Answer:

Explanation:
The maximum velocity of an object moving in a curve beyond which it will slide off the curve is given by the relationship in equation (1);

where
is the coefficient of friction between the object and the surface of the curve, g is acceleration due to gravity and r is the radius of the curve.
Given;
v = 0.8m/s
g = 
r = ?

In order to solve for
, we can simply make it the subject of formula from equation (1) as follows;

since we were not given the value of r, we can just substitute other known values, then solve and leave the answer in terms of r.
Therefore;


The gravitational force experienced by Earth due to the Moon is <u>equal to </u>the gravitational force experienced by the Moon due to Earth.
<u>Explanation</u>:
The force that attracts any two objects/bodies with mass towards each other is defined as gravitational force. Generally the gravitational force is attractive, as it always pulls the masses together and never pushes them apart.
The gravitational force can be calculated effectively using the following formula: F=GMmr^2
where “G” is the gravitational constant.
Though gravity has the ability to pull the masses together, it is the weakest force in the nature.
The mass of the Earth and moon varies, but still the gravitational force felt by the Earth and Moon are alike.
Answer:
a)
= 0.25 m / s b) u = 0.25 m / s
Explanation:
a) To solve this problem let's start with the conservation of the moment, for this we define a system formed by the ball plus the dog, in this case all the forces are internal and the moment is conserved
We will write the data
m₁ = 0.40 kg
v₁₀ = 9.0 m / s
m₂ = 14 kg
v₂₀ = 0
Initial
po = m₁ v₁₀
Final
= (m₁ + m₂) vf
po = pf
m₁ v₁₀ = (m₁ + m₂) 
= v₁₀ m₁ / (m₁ + m₂)
= 9.0 (0.40 / (0.40 +14)
= 0.25 m / s
b) This is the reference frame of the center of mass of the system in this case the speed of this frame is the speed of the center of mass
u = 0.25 m / s
In the direction of movement of the ball
c) Let's calculate the kinetic energy in both moments
Initial
K₀ = ½ m₁ v₁₀² +0
K₀ = ½ 0.40 9 2
K₀ = 16.2 J
Final
= ½ (m₁ + m₂)
2
= ½ (0.4 +14) 0.25 2
= 0.45 J
ΔK = K₀ - 
ΔK = 16.2-0.445
ΔK = 1575 J
These will transform internal system energy
d) In order to find the kinetic energy, we must first find the velocities of the individual in this reference system.
v₁₀’= v₁₀ -u
v₁₀’= 9 -.025
v₁₀‘= 8.75 m / s
v₂₀ ‘= v₂₀ -u
v₂₀‘= - 0.25 m / s
‘=
- u
= 0
Initial
K₀ = ½ m₁ v₁₀‘² + ½ m₂ v₂₀‘²
Ko = ½ 0.4 8.75² + ½ 14.0 0.25²
Ko = 15.31 + 0.4375
K o = 15.75 J
Final
= ½ (m₁ + m₂) vf’²
= 0
All initial kinetic energy is transformed into internal energy in this reference system
<h2>
Depth of river is 4.48 m</h2>
Explanation:
Discharge = Area x Velocity
Discharge of river = Discharge of stream 1 + Discharge of stream 2

Area of stream 1 = w₁ x d₁ = 8.9 x 3.8 = 33.82 m²
Area of stream 2 = w₂ x d₂ = 6.5 x 3.9 = 25.35 m²
Velocity of stream 1 = 2.3 m/s
Velocity of stream 2 = 2.6 m/s
Velocity of river = 3 m/s

Substituting

Depth of river = 4.48 m