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laiz [17]
3 years ago
10

The length of second hand of clock is 14cm, an ant sits on the top of second hand. find the following

Physics
1 answer:
Anastaziya [24]3 years ago
4 0

Answer:

The answer is below

Explanation:

i) Since the length of the second clock (radius) is 14 cm = 0.14 m, the distance covered by the second hand in one revelution is:

Distance covered = 2πr = 2π(0.14) = 0.88 m

The time taking to complete one revolution = 60 seconds, hence;

Speed = distance covered in one revolution / time take o complete a revolution

Speed = 0.88 m / 60 s = 0.0147 m/s

ii) Distance covered in 150 s = speed * 150 s = 0.0147 * 150 = 2.2 m

iii) Displacement in 150 seconds = distance from initial position to final position

At 150 s, the hand has covered 2 revolutions and moved 30 s. Hence:

Displacement in 150 seconds = speed * 30 s = 0.0147 * 30 = 0.44 m

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\textbf{2500 }\dfrac{\textbf{kg}}{\textbf{s}^{\textbf{2}}}

Explanation:

       Natural length of a spring is 0.5\text{ }m. The spring is streched by 0.02\text{ }m. The resultant energy of the spring is 0.5\text{ }J.

       The potential energy of an ideal spring with spring constant k and elongation x is given by \dfrac{1}{2}kx^{2}.

       So, in the current problem, the natural length of the spring is not required to find the spring constant k.

       \text{Potential Energy in the spring = }\dfrac{1}{2}kx^{2}\\0.5\text{ }J\text{ }=\text{ }\dfrac{1}{2}k(0.02\text{ }m)^{2}\\k\times0.0004\text{ }m^{2}\text{ }=\text{ }1\text{ }J\text{ }=\text{ }1\text{ }kg\frac{m^{2}}{s^{2}}\\k\text{ }=\text{ }\dfrac{1\text{ }kg\dfrac{m^{2}}{s^{2}}}{0.0004\text{ }m^{2}}\text{ }=\text{ }2500\text{ }\frac{kg}{s^{2}}

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4 0
4 years ago
Two blocks of clay, one of mass 1.00 kg and one of mass 7.00 kg, undergo a completely inelastic collision. Before the collision
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Answer:8 m/s

Explanation:

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kinetic Energy of m_1 is =\frac{m_1u^2}{2}=32

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In Completely inelastic collision both mass stick together and move with common velocity

Suppose v is the common velocity

m_1u+0=(m_1+m_2)v

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