3 years ago

The value of g is greater at the poles than at the equator why

Physics

1 answer:

babunello [35]3 years ago

6 0

Answer:

because the gravitational pull is maximum at the poles and decreases as it comes down toward the equator.

You might be interested in

Which of the following should you always use to protect yourself from hazardous moving parts of power tools and equipment

Andrew [12]

What are the following answers?

7 0

3 years ago

Vector question,university physics zemanski

just olya [345]

Answer:

yes This is correct Answer

3 0

3 years ago

A ball is thrown vertically downward at 10 m/s. what is it’s speed 1s and 2s later

soldi70 [24.7K]

Answer:

20 m/s

30 m/s

Explanation:

Given:

v₀ = -10 m/s

a = -9.8 m/s²

When t = 1 s:

v = v₀ + at

v = (-10 m/s) + (-9.8 m/s²) (1 s)

v = -19.8 m/s

When t = 2 s:

v = v₀ + at

v = (-10 m/s) + (-9.8 m/s²) (2 s)

v = -29.6 m/s

Rounded to one significant figures, the speed of the ball at 1 s and 2 s is 20 m/s and 30 m/s, respectively.

3 0

3 years ago

If a cart of 4 kg mass has a force of 8 newtons exerted on it, what is its acceleration?

dybincka [34]

M = 4kg
F =8N
a..?

F =m.a
8 = 4.a
a = 2m/s^2

6 0

3 years ago

Read 2 more answers

A 1.2 kg block is held at rest against the spring with a force constant k= 730 N/m. Initially, the spring is compressed a distan

mariarad [96]

Answer:

d = 9.69 cm

Explanation:

given,

mass of the block = 1.2 Kg

spring force constant(k) = 730 N/m

spring is compressed = d = ?

rough patch width = 5 cm

μ_k = 0.44

work done by friction = energy lost

$\mu_k mg \times x = \dfrac{1}{2}kd^2-\dfrac{1}{2}mv^2$

$0.44\times 1.2 \times 9.8 \times 0.05 = \dfrac{1}{2}\times 730 \times d^2-\dfrac{1}{2}\times 1.2 \times 2.3^2$

$3.432 = 365 d^2$

$d = \dfrac{3.432}{365}$

d = 0.0969 m

d = 9.69 cm

5 0

3 years ago

The value of g is greater at the poles than at the equator why ​

The value of g is greater at the poles than at the equator why