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3 years ago

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A vehicle of mass 1600kg take off from rest and coveres a distance of 280m in 60s. It continues at this speed for 98seconds and

comes to rest after 106s. i. Determine the total distance covered. ii. Determine the amount of force exerted as it slow to a stop. iii. If the road is inclined at 30degrees to the horizontal,what is the horizontal component of its velocity?

1 answer:

Lady bird [3.3K]3 years ago

4 0

1. V = Vo + a*t.
<span>15 = 0 + a*5 a = 3 m/s^2. </span>

<span>2. V = Vo + a*t. </span>
<span>0 = 15 - 5*t, t = 3 s. </span>

<span>3. d1 = Vo*t + 0.5a*t^2. </span>
<span>Vo = 0, t = 5, a = 3 m/s^2, d1 = ?. </span>

<span>d2 = Vo*t. Vo = 15 m/s, t = 15 s, </span>
<span>d2 = ?. </span>

<span>V^2 = Vo^2 + 2a*d3. </span>
<span>V = 0, Vo = 15 m/s, a = -5 m/s^2, d3 = ?. </span>

<span>d1+d2+d3 = Total dist.</span>

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ρ(solution)= m(solution) /v(solution)

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total moles of solution = n(H2SO4) + n(H2O)

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<h3>What is a solution?</h3>

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Solvent
Solute

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Physics B 2020 Unit 3 Test

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Answer:

1)

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where here:

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2)

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Therefore, this means that the magnetic field of a magnet is stronger near the North and South Pole.

3)

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It can be applied as follows:

- Direction of index finger = direction of motion of the charge

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5)

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6)

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7)

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8)

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See the last 7 answers in the attached document.

<span class="sg-text sg-text--link sg-text--bold sg-text--link-disabled sg-text--blue-dark"> docx </span>

<span class="sg-text sg-text--link sg-text--bold sg-text--link-disabled sg-text--blue-dark"> pdf </span>

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