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3 years ago

What shape is the Milky way galaxy

Physics

2 answers:

gizmo_the_mogwai [7]3 years ago

5 0

It appears to be a spiral shape.

podryga [215]3 years ago

5 0

The shape of the Milky Way Galaxy is spiral.

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A sample of helium gas has a volume of 900. milliliters and a pressure of 2.50 atm at 298 K. What is the new pressure when the t

Sladkaya [172]

Answer:

The new pressure = 5.64 atm.

Explanation:

Using The general gas equation,

P₁V₁/T₁ = P₂V₂/T₂..................... Equation 1

making P₂ the subject of the equation

P₂ = P₁V₁T₂/V₂T₁ ...................... Equation 2

Where P₁ = initial pressure, V₁ = initial Volume, T₁ = Initial temperature, P₂ = final pressure, V₂ = final volume, T₂ = final Temperature.

Given: P ₁= 2.5 atm, T₁ = 298 K, V₁= 900 milliliters, T ₂= 336 K, V₂ = 450 milliliters

Substituting these values into equation 2,

P₂ = (2.5×900×336)/(298×450)

P₂ = 756000/134100

P₂ = 5.64 atm.

Thus the new pressure = 5.64 atm.

5 0

2 years ago

A potter’s wheel moves from rest to an angular speed of 0.50 rev/s in 28.9 s. Assuming constant angular acceleration, what is it

Fudgin [204]

Answer:

0.108 rad/s².

Explanation:

Given that

Initial angular velocity ,ωi = 0 rad/s

Final angular velocity ωf= 0.5 rev/s

We know that

1 rev/s = 6.28 rad/s

ωf= 3.14 rad/s

t= 28.9 s

We know that (if acceleration is constant)

ωf=ωi + α t

α=Angular acceleration

3.14 = 0 + α x 28.9

$\alpha=\dfrac{3.14}{28.9}\ rad/s^2\\\alpha=0.108\ rad/s^2$

Therefore the acceleration will be 0.108 rad/s².

Therefore the answer will be 0.108 rad/s².

5 0

3 years ago

Is this a boy or a girl

ivanzaharov [21]

I believe that it is a boy but if they would like to be classified as a girl or a boy their decision needs to be respected

3 0

3 years ago

A driver travels 4.1 km west, 17.3 km north, and finally 1.2 km at an angle of 65.4 degree north of west. What is the driver’s d

jek_recluse [69]

Answer:

Explanation:

We shall represent each displacement in vector form .

i will represent east , j will represent north .

D₁ = 4.1 west = - 4.1 i

D₂ = 17.3 north = 17.3 j

D₃ = - 1.2 cos65.4 i + 1.2 sin65.4 j

= - .5 i + 1.09 j

Total displacement

= D₁ + D₂ + D₃

= - 4.1 i + 17.3 j - .5 i + 1.09 j

D = - 4.6 i + 18.39 j

magnitude of D

= √ ( 4.6² + 18.39² )

= √ (21.16 + 338.2 )

= √359.36

= 18.95 km .

Final displacement = 18.95 km .

5 0

3 years ago

A 2530-kg test rocket is launched vertically from the launch pad. Its fuel (of negligible mass) provides a thrust force so that

gavmur [86]

Answer:

A = 1.4 m/s²

B = -0.10493 m/s³

a = 1.29507 m/s²

T = 28095.8271 N

T = 1.13198 W

Explanation:

t = Time taken

g = Acceleration due to gravity = 9.81 m/s²

The equation

$v(t)=At+Bt^2$

Differentiating with respect to time

$\frac{dv}{dt}=\frac{d(At+Bt^2)}{dt}\\\Rightarrow 1.4=A+2Bt$

At t = 0

$1.4=A$

Hence, A = 1.4 m/s²

$B=\frac{v-At}{t^2}\\\Rightarrow B=\frac{2.18-1.4\times 1.8}{1.8^2}\\\Rightarrow B=-0.10493\ m/s^3$

B = -0.10493 m/s³

At t = 5 seconds

$a=1.4+2\times -0.010493\times 5=1.29507\ m/s^2$

a = 1.29507 m/s²

$T=m(a+g)\\\Rightarrow T=2530(1.29507+9.81)\\\Rightarrow T=28095.8271\ N$

T = 28095.8271 N

Weight of rocket

$W=2530\times 9.81=24819.9\ N$

$\frac{T}{W}=\frac{28095.8271}{24819.9}\\\Rightarrow \frac{T}{W}=1.13198\\\Rightarrow T=1.13198W$

T = 1.13198 W

3 0

3 years ago