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miss Akunina [59]
2 years ago
15

The mass of 04kg is rotated by a spring at a constant speed V . If the length of the string is 1m and the minimum tension is 3N,

what is value of V​
Physics
1 answer:
wlad13 [49]2 years ago
3 0

I think you mean "rotated by a string", which is to say a 0.4-kg mass is suspended from a string and swung around in a circle with constant speed.

The centripetal acceleration a of the object is

a = V² / R

where R is the length of the string, 1 m, so that

a = V² / (1 m) = V² m⁻¹

As the mass goes through its uniform circular motion, the only relevant force acting on it is tension with magnitude T, pointing in the direction in which the mass is accelerating (the center of its circular path). By Newton's second law,

T = ma

where m is the object's mass, 0.4 kg. Then

3 N = (0.4 kg) (V² m⁻¹)

⇒   V² = (3 N) / (0.4 kg) • (1 m) = 7.5 m²/s²

⇒   V ≈ 2.74 m/s

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Answer:

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Explanation:

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a)

The final velocity of cart 1 after collision is given as:

v_1=(\frac{m_1-m_2}{m_1+m_2})u_1+\frac{2m_2}{m_1+m_2}u_2\\  Substituting:\\v_1=\frac{0.6-0.8}{0.6+0.8} (-5)+\frac{2*0.8}{0.6+0.8}(2)= 5/7+16/7=3\ m/s

The final velocity of cart 2 after collision is given as:

v_2=(\frac{m_2-m_1}{m_1+m_2})u_2+\frac{2m_1}{m_1+m_2}u_1\\  Substituting:\\v_1=\frac{0.8-0.6}{0.6+0.8} (2)+\frac{2*0.6}{0.6+0.8}(-5)= 2/7-30/7=-4\ m/s

b) Using the law of conservation of energy:

\frac{1}{2}m_1u_1+ \frac{1}{2}m_2u_2=\frac{1}{2}m_1v_1+\frac{1}{2}m_2v_2+\frac{1}{2}kx^2\\x=\sqrt{\frac{m_1u_1+m_2u_2-m_1v_1-m_2v_2}{k}}\\ Substituting\ gives:\\x=\sqrt{\frac{0.6*(-5)^2+0.8*2^2-(0.6*3^2)-(0.8*(-4)^2)}{1200}}=\sqrt{0}=0\ cm

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