If a spring is stretched from its equilibrium position, then a force with magnitude proportional to the increase in length from the equilibrium length is pulling each end.

F = kx

where k is the proportionality constant called the spring constant or force constant.

Up to a point, the elongation of a spring is directly proportional to the force applied to it. Once you extend the spring more than 10.0 centimeters, however, it no longer follows that simple linear rule.

Let the spring constant be very low 0.04N/m

The force applied is

F = 10 cm / 0.04

F = 0.1 m / 0.04

F = 2.5 N

Thus, the force result in stretching the spring 10cm is 2.5 N.

Learn more about hooke's law.

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2 years ago

A skydiver of 75 kg mass has a terminal velocity of 60 m/s. At what speed is the resistive force on the skydiver half that when

ankoles [38]

Answer:

The speed of the resistive force is 42.426 m/s

Explanation:

Given;

mass of skydiver, m = 75 kg

terminal velocity, $V_T = 60 \ m/s$

The resistive force on the skydiver is known as drag force.

Drag force is directly proportional to square of terminal velocity.

$F_D = kV_T^2$

Where;

k is a constant

$k = \frac{F_D_1}{V_{T1}^2} = \frac{F_D_2}{V_{T2}^2}$

When the new drag force is half of the original drag force;

$F_D_2 = \frac{F_D_1}{2} \\\\\frac{F_D_1}{V_{T1}^2} = \frac{F_D_2}{V_{T2}^2} \\\\\frac{F_D_1}{V_{T1}^2} = \frac{F_D_1}{2V_{T2}^2} \\\\\frac{1}{V_{T1}^2} = \frac{1}{2V_{T2}^2}\\\\2V_{T2}^2 = V_{T1}^2\\\\V_{T2}^2= \frac{V_{T1}^2}{2} \\\\V_{T2}= \sqrt{\frac{V_{T1}^2}{2} } \\\\V_{T2}= \frac{V_{T1}}{\sqrt{2} } \\\\V_{T2}= 0.7071(V_{T1})\\\\V_{T2}= 0.7071(60 \ m/s)\\\\V_{T2}= 42.426 \ m/s$