What is the maximum value the string tension can have before the can slips? The coefficient of static friction between the can a

Answer:

The response to your inquiry is: 15 m/s²

Explanation:

Equation    x = at³ - bt² + ct

a = 4.1 m/s³

b = 2.2 m/s²

c = 1.7 m/s

First we calculate x at t = 4.1 s

x = 4.1(4.1)³ - 2.2(4.1)² + 1.7(4.1)

x = 4.1(68.921) - 2.2(16.81) + 6.97

x = 282.58 - 36.98 + 6.98

x = 252.58 m

Now we calculate speed

v = x/t = 252.58/ 4.1 = 61.6 m/s

Finally

acceleration = v/t = 61.6/4.1 = 15 m/s²

The calculation for the horizontal component is performed as follows:
Vhorizontal = V · cos(angle)

For your instance, Vhorizontal = 16 · cos(40) equates to 12.3 m/s

Conclusion: 12.3 m/s

The initial description is the accurate one.

Answer:

155.38424 K

2.2721 kg/m³

Explanation:

= Reservoir pressure = 10 atm

= Reservoir temperature = 300 K

= Exit pressure = 1 atm

= Exit temperature

= Specific gas constant = 287 J/kgK

= Specific heat ratio = 1.4 for air

Assuming isentropic flow

Flow temperature at exit is 155.38424 K

Density at exit can be derived using the ideal gas equation

Flow density at exit measures 2.2721 kg/m³

Response:

D: The distance among the particles diminishes

Clarification:

Removing energy reduces the activity of molecules, similar to how one slows down in cold temperatures (I believe).