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3 years ago

7

When people do a bungee jump theyare asked how much they weigh. Why?

1 answer:

Katena32 [7]3 years ago

3 0

Answer:

The elasticity of the bungee cord reduces the gravitational forces applied on the body during bungee jumping. For example, if a 100-pound individual jumps from a building and encounters 900 pounds of deceleration force, they will feel 9 "G's" of force.

hopefully this'll help

have a nice day!!! :D

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an object is thrown with an initial horizontal velocity of 10 meters per second and take approximately 9 seconds to reach the gr

pantera1 [17]

The horizontal velocity was constant, so:

$s = vt$

$s = 10\cdot9$

$s = 90$

it traveled 90meters

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3 years ago

What is the fundamental frequency (in Hz) of a 0.632 m long tube, open at both ends, on a day when the speed of sound is 344 m/s

Greeley [361]

Answer:

$f=272.15Hz$

Explanation:

Given data

Length of tube L=0.632 m

Speed of sound v=344 m/s

To find

Fundamental frequency f

Solution

The fundamental frequency of the tube can be given as:

$f=\frac{v}{2L}\\ f=\frac{344m/s}{2(0.632m)}\\ f=272.15Hz$

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4 years ago

Which statement best describes the energy changes that occur while a child is riding on a sled down a steep, snow-covered hill?

Degger [83]

B. kinetic energy increases and potential energy decreases

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4 years ago

A wire of length 5mm and Diameter 2m extends by 0.25 when a force of 50N was use. calculate the

bazaltina [42]

Answer and Explanation:

Data provided in the question

Force = 50N

Length = 5mm

diameter = 2.0m = $2\times 10^{-3}$

Extended by = 0.25mm = $0.25\times 10^{-3}$

Based on the above information, the calculation is as follows

a. The Stress of the wire is

$= \frac{force\ applied}{area\ of \ circle}$

here area of circle = perpendicular to the are i.e cross-sectional i.e

= $\frac{\pi d^{2}}{4}$

= $\frac{\pi(2\times 10^{-3})^2}{4}$

Now place these above values to the above formula

$= \frac{4\times 50}{\pi\times 4 \times 10^{-6}} \\\\ = \frac{50}{\pi}$

= 15.92 MPa

As 1Pa = 1 by N m^2

So,

MPa = 10^6 N m^2

b. Now the strain of the wire is

$= \frac{Change\ in\ length}{initial\ length} \\\\ = \frac{0.25\times 10^{-3}}{5}$

= $5 \times 10^{-5}$

3 0

4 years ago

What are the wavelength ranges for the following? (a) the AM radio band (540–1600 kHz) maximum wavelength m minimum wavelength m

Pie

Answer:

Explanation:

a ) AM radio band (540–1600 kHz)

frequency = 540 kHz = 540 x 10³ Hz

wave length = velocity of light / frequency

= 3 x 10⁸ / 540 x 10³

= 555.55 m

frequency = 1600 kHz = 1600 x 10³ Hz

wave length = velocity of light / frequency

= 3 x 10⁸ / 1600 x 10³

= 187.5 m

maximum wavelength = 555.55 m

minimum wavelength = 187.5 m

b )

AM radio band (88 - 108 MHz)

frequency = 88 MHz = 88 x 10⁶ Hz

wave length = velocity of light / frequency

= 3 x 10⁸ / 88 x 10⁶

= 3.41 m

frequency = 108 MHz = 108 x 10⁶ Hz

wave length = velocity of light / frequency

= 3 x 10⁸ / 108 x 10⁶

= 2.78 m

maximum wavelength = 3.41 m

minimum wavelength = 2.78 m

3 0

3 years ago