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Over [174]
3 years ago
15

A gas heated to millions of degrees would emit

Physics
1 answer:
makkiz [27]3 years ago
4 0

A gas heated to millions of degrees would emit mostly x-rays.

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photoshop1234 [79]

Answer:-2

Explanation:

7 0
2 years ago
The figure shows two springs (k1 = 10 N/m and k2 = 20 N/m ) attached to a block that can slide on a frictionless surface. In the
Rainbow [258]

Answer:

Explanation:

a )

Energy stored by left spring when compressed = 1/2 k x²

= .5 x 10 x .02² = .002 J .

Let compression in right spring = y

energy stored to right spring = 1/2 k y²

1/2 k y² = 0.002

.5 x 20 x y² = 0.002

y = .01414 m

= 1.4  cm

7 0
2 years ago
The largest graduated cylinder in my lab holds 2 L and has an inner diamter (the part that holds the water) of 8 cm. When it is
mestny [16]

Answer:

<em>3924 Pa</em>

<em></em>

Explanation:

Volume of cylinder = 2 L = 0.002 m^3  (1000 L = 1 m^3)

diameter of the inner cylinder = 8 cm = 0.08 m  (100 cm = 1 m)

radius of the inner cylinder = diameter/2 = 0.08/2 = 0.04 m

area of the inner cylinder = \pi r^{2}

where \pi = 3.142,

and r = radius = 0.04 m

area of inner cylinder = 3.142 x 0.04^{2} = 0.005 m^2

<em>height h of the water in this cylinder = volume/area</em>

h = 0.002/0.005 = 0.4 m

<em>pressure at the bottom of the cylinder due to the height of water = pgh</em>

where

p = density of water = 1000 kg/m^3

g = acceleration due to gravity = 9.81 m/s^2

h = height of water within this cylinder = 0.4 m

pressure = 1000 x 9.81 x 0.4 = <em>3924 Pa</em>

3 0
3 years ago
when a metal ball is heated through 30°c,it volume becomes 1.0018cm^3 if the linear expansivity of the material of the ball is 2
Vlad1618 [11]

Answer:

The original volume of the metal sphere is approximately 1 cm³

Explanation:

The given parameters are;

The temperature change of the metal ball, ΔT = 30°C = 30 K

The new volume of the metal ball, V₂  = 1.0018 cm³

The linear expansivity of the material ball, α = 2.0 × 10⁻⁵ K⁻¹

We have;

d₂ = d₁·(1 + α·ΔT)

Where;

d₁ = The original diameter of the metal ball

d₂ = The new diameter of the ball

From the volume of the ball, V₂, we have;

V₂ = 1.0018 cm³ = (4/3)×π×r₂³

Where;

r₂ = The new radius = d₂/2

∴ V₂ = 1.0018 cm³ = (4/3)×π×(d₂/2)³

∴ d₂ = ∛(2³ × 1.0018 cm³/((4/3) × π)) ≈ 1.241445 cm

d₁ = d₂/(1 + α·ΔT)

∴ d₁ ≈ 1.241445 cm/(1 + 2.0 × 10⁻⁵·K × 30 K) ≈ 1.24070058 cm

The original volume of the metal ball, V₁ = (4/3)×π×(d₁/2)³

∴ V₁ = (4/3)×π×(1.24070058/2)³ ≈ 0.99999902845 cm³ ≈ 1 cm³

The original volume of the metal sphere, V₁ ≈ 1 cm³.

8 0
3 years ago
Assume it takes 5.00 minutes to fill a 50.0-gal gasoline tank. (1 U.S. gal = 231 in.3).
dolphi86 [110]

               (50 gal / 5 min) x (.0037854 m³/gal) x (1 min / 60 sec)

         =    (50 · 0.0037854 · 1) / (5 · 60)          m³/sec

         =                  0.000631  m³/sec
7 0
3 years ago
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