Which of earth sphere have the most mass

A capacitor is made from two hollow, coaxial, iron cylinders, one inside the other. The inner cylinder is negatively charged and

notsponge [240]

Answer: 1) 17.65 * 10^-12 C/V; 2) 0.68 V

Explanation: In order to calculate the capacitance of one cylinder capacitor we have to use the following expression:

$C=\frac{2\pi \epsilon o L}{ln (b/a)}$ where and b are the inner and outer radius of teh cylinder, respectively. L is length of the cylinder.

Finally we also kwn that C=Q/ΔV

then we have

ΔV =Q/C

ΔV = 12 * 10^-12/17.65*10^-12= 0.68V

3 0

3 years ago

Which instrument is used to measure the height of a 20 litre jerrican

Lynna [10]

tape measure?

Hope that helped!!! k

7 0

4 years ago

Read 2 more answers

Two protons are aimed directly toward each other by a cyclotron accelerator with speeds of 1200 km/s , measured relative to the

Radda [10]

Answer:

The maximum electrical force is $2.512\times10^{-2}\ N$ .

Explanation:

Given that,

Speed of cyclotron = 1200 km/s

Initially the two protons are having kinetic energy given by

$\dfrac{1}{2}mv^2=\dfrac{1}{2}mv^2$

When they come to the closest distance the total kinetic energy is converts into potential energy given by

Using conservation of energy

$mv^2=\dfrac{kq^2}{r}$

$r=\dfrac{kq^2}{mv^2}$

Put the value into the formula

$r=\dfrac{8.99\times10^{9}\times(1.6\times10^{-19})^2}{1.67\times10^{-27}\times(1200\times10^{3})^2}$

$r=9.57\times10^{-14}\ m$

We need to calculate the maximum electrical force

Using formula of force

$F=\dfrac{kq^2}{r^2}$

$F=\dfrac{8.99\times10^{9}\times(1.6\times10^{-19})^2}{(9.57\times10^{-14})^2}$

$F=2.512\times10^{-2}\ N$

Hence, The maximum electrical force is $2.512\times10^{-2}\ N$ .

8 0

3 years ago

What is the mass of mars

tiny-mole [99]

the mass of mars is

6.39 × 10^23 kg

3 0

2 years ago

Suppose you have two solid bars, both with square cross-sections of 1 cm2. They are both 24.6 cm long, but one is made of copper

vodka [1.7K]

Explanation:

Expression to calculate thermal resistance for iron ( $R_{I}$ ) is as follows.

$R_{I} = \frac{L_{I}}{k_{I} \times A_{I}}$

where, $L_{I}$ = length of the iron bar

$k_{I}$ = thermal conductivity of iron

$A_{I}$ = Area of cross-section for the iron bar

Thermal resistance for copper ( $R_{c}$ ) = \frac{L_{c}}{k_{c} \times A_{c}}[/tex]

where, $L_{c}$ = length of copper bar

$k_{c}$ = thermal conductivity of copper

$A_{c}$ = Area of cross-section for the copper bar

Now, expression for the transfer of heat per unit cell is as follows.

Q = $\frac{(100^{o} - 0^{o}}{\frac{L_{I}}{k_{I}.A_{I}} + \frac{L_{c}}{k_{c}.A_{c}}}$

Putting the given values into the above formula as follows.

Q = $\frac{(100^{o} - 0^{o})}{\frac{L_{I}}{k_{I}.A_{I}} + \frac{L_{c}}{k_{c}.A_{c}}}$

= $\frac{(100^{o} - 0^{o})}{21 \times 10^{-2} m[\frac{1}{73 \times 10^{-4}m^{2}} + \frac{1}{386 \times 10^{-4}m^{2}}}$

= 2.92 Joule

It is known that heat transfer per unit time is equal to the power conducted through the rod. Hence,

P = $\frac{Q}{T}$

Here, T is 1 second so, power conducted is equal to heat transferred.

So, P = 2.92 watt

Thus, we can conclude that 2.92 watt power will be conducted through the rod when it reaches steady state.

7 0

3 years ago