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1 year ago

When the leveling bulb is higher than the water level, the pressure in the system is greater than atmospheric pressure.

1 answer:

4 0

When the leveling bulb is higher than the water level, the pressure in the system is greater than atmospheric pressure. This statement is true.

In the physical sciences, pressure is defined as the stress at a point within a confined fluid or the perpendicular force per unit area. A 42-pound box with a bottom area of 84 square inches will impose pressure on a surface equal to the force divided by the area it is applied to, or half a pound per square inch.

Atmospheric pressure, which is roughly 15 pounds per square inch at sea level, is the weight of the atmosphere pressing down on each unit area of the Earth's surface. Pascals are used to express pressure in SI units; one pascal is equivalent to one newton per square meter. Almost 100,000 pascals of atmospheric pressure are present.

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Compare animal and plant cell

vagabundo [1.1K]

Answer:

A plant cell contains a large, singular vacuole that is used for storage and maintaining the shape of the cell. In contrast, animal cells have many, smaller vacuoles. Plant cells have a cell wall, as well as a cell membrane. Animal cells simply have a cell membrane, but no cell wall.

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8 0

2 years ago

Read 2 more answers

When light of wavelength 240 nm falls on a cobalt surface, electrons having a maximum kinetic energy of 0.17 eV are emitted. Fin

dusya [7]

Answer:

(a) 5.04 eV (B) 248.14 nm (c) $1.21\times 10^{15}Hz$

Explanation:

We have given Wavelength of the light \lambda = 240 nm

According to plank's rule ,energy of light

$E = h\nu = \frac{hc}{}\lambda$

$E = h\nu = \frac{6.67\times 10^{-34} J.s\times 3\times 10^{8}m/s}{ 240\times 10^{-9} m\times 1.6\times 10^{-19}J/eV}= 5.21 eV$

Maximum KE of emitted electron i= 0.17 eV

Part( A) Using Einstien's equation

$E = KE_{max}+\Phi _{0}$ , here $\Phi _0$ is work function.

$\Phi _{0}=E - KE_{max}$ = 5.21 eV-0.17 eV = 5.04 eV

Part( B) We have to find cutoff wavelength

$\Phi _{0} = \frac{hc}{\lambda_{cuttoff}}$

$\lambda_{cuttoff}= \frac{hc}{\Phi _{0} }$

$\lambda_{cuttoff}= \frac{6.67\times 10^{-34} J.s\times 3\times 10^{8}m/s}{5.04 eV\times 1.6\times 10^{-19}J/eV }=248.14 nm$

Part (C) In this part we have to find the cutoff frequency

$\nu = \frac{c}{\lambda_{cuttoff}}= \frac{3\times 10^{8}m/s}{248.14 \times 10^{-19} m }= 1.21\times 10^{15} Hz$

5 0

2 years ago

A circular parallel plate capacitor is constructed with a radius of 0.52 mm, a plate separation of 0.013 mm, and filled with an

olya-2409 [2.1K]

Answer:

289282

Explanation:

r = Radius of plate = 0.52 mm

d = Plate separation = 0.013 mm

A = Area = $\pi r^2$

V = Potential applied = 2 mV

k = Dielectric constant = 40

$\epsilon_0$ = Electric constant = $8.854\times 10^{-12}\ \text{F/m}$

Capacitance is given by

$C=\dfrac{k\epsilon_0A}{d}$

Charge is given by

$Q=CV\\\Rightarrow Q=\dfrac{k\epsilon_0AV}{d}\\\Rightarrow Q=\dfrac{40\times 8.854\times 10^{-12}\times\pi \times (0.52\times 10^{-3})^2\times 2\times 10^{-3}}{0.013\times 10^{-3}}\\\Rightarrow Q=4.6285\times 10^{-14}\ \text{C}$

Number of electron is given by

$n=\dfrac{Q}{e}\\\Rightarrow n=\dfrac{4.6285\times 10^{-14}}{1.6\times10^{-19}}\\\Rightarrow n=289281.25\ \text{electrons}$

The number of charge carriers that will accumulate on this capacitor is approximately 289282.

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2 years ago

Doubling the frequency of a wave source doubles the speed of the wave assuing that wavelength remains the same? True or false? W

SVETLANKA909090 [29]

Answer:true,because velocity is directly proportional to speed or velocity

Explanation:

Velocity = frequency x wavelength

The velocity or speed varies directly with the frequency, so as the frequency is increased, the velocity or speed is also increased

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2 years ago

The stone falls a further 25 m in the next 1.0 s of its fall. Calculate the stone's average speed during the 3 s of its fall.

Angelina_Jolie [31]

Answer:

5 minutes sweetheart

Explanation:

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2 years ago