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3 years ago

A 40 kg box has a surface area of 0.2 m2, determine its pressure.

Physics

1 answer:

vesna_86 [32]3 years ago

6 0

Answer:

P = 1960 Pa

Explanation:

The pressure is computed as the force acting upon a surface divided by the area of the surface

$P=\frac{F}{A}$

The box has 40 kg, it presses on its own surface when placed over a table or any flat area. The acting force is its weight.

$F = W = m.g = 40 kg \ 9.8m/sec^2 = 392 N$

The pressure is:

$P=\frac{392N}{0.2m^2}=1960 \ Pa$

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3 years ago

Need some help with these two physics problems!

Juliette [100K]

The force that keeps the puck moving is 0.25 N while the velocity of the puck is 3.7 m/s.

<h3>What is the centripetal force?</h3>

We know that the centripetal force is the force that acts on a body that is moving along a circular path. In this case, we are told that the puck is moving along a circular path hence it is acted upon by the centripetal force that acts on it.

The centripetal force in this case would be supplied by the weight of the object that is moving in the circular path. Thus we can write in our equation that;

Centripetal force = Weight of object = mg

m = mass of the object

g = acceleration due to gravity

Then;

W = 0.026 Kg * 9.8 m/s^2

W = 0.25 N

To obtain the velocity of the object;

FT = mv^2/r

v = √ FT r/m

v = √0.25 * 1.4/0.026

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1 year ago

A particular coaxial cable is comprised of inner and outer conductors having radii 1 mm and 3 mm respectively, separated by air.

noname [10]

Answer:

The value is $\rho_s = 4.026 *10^{-6} \ C/m^2$

Explanation:

From the question we are told that

The radius of the inner conductor is $r_1 = 1 \ mm = 0.001 \ m$

The radius of the outer conductor is $r_2 = 3 \ mm = 0.003 \ m$

The potential at the outer conductor is $V = 1.5 kV = 1.5 *10^{3} \ V$

Generally the capacitance per length of the capacitor like set up of the two conductors is

$C= \frac{2 * \pi * \epsilon_o }{ ln [\frac{r_2}{r_1} ]}$

Here $\epsilon_o$ is the permitivity of free space with value $\epsilon_o = 8.85*10^{-12} C/(V \cdot m)$

=> $C= \frac{2 * 3.142 * 8.85*10^{-12} }{ ln [\frac{0.003}{0.001} ]}$

=> $C= 50.6 *10^{-12} \ F/m$

Generally given that the potential of the outer conductor with respect to the inner conductor is positive it then mean that the outer conductor is positively charge

Generally the line charge density of the outer conductor is mathematically represented as

$\rho_l = C * V$