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2 years ago

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A student is bouncing on a trampoline. at her highest point, her feet are 65 cm above the trampoline. when she lands, the trampo

line sags 15 cm before propelling her back up. part a for how long is she in contact with the trampoline?

1 answer:

Ede4ka [16]2 years ago

3 0

<span>she is in contact with the trampoline for less than one second.</span>

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An object blocks the path of a wave causing the wave to change direction is called what

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"Reflection" is the change direction of waves when blocked by obstacle

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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

Agata [3.3K]

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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Cause and Effect: What would happen to a space vehicle in orbit around Earth if it sped up?

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Answer:

G M m / R^2 = m v^2 / R gravitational force = centripetal force

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As v increases R will must decrease

Take the moon as an example

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S / 1 day = 54,000 miles/day for a 28 day circuit

S / 1 hr = 54000 / 24 = 2200 mph which is much less than a satellite in orbit

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Assume north and east as two sides of a right angled triangle. magnitude of the distance is then given by the length of the hypotenuse which is $\sqrt{a^2 + b^2}$

where a = 1.2 km north

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magnitude = 2 km

Direction is given by the angle between them, that is atan(a/b) = 36.86 deg north of east = 53.1 deg east of north.

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