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sveticcg [70]
3 years ago
12

A pen rolls off a 0.55–meter high table with an initial horizontal velocity of 1.20 meters/second. At what horizontal distance f

rom the base of the table will the pen land? A.
0.18 meters
B.
0.20 meters
C.
0.40 meters
D.
0.62 meters
Physics
2 answers:
Evgesh-ka [11]3 years ago
3 0
Whats the weight of the pen
LiRa [457]3 years ago
3 0

Answer:

s_x=0.40\ meters

Explanation:

Given that,

Height of the table,s_y = 0.55 m

Initial horizontal velocity, u = 1.2 m/s

We need to find the horizontal distance from the base of the table will the pen land. Let t is time taken by the pen to land. Let s_x is the horizontal distance. It is given by :

s_x=ut...........(1)

Using second equation of motion to find time t.

s_y=u_yt+\dfrac{1}{2}gt^2

Since, u_y=0

t=\sqrt{\dfrac{2s_y}{g}}

t=\sqrt{\dfrac{2\times 0.55}{9.8}}

t = 0.33 s

Use the value of t in equation (1) as :

s_x=1.2\times 0.33

s_x=0.396\ m

or

s_x=0.40\ meters

So, the horizontal distance from the base of the table 0.4 meters. Hence, this is the required solution.      

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RoseWind [281]

M = mass of the girl = 53.4 kg

m = mass of skateboard = 3.55 kg

V' = velocity of the combination of girl and skateboard before she jumps forward = 0 \frac{m }{s}

V = velocity of the girl forward = 1.32 \frac{m }{s}

v = velocity of the skateboard afterward = ?

Using conservation of momentum

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inserting the values

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v = - 19.86 \frac{m }{s}

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Put these value in the relation above

[tex]\frac{210}{221} = \frac{343-v}{343+v}[/tex]

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sashaice [31]
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