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Alex73 [517]
3 years ago
9

What is the value of (g) at north pole and at equator?

Physics
1 answer:
iVinArrow [24]3 years ago
5 0
At the north pole:
       9.832 m/s²
     32.258 ft/s²

At the equator:
      9.780 m/s²
    32.088 ft/s² 

It's about  0.532%  greater at the pole than at the equator.

A large person who weighs 200 pounds on the equator
would weigh  1 pound 1 ounce more at the pole.
(Even more if he was properly dressed for it !)
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Cosine law can be applied when .....................
jarptica [38.1K]

Answer:

<em>Two</em><em> </em><em>sides</em><em> </em><em>and</em><em> </em><em>angle</em><em> </em><em>between</em><em> </em><em>them</em><em> </em><em>is</em><em> </em><em>given</em><em> </em>

3 0
2 years ago
An ice cube measures 1.38 in. on each edge and weighs 1.39 oz. What is the density of the ice cube in g/cm^3
brilliants [131]

Answer:

0.91 g/cm³

Explanation:

Density: This can be defined as the ratio of mass of an object to its volume.

The mathematical expression for density is given as,

D = m/v .............................. Equation 1

Where D = Density of the ice cube, m = mass of the ice cube, v = volume of the ice cube.

v = l³

Where l = length of each side of the cube.

l = 1.38 in = (1.38×2.54) cm = 3.51 cm

v = 3.51³

v = 43.24 cm³

Given: m = 1.39 oz = (1.39×28.35) g = 39.41 g

Substitute into equation 1

D = 39.41/43.24

D = 0.91 g/cm³

Hence the density of the ice cube = 0.91 g/cm³

4 0
4 years ago
A 75.0 kg students sits 1.15 m away from a 68.4 kg student. What is the force of gravitational attraction between them?
AVprozaik [17]

Answer:

F=2.589×10⁻⁷ Newtons

Explanation:

The gravitational force of attraction between two bodies is given by

F= Gm₁m₂/r²

Where m₁ and m₂ are the masses of the two bodies, G is the universal gravitational constant while r is the distance between the two bodies.

G=6.67408 × 10⁻¹¹m³kg⁻¹s⁻²

r=1.15m

m₁=75.0kg

m₂=68.4kg

Therefore F= (6.67408×10⁻¹¹m³kg⁻¹s⁻²×75.0kg×68.4kg)/(1.15m)²

F=2.589×10⁻⁷ Newtons

7 0
4 years ago
Help me pretty please
aleksklad [387]

Answer:a

Explanation:

4 0
3 years ago
Charge q is 1 unit of distance away from the source charge S. Charge p is four times further away. The force exerted between S a
zhenek [66]

The correct answer is:

sixteen times


In fact, the distance between charge q and the source S is 1 unit. Instead, the distance between charge p and the source S is 4 units. The magnitude of the electrostatic force is inversely proportional to the square of the distance between the charge and the source:

E \sim \frac{1}{r^2}

where r is the distance. If we take the force between q and S as reference, we have r=1, so that

E_{qS}=\frac{1}{1^2}=1

while the force between p and S is

E_{pS}=\frac{1}{4^2}=\frac{1}{16}

Therefore, we see that the force exerted between q and S is 16 times the force exerted between p and S.

4 0
3 years ago
Read 2 more answers
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