Gravity is affected by

The first law of thermodynamics states that energy can neither be created nor destroyed. If this is true then why are we always

riadik2000 [5.3K]

Answer:

The second law of thermodynamics states in an isolated system, the entropy (the amount of thermal energy that cannot be converted into mechanical work, also known as the amount of disorder) always increases, therefore, an isolated system always require an external input (new sources) of energy for there to be orderliness or for the available energy of the system to remain constant or increase

Explanation:

5 0

2 years ago

Where in the motion is the magnitude of the force from the spring on the object zero? Where in the motion is the magnitude of th

kkurt [141]

Answer:

1. The magnitude of the force from the spring on the object is zero on Equilibrium.

2. The magnitude of the force from the spring on the object is a maximum on The top and bottom.

3. The magnitude of the net force on the object is zero on The Bottom.

4. The magnitude of the force on the object is a maximum on the Top.

Explanation:

1. Because the change in position delta X is zero.

2. Because of delta X.

3. Beacuse, the force of gravity and the force of the spring oppose each other to keep the block at rest, away from the equilibrium position.

4. Because, the force of the spring from compressiom and the force of gravity both act on the mass.

8 0

3 years ago

A vertical spring has a spring constant of 2900 N/m. The spring is compressed 80 cm and a 8 kg spider is placed on the spring. T

Serga [27]

Answer:

a) $k_{e}$ = 928 J , b)U = -62.7 J , c) K = 0 , d) Y = 11.0367 m, e) v = 15.23 m / s

Explanation:

To solve this exercise we will use the concepts of mechanical energy.

a) The elastic potential energy is

$k_{e}$ = ½ k x²

$k_{e}$ = ½ 2900 0.80²

$k_{e}$ = 928 J

b) place the origin at the point of the uncompressed spring, the spider's potential energy

U = m h and

U = 8 9.8 (-0.80)

U = -62.7 J

c) Before releasing the spring the spider is still, so its true speed and therefore the kinetic energy also

K = ½ m v²

K = 0

d) write the energy at two points, maximum compression and maximum height

Em₀ = ke = ½ m x²

$E_{mf}$ = mg y

Emo = $E_{mf}$

½ k x² = m g y

y = ½ k x² / m g

y = ½ 2900 0.8² / (8 9.8)

y = 11.8367 m

As zero was placed for the spring without stretching the height from that reference is

Y = y- 0.80

Y = 11.8367 -0.80

Y = 11.0367 m

Bonus

Energy for maximum compression and uncompressed spring

Emo = ½ k x² = 928 J

$E_{mf}$ = ½ m v²

Emo = $E_{mf}$

Emo = ½ m v²

v =√ 2Emo / m

v = √ (2 928/8)

v = 15.23 m / s

8 0

3 years ago

A drone flies 8 m/s due East with respects to the wind. The wind is blowing 6 m/s due North with respects to the ground. What is

vekshin1

Answer:

B. 10m/s

Explanation:

If a drone flies 8 m/s due East with respects to the wind and the wind is blowing 6 m/s due North, the speed of the drone with respect to the ground is its displacement.

Displacement is calculated using Pythagoras theorem.

d² = 8²+6²

d² = 64+36

d² = 100

Square root both sides

√d² = √100

d = 10m/s

Hence the distance of the drone with respect to the ground is 10m/s

Option B is correct

8 0

3 years ago

A 1200-kg car accelerates it’s speed from 4 m/s to 10 m/s in 3 seconds. Find the car average speed , the car acceleration , the

dsp73

Answer:

Given: mass 1200kg

initial velocity: 4m/s

finial velocity: 10 m/s

time 3 sec

then

speed; initial velocity + final velocity/2

4+10/3

: 4.66m/s2

8 0

2 years ago