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balandron [24]
3 years ago
12

1. A 70-kg swimmer dives horizontally off a 500-kg raft. The diver's speed immediately after leaving the raft is 6.0 m/s. A micr

o-sensor system attached to the edge of the raft measures the time interval during which the diver applies an impulse to the raft just prior to leaving the raft surface. If the time interval is read as 0.25 s, what is the magnitude of the average horizontal force by diver on the raft?
Physics
1 answer:
MA_775_DIABLO [31]3 years ago
6 0

To solve this problem it is necessary to apply the concepts related to momentum theorem.

The equation for impulse is given as

I = Ft

Where

I = Force

t = Time

At the same time we have the equation for momentum is given as

p = mv

The impulse momentum theorem states that the change in momentum of an object is equal to the impulse applied to it. Therefore

I = p

Ft = mv

Solving to find the force

F = \frac{mv}{t}

F = \frac{(70)(6)}{0.25}

F = 1680N

Therefore the magnitude of the average horizontal force by diver on the raft is 1680N

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To practice Problem-Solving Strategy 11.1 Equilibrium of a Rigid Body. A horizontal uniform bar of mass 2.7 kg and length 3.0 m
Zanzabum

Answer:

T_{1} = 14.88 N

Explanation:

Let's begin by listing out the given variables:

M = 2.7 kg, L = 3 m, m = 1.35 kg, d = 0.6 m,

g = 9.8 m/s²

At equilibrium, the sum of all external torque acting on an object equals zero

τ(net) = 0

Taking moment about T_{1} we have:

(M + m) g * 0.5L - T_{2}(L - d) = 0

⇒ T_{2} = [(M + m) g * 0.5L] ÷ (L - d)

T_{2} = [(2.7 + 1.35) * 9.8 * 0.5(3)] ÷ (3 - 0.6)

T_{2}= 59.535 ÷ 2.4

T_{2} = 24.80625 N ≈ 24.81 N

Weight of bar(W) = M * g = 2.7 * 9.8 = 26.46 N

Weight of monkey(w) = m * g = 1.35 * 9.8 = 13.23 N

Using sum of equilibrium in the vertical direction, we have:

T_{1} + T_{2} = W + w   ------- Eqn 1

Substituting T2, W & w into the Eqn 1

T_{1} + 24.81 = 26.46 + 13.23

T_{1} = <u>14.88</u> N

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3 years ago
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