(10 points) A spring with a 7-kg mass and a damping constant 12 can be held stretched 1 meters beyond its natural length by a fo

rce of 4 newtons. Suppose the spring is stretched 2 meters beyond its natural length and then released with zero velocity. In the notation of the text, what is the value c2−4mk? m2kg2/sec2 Find the position of the mass, in meters, after t seconds. Your answer should be a function of the variable t of the form c1eαt+c2eβt where α= (the larger of the two) β=

Physics

You might be interested in

(a) Two point charges totaling 8.00 μC exert a repulsive force of 0.150 N on one another when separated by 0.500 m. What is the

Softa [3030]

3 is jxhuneuxndnzixbf

7 0

1 month ago

Read 2 more answers

A simple watermelon launcher is designed as a spring with a light platform for the watermelon. When an 8.00 kg watermelon is put

Yuliya22 [3333]

To tackle this problem, it's essential to employ concepts associated with force as per Hooke's law, alongside the forces described by Newton's second law and the concept of potential elastic energy. Since the forces are in equilibrium, the spring force matches the gravitational force. To find the spring constant k, we recognize the compression is 40cm at launch, hence applying the potential elastic energy formula results in determining the energy stored in the spring as 63.72 Joules.

6 0

2 months ago

A gas is compressed in a cylinder from a volume of 20.0 l to 2.0 l by a constant pressure of 10.0 atm. calculate the amount of w

Maru [3345]

First, we must transform the pressure into SI units, considering that $1 atm = 1.01 \cdot 10^5 Pa$ :
$p=10.0 atm =1.01 \cdot 10^6 Pa$

The starting and ending volumes of the gas will be as follows (keeping in mind that $1.0 L = 0.001 m^3$ ):
$V_i = 20.0 L=0.020 m^3$
$V_f = 2.0 L=0.002 m^3$

Thus, the work performed on the gas by its surroundings is
$W= -p \Delta V=-p(V_f-V_i)=-(1.01 \cdot 10^6 Pa)(0.002 m^3-0.020 m^3)=18180 J$
The positive outcome indicates that this work leads to a rise in the gas's internal energy.

8 0

2 months ago

A mouse runs along a baseboard in your house. The mouse's position as a function of time is given by x(t)=pt2+qt, with p = 0.36

ValentinkaMS [3465]

Response:

0.60 m/s

Details:

The average speed between times t = a and t = b can be expressed as:

v_avg = (x(b) − x(a)) / (b − a)

Given the function x(t) = 0.36t² − 1.20t, and considering the interval from 1.0 to 4.0:

v_avg = (x(4.0) − x(1.0)) / (4.0 − 1.0)

v_avg = [(0.36(4.0)² − 1.20(4.0)) − (0.36(1.0)² − 1.20(1.0))] / 3.0

v_avg = [(5.76 − 4.8) − (0.36 − 1.20)] / 3.0

v_avg = [0.96 − (-0.84)] / 3.0

v_avg = 0.60

The average speed calculated is 0.60 m/s.

5 0

3 months ago