Physics, 31.03.2020 00:43, shaelynwolf326

A 1.60-kg object is held 1.05 m above a relaxed, massless vertical spring with a force constant of 315 N/m. The object is dropped onto the spring. (a) How far does the object compress the spring? .377 Correct: Your answer is correct. m (b) Repeat part (a), but this time assume a constant air-resistance force of 0.750 N acts on the object during its motion.

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answered: ashylynn2423

a) the distance up to which the object compress the spring is x = 0.378m

b),the distance up to which the object compress the spring is x= 0.359

Explanation:

Given that,

Mass , m = 1.6kg

spring at vertical height ,h = 1.05m

Spring constant ,k = 315 N/m

Air resistance force = 0.750N

Let assume that the lowest point is the origin

Therefore, initial displacement is

y_i = 1.05 + x

And final displacement is

y_f = 0

Apply the conservation of energy

$\frac{mv_i^2}{2} +\frac{kx_i^2}{2} +mgy_i=\frac{mv_f^2}{2} +\frac{kx^2}{2} +mgy_f$

$\frac{m(0)^2}{2} +\frac{k(0)^2}{2} +mg(1.05+x)=\frac{m(0)^2}{2} +\frac{kx^2}{2} +mg(0)$

$(1.60)(9.81)(1.05 + x)^2=\frac{(315)}{2} x^2\\\\16.4808+15.969x=157.5x^2$

x^2-0.1014x-0.1046=0

solve the equation using quadratic equation

$x = \frac{-(-0.1014\pm \sqrt{(-0.1014)^2-4(1)(-1046)} )}{2}$

$x = \frac{0.1014\pm\sqrt{0.42868} }{2} \\\\x = \frac{0.1014\pm0.6547}{2} \\\\x = \frac{0.1014+0.6547}{2} or \frac{0.1014-0.6547}{2} \\\\x = 0.3781 or -0.2767$

Take the positive value

Hence , the distance up to which the object compress the spring is x = 0.378m

b)If the air resistance is taken into cosideration

therefore apply conservation of energy

mg(1.05 + x) - fd = kx²/2

mg(1.05 +x) - (0.750N) (1.05 + x) = kx²/2

((1.60× 9.81) - 0.750)(1.05 + x) = (315/2)x²

14.946 (1.05 + x) = 157.5x²

15.6933 + 14.946x = 157.5x²

x² - 0.0949x - 0.00996

solve the equation using quadratic equation

x = 0.359 or -0.264

Hence ,the distance up to which the object compress the spring is x= 0.359

answered: Guest

electromagnetic shielding is the practice of reducing the electromagnetic field in a space by blocking the field with barriers made of conductive or magnetic materials. shielding is typically applied to enclosures to isolate electrical devices from their surroundings, and to cables to isolate wires from the environment through which the cable runs. electromagnetic shielding that blocks radio frequency electromagnetic radiation is also known as rf shielding.

the shielding can reduce the coupling of radio waves, electromagnetic fields and electrostatic fields. a conductive enclosure used to block electrostatic fields is also known as a faraday cage. the amount of reduction depends very much upon the material used, its thickness, the size of the shielded volume and the frequency of the fields of interest and the size, shape and orientation of apertures in a shield to an incident electromagnetic

from wikipedia

answered: Guest

You would need to take a picture so i can see

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A 1.60-kg object is held 1.05 m above a relaxed, massless vertical spring with a force constant of 3...