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bagirrra123 [75]
3 years ago
11

Help quick

Physics
1 answer:
Yanka [14]3 years ago
8 0

C it reduces the amount of useful work done on objects move it up the ramp

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For a star with a parallax angle of 1/2 of an at arcsecond, what will be its distance in parsec?
avanturin [10]

(1 parsec) is the distance at which an object has a parallax of 1 arcsecond. The distance is about 3.26 light years.

Another way to understand it is:  The distance from which the Earth's orbit appears 1 arcsecond across.

For a parallax angle of 1/2 arcsecond, the distance is <em>2 parsecs </em>(about 6.52 light years).

1 arcsecond is 1/3600 of a degree, 0.00028 degree.  

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3 years ago
The rebirth of science and learning during the fifteenth century is termed the
Pepsi [2]
The Renaissance, was the rebirth to science as well as many other advancements
5 0
4 years ago
supose at 20 degree celsius the resistance of Tungsten thermometer is 154.9. WHen placed in a particular solution , the resistan
saw5 [17]

Answer:

T₂ = 95.56°C

Explanation:

The final resistance of a material after being heated is given by the relation:

R' = R(1 + αΔT)

where,

R' = Final Resistance = 207.4 Ω

R = Initial Resistance = 154.9 Ω

α = Temperature Coefficient of Resistance of Tungsten = 0.0045 °C⁻¹

ΔT = Change in Temperature = ?

Therefore,

207.4 Ω = 154.9 Ω[1 + (0.0045°C⁻¹)ΔT]

207.4 Ω/154.9 Ω = 1 + (0.0045°C⁻¹)ΔT

1.34 - 1 = (0.0045°C⁻¹)ΔT

ΔT = 0.34/0.0045°C⁻¹

ΔT = 75.56°C

but,

ΔT = Final Temperature - Initial Temperature

ΔT = T₂ - T₁ = T₂ - 20°C

T₂ - 20°C = 75.56°C

T₂ = 75.56°C + 20°C

<u>T₂ = 95.56°C</u>

7 0
3 years ago
Where can volcanoes form? A. Where there are cracks in the crust B. Along fault lines C. Where the crust is thin and can be rupt
schepotkina [342]
The answer is all of the above
8 0
4 years ago
Read 2 more answers
A 50.0-kg block is being pulled up a 16.0° slope by a force of 250 n that is parallel to the slope. the coefficient of kinetic
Drupady [299]
When dealing with multiple forces acting on a body, it is advisable to draw a free-body diagram like that shown in the picture. There are four forces acting on the box: weight (W) pointing straight down, normal force perpendicular to the slope denoted as Fn, force used to push the box upwards along the slope and the frictional force acting opposite to the direction of motion of the box denoted as Ff. Frictional force is equal to coefficient of kinetic friction (μk) multiplied with Fn.

∑Fy = Fn - mgcos30° = 0
           Fn = (50)(9.81)(cos 16) = 471.5 N

When in motion, the net force is equal to mass times acceleration according to Newton's 2nd Law of Motion:

Fnet = F - μk*Fn - mgsin30° = ma
          250 - (0.2)(471.5 N) - (50)(sin 16°) = (50)(a)
          a = 2.84 m/s²

8 0
3 years ago
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