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3 years ago

Difference between universal law of gravitation and acceleration due to gravity

Physics

1 answer:

marishachu [46]3 years ago

7 0

Explanation:

The acceleration on an object due to the gravity of any massive body is represented by g (small g). The force of attraction between any two unit masses separated by unit distance is called universal gravitational constant denoted by G(capital g). The relation between G and g is not proportional. That means they are independent entities.

G and g

In physics, G and g can be related mathematically as –

$g=\frac{GM}{R^{2}}$

Where,

1=g is the acceleration due to the gravity of any massive body measured in m/s2.

2=G is the universal gravitational constant measured in Nm2/kg2.

3=R is the radius of the massive body measured in km.

4=M is the mass of the massive body measured in Kg.

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Jane, looking for tarzan, is running at top speed 5.3 m/s and grabs a vine hanging vertically from a tall tree in the jungle. ho

lutik1710 [3]

v₀ = initial speed as tarzan grabs the vine = 5.3 m/s

v = final speed as the tarzan reach the maximum height = 0 m/s

h = maximum height gained by the tarzan

m = mass of tarzan

using conservation of energy

initial kinetic energy = final kinetic energy + potential energy

(0.5) m v²₀ = (0.5) m v² + m g h

(0.5) v²₀ = (0.5) v² + g h

(0.5) (5.3)² = (0.5) (0)² + (9.8) h

h = 1.43 m

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3 years ago

The best rebounders in basketball have a vertical leap (that is, the vertical movement of a fixed point on their body) of about

nadya68 [22]

Answer:

a) 4.45 m/s

b) 0.9 seconds

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration due to gravity = 9.81 m/s²

$v^2-u^2=2as\\\Rightarrow -u^2=2as-v^2\\\Rightarrow u=\sqrt{v^2-2as}\\\Rightarrow u=\sqrt{0^2-2\times -9.81\times 1}\\\Rightarrow u=4.45\ m/s$

a) The vertical speed when the player leaves the ground is 4.45 m/s

$v=u+at\\\Rightarrow t=\frac{v-u}{a}\\\Rightarrow t=\frac{0-4.45}{-9.81}\\\Rightarrow t=0.45\ s$

Time taken to reach the maximum height is 0.45 seconds

$s=ut+\frac{1}{2}at^2\\\Rightarrow 1=0t+\frac{1}{2}\times 9.81\times t^2\\\Rightarrow t=\sqrt{\frac{1\times 2}{9.81}}\\\Rightarrow t=0.45\ s$

Time taken to reach the ground from the maximum height is 0.45 seconds

b) Time the player stayed in the air is 0.45+0.45 = 0.9 seconds

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3 years ago

A large, cylindrical water tank with diameter 3.60 m is on a platform 2.00 m above the ground. The vertical tank is open to the

zysi [14]

To solve this problem it is necessary to apply the concepts related to the geometry of a cylindrical tank and its respective definition.

The volume of a tank is given by

$V = \frac{\pi d^2}{4}h$

Where

d = Diameter

h = Height

Considering that there are two stages, let's define the initial and final volume as,

$V_0 = \frac{\pi d^2}{4}H$

$V_f = \frac{\pi d^2}{4}h$

We know as well by definition that

$1gal = 3.785*10^{-3}m^3$

Then we have for the statement that

$V_f = V_0 -1gal$

$V_f = V_0 - 3.785*10^{-3}$

Replacing the previous data

$\frac{\pi d^2}{4}h = \frac{\pi d^2}{4}H- 3.785*10^{-3}$

$\frac{\pi (3.6)^2}{4}h = \frac{\pi (3.6)^2}{4}(2)- 3.785*10^{-3}$

Solving to get h,

$h = 1.99963m$

Therefore the change is

$\Delta h = H-h$

$\Delta h = 2- 1.99963$

$\Delta h = 3.7*10^{-4}m=0.37mm$

Therefore te change in the height of the water in the tank is 0.37mm

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3 years ago

Assuming that the circuit is in the steady state, what is the total energy stored in the two capacitors? j

Neporo4naja [7]

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4 years ago

How to find the density of air?

nikitadnepr [17]

You have to divide the pressure exerted by the air into two partial pressures: of the dry air and of the water vapor. Combining these two values gives you the parameter.

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3 years ago

Difference between universal law of gravitation and acceleration due to gravity​

Difference between universal law of gravitation and acceleration due to gravity