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ddd [48]
3 years ago
8

Suppose you are standing on the edge of a dock and jump straight down. If you land on sand your stopping time is much shorter th

an if you land on water. Using the impulse-momentum theorem as a guide, determine which one of the following statements is correct.
(a) In bringing you to a halt, the sand exerts a greater impulse on you than does the water.

(b) In bringing you to a halt, the sand and the water exert the same impulse on you, but the sand exerts a greater average force.

(c) In bringing you to a halt, the sand and the water exert the same impulse on you, but the sand exerts a smaller average force.
Physics
1 answer:
denis-greek [22]3 years ago
7 0

Answer:

In bringing you to a halt, the sand and the water exert the same impulse on you, but the sand exerts a greater average force

Explanation:

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In a displacement versus time graph, what does the slope of a line at any point indicate?
Rom4ik [11]

In a displacement versus time graph, the slope of the line at any point on the graph indicates the <em>magnitude of velocity</em>.  

(It can't indicate velocity completely, because the graph shows nothing about the direction of the motion.)

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3 years ago
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What is the net charge of a metal ball if there are 21,749 extra electrons in it?
pickupchik [31]

Answer:

Q=3.47\times 10^{-15}\ C

Explanation:

Given that,

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We need to find the net charge on the metal ball. Let Q is the net charge.

We know that the charge on an electron is q=1.6\times 10^{-19}\ C

To find the net charge if there are n number of extra electrons is :

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3 years ago
How can understanding velocity help to prevent a mid-air collision
dsp73

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To maintain enough time to prevent a collision, a system operating in air traffic where aircraft speed does not

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3 years ago
An electron is moving east in a uniform electric field of 1.47 {\rm N/C} directed to the west. At point A, the velocity of the e
zimovet [89]

Answer:

a) v = 6.25\times 10^{5} m/s

b) v = 1.73\times 10^{4} m/s

Explanation:

Given data:

Electric field = 1.47 N/C

velocity of electron is 4.55\times 10^5 m/s

distance of point b from point A is 0.55 m

we know that acceleration of particle is given as

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a =\frac{q E}{m}

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b) for proton

a = \frac{1.6\times 10^{-19} \times -1.47}{1.6\times 10^{-27}}

a = -1.41\times 10^{8} m/s^2

from equation of motion we have

v^2 = u^2 + 2aS

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