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Olin [163]
2 years ago
11

What is the value of acceleration due to gravity at the pole, equator and the centre of the earth

Physics
1 answer:
fgiga [73]2 years ago
7 0

Answer:

In combination, the equatorial bulge and the effects of the surface centrifugal force due to rotation mean that sea-level gravity increases from about 9.780 m/s2 at the Equator to about 9.832 m/s2 at the poles, so an object will weigh approximately 0.5% more at the poles than at the Equator.

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If a species experiences a helpful mutation, like camouflage, explain how that mutation would help the species to better survive
ExtremeBDS [4]

Answer:

A helpful mutation like camoflage would immensely help a species survive and flourish. The reason being the appliances of such a mutation. Camoflage is one of the most apllicable of abilities. It can be used as a defense mechanism, to hide in plain sight, it can be used offensively to avoid being seen while hunting and it can even be used as a form of communication between the species. All of these appliances would greatly benefit the species and would most definitely help it survive and flourish.

4 0
2 years ago
What is momentum in physics?
Shalnov [3]

Answer:

mv

Explanation:

momentum = mass * velocity

7 0
3 years ago
Read 2 more answers
A 6.0-cm-diameter horizontal pipe gradually narrows to 4.5 cm. When water flows through this pipe at a certain rate, the gauge p
Rufina [12.5K]

Answer:

0.0072 m³/s

Explanation:

Using Bernoulli's law

P₁ + 1/2ρv₁² = P₂ + 1/2ρv₂ since the pipe is horizontal

1/2ρv₂² - 1/2ρv₁² = P₁ - P₂

flow rate is constant

A₁v₁ = A₂v₂

A₁ = πr₁² = π (0.06/2)² = 0.0028278 m²

A₂ = πr₂² = π (0.0225)² = 0.00159 m²

v₁  = (A₂ / A₁)v₂

v₁  = (0.00159 m²/ 0.0028278  m²) v₂ = 0.562  v₂

substitute v₁  into the Bernoulli's equation

1/2ρv₂² - 1/2ρv₁² = P₁ - P₂

500 ( 1 - 0.3161 ) v₂²  = (31.0 - 24 ) × 10³ Pa

341.924 v₂² = 7000

v₂² = 20.472

v₂ = √ 20.472 = 4.525 m/s

volume follow rate = 0.00159 m² ×  4.525 m/s = 0.0072 m³/s

8 0
3 years ago
What is the resistance of resistor R3?
Vsevolod [243]

Answer: 2.0

Explanation: I KNOW !

5 0
3 years ago
Gold has a density of 19300 kg/m3 calculate the mass of 0.02m3 of gold in kilograms​
aev [14]

Answer:

The mass of 0.02 m³ of gold is 386 kilograms

Explanation:

Given:

The density of the gold = 19300 kg/m³.

The volume of gold = 0.02 m³

To Find:

The mass of gold = ?

Solution:

We know that density is mass divided per unit volume.

Thus mathematically

Density = \frac{mass}{volume}Density=

volume

mass

Rewriting in terms of mass ,

Mass = density * volume

On substituting the known values

Mass = 19300 kg/m³ * 0.02 m³

Mass = 386 kilograms

Learn more about Mass and Density:

Mass=?,volume=190,density=4

Mass 350 kg volume 175 density ans

This is not my answer I copied it but hope it helps:)

5 0
3 years ago
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