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Sauron [17]
3 years ago
13

Use Hooke's Law, which states that the distance a spring stretches (or compresses) from its natural, or equilibrium, length vari

es directly as the applied force on the spring. A force of 265 newtons stretches a spring 0.15 meter. (a) What force stretches the spring 0.4 meter
Physics
1 answer:
Phantasy [73]3 years ago
6 0

Answer:

706.68 N

Explanation:

By Hooke's law,

F = ke

k=\dfrac{F}{e}

Using the values in the question,

k=\dfrac{265\text{ N}}{0.15 \text{ m}}=1766.7\text{ N/m}

When e = 0.4 m,

F = 1766.7\text{ N/m}\times0.4\text{ m}=706.68\text{ N}

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Assuming that Bernoulli's equation applies, compute the volume of water ΔV that flows across the exit of the pipe in 1.00 s . In
OLEGan [10]

Answer:

discharge rate (Q) = 0.2005 m^{3} / s

Explanation:

if you read the question you would see that some requirements are missing, by using search engines, you can get the complete question as stated below:

Water flows steadily from an open tank as shown in the figure. (Figure 1) The elevation of point 1 is 10.0m , and the elevation of points 2 and 3 is 2.00 m . The cross-sectional area at point 2 is 4.80x10-2m ; at point 3, where the water is discharged, it is 1.60x10-2m. The cross-sectional area of the tank is very large compared with the cross-sectional area of the pipe. Part A Assuming that Bernoulli's equation applies, compute the volume of water DeltaV that flows across the exit of the pipe in 1.00 s . In other words, find the discharge rate \Delta V/Delta t. Express your answer numerically in cubic meters per second.

solution:

time = 1 s

elevation of point 1 (z1) = 10 m

elevation of point 2 (z2) = 2 m

elevation of point 3 (z3) = 2 m

cross section area of point 2 = 4.8 x 10^{2} m

cross section area of point 3 = 1.6 x 10^{2} m

g

acceleration due to gravity (g) = 9.8 m/s^{2}

find the discharge rate at point 3 which is the exit pipe.

discharge rate (Q) = A3 x V3

where A3 is the cross sectional area at point 3 and V3 is the velocity of the fluid and can be gotten by applying Bernoulli's equation below

\frac{P1}{ρg} +  \frac{V1^{2} }{2g} + Z1 =  \frac{P3}{ρg} + \frac{V3^{2} }{2g} + Z3

pressure at point 1 (P1) is the same as pressure at point 3 (P3), and at point 1, the velocity (V1) = 0. therefore the equation now becomes

\frac{P1}{ρg} + Z1 =  \frac{P1}{ρg} + \frac{V3^{2} }{2g} + Z3

Z1 = \frac{V3^{2} }{2g} + Z3

V3 = \sqrt{2g(Z1-Z3)}

V3 = \sqrt{2 x 9.8 x (10 - 3)}

V3 = 12.53 m/s

discharge rate (Q) = A3 x V3 = 1.6 x 10^{-2} x 12.53

discharge rate (Q) = 0.2005 m^{3} / s

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The woman on the platform is correct because it is the pace of the man moving on the train not walking.

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You have a two-wheel trailer that you pull behind your ATV. Two children with a combined mass of 77.2 kg hop on board for a ride
lions [1.4K]

To solve this problem it is necessary to apply the kinematic equations of motion and Hook's law.

By Hook's law we know that force is defined as,

F= kx

Where,

k = spring constant

x = Displacement change

PART A) For the case of the spring constant we can use the above equation and clear k so that

k= \frac{F}{x}

k = \frac{mg}{x}

k= \frac{77.2*9.8}{0.0637}

k = 11876.92N/m

Therefore the spring constant for each one is 11876.92/2 = 5933.46N/m

PART B) In the case of speed we can obtain it through the period, which is given by

T = \frac{2\pi}{\omega}

Re-arrange to find \omega,

\omega = \frac{2\pi}{T}

\omega = \frac{2\pi}{2.14}

\omega = 2.93rad/s

Then through angular kinematic equations where angular velocity is given as a function of mass and spring constant we have to

\omega^2 = \frac{k}{m}

m = \frac{k}{\omega^2}

m = \frac{ 11876.92}{2.93}

m = 4093.55Kg

Therefore the mass of the trailer is 4093.55Kg

PART C) The frequency by definition is inversely to the period therefore

f = \frac{1}{T}

f = \frac{1}{2.14}

f = 0.4672 Hz

Therefore the frequency of the oscillation is 0.4672 Hz

PART D) The time it takes to make the route 10 times would be 10 times the period, that is

t_T = 10*T

t_T = 10 *2.14s

t_T = 21.4s

Therefore the total time it takes for the trailer to bounce up and down 10 times is 21.4s

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3 years ago
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