A body of mass 5.0 kg is suspended by a spring which stretches 10 cm when the mass is attached. It is then displaced downward an additional 5.0 cm and released. Its position as a function of time is approximately what? Group of answer choices

0.05cos10t

Explanation:

X(t) = Acos(wt+φ)

The oscillation angular frequency can be calculated using below formula

w = √(k/M)

Where K is the spring constant

But we were given body mass of 5.0 kg

We know acceleration due to gravity as 9.8m)s^2

The lenghth of spring which stretches =10 cm

Then we can calculate the value of K

k = (5.0kg*9.8 m/s^2)/0.10 m

K= 490 N/m

Then if we substitute these values into the formula above we have

w = √(k/M)

w = √(490/5)

= 9.90 rad/s=10rads/s(approximately)

Its position as a function of time can be calculated using the below expresion

X(t) = Acos(wt+φ)

We were given amplitude of 5 cm , if we convert to metre = 0.05m

w=10rads/s

Then if we substitute we have

X(t)=0.05cos(10×t)

X(t)= 0.05cos10t

Therefore,Its position as a function of time=

X(t)= 0.05cos10t

c. 7/3

step-by-step explanation:

a rational number is like pi- it is a repeating decimal.

a. is -1.8

b. 2.236068

c. 2.3333333333333333333

d.   is 3

i hope that ! =)

the tires are rotating at 0.434 rpmlinear speed=45 m/sdiameter=0.550 mradius=0.275 mnow v= rωω= v/rω=45/0.275ω=163.6 rad/sω=163.6 rad/s (1rev/2π rad) (60s/1m)ω=1563 rpm