As compared to developing countries, developed countries have a

A uniform rod of length 50cm and mass 0.2kg is placed on a fulcrum at a distance of 40cm from the left end of the rod. At what d

oksano4ka [1.4K]

Answer:

x = 45 cm

Explanation:

Given that,

The length of a rod, L = 50 cm

Mass, m₁ = 0.2 kg

It is at 40cm from the left end of the rod.

We need to find the distance from the left end of the rod should a 0.6kg mass be hung to balance the rod.

The centre of mass of the rod is at 25 cm.

Taking moments of both masses such that,

$15\times 0.2=x\times 0.6\\\\x=\drac{3}{0.6}\\\\x=5\ cm$

The distance from the left end is 40+5 = 45 cm.

Hence, at a distance of 45 cm from the left end it will balance the rod.

5 0

3 years ago

A desk was moved 12 m using 280 j of work. how much force was used to move the desk?

Oksana_A [137]

Let F = required force, N

Given:
d = 12 m, distance
W = 280 J, work done

By definition,
W = F*d,
therefore
(F N)*(12 m) = (280 J)
F = 280/12 = 23.33 N

Answer: The force is 23.3 N (nearest tenth)

4 0

3 years ago

To what volume will a sample of gas expand if it is heated from 50.0°c and 2.33 l to 500.0°c?

Lena [83]

The volume of the gas at $500^{\circ} \text{C}$ is $\fbox{\begin\ 5.576\,{\text{L}}\\\end{minispace}}$ .

Further Explanation:

Consider the pressure of the gas to be constant.

The change in the volume of the gas when the temperature of the gas is varied by keeping the pressure of the gas at a constant value is defined by the Charles' Law.

Concept:

According to the Charles law, the volume of the gas is directly proportional to the temperature of the gas at constant pressure.

The Charles' law can be stated as:

$\fbox{\begin\\V\propto T\\\end{minispace}}$

The above expression can we written as.

$\dfrac{V_1}{T_1}=\dfrac{V_2}{T_2}$

Convert the temperature of the gas into kelvin.

$T=273+T^\circ\text{C}}}$

Here, T is the temperature in kelvin and $T^\circ\text{C}}}$ is the temperature in centigrade.

The initial temperature of the gas is $50^\circ\text{C}$ . The temperature of the gas in kelvin is.

$\begin{aligned}{T_1}&=273+50\\&=323\,{\text{K}}\\\end{aligned}$

The final temperature of the gas is $500^\circ\text{C}$ . The temperature in kelvin is.

$\begin{aligned}{T_2}&=273+500\\&=773\,{\text{K}}\\\end{aligned}$

Substitute the values of temperature and volume in the expression of the Charles' Law.

$\begin{aligned}{V_2}&=\frac{{{T_2}}}{{{T_1}}}{V_1}\\&=\frac{{773\,{\text{K}}}}{{323\,{\text{K}}}}\left({2.33\,{\text{L}}}\right)\\&=5.576\,{\text{L}}\\\end{aligned}$