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viktelen [127]
3 years ago
12

Assume that the Deschutes River has straight and parallel banks and that the current is 0.75 m/s. Drifting down the river, you f

all out of your boat and immediately grab a piling of the Warm Springs Bridge. You hold on for 40 s and then swim after the boat with a speed relative to the water of 0.95 m/s. The distance of the boat downstream from the bridge when you catch it is______________.
Physics
1 answer:
DedPeter [7]3 years ago
3 0

Answer:

    d = 142.5 m

Explanation:

This is a vector exercise. Let's calculate how much the boat travels in the 40s

     d₀ = v_{b} t

    d₀ = 0.75 40

    d₀ = 30 m

Let's write the kinematic equations

Boat

     x = d₀  +  v_{b} t

     x = 0 +  v_{h} t

At the meeting point the coordinate is the same for both

    d₀  +  v_{b} t =  v_{h} t

    t ( v_{h} -  v_{b}) = d₀  

    t = d₀  / ( v_{b}-  v_{h})

The two go in the same direction therefore the speeds have the same sign

     t = 30 / (0.95-0.775)

     t = 150 s

The distance traveled by man is

     d =  v_{h} t

     d = 0.95 150

     d = 142.5 m

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A large fake cookie sliding on a horizontal surface is attached to one end of a horizontal spring with spring constant k = 440 N
irinina [24]

Answer:

a) 0.275 m b) 13.6 J

Explanation:

In absence of friction, the energy is exchanged between the spring (potential energy) and the cookie (kinetic energy), so at any point, the sum of both energies must be the same:

E = ½ kx2 + ½ mv2

If we take as initial state, the instant when the cookie is passing through the spring’s equilibrium position, all the energy is kinetic, and we know that is equal to 20.0 J.

After sliding to the right, while is being acted on by a friction force, it came momentarily at rest. At this point, the initial kinetic energy, has become potential elastic energy, in part, and in thermal energy also, represented by the work done by the friction force.

So, for this state, we can say the following:

Ki = Uf + Eth = ½* k*d2 + Ff*d

20.0J = ½ *440 N/m* d2 + 11.0 *d, where d is the compressed length of the spring, which is equal to the distance travelled by the cookie before coming momentarily at rest.

We have a quadratic equation, that, after simplifying terms, can be solved as follows, applying the quadratic formula:

d = -0.05/2 +/- √0.090625 = -0.025 +/- 0.3 = 0.275 m (we take the positive root)

b) If we take as our new initial status the moment at which the spring is compressed, and the cookie is at rest, all the energy is potential:

E = Ui = 1/2 k d²

In this case, d is the same value that we got in a), i.e., 0.275 m (as the distance travelled by the cookie after going through the equilibrium point is the same length that the spring have been compressed).

E= 1/2 440 N/m . (0.275)m² = 16.6 J

When the cookie passes again through the equilibrium position, the energy will be in part kinetic, and in part, it will have become thermal energy again.

So, we can write the following equation:

Kf = Ui - Ff.d = 16.6 J - 11.0 (0.275) m = 16.6 J - 3.03 J = 13.6 J

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