A car travels 1000 meters in 5 seconds. What is its average velocity?

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

For a cylindrical capacitor, the capacitance does not depend on which of the following values?

IgorC [24]

Answer:

Capacitance of cylindrical capacitor does not depends on the amount of charge on the conductors

Explanation:

Consider a cylindrical capacitor of length L, inner radius R₁ and outer radius R₂, permitivity ε₀ constant then capacitance of cylindrical capacitor is given by:

$C=\frac{2\pi \epsilon_{o}L}{ln\frac{R_{2} }{R_{1}} }$

From this equation it is clear that capacitance of cylindrical capacitor is independent of the amount of charge on the conductors where as directly proportional permitivity constant and length of cylinder where as inversely proportional to natural log of ratio of R₂ and R₁

5 0

3 years ago

The astronomer who discovered the dwarf planet Eris suggests there might be another object far beyond the Kupier belt. If this P

PolarNik [594]

These objects would be classified as extreme trans Neptunian object (ETNO).

Explanation:

ETNO’s are the objects lying beyond the planet Neptune and orbiting the Sun. They follow a highly eccentric path which is tilted. ETNO has been grouped into three major according to their respective perihelia.

Within this region (beyond Neptune’s orbit), a hypothetical planet has been discovered. It was discovered following its gravitational effect on the other objects of Kuiper Belt (region beyond the orbit of the Neptune- the last planet of our Solar system)

The Planet is assumed to be around 2 times the Earth’s size and around 10 times heavier than Earth.

4 0

3 years ago

Please help

Snezhnost [94]

Answer:

Range = 22.61 m

Explanation:

We can use the formula for the Range in flat ground, given by:

$Range=v_i^2\frac{sin(2\theta)}{g}$

which for our case renders:

$Range=15^2\frac{sin(80^o)}{9.8} \approx 22.61\,\,m$

4 0

2 years ago

Light bulb 1 operates with a filament temperature of 3000 K, whereas light bulb 2 has a filament temperature of 2000 K. Both fil

MatroZZZ [7]

Answer:

A₁/A₂ = 0.44

Explanation:

The emissive power of the bulb is given by the formula:

P = σεAT⁴

where,

P = Emissive Power

σ = Stefan-Boltzman constant

ε = Emissivity

A = Surface Area

T = Absolute Temperature of Surface

FOR BULB 1:

Since, emissivity and emissive power are constant.

Therefore,

P = σεA₁T₁⁴ ----------- equation 1

where,

A₁ = Surface Area of Bulb 1

T₁ = Temperature of Bulb 1 = 3000 k

FOR BULB 2:

Since, emissivity and emissive power are constant.

Therefore,

P = σεA₂T₂⁴ ----------- equation 2

where,

A₂ = Surface Area of Bulb 2

T₂ = Temperature of Bulb 1 = 2000 k

Dividing equation 1 by equation 2, we get:

P/P = σεA₁T₁⁴/σεA₂T₂⁴

1 = A₁(3000)²/A₂(2000)²

A₁/A₂ = (2000)²/(3000)²

A₁/A₂ = 0.44

6 0

3 years ago