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3 years ago

The pressure at the top of a liquid

Physics

1 answer:

attashe74 [19]3 years ago

7 0

Answer:

$\displaystyle \rho=1,024.74\ kg/m^3$

Explanation:

Pressure

$P=P_o+\rho gh$

Where $P_o$ is the atmospheric pressure, $\rho$ is the density of the fluid, and h is the depth

The values are :

$P_o= 1\ atm= 101,325\ Pa$

$P=261,000\ Pa$

$g=9.8\ m/s^2$

$h=15.9\ m$

Solving the above equation for $\rho$

$\displaystyle \rho=\frac{P-P_o}{gh}$

$\displaystyle \rho=\frac{261,000-101,325}{9.8(15.9)}$

$\displaystyle \rho=\frac{159,675}{155.82}$

$\boxed{ \rho=1,024.74\ kg/m^3}$

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Two parallel-plate capacitors have the same plate area. Capacitor 1 has a plate separation twice that of capacitor 2, and the qu

Luba_88 [7]

Answer:

$V_1=8 V_2$

Explanation:

Given that:

Area of the plate of capacitor 1= Area of the plate of capacitor 2=A

separation distance of capacitor 2, $d_2=d$

separation distance of capacitor 1, $d_1=2d$

quantity of charge on capacitor 2, $Q_2=Q$

quantity of charge on capacitor 1, $Q_1=4Q$

We know that the Capacitance of a parallel plate capacitor is directly proportional to the area and inversely proportional to the distance of separation.

Mathematically given as:

$C=\frac{k.\epsilon_0.A}{d}$ .....................................(1)

where:

k = relative permittivity of the dielectric material between the plates= 1 for air

$\epsilon_0 = 8.85\times 10^{-12}\,F.m^{-1}$

From eq. (1)

For capacitor 2:

$C_2=\frac{k.\epsilon_0.A}{d}$

For capacitor 1:

$C_1=\frac{k.\epsilon_0.A}{2d}$

$C_1=\frac{1}{2} [ \frac{k.\epsilon_0.A}{d}]$

We know, potential differences across a capacitor is given by:

$V=\frac{Q}{C}$ ..........................................(2)

where, Q = charge on the capacitor plates.

for capacitor 2:

$V_2=\frac{Q}{\frac{k.\epsilon_0.A}{d}}$

$V_2=\frac{Q.d}{k.\epsilon_0.A}$

& for capacitor 1:

$V_1=\frac{4Q}{\frac{k.\epsilon_0.A}{2d}}$

$V_1=\frac{4Q\times 2d}{k.\epsilon_0.A}$

$V_1=8\times [\frac{Q.d}{k.\epsilon_0.A}]$

$V_1=8 V_2$

6 0

3 years ago

you read a primary source and a secondary source that discuss the same experiment. There is a difference in the conclusion made

Nuetrik [128]

You should trust the primary source more.

This is because the primary source is make its conclusion from direct observation, while the secondary source is possibly making reference to another secondary source or to another primary.

The primary source should be trusted more because it is from direct observation.

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3 years ago

a snail can move approximately 0.30 meters per minute.How many meters can the snail cover in 15 minutes ?

hichkok12 [17]

Sometimes if you have the word per, you need to multiply. In this problem, they are asking you to find the amount of meters a snail can travel per minute.They are giving you the amount of minutes, so in this case you multiply 15 times 0.30. The answer is 4.5 meters.

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3 years ago

1. A car starting from rest accelerates uniformly at 3

amid [387]

The car's final speed in m/s after the acceleration is 30.

Given the following data:

Initial velocity, U = 0 m/s (since the cars starts from rest)

Time, t = 10 seconds

Acceleration, a = 3 meter per seconds square.

To find the car's final speed in m/s after the acceleration, we would use the first equation of motion;

Mathematically, the first equation of motion is given by the formula;

$V = U + at\\\\V = 0 + 3(10)$

Final speed, V = 30 m/s

Therefore, the car's final speed in m/s after the acceleration is 30.