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3 years ago

How do you know the earth is rotating on its axis

Physics

1 answer:

cupoosta [38]3 years ago

4 0

Answer:

Read below!

Explanation:

You can watch the sun wheel across the sky during the day, and the stars at night. Focus a telescope on any star besides the north star--especially southern stars--and you can watch them drift across your field of view.

An alternative explanation is that all the stars are painted on (or holes in) some canopy that rotates around the earth. This explanation does not account for the motion of the "wanderers," or planets, as the Greeks called them, or for the path of the moon among the stars.

As we know the stars are massive bodies of significant and varying distance to the earth, the notion they all swing around us in unison seems highly implausible

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1. Mike wears an 8.50 kg backpack while scaling a cliff. After 90 minutes, Mike is 11.2 m above the starting point.

svet-max [94.6K]

Answer:

a. 952Joules

b. 8008 joules

c. 1.48 watts

8 0

2 years ago

g The electric power needs of a community are to be met by windmills with 40-m-diameter rotors. The windmills are to be located

Ksenya-84 [330]

Answer:

Explanation:

Given Data

The diameter of the wind mills is d = 40m

Velocity of the air is V = 6 m / s

Required power output is: P ₀ = 2100 k W

Expression to calculate the exergy of the air is

E = V ² / 2

Substitute the value in above expression

E = ( 6 m / s ) ² / 2

E = 18 m ² / s ² x (1kJ/kg / 1000m²/s²)

E = 0.018 k J / k g

Expression to calculate the density of the air is

P v =m R T

m /v = P /RT ⋯ ⋯( I )

Here

m is the mass of the air,

v is the volume of the air,

P is the atmospheric pressure,

T is the standard temperature at the atmospheric pressure and

R is the gas constant

As the density is

ρ = m /V

Substitute the value in expression (I)

ρ = 101 kP a /( 0.287 k J / k g ⋅ K ) ( 298 K )

ρ = 1.180 k g / m ²

Expression to calculate the mass flow rate is

m = ρ A V ⋯ ⋯ ( I I )

Here A is the area of the windmill

Expression to calculate the A is

A = π /4 d ²

Substitute the value in above expression

A = π /4 ( 40 m ²)

A = 1256.63 m ²

Substitute the value in expression (II)

m = ( 1.180 k g / m ³) ( 1256.63 m ²) ( 6 m / s )

m = 8896.94 k g / s

Expression to calculate the maximum power available to the windmill is

P w = m ( V ² /2 )

Substitute the value in above expression

P w = 8896.94 k g / s ( (6m/s)²/2 )

P w = 160144.92 W × ( 1 W /1000 k W )

P w = 160.144 k W

Expression to calculate the number of windmills required is

n = P o /P w

Substitute the value in above expression

n=2100kw/160.144kw

n=13.11

8 0

3 years ago

Multidimensional development means that:

Snezhnost [94]

Answer:

option D

Explanation:

this is because it occurs in many different dimensions, including biological, cognitive and socioemotional. this is also the answer on apex.

5 0

3 years ago

An electron is initially at rest in a uniform electric field having a strength of 1.85 × 106 V/m. It is then released and accele

kirza4 [7]

Answer:

$W = 462.5 keV$

Explanation:

As we know that when electron moved in electric field then work done by electric field must be equal to the change in kinetic energy of the electron

So here we have to find the work done by electric field on moving electron

So we have

$F = qE$

$F = (1.6 \times 10^{-19})(1.85 \times 10^6)$

$F = 2.96 \times 10^{-13} N$

now the distance moved by the electron is given as

$d = 0.25 m$

so we have

$W = F.d$

$W = (1.6 \times 10^{-19})(1.85 \times 10^6)(0.25)$

$W = 7.4 \times 10^{-14} J$

now we have to convert it into keV units

so we have

$1 keV = 1.6 \times 10^{-16} J$

$W = 462.5 keV$

5 0

3 years ago

A 30-mm-diameter copper rod is 1 m long with a yield strength of 70 MPa. Determine the axial force necessary to cause the diamet

ivolga24 [154]

Explanation:

Given data:

d = 30 mm = 0.03 m

L = 1m

S $_{y}$ = 70 Mpa

Δd = -0.0001d

Axial force = ?

validity of elastic deformation assumption.

Solution:

O'₂ = Δd/d = (-0.0001d)/d = -0.0001

For copper,

v = 0.326 E = 119×10³ Mpa

O'₁ = O'₂/v = (-0.0001)/0.326 = 306×10⁶

∵δ = F.L/E.A and σ = F/A so,

σ = δ.E/L = O'₁ .E = (306×10⁻⁶).(119×10³) = 36.5 MPa

F = σ . A = (36.5 × 10⁻⁶) . (π/4 × (0.03)²) = 25800 KN

S $_{y}$ = 70 MPa > σ = 36.5 MPa

∵ elastic deformation assumption is valid.

so the answer is

F = 25800 K N and S $_{y}$ > σ

3 0

3 years ago