How is the pressure exerted and the area on which it is exerted are related to each other?

is weigh-in time for the local under-85-kg rugby team. The bathroom scale used to assess eligibility can be described by Hooke’s

yulyashka [42]

Answer:

a) 156960 N/m

b) Yes the player can play on the team.

Explanation:

F = Force

a = Acceleration due to gravity = 9.81 m/s²

m = Mass

k = Spring constant

x = Deformation length of spring

F = ma

From Hooke's law

$F=kx\\\Rightarrow ma=kx\\\Rightarrow k=\frac{ma}{x}\\\Rightarrow k=\frac{120\times 9.81}{0.0075}\\\Rightarrow k=156960\ N/m$

The effective spring constant is 156960 N/m

When x = 0.48 cm = 0.0048 m

$m=\frac{kx}{a}\\\Rightarrow m=\frac{156960\times 0.0048}{9.81}\\\Rightarrow m=76.8\ kg$

The mass of the player is 76.8 kg which is less than the required mass limit of the players' which is 85 kg. So, the player is eligible to play.

5 0

3 years ago

A material that causes a wave to bounce off of it

Deffense [45]

the answer would be white body

7 0

3 years ago

Read 2 more answers

Did the displacement at this point reach its maximum of 2 mm before or after the interval of time when the displacement was a co

Igoryamba

Answer:

a. before

Explanation:

Did the displacement at this point reach its maximum of 2 mm before or after the interval of time when the displacement was a constant 1 mm?

from the graph given from a source. the vertical axis represents the displacement of the graph motion, whilst the horizontal side is representing the time variable of the motion .

displacement is distance in a specific direction.

before the displacement was maximum at 2mm was instant at time=0.04s.

But later was constant at 0.06s at a displacement point of 1mm

4 0

4 years ago

A spaceship maneuvering near planet zeta is located at r⃗ =(600i^−400j^+200k^)×103km, relative to the planet, and traveling at v

slava [35]

<u>Answer:</u>

The spaceship's position when the engine shuts off = $(708.15 i - 444.1 j + 200 k)*10^3km$

<u>Explanation:</u>

Initial location of spaceship = (600 i - 400 j + 200 k)* $10^3km$ = (600 i - 400 j + 200 k)* $10^6m$

Initial velocity = 9500 i m/s

Acceleration = (40 i - 20 k) $10^3m/s^2$

Time = 35 minute = 35 * 60 = 2100 seconds

We have equation of motion , $s= ut+\frac{1}{2} at^2$ , s is the displacement, u is the initial velocity, a is the acceleration and t is the time.

Substituting

$s= (9500 i)*2100+\frac{1}{2}*(40 i - 20 k)*2100^2\\ \\ s=9500*2100 i+20*2100^2i-10*2100^2j\\ \\ s=19.95*10^6i+88.2*10^6i-44.1*10^6j\\ \\ \\s=(108.15i-44.1j)*10^6m$