In science, all concepts are

A raft with the area A , thickness= h and the mass 600 kg, Floats in still water with 7 cm

elena55 [62]

In this problem, we need to solve for Bubba’s mass. To do this, we let A be the area of the raft and set the weight of the displaced fluid with the raft alone as ρwAd1g and ρwAd2g with the person on the raft, where ρw is the density of water, d1 = 7cm, and d2= 8.4 cm. Set the weight of displaced fluid equal to the weight of the floating objects to eliminate A and ρw then solve for m.

ρwAd1g = Mg

ρwAd2g = (M + m) g

d2∕d1 = (M + m)/g

m = [(d2∕d1)-1] M = [(8.4 cm/7.0 cm) - 1] (600 kg) =120 kg

This means that Bubba’s mass is 120 kg.

7 0

3 years ago

What is cartilage?

Sphinxa [80]

"The connective tissue that's found between bones."

An example would be between the vertabrae in the spine.

Like the other (inappropriate) answer said, a ligament is what attches muscles and bones.

Also, the tissue inside of bones is bone marrow.

If you know that, you can take out those two options (Option 1 and 3)

5 0

2 years ago

Show all work.

lys-0071 [83]

The new gravitation force at the new location is 40 N

Explanation:

The weight of the astronaut is given by the equation

$F=mg$ (1)

where

m is the mass of the astronaut

g is the acceleration of gravity

The acceleration of gravity at a certain distance $r$ from the centre of the Earth is given by

$g=\frac{GM}{r^2}$

where G is the gravitational constant and M is the Earth's mass. So we can rewrite eq.(1) as

$F=\frac{GMm}{r^2}$

When the astronaut is on the Earth's surface, $r=R$ (where R is the Earth's radius), so his weight is

$F=\frac{GMm}{R^2}=640 N$

Later, he moves to another location where his distance from the Earth's surface is 3 times the previous distance, so the new distance from the Earth's centre is

$r'=3R+R=4R$