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mixer [17]
1 year ago
6

In the Atwood machine shown below, m1 = 2.00 kg and m2 = 6.05 kg. The masses of the pulley and string are negligible by comparis

on. The pulley turns without friction and the string does not stretch. The lighter object is released with a sharp push that sets it into motion at vi = 3.00 m/s downward.. (a) How far will m1 descend below its initial level?
Physics
1 answer:
Rus_ich [418]1 year ago
5 0
M1 descending
−m1g + T = m1a 

m2 ascending
m2g − T = m2a

this gives :
(m2 − m1)g = (m1 + m2)a 

a = (m2 − m1)g/m1 + m2
   = (5.60 − 2)/(2 + 5.60) x 9.81 
   = = 4.65m/s^2
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A rotating beacon is located 2 miles out in the water. Let A be the point on the shore that is closest to the beacon. As the bea
Mice21 [21]

Answer:

v = 3369.2 m/s

Explanation:

As we know that Beacon is rotating with angular speed

f = 10 rev/min

so we have

\omega = 2\pi f

\omega = 2\pi(\frac{10}{60})

\omega = 1.047 rad/s

now we know that

v = r \omega

here we will have

r = 2 miles = 2(1609 m)

r = 3218 m

so we have

v = 3218(1.047)

v = 3369.2 m/s

6 0
1 year ago
Consider a steel guitar string of initial length l=1.00m and cross-sectional area a=0.500mm2. the young's modulus of the steel i
laiz [17]
L = 1.00 m, the original length
A = 0.5 mm² = 0.5 x 10⁻⁶ m², the cross sectional area
E = 2.0 x 10¹¹ n/m², Young's modulus
P = 1500 N, the applied tension

Calculate the stress.
σ = P/A = (1500 N)/(0.5 x 10⁻⁶ m²) = 3 x 10⁹ N/m²

Let δ =  the stretch of the string.
Then the strain is
ε = δ/L

By definition, the strain is
ε = σ/E = (3 x 10⁹ N/m²)/(2 x 10¹¹ N/m²) = 0.015
Therefore
δ/(1 m) = 0.015
δ = 0.015 m = 15 mm

Answer:  15 mm
4 0
2 years ago
Process in which permanent deformation of metals occurs due to applied stress and results in breaking of bonds and then reformin
luda_lava [24]

Answer:

This process involves the motion of dislocations and is termed slip (or glide in some textbooks)

Explanation:

Plastic deformation of metals (and other crystalline materials) usually occurs by slip, which is the sliding of planes of atoms over one another by dislocation movements.

On a microscopic scale, stress causes planes of crystalline objects to leave their original position and slide over other planes into new positions, these microscopic movements manifest as a slip on a macroscopic scale. And the planes do not return back to their original position after the removal of the dislocation-causing stress.

6 0
2 years ago
You drop your keys in a high-speed elevator going up at a constant speed. Part APart complete Do the keys accelerate faster towa
anzhelika [568]

Answer:

Explained

Explanation:

a) No, the keys were initially moving upward in the elevator only effects the initial velocity of the key and not the rate of change of velocity that is acceleration. So, the keys accelerate with the same acceleration as before.

b)Yes, keys will accelerate towards the floor faster if it is a constant speed than it is moving downward because if the elevator is accelerating downward, the downward change in velocity of the keys is at least partially matched by a downward change in the velocity of the of the elevator.

5 0
2 years ago
Little Tammy lines up to tackle Jackson to (unsuccessfully) prove the law of conservation of momentum. Tammy’s mass is 34.0 kg a
Naily [24]

Answer:

So Tammy must move with speed 4.76 m/s in opposite direction of Jackson

Explanation:

As per law of conservation of momentum we know that there is no external force on it

So here we can say that initial momentum of the system must be equal to the final momentum of the system

now we have

m_1v_1 + m_2v_2 = 0

final they both comes to rest so here we can say that final momentum must be zero

now we have

34 v + 54 (3 m/s) = 0

v = -4.76 m/s

8 0
1 year ago
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