Ask question

Ask question

All categories

3 years ago

7

A man on the moon with a mass of 90 kilograms weighs 146 newtons. The radius of the moon is 1.74 x 10^6

1 answer:

Anestetic [448]3 years ago

4 0

Answer:

7.36 × 10^22 kg

Explanation:

You might be interested in

2. What is meant by the statement that the MA of a simple machine is 3<br>

Blababa [14]

Explanation:

MA of a machine is its mechanical advantage.

Mechanical advantage is the ability of a machine to multiply force so as to get a work done.

This is done by minimizing effort and maximizing the load or output.
A MA of a simple machine with a value of 3 suggests that such a machine will multiply input force by a factor of 3.
Therefore, such a machine will amplify the force input into the system.

4 0

3 years ago

Why it is important to have a regular

frozen [14]

Answer:

https://magazine.com/an

Explanation:

that's answer

3 0

3 years ago

Read 2 more answers

Reading accuracy of stop watch

Mice21 [21]

Answer:

Is there a more speciific question?

Explanation:

3 0

3 years ago

A copper wire and a tungsten wire of the same length have the same resistance. What is the ratio of the diameter of the copper w

spayn [35]

Answer:

Therefore the ratio of diameter of the copper to that of the tungsten is

$\sqrt{3} :\sqrt{10}$

Explanation:

Resistance: Resistance is defined to the ratio of voltage to the electricity.

The resistance of a wire is

directly proportional to its length i.e $R\propto l$
inversely proportional to its cross section area i.e $R\propto \frac{1}{A}$

Therefore

$R=\rho\frac{l}{A}$

ρ is the resistivity.

The unit of resistance is ohm (Ω).

The resistivity of copper(ρ₁) is 1.68×10⁻⁸ ohm-m

The resistivity of tungsten(ρ₂) is 5.6×10⁻⁸ ohm-m

For copper:

$A=\pi r_1^2 =\pi (\frac{d_1}{2} )^2$

$R_1=\rho_1\frac{l_1}{\pi(\frac{d_1}{2})^2 }$

$\Rightarrow (\frac{d_1}{2})^2=\rho_1\frac{l_1}{\pi R_1 }$ ......(1)

Again for tungsten:

$R_2=\rho_2\frac{l_2}{\pi(\frac{d_2}{2})^2 }$

$\Rightarrow (\frac{d_2}{2})^2=\rho_2\frac{l_2}{\pi R_2 }$ ........(2)

Given that $R_1=R_2$ and $l_1=l_2$

Dividing the equation (1) and (2)

$\Rightarrow\frac{ (\frac{d_1}{2})^2}{ (\frac{d_2}{2})^2}=\frac{\rho_1\frac{l_1}{\pi R_1 }}{\rho_2\frac{l_2}{\pi R_2 }}$

$\Rightarrow( \frac{d_1}{d_2} )^2=\frac{1.68\times 10^{-8}}{5.6\times 10^{-8}}$ [since $R_1=R_2$ and $l_1=l_2$ ]

$\Rightarrow( \frac{d_1}{d_2} )=\sqrt{\frac{1.68\times 10^{-8}}{5.6\times 10^{-8}}}$

$\Rightarrow( \frac{d_1}{d_2} )=\sqrt{\frac{3}{10}}$

$\Rightarrow d_1:d_2=\sqrt{3} :\sqrt{10}$

Therefore the ratio of diameter of the copper to that of the tungsten is

$\sqrt{3} :\sqrt{10}$

8 0

4 years ago

A gas has a volume of 60dm³ when it is under a pressure of 30kpa. Calculate it's volume when pressure changes

vesna_86 [32]

Explanation:

Given that,

Initial volume = 60dm³

Initial pressure = 30 kPa

We need to find the new volume when pressure changes to 40 kPa (say).

We know that the mathematical relation between volume and pressure is given by :

$\dfrac{P_1}{P_2}=\dfrac{V_2}{V_1}$

Put all the values,

$V_2=\dfrac{P_1V_1}{P_2}\\\\V_2=\dfrac{30\times 60}{40}\\\\V_2=45\ dm^3$

So, when pressure of the gas increases, its volume get decreased.

7 0

3 years ago