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alexandr402 [8]

2 years ago

13

A girl, standing on a bridge, throws a stone vertically downward with an initial velocity of 12.0 m/s, into the river below. If

the stone hits the water 2.50 seconds later, what is the height of the bridge above the water?
Remember to identify all your data, write the equation, and show your work.

1 answer:

melamori03 [73]2 years ago

3 0

Answer: 60.65 m

Explanation

Use eqation for distance for vertical throw( down)

Vo=12m/s

t=2.5

S=Vot +gt^2/2

S=12*2.5+9.81*2.5^2/2

S=30+30.65

S=60.65m

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A spring with a spring constant of 2500 n/m. is stretched 4.00 cm. what is the work required to stretch the spring?

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A 3.00-kg model airplane has velocity components of 5.00 m/s due east and 8.00 m/s due north. What is the plane’s kinetic energy

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A plate drops onto a smooth floor and shatters into three pieces of equal mass.Two of the pieces go off with equal speeds v at r

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Answer:

Speed of the this part is given as

$v_3 = \sqrt2 v$

Also the direction of the velocity of the third part of plate is moving along 135 degree with respect to one part of the moving plate

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As we know by the momentum conservation of the system

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here we know that

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so we will have

$P_1 + P_2 = \sqrt{P^2 + P^2}$

$P_1 + P_2 = \sqrt2 mv$

now from above equation we have

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2 years ago

While ice skating, you unintentionally crash into a person. Your mass is 60 kg, and you are traveling east at 8.0 m/s with respe

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Answer:

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Roller skate collision

The final direction of the system (me=M + person=P) velocity vector is at an angle; Ф, to the direction running south to north. Apply the component form of the impulse-momentum equation, firstly;

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$P_{Miy}$ + $p_{Piy}$ + $j_{y}$ = $P_{Mfy}$ + $P_{pfy}$

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y-axis component form (+y north);

$P_{Mix}$ + $p_{Pix}$ + $j_{x}$ = $P_{Mfx}$ + $P_{pfx}$

$m_{Mu_{Mix}+ m_{pu_{pix}}+0=(m_{M}+m_{p})V_{f} cos$ Ф

0 + 80.9 = (60 + 80) $V_{f}cos$ Ф

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140Vf= $\frac{720}{cos}$ Ф......................................(2)

Substituting (2) into (1) to give the angle;

480 = 720tan Ф

Ф = arctan(0.67) =33.69°.......................(3)

Evaluating (1) with (3) gives the velocity magnitude

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Vf=6.18 m/s

note 1:

This angle corresponds to a direction; 90° - 33.69° = 56.31° north of east.

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