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3 years ago

Which is the correct equation for the force applied by a spring?

Physics

1 answer:

slava [35]3 years ago

3 0

Answer:

B. $F = -k*x$

Explanation:

The force of a spring can be easily calculated using Hooke's law which tells us that the force applied on a spring is equal to the product of the constant of a spring by the compressed or stretched distance of the spring.

That is:

$F = k*x$

k = spring constant [N/m]

x = distance [m]

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A car comes to a bridge during a storm and finds the bridge washed out. The driver must get to the other side, so he decides to

aksik [14]

Answer:

The answer is below

Explanation:

a) The vertical displacement = Δy = 21.5 m - 1.5 m = 20 m

The horizontal displacement = Δx = 69 m wide

Using the formula:

$\Delta y = u_yt+ \frac{1}{2}a_yt^2\\ \\u_y=initial\ velocity\ of \ car\ in\ y\ direction = 0,a_y=g=acceleration\ due\ to\ gravity\\=10m/s^2\\\\\Delta y = \frac{1}{2}a_yt^2\\\\\Delta y=\frac{1}{2}a_yt^2\\\\t=\sqrt{\frac{2\Delta y}{a_y} }=\sqrt{\frac{2*20}{10} } =2\ m/s$

Also:

$\Delta x = u_xt+ \frac{1}{2}a_xt^2\\ \\u_x=initial\ velocity\ of \ car\ in\ x\ direction = 0,a_x=acceleration=0\\\\\Delta x = u_xt\\\\u_x=\frac{\Delta x}{t}=\frac{69}{2} =34.5\ m/s$

b)The car is moving at a constant speed in the horizontal direction, hence the initial velocity = final velocity

$v_x=u_x=34.5\ m/s\\\\v_y=u_y+a_yt\\\\v_y=0+gt\\\\v_y=10(2)=20\ m/s\\\\v=\sqrt{v_x^2+v_y^2}=\sqrt{34.5^2+20^2}=39.9\ m/s\\ v=39.9\ m/s$

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4 years ago

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SIZIF [17.4K]

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ycow [4]

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