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siniylev [52]
3 years ago
10

A boat moves ar 10.0 m/s relative to

Physics
1 answer:
Charra [1.4K]3 years ago
5 0

Answer: 208.3 s

Explanation:

Hi!

You need to calculate the velocity of the boat relative to the shore, in each trip.

Relative velocities are transformed according to:

V_{b,s} = V_{b,w} + V_{w,s}\\ V_{b,s} = \text{velocity of boat relative to shore}\\V_{b,w} = \text{velocity of boat relative to water} \\V_{b,w} = \text{velocity of water relative to shore}\\

Let's take the upstram direction as positive. Water flows downstream, so it's velocity relative to shore is negative , -2 m/s

In the upstream trip, velocity of boat relative to water is positive: 10m/s. But in the downstream trip it is negatoive: -10m/s

In upstream trip we have:

V_{b,s} = (10 - 2)\frac{m}{s} = 8 ms

In dowstream we have:

V_{b,s} = (-10 - 2)\frac{m}{s} = -12 ms

In both cases the distance travelled is 1000m. Then the time it takes the round trip is:

T = T_{up} + T_{down} = \frac{1000}{8}s + \frac{1000}{12}s = 208.3 s

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A boy pulls a 28.0-kg box with a 230-N force at 35° above a horizontal surface. If the coefficient of kinetic friction between t
ddd [48]

Answer:

1977.696 J

Explanation:

Given;

Weight of the box = 28.0 kg

Force applied by the boy = 230 N

angle between the horizontal and the force = 35°

Therefore,

the horizontal component of the force = 230 × cosθ

= 230 × cos 35°

= 188.405 N

Coefficient of kinetic friction, μ = 0.24

Force by friction, f = μN

here,

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therefore,

f = 0.24 × 274.68

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= 5652.15 J

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the

work done by the friction, W₂ = - 65.9232 N × Displacement  

= - 65.9232 N × 30 m

= - 1977.696 J

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