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3 years ago

To which category does galaxy #1 belong? Why does it belong in this category?

Physics

1 answer:

vodka [1.7K]3 years ago

5 0

Answer:

This galaxy belongs to the elliptical galaxy category. This is because it does not have spiral arms.

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The density of aluminum is 2.7 × 103 kg/m3 . the speed of longitudinal waves in an aluminum rod is measured to be 5.1 × 103 m/s.

andrey2020 [161]

<span>The speed of longitudinal waves, S, in a thin rod = âšYoung modulus / density , where Y is in N/m^2. So, S = âšYoung modulus/ density. Squaring both sides, we have, S^2 = Young Modulus/ density. So, Young Modulus = S^2 * density; where S is the speed of the longitudinal wave. Then Substiting into the eqn we have (5.1 *10^3)^2 * 2.7 * 10^3 = 26.01 * 10^6 * 2.7 *10^6 = 26.01 * 2.7 * 10^ (6+3) = 70.227 * 10 ^9</span>

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3 years ago

Which type of matter is gold an element used in making jewelry A.) A heterogeneous mixture B.) A pure substance C.) A crystallin

Mnenie [13.5K]

Answer:

Explanation:

<em>The gold used in the making of jewelry is usually not pure but a heterogeneous mixture of metals. Pure gold is quite soft and even though it may look better in appearance compared to those made using heterogeneous mixtures, it usually bends easily. Hence, in order to make the jewelry more durable, gold is usually mixed with other metals to form a heterogeneous mixture. </em>

The correct option is A.

7 0

3 years ago

Read 2 more answers

A freight train rolls along a track with considerable momentum. If it were to roll at the same speed but had twice as much mass,

fomenos

Answer:

The momentum would be doubled

Explanation:

The magnitude of the momentum of the freight train is given by:

$p=mv$

where

m is the mass of the train

v is its speed

In this problem, we have that the speed of the train is unchanged, while the mass of the train is doubled:

$m'=2m$

therefore, the new momentum is

$p'=m'v=(2m)v=2(mv)=2p$

so, the momentum has also doubled.

7 0

3 years ago

When light propagates from a material with a given index of refraction into a material with a smaller index of refraction, the s

Ket [755]

Hope this can help u

7 0

3 years ago

Needddd helppppppp!!!

yulyashka [42]

Answer:

2/9 times as strong.

Explanation:

From the question given above, the following assumptions were made:

Initial mass of 1st planet (M₁ ) = M

Initial mass of 2nd planet (m₁ ) = m

Initial distance apart (r₁) = r

Initial Force of attraction (F₁) = F

Final mass of 1st planet (M₂) = 2M

Final mass of 1st planet (m₂) = constant = m

Final distance apart (r₂) = 3r

Final force of attraction (F₂) =?

Next, we shall obtain an expression to determine the new force. This can be obtained as follow:

F = GMm / r²

Cross multiply

Fr² = GMm

Divide both side by Mn

G = Fr² / Mm

Since G is constant, then we have

F₁r₁² / M₁m₁ = F₂r₂² / M₂m₂

Finally, we shall determine the new force as follow:

Initial mass of 1st planet (M₁ ) = M

Initial mass of 2nd planet (m₁ ) = m

Initial distance apart (r₁) = r

Initial Force of attraction (F₁) = F

Final mass of 1st planet (M₂) = 2M

Final mass of 1st planet (m₂) = constant = m

Final distance apart (r₂) = 3r

Final force of attraction (F₂) =?

F₁r₁² / M₁m₁ = F₂r₂² / M₂m₂

Fr² / Mm = F₂ × (3r)² / 2M × m

Fr² / Mm = F₂ × 9r² / 2Mm

Cross multiply

Fr² × 2Mm = F₂ × 9r² × Mm

Divide both side by 9r² × Mm

F₂ = Fr² × 2Mm / 9r² × Mm

F₂ = F × 2 / 9

F₂ = 2/9 F

Thus, the new force is 2/9 times the original force i.e 2/9 times as strong.

4 0

3 years ago