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beks73 [17]
3 years ago
9

The force required to stretch a Hooke’s-law spring varies from 0 N to 63.5 N as we stretch the spring by moving one end 5.31 cm

from its unstressed position. Find the force constant of the spring. Answer in units of N/m. Find the work done in stretching the spring. Answer in units of J.
Physics
1 answer:
Alika [10]3 years ago
8 0

Answer:

Force constant will be 1195.85 N/m

Work done will be 1.6859 J

Explanation:

We have given the force,  F = 63.5 N

Spring is stretched by 5.31 cm

So x = 0.0531 m

Force is given , F = 63.5 N

We know that force is given by F=kx

So 63.5=k\times 0.0531

k = 1195.85 N/m

Now we have to find the work done

We know that work done is given by

W=\frac{1}{2}kx^2=\frac{1}{2}\times 1195.85\times 0.0531^2=1.6859J

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An airplane with a speed of 92.3 m/s is climbing upward at an angle of 51.1 ° with respect to the horizontal. When the plane's a
devlian [24]

Answer:

a

D =  1162.7 \  m

b

\beta =- 65.55^o

Explanation:

From the question we are told that

  The speed of the airplane is  u  =  92.3 \ m/s

   The  angle is  \theta = 51.1^o

    The altitude of the plane is  d =  532 \  m

Generally the y-component of the airplanes velocity is  

       u_y  =  v *  sin (\theta )

=>     u_y  =   92.3 *  sin ( 51.1 )

=>     u_y  =  71.83  \ m/s

Generally the displacement  traveled by the package in the vertical direction is

       d =  (u_y)t +  \frac{1}{2}(-g)t^2

=>       -532  = 71.83 t +  \frac{1}{2}(-9.8)t^2

Here the negative sign for the distance show that the direction is along the negative y-axis

 =>   4.9t^2 - 71.83t - 532 = 0

Solving this using quadratic formula we obtain that

    t =  20.06 \  s

Generally the x-component of the velocity is  

     u_x  =  u  *  cos (\theta)

=>    u_x  =   92.3  *  cos (51.1)

=>   u_x  =   57.96 \ m/s

Generally the distance travel in the horizontal  direction is    

     D =  u_x  *  t

=>   D =  57.96  *   20.06

=>    D =  1162.7 \  m

Generally the angle of the velocity vector relative to the ground is mathematically represented as

       \beta  =  tan ^{-1}[\frac{v_y}{v_x } ]

Here v_y is the final  velocity of the package along the vertical  axis and this is mathematically represented as  

     v_y  =  u_y  -   gt

=>  v_y  =  71.83  -    9.8 *  20.06

=>  v_y  =  -130.05 \  m/s  

and  v_x is the final  velocity of the package which is equivalent to the initial velocity u_x

So

       \beta  =  tan ^{-1}[-130.05}{57.96 } ]

       \beta =- 65.55^o

The negative direction show that it is moving towards the south east direction

   

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Answer:

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Explanation:

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Voltage of primary coil, V_p=120\ V

Current in the secondary coil, I_s=1.8\times 10^{-3}\ A

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P_s=I_s\times V_s...........(1)

For a transformer, \dfrac{N_s}{N_p}=\dfrac{V_s}{V_p}

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P_s=1.8\times 10^{-3}\times (\dfrac{46}{1})\times 120

P_s=9.936\ Watts

So, the power consumed by the air filter is 9.936 watts. Hence, this is the required solution.                

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