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drek231 [11]
3 years ago
9

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.
Physics
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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C) Mass

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3 years ago
Help me to solve it . It’s urgent
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Answer: 0°

Explanation:

Step 1: Squaring the given equation and simplifying it

Let θ be the angle between a and b.

Given: a+b=c

Squaring on both sides:

... (a+b) . (a+b) = c.c

> |a|² + |b|² + 2(a.b) = |c|²

> |a|² + |b|² + 2|a| |b| cos 0 = |c|²

a.b = |a| |b| cos 0]

We are also given;

|a+|b| = |c|

Squaring above equation

> |a|² + |b|² + 2|a| |b| = |c|²

Step 2: Comparing the equations:

Comparing eq( insert: small n)(1) and (2)

We get, cos 0 = 1

> 0 = 0°

Final answer: 0°

[Reminders: every letters in here has an arrow above on it]

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