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GalinKa [24]
3 years ago
11

A life preserver is thrown from an helicopter straight down to a person in distress. The initial velocity of the life preserver

is 1.60 m/s and it takes 2.3 s to reach the water. (a) List the knowns in this problem. (b) How high above the water was the preserver released? Note Ignore air resistance.
Physics
1 answer:
Leno4ka [110]3 years ago
6 0

Answer:29.627 m

Explanation:

Given

Initial velocity of life preserver(u) is 1.6 m/s

it takes 2.3 s to reach the water

using equation of motion

v=u+at

v=1.6+9.81\times 2.3

v=24.163 m/s

Let s be the height of life preserver

v^2-u^2=2gs

24.163^2-1.6^2=2\times 9.81\times s

s=\frac{581.29}{2\times 9.81}

s=29.627 m

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Answer:

the force P required for impending motion is 132.3 N

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Explanation:

Given that:

mass of the cabinet  m = 45 kg

coefficient of static friction μ =  0.30

A free flow body diagram illustrating what the question represents is attached in the file below;

The given condition from the question let us realize that ; the casters are locked to prevent the tires from rotating.

Thus; considering the forces along the vertical axis ; we have :

\sum f_y =0

The upward force and the downward force is :

N_A+N_B = mg

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Considering the forces on the horizontal axis:

\sum f_x = 0

F_A +F_B  = P

where ;

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which can be written also as:

\mu_s N_A + \mu_s N_B  = P

\mu_s( N_A +  N_B)  = P

replacing our value from equation (1)

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b) Since the horizontal distance between the casters A and B is 480 mm; Then half the distance = 480 mm/2 = 240 mm = 0.24 cm

the largest value of "h" allowed for  the cabinet is not to tip over is calculated by determining the limiting condition  of the unbalanced torque whose effect is canceled by the normal reaction at N_A and it is shifted to N_B:  

Then:

\sum M _B = 0

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Answer:

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E_n=-13.6\times \frac{Z^2}{n^2}ev

Formula used for the radius of the n^{th} orbit will be,

where,

E_n = energy of n^{th} orbit

n = number of orbit

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E=E_3-E_2

E=(-1.51 eV) -(-3.40 eV)=1.89 eV

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E'=E_6-E_5

E'=(-0.378 eV)-(-0.544 eV) =0.166 eV

1.89 eV > 0.166 eV

E> E'

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