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-BARSIC- [3]
2 years ago
12

A. A cord passing over an easily turned pulley (one that is both massless and frictionless) has 7kg mass hanging from one end an

d a 9kg mass hanging from the other end as shown below. G 7.0 kg 9.0 ks This arrangement is known as Atwood's machine. Calculate the (i). acceleration of the masses (ii). tension in the cord. [2.5 ​
Physics
1 answer:
Zepler [3.9K]2 years ago
6 0

Answer:

M g = M a = a (9 - 7) kg      with the two masses going in different directions

(7 + 9) kg * g = a * 2 kg       2 kg effective in prducing acceleration

a = 16 / 2 * g = 9.8 m/s^2 / 8 = 1.23 m/s^2

M1 a = T - M1 * g        T producing upward acceleration of 1.23 m/s^2

T = 7 (a + g) = 7 * 11 = 77 N

Check:

M2 a =  M2 * g - T          M2 producing downward acceleration

T = M2 (g - a) = 9 * (9.8 - 1.23) = 77 N

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3 years ago
How do sea surface temperatures affect evaporation rate?
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8 0
3 years ago
The lawn sprinkler consists of four arms that rotate in the horizontal plane. The diameter of each nozzle is 8 mm, and the water
sashaice [31]

Answer: the constant angular velocity of the arms is 86.1883 rad/sec

Explanation:

First we calculate the linear velocity of the single sprinkler;

Area of the nozzle = π/4 × d²

given that d = 8mm = 8 × 10⁻³

Area of the nozzle = π/4 × (8 × 10⁻³)²

A = 5.024 × 10⁻⁵ m²

Now total discharge is dived into 4 jets so discharge for single jet will be;

Q_single = Q / n = 0.006 / 4 = 1.5 × 10⁻³ m³/sec

So using continuity equation ;

Q_single = A × V_single

V_single = Q_single/A

we substitute

V_single = (1.5 × 10⁻³) / (5.024 × 10⁻⁵)

V_single = 29.8566 m/s

Now resolving the forces as shown in the second image,

Vt = Vcos30°

Vt = 29.8566 × cos30°

Vt = 25.8565 m/s

Finally we calculate the angular velocity;

Vt = rω

ω_single = Vt / r

from the given diagram, radius is 300mm = 0.3m

so we substitute

ω_single = 25.8565 / 0.3

ω_single = 86.1883 rad/sec

Therefore the constant angular velocity of the arms is 86.1883 rad/sec

7 0
3 years ago
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