Gravitational force is

GmM / r^2

Force is also m x a

When equating both of them, m cancels out, which leaves us with GM/r^2

Hope this helps

**The force on one end of the trough is 5.4 X 10⁵ N**

<u>**Explanation:**</u>

The triangle is equilateral which means all the interior angles are 60° and the sides are 6m long.

According to the figure,

AI / 8 = sin (60) = √3/2

AI = 4√3

The depth of the water is AI = 4√3

The interval becomes, | 0 , 4√3|

w = 2JK

(the hydrostatic force acting on the strip is the product of the pressure and the area)

where.

ρ = 875 kg/m³

g = 9.8m/s²

d = depth ( d = y')

limit is 0 → 4√3

On solving the equation, we get the value of limit as 32√3

**Therefore, the force on one end of the trough is 5.4 X 10⁵ N**

**Answer:**

10.28571 N

**Explanation:**

m = Mass of toboggan = 15 kg

u = Initial velocity = 4.8 m/s

v = Final velocity = 0

t = Time taken = 7 seconds

Friction force is given by the change in momentum over time

**The magnitude of the average friction force acting on the toboggan while it was moving is 10.28571 N**

I think its suicidal ideation......

I think

**Answer:264.58 mm of Hg**

**Explanation:**

Given

Volume of gas

Pressure

If the Plunger is pressed volume is reduced to

As temperature of the gas does not change so it follows boyle's law

According to which