s what is the magnitude of the force
Physics
1 answer:
vladimir1956 [14]2 years ago
3 0

The car accelerates with magnitude <em>a</em> such that

31.3 m/s = 25.6 m/s + <em>a</em> (5.4 s)

→   <em>a</em> = (31.3 m/s - 25.6 m/s) / (5.4 s) ≈ 1.056 m/s²

Then the applied force has a magnitude <em>F</em> of

<em>F</em> = (1430 kg) <em>a</em> ≈ 1500 N

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Which of the following distinguishes the isotope uranium-238 from the isotope uranium-235?
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Answer:

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also we know that

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3 years ago
A 2,000 kg rocket is launched 12 km straight up at a constant acceleration into the sky at which point the rocket is travelling
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Answer:

797700000 J

Explanation:

From the question,

The work done by the rocket, is given as,

W = Ek+Ep............. Equation 1

Where Ek and Ep are the potential and the kinetic energy of the rocket respectively.

Ep = mgh............ Equation 2

Ek = 1/2mv²............. equation 3

Substitute equation 2 and equation 3 into equation 1

W = mgh+1/2mv².............. Equation 4

Where m = mass of the rocket, h = height, v = velocity of the rocket, g = acceleration due to gravity.

Given: m = 2000 kg, h = 12 km = 12000 m, v = 750 m/s, g = 9.8 m/s²

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W = 235200000+562500000

W = 797700000 J

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3 years ago
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