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3 years ago

A merry-go-round rotates at the rate of

Physics

1 answer:

Mila [183]3 years ago

6 0

Explanation:

Angular momentum is conserved.

I₁ ω₁ = I₂ ω₂

(½ Mr² + md²) ω₁ = (½ Mr²) ω₂

(½ (35 kg) (2.3 m)² + (84 kg) (2.3 m)²) (0.28 rev/s) = (½ (35 kg) (2.3 m)²) ω

ω = 1.624 rev/s

ω = 10.2 rad/s

Round as needed.

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What information is given by the formula of an ionic compound?

Sergio [31]

MgCl2 is ionic compound.........Mg +2 and Cl -1
both charges are cross multiplied to each element......formula tells us that to balance the positive and negative charges on both sides they are cross multiplied........MgCl2......meaning there is one atom of Mg and 2 atoms of Cl.......

HOPE IT HELPS !!!

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3 years ago

Explain how climbing a mountain is similar to hiking from the equator to one of the poles

4vir4ik [10]

Answer:

Explanation:

SRRY God bless

7 0

3 years ago

Read 2 more answers

The number of confirmed exoplanets is (a) less than 10; (b) roughly 50; (c) more than 500; (d) more than 5000.

Bad White [126]

Answer:

Explanation:

Until 2019, more than 3000 planetary systems have been discovered that contain more than 4000 exoplanets, since some of these systems contain multiple planets. Most known extrasolar planets are gas giants equal to or more massive than the planet Jupiter, with orbits very close to its star.

8 0

3 years ago

Water flows through a horizontal pipe of varying cross-section. In the first section, the cross-sectional area is 10 cm2 and flo

Stels [109]

Answer:

(a) the flow speed of the second section is 11 m/s

(b) the pressure of the second section is 6.33 x 10⁴ Pa

Explanation:

Given;

flow rate in the first section, Q₁ = 2750 cm³/sec

area of the first cross section, A₁ = 10 cm²

pressure in the first cross section, P₁ = 1.2 x 10⁵ Pa

area of the second section, A₂ = 2.5 cm²

(a) the flow speed of the second section (V₂)

Apply continuity equation;

Q₁ = Q₂

Q₁ = A₂V₂

V₂ = Q₁ / A₂

V₂ = (2750) / (2.5)

V₂ = 1100 cm/s = 11 m/s

(b) the pressure of the second section (P₂)

Apply Bernoulli's equation;

P₁ + ¹/₂ρV₁² = P₂ + ¹/₂ρV₂²

where;

ρ is density of water = 1000 kg/m³

V₁ is the speed of water in the first section;

Q₁ = A₁V₁

V₁ = Q₁ / A₁

V₁ = (2750) / (10)

V₁ = 275 cm/s = 2.75 m/s

P₂ = P₁ + ¹/₂ρV₁² - ¹/₂ρV₂²

P₂ = P₁ + ¹/₂ρ(V₁² - V₂²)

P₂ = 1.2 x 10⁵ Pa + ¹/₂ x 1000 (2.75² - 11²)

P₂ = 1.2 x 10⁵ Pa + 500(-113.438)

P₂ = 1.2 x 10⁵ Pa - 0.567 x 10⁵ Pa

P₂ = 0.633 x 10⁵ Pa

P₂ = 6.33 x 10⁴ Pa

8 0

2 years ago

An empty beaker is placed on a top-pan balance. Some water is now poured into the beaker.What is the weight of the water? A. 0.0

mario62 [17]

Answer:

A. 0.044 kg

Explanation:

We need to subtract the sum of (beaker+water - empty beaker) which is 0.106 kg - 0.062 kg = 0.044 kg. The answer will not be written in Newton because this unit is used for force only and in this question w have to find the weight.

Hope it is enough.

Please mark me as brainliest.

6 0

3 years ago