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N76 [4]
2 years ago
14

A high-jumper, having just cleared the bar, lands on an air mattress and comes to rest. Had she landed directly on the hard grou

nd, her stopping time would have been much shorter. Using the impulse-momentum theorem as your guide, determine which one of the following statements is correct.a. the air mattress exerts the same impulse, but a greater net average force, on the high-jumper than does the hard ground
b. the air mattress exerts a greater impulse, and a greater net average force, on the high-jumper than does the hard ground
c. the air mattress exerts a smaller impulse, and a smaller net average force, on the high-jumper than does the hard ground
d. the air mattress exerts a greater impulse, but a smaller net average force, on the high-jumper than does the hard ground
e. the air mattress exerts the same impulse, but a smaller net avg force, on the hj than hg
Physics
1 answer:
attashe74 [19]2 years ago
8 0

Answer:

Explanation:

Impulse = Force x time = change in momentum

F x t = m ( v - u )

In both cases, u are same and v=0

So change in momentum is same

hence , impulse is same.

F = Change in momentum / time

In case of air mattress , time increases

Hence average force decreases .

Option e is correct .

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V = (14 ft)(15 ft)(8 ft)(30.48 cm/1 ft)³ = 0.0593 cm³

The mass is equal to:
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It has been proposed that extending a long conducting wire from a spacecraft (a "tether") could be used for a variety of applica
denis23 [38]

Complete Question

The complete question is shown on the first uploaded image

Answer:

The angle between shuttle's velocity and the Earth's field is  \theta =   24.2^o

Explanation:

From the question we are told that

     The length of eire let out is  L = 250 \ m

      The emf generated is \epsilon = 40 V

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The emf generated is mathematically represented as

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making \theta  the subject of the formula

                        \theta =   sin ^{-1}[ \frac{\epsilon}{L  * B  *v} ]

substituting values

                        \theta =   sin ^{-1}[ \frac{40}{250  * (5*10^{-5})  *(7.80 *10^{3})} ]

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6 0
2 years ago
In ideal flow, a liquid of density 850 kg/m3 moves from a horizontal tube of radius 1.00 cm into a second horizontal tube of rad
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Answer:

a)   Q = π r₁ √ 2ΔP / rho [r₁² / r₂² -1] , b) Q = 3.4 10⁻² m³ / s , c)      Q = 4.8 10⁻² m³ / s

Explanation:

We can solve this fluid problem with Bernoulli's equation.

         P₁ + ½ ρ v₁² + ρ g y₁ = P₂ + ½ ρ v₂² + ρ g y₂

With the two tubes they are at the same height y₁ = y₂

        P₁-P₂ = ½ ρ (v₂² - v₁²)

The flow rate is given by

         A₁ v₁ = A₂ v₂

         v₂ = v₁ A₁ / A₂

We replace

         ΔP = ½ ρ [(v₁ A₁ / A₂)² - v₁²]

         ΔP = ½ ρ v₁² [(A₁ / A₂)² -1]

Let's clear the speed

         v₁ = √ 2ΔP /ρ[(A₁ / A₂)² -1]

The expression for the flow is

           Q = A v

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The areas are

            A₁ = π r₁

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We replace

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Therefore, the magnitude of ship's momentum is 6.33\times 10^8\ kg\cdot m/s.

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8 0
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