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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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2. Two people, one on Earth and the other on the Moon, try and lift 200 kg blocks. Which person

Art [367]

Answer:

The person on Earth will have to exert more force to lift their block

Explanation:

The mass of the blocks to be lifted = 200 kg

The location of the first person = On Earth

The location of the second person = On the Moon

The force a person will have to exert to lift the block = The weight of the block = The gravitational force, F, on the block which is given as follows;

$F = \dfrac{G \times M_1}{R^2} \times m_2$

Where;

$\dfrac{G \times M_1}{R^2}$ = The acceleration due to gravity on the Earth or the Moon, depending on the location of the block

m₂ = The mass of the block

Therefore, given that the acceleration due to gravity on the Earth is larger than the acceleration due to gravity on the Moon, the weight of the block on the Earth is larger than the weight of the block on the Moon, and the person on Earth have to exert more force to lift the heavier weight of the block on Earth than the person on the Moon will have to exert to lift the same block as the block has a lower weight on the Moon due to lower acceleration due to gravity on the Moon.

6 0

3 years ago

A sled whose total mass with cargo is 30.0 kg rests on ice. The coefficient of static friction is 0.20 and the coefficient of ki

romanna [79]

when sled just start to move the force on the sled will be equal to the static friction

So here we need to find the value of static friction

We know that

m = 30 kg

$\mu_s = 0.20$

now we know that normal force on the block is counterbalanced by weight of the block

$F_n = mg$

$F_n = 30*9.8$

now in order to find the friction force we can use

$F_s = \mu_s * F_n$

$F_s = 0.20 * 30 * 9.8$

$F_s = 59 N$

so it requires 59 N of force to move the sled

5 0

3 years ago

Please Help. I would Really Appreciate it

Readme [11.4K]

Answer:

a) 0.60 kg cart has final velocity 3.0 m/s [E]

0.80 kg cart has final velocity 4.0 m/s [W]

b) 0.12 m

Explanation:

Take east to be positive.

a) Momentum is conserved.

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

(0.60) (-5.0) + (0.80) (2.0) = (0.60) v₁ + (0.80) v₂

-1.4 = 0.6 v₁ + 0.8 v₂

Kinetic energy is conserved in elastic collisions.

½ m₁u₁² + ½ m₂u₂² = ½ m₁v₁² + ½ m₂v₂²

m₁u₁² + m₂u₂² = m₁v₁² + m₂v₂²

(0.60) (-5.0)² + (0.80) (2.0)² = (0.60) v₁² + (0.80) v₂²

18.2 = 0.6 v₁² + 0.8 v₂²

Solve the system of equations.

-1.4 = 0.6 v₁ + 0.8 v₂

-1.4 − 0.6 v₁ = 0.8 v₂

-1.75 − 0.75 v₁ = v₂

18.2 = 0.6 v₁² + 0.8 (-1.75 − 0.75 v₁)²

18.2 = 0.6 v₁² + 0.8 (3.0625 +2.625 v₁ + 0.5625 v₁²)

182 = 6 v₁² + 8 (3.0625 + 2.625 v₁ + 0.5625 v₁²)

182 = 6 v₁² + 24.5 + 21 v₁ + 4.5 v₁²

0 = 10.5 v₁² + 21 v₁ − 157.5

0 = v₁² + 2 v₁ − 15

0 = (v₁ − 3) (v₁ + 5)

v₁ = 3 or -5

Since u₁ = -5.0 m/s, v₁ must be 3.0 m/s.

Solving for v₂:

v₂ = -0.75 v₁ − 1.75

v₂ = -4.0 m/s

b) The compression of the spring is a maximum when the carts have the same velocity.

Momentum is conserved.

m₁u₁ + m₂u₂ = (m₁ + m₂) v

(0.60) (-5.0) + (0.80) (2.0) = (0.60 + 0.80) v

-1.4 = 1.4 v

v = -1.0

Energy is conserved.

½ m₁u₁² + ½ m₂u₂² = ½ (m₁ + m₂) v² + ½ kx²

m₁u₁² + m₂u₂² = (m₁ + m₂) v² + kx²

(0.60) (-5.0)² + (0.80) (2.0)² = (0.60 + 0.80) (-1.0)² + (1200) x²

18.2 = 1.4 + 1200 x²

16.8 = 1200 x²

x² = 0.014

x = 0.12

5 0

4 years ago

What are parasites? Give some example

Inga [223]

Answer:

parasites are creatures the gain benefit off of other animals usually harming them eg:ticks on dogs

4 0

3 years ago

A conducting coil of 1470 turns is connected to a galvanometer, and the total resistance of the circuit is 56.0 . The area of ea

Damm [24]

Answer:

B = - 1.51 10⁻⁷ T

Explanation:

For this exercise we can use Faraday's law of induction

E = - $- \frac{d \phi _B}{dt} = - \frac{d (B A cos \theta )}{dt}$

In this case, they indicate that the normal and the magnetite field are in the same direction, so the angle is zero (cos 0 = 1), they give the area of the loop A = 4.32 10⁻⁴ m² and since we have N = 1470 turns in each one a voltage is induced

E = - N B A

B = -E A / N (1)

we find the induced voltage with ohm's law

V = i R

where the current is defined by

i = Q / t

we substitute

V = Q R / t

let's calculate

V = 9.18 10-3 56.0 / t

We must assume a time normally is t = 1 s

V = 0.514 V

this is the voltage in the circuit which must be the induced voltage V = E

we substitute in 1

B = - 0.514 4.32 10⁻⁴ / 1470

B = - 1.51 10⁻⁷ T

3 0

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