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3 years ago

A __ is used to measure air pressure.

Physics

2 answers:

larisa86 [58]3 years ago

8 0

A barometer is used to measure air pressure!

PSYCHO15rus [73]3 years ago

3 0

Answer:

Barometer is used to measure air pressure.

Explanation:

Barometer :

Barometer used to measure the air pressure.

It is the scientific device which is used to measure the atmospheric pressure.

Atmospheric pressure change with the distance above and below the sea level.

When the air pressure is greater, then the mercury rises the higher.

Air pressure :

The weight of the air molecules pressing down on the earth it is called air pressure.

The density of air molecules is higher at sea level where the air pressure will be higher.

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A spring with spring-constant of 100 n/m is compressed 0.10 m. what is its maximum stored elastic potential energy

Ganezh [65]

Elastic potential energy stored in a spring is

(1/2) · (spring constant) · (stretch or compress)² .

PE = (1/2) · (100 N/m) · (0.1 m)²

PE = (50 N/m) · (0.01 m²)

PE = (50 · 0.01) (N · m / m²)

PE = 0.5 N · m

PE = 0.5 Joule

5 0

3 years ago

Consider two points in an electric field. The potential at point 1, V1, is 33 V. The potential at point 2, V2, is 175 V. An elec

Mnenie [13.5K]

Answer:

ΔU = e(V₂ - V₁) and its value ΔU = -2.275 × 10⁻²¹ J

Explanation:

Since the electric potential at point 1 is V₁ = 33 V and the electric potential at point 2 is V₂ = 175 V, when the electron is accelerated from point 1 to point 2, there is a change in electric potential ΔV which is given by ΔV = V₂ - V₁.

Substituting the values of the variables into the equation, we have

ΔV = V₂ - V₁.

ΔV = 175 V - 33 V.

ΔV = 142 V

The change in electric potential energy ΔU = eΔV = e(V₂ - V₁) where e = electron charge = -1.602 × 10⁻¹⁹ C and ΔV = electric potential change from point 1 to point 2 = 142 V.

So, substituting the values of the variables into the equation, we have

ΔU = eΔV

ΔU = -1.602 × 10⁻¹⁹ C × 142 V

ΔU = -227.484 × 10⁻¹⁹ J

ΔU = -2.27484 × 10⁻²¹ J

ΔU ≅ -2.275 × 10⁻²¹ J

So, the required equation for the electric potential energy change is

ΔU = e(V₂ - V₁) and its value ΔU = -2.275 × 10⁻²¹ J

5 0

3 years ago

A pressure that will support a column of Hg to a height of 256 mm would support a column of water to what height? The density of

Paul [167]

Answer:

The height of water in the column = 348.14 cm

Explanation:

Pressure:This is defined as the ratio of the force acting normally ( perpendicular) to the area of surface in contact. The S.I unit of pressure is N/m²

p = Dgh............... Equation 1

Where p = pressure, D = density, g = acceleration due to gravity, h = height.

From the question, the same pressure will support the column of mercury and water.

p₁ = p₂

Where p₁ = pressure of mercury, p₂ = pressure of water

D₁gh₁ = D₂gh₂.................. Equation 2

making h₂ the subject of equation 2

h₂ = D₁gh/D₂g............... Equation 3

Where D₁ and D₂ = Density of mercury and water respectively, h₁ and h₂ = height of mercury and water respectively

Given: D₁ = 13.6 g/cm³, D₂ = 1.00 g/cm³, h₁ = 256 mm = 25.6 cm.

Constant: g = 9.8 m/s²

Substituting these values into Equation 3,

h₂ = (13.6×9.8×25.6)/1×9.8

h₂ = 348.14 cm

The height of water in the column = 348.14 cm

6 0

3 years ago

The block in the figure below has a mass of 5.1 kg and it rests on an incline of angle . You pull on the rope with a force F = 3

viktelen [127]

42.9°

Explanation:

Let's assume that the x-axis is aligned with the incline and the positive direction is up the incline. We can then apply Newton's 2nd law as follows:

$x:\;\;\;\;F - mg\sin{\theta} = 0\;\;\;\;$

$\Rightarrow mg\sin{\theta} = F$

Note that the net force is zero because the block is moving with a constant speed when the angle of the incline is set at $\theta.$ Solving for the angle, we get

$\sin{\theta} = \dfrac{F}{mg}$

$\theta = \sin^{-1}\left(\dfrac{F}{mg}\right)$

$\;\;\;= \sin^{-1}\left[\dfrac{34\:\text{N}}{(5.1\:\text{kg})(9.8\:\text{m/s}^2)}\right]$

$\;\;\;=42.9°$

6 0

2 years ago

house current is 120 volts. if a light bulb runs a current of 0.5 amps, what is the resistance of the bulb

Sauron [17]

Ohm's Law tells the relationship between voltage, current, and resistance.
It can be written in three different ways, depending on which ones you know,
and which one you want to find.

Here's the one we need:

   Resistance = (voltage) divided by (current)

   = (120 V) / (0.5 Amp)

   =    240 ohms .

6 0

3 years ago