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3 years ago

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A 0.35-kgkg cord is stretched between two supports, 7.4 mm apart. When one support is struck by a hammer, a transverse wave trav

els down the cord and reaches the other support in 0.88 ss .What is the tension in the cord? (Express your answer to two significant figures and include the appropriate units.

1 answer:

Agata [3.3K]3 years ago

4 0

Answer:

The tension in the string is $T = 1.49*10^{-6}N$ .

Explanation:

For a string with tension $T$ and linear density $\mu_d$ carrying a transverse wave at speed $v$ it is true that

$v = \sqrt{\dfrac{T}{\mu_d} }$

solving for $T$ we get:

$T = \dfrac{v^2}{\mu_d}.$

Now, the transverse wave covers the distance of 7.4mm in 0.88s, which means it's speed is

$v =\dfrac{7.4*10^{-3}m}{0.88s} \\\\v = 8.4*10^{-3}s$

And it's linear density (mass per unit length) is

$\mu_d = \dfrac{0.35kg}{7.4*10^{-3}m} \\\\\mu_d = 47.3kg/m$

Therefore, the tension in the cord is

$T = \dfrac{(8.4*10^{-3}m/s^2)^2}{47.3kg/m}.$

$\boxed{T = 1.5*10^{-6}N}$

or in micro newtons

$T =1.5\mu N$

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Water is drawn from a well in a bucket tied to the end of a rope whose other end wraps around a solid cylinder of mass 50 kg and

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Answer:

$\alpha=78.4\ rad.s^{-2}$

Explanation:

Given:

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diameter of cylinder, $d=0.25\ m$

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<em>When the bucket is released to fall in the well, it fall under the acceleration due to gravity.</em>

We have formula for angular acceleration as:

$\alpha=\frac{g}{r}$

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3 years ago

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Answer:

Part a)

$a= 0.32 m/s^2$

Part b)

$F_c = 3.6 N$

Part c)

$F_c = 5.5 N$

Explanation:

Part a)

As we know that the friction force on two boxes is given as

$F_f = \mu m_a g + \mu m_b g$

$F_f = 0.02(10.6 + 7)9.81$

$F_f = 3.45 N$

Now we know by Newton's II law

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$a = \frac{5.65}{17.6}$

$a= 0.32 m/s^2$

Part b)

For block B we know that net force on it will push it forward with same acceleration so we have

$F_c - F_f = m_b a$

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If Alex push from other side then also the acceleration will be same

So for box B we can say that Net force is given as

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$9.1 - 0.02(7)(9.8) - F_c = 7(0.32)$

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$F_c = 5.5 N$

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3 years ago

ASTM A229 oil-tempered carbon steel is used for a helical coil spring. The spring is wound with D = 50 mm d = 10.0 mm, and a pit

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Answer

given,

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$\tau_s \leq 0.45 \times 1300$

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since corrected stress is less than the $\tau_s$

hence, spring will return to its original shape.

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