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sasho [114]
3 years ago
11

Work of 5 Joules is done in stretching a spring from its natural length to 19 cm beyond its natural length. What is the force (i

n Newtons) that holds the spring stretched at the same distance (19 cm)? Don't forget to enter the correct units.
Physics
1 answer:
densk [106]3 years ago
6 0

Answer:

Explanation:

Given that,

5J work is done by stretching a spring

e = 19cm = 0.19m

Assuming the spring is ideal, then we can apply Hooke's law

F = kx

To calculate k, we can apply the Workdone by a spring formula

W=∫F.dx

Since F=kx

W = ∫kx dx from x = 0 to x = 0.19

W = ½kx² from x = 0 to x = 0.19

W = ½k (0.19²-0²)

5 = ½k(0.0361-0)

5×2 = 0.0361k

Then, k = 10/0.0361

k = 277.008 N/m

The spring constant is 277.008N/m

Then, applying Hooke's law to find the applied force

F = kx

F = 277.008 × 0.19

F = 52.63 N

The applied force is 52.63N

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In the voltage multiplier experiment, why not use a zener diode?​
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Copper is used in wiring because its metallic bonding gives its _______ a high degree of freedom and the material a high conduct
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Read 2 more answers
A truck on a straight road starts from rest, accelerating at 2.00 m/s2 until it reaches a speed of 20.0 m/s. Then the truck trav
aniked [119]
<span>You can use the equation
V_xf = V_xi + a_x(t)

V_xf = 20.0m/s
V_xi = 0m/s
ax = 2.0 t

Thus, solve for t and get 10seconds and then take 5 seconds to break after 20 seconds of driving so for

a) 10 + 20 + 5 = 35 seconds

</span><span>for part b)
You can use the formula
Delta x/Delta t = average velocity
 
Need to find xf, knowing xi = 0

Thus, use the formula
 x_f = x_i + V_xi(t) + (1/2)a_x(t)^(2)
x_f = 0 + 0(10) + (1/2)(2.0)(10)^(2)
 x_f = 100m
 
so for the first 10 seconds the truck traveled 100ms At a speed of 20m/s

20m/s = xm/20s 20*20 = x
x = 400
 
thus we have 100+400 = 500m then it slows down from 500m to x_f
 
thus I use the equation
x_f = x_i + (1/2)(V_xf + V_xi)t
x_f = 500 + (1/2)(0 + 20)(5) x_f = 500 + 50
x_f = 550
 
therefore the total distance traveled is 550m
</span>
<span>to calculate average velocity
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thus V_xavg = 16m/s</span>
8 0
3 years ago
Water, which we can treat as ideal and incompressible, flows at 12 m/s in a horizontal pipe with a pressure of 3.0 × 10^4 Pa.
Triss [41]

Answer:

9.8 × 10⁴Pa

Explanation:

Given:

Velocity V₁ = 12m/s

Pressure P₁ = 3 × 10⁴ Pa

From continuity equation we have

                              ρA₁V₁ = ρA₂V₂

                                 A₁V₁ = A₂V₂

making V₂ the subject of the equation;

                               V_{2} = \frac{A_{1}V_{1}}{A_{2}}

the pipe is widened to twice its original radius,

                                r₂ = 2r₁          

then the cross-sectional area A₂ = 4A₁

                           ⇒  V_{2}= \frac{A_{1}V_{1}}{4A_{1}}

                                  V_{2}= \frac{V_{1}}{4}

This implies that the water speed will drop by a factor of  \frac{1}{4} because of the increase the pipe cross-sectional area.  

 The Bernoulli Equation;

     Energy per unit volume before = Energy per unit volume after    

        p₁ + \frac{1}{2}ρV₁²  + ρgh₁ = p₂ + \frac{1}{2}ρV₂²  + ρgh₂  

Total pressure is constant and P_{T} = P = \frac{1}{2}ρV₂²ρV²  

        p₁ + \frac{1}{2}ρV₁²  = p₂ + \frac{1}{2}ρV₂²

Making p₂ the subject of the equation above;

        p₂ = p₁ + \frac{1}{2}ρV₁² - \frac{1}{2}ρV₂²

But V_{2}  = \frac{V_{1}}{4} so,

        p₂ = p₁ + \frac{1}{2}ρV₁² - \frac{1}{2}ρ\frac{V_{1}^{2}}{4^{2}}      

       p₂ = 3.0 x 10⁴ + (\frac{1}{2} × 1000 × 12²) - ( \frac{1}{2} × 1000 × 12²/4² )

      P₂ = 3.0 x 10⁴ + 7.2 × 10⁴ - 4.05 x 10³    

       P₂ = 9.79 × 10⁴Pa      

      P₂ = 9.8 × 10⁴Pa                      

4 0
3 years ago
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