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3 years ago

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It took 3.5 hours for a train to travel the distance between two cities at a velocity of 120mi/h. How many miles lie between the

two cities?

1 answer:

erma4kov [3.2K]3 years ago

3 0

Answer:

420 miles

Explanation:

3.5 * 120 mi/h = 420 miles

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A 15 kg bucket of water is suspended by a very light ropewrapped around a solid uniform cylinder 0.300 m in diamter withmass 12.

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Answer:

Part a)

$T = 42 N$

Part b)

$v_f = 11.8 m/s$

Part c)

$t = 1.7 s$

Part d)

$F = 159.7 N$

Explanation:

Part a)

While bucket is falling downwards we have force equation of the bucket given as

$mg - T = ma$

for uniform cylinder we will have

$TR = I\alpha$

so we have

$T = \frac{1}{2}MR^2(\frac{a}{R^2})$

$T = \frac{1}{2}Ma$

now we have

$mg = (\frac{M}{2} + m)a$

$a = \frac{mg}{(\frac{M}{2} + m)}$

$a = \frac{15 \times 9.81}{(6 + 15)}$

$a = 7 m/s^2$

now we have

$T = \frac{12 \times 7}{2}$

$T = 42 N$

Part b)

speed of the bucket can be found using kinematics

so we have

$v_f^2 - v_i^2 = 2 a d$

$v_f^2 - 0 = 2(7)(10)$

$v_f = 11.8 m/s$

Part c)

now in order to find the time of fall we can use another equation

$v_f - v_i = at$

$11.8 - 0 = 7 t$

$t = 1.7 s$

Part d)

as we know that cylinder is at rest and not moving downwards

so here we can use force balance

$F = T + Mg$

$F = 42 + (12 \times 9.81)$

$F = 159.7 N$

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An 8.0 kg object starts from rest and is pushed with a horizontal 24.N force for 10. s on a surface with negligible friction. Th

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The 24 N force applies an acceleration <em>a</em>₁ such that

24 N = (8.0 kg) <em>a</em>₁

<em>a</em>₁ = (24 N) / (8.0 kg)

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Starting from rest, the object accelerates at this rate for 10 s, so that it reaches a velocity <em>v</em> of

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<em>v</em> = 30 m/s

and in these 10 s, the object travels a distance <em>d</em>₁ of

<em>d</em>₁ = 1/2 (3.0 m/s²) (10 s)²

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When it hits the rough surface, it slows to a stop in 10 s, so that its accleration <em>a</em>₂ during this time is such that

0 = 30 m/s + <em>a</em>₂ (10 s)

<em>a</em>₂ = - (30 m/s) / (10 s)

<em>a</em>₂ = -3.0 m/s²

so that it covers an additional distance <em>d</em>₂ such that

<em>d</em>₂ = 1/2 (-3.0 m/s²) (10 s)²

<em>d</em>₂ = 150 m

So the object travels a <em>total</em> distance of <em>d</em>₁ + <em>d</em>₂ = 300 m.

Alternatively, once we find the accelerations during both 10 s intervals and the velocity after the first 10 s, we get

(30 m/s)² - 0² = 2 (3.0 m/s²) <em>d</em>₁

<em>d</em>₁ = (30 m/s)² / (6.0 m/s²)

<em>d</em>₁ = 150 m

and

0²- (30 m/s)² = 2 (-3.0 m/s²) <em>d</em>₂

<em>d</em>₂ = (30 m/s)² / (6.0 m/s²)

<em>d</em>₂ = 150 m

so that, again, <em>d</em>₁ + <em>d</em>₂ = 300 m.

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