Question

A hard wood cylinder of radius 2.0 cm and length 75 cm weighs 800 g. Calculate the density of wood in kg/m³.

Pls help..​

Answer 1

Answer:

Given :

Radius(r) = 2.0cm

Length(l) = 75cm

Mass(m) = 800g

To find :

Density of wood

Solution :

\large\therefore∴ Volume (V) = πr²l

= \sf{\dfrac{22}{7}\times (2.0)^2 \times 75 = \dfrac{6600}{7}g/cm^3}

7

22

×(2.0)

2

×75=

7

6600

g/cm

3

Finding the density now,

\therefore∴ Density(D) = \sf\dfrac{Mass(M)}{Volume(V)}

Volume(V)

Mass(M)

= \sf{\dfrac{800}{\frac{6600}{7}} = \dfrac{800 \times 7}{6600} = \dfrac{28}{33}g/cm^3}

7

6600

800

=

6600

800×7

=

33

28

g/cm

3

= \sf{\dfrac{28}{33} \times 1000kg/m^3 = 848.5 kg/m^3}

33

28

×1000kg/m

3

=848.5kg/m

3

Hence , Density of the wood is 848.5 kg/m³.

Step-by-step explanation:

hope it helps!!!

is your brainly bio visible??

can u tell me there is ur old I'd is ||Dishant||

Answer 2

given :-

• radius - 2.0cm

• length - 75cm

• mass - 800g

To find :-

• density of wood

formula :-

• volume = πr²

• density = mass/volume

solution :-

$\dashrightarrow\sf \: \frac{22}{7} \times (2.0)^{2} \times 75 = \frac{6600}{7}kgcm^{3}$

now using 2nd formula -

$\dashrightarrow\sf \: density \: = \frac{mass}{volume}$

$\dashrightarrow\sf \: \frac{800}{ \frac{6600}{7} } = \frac{800 \times 7}{6600} = \frac{28}{33}kgcm^{3}$

$\dashrightarrow\sf \: \frac{28}{33} \times 1000kgm^{3} = 848.5kgm^{3}$

so therefore 848.5kgm³ is the density of wood