Refer to the diagram shown below.
W₁ = (4 kg)*(9.8 m/s²) = 39.2 N
W₂ = (1 kg)*(9.8 m/s²) = 9.8 N
The normal reaction on the 4-kg mass is
N = (39.2 N)*cos(25°) = 35.5273 N
The force acting down the inclined plane due to the weight is
F = (39.2 N)*sin(25°) = 16.5666 N
The net force that accelerates the 4-kg mass at a m/²s down the plane is
F - W₂ = (4 kg)*(a m/s²)
4a = 16.5666 - 9.8
a = 1.6917 m/s²
Answer: 1.69 m/s² (nearest hundredth)
Answer:
The speed of the large cart after collision is 0.301 m/s.
Explanation:
Given that,
Mass of the cart, 
Initial speed of the cart, 
Mass of the larger cart, 
Initial speed of the larger cart, 
After the collision,
Final speed of the smaller cart,
(as its recolis)
To find,
The speed of the large cart after collision.
Solution,
Let
is the speed of the large cart after collision. It can be calculated using conservation of momentum as :





So, the speed of the large cart after collision is 0.301 m/s.
1) 3 miles/Hour
The speed is defined as the distance covered divided by the time taken:

where
d = 1.5 mi is the distance
t = 0.5 h is the time taken
Substituting,

2) 1.34 m/s south
Velocity, instead, is a vector, so it has both a magnitude and a direction. We have:
is the displacement in meters
is the time taken in seconds
Substituting,

And the direction of the velocity is the same as the displacement, so it is south.
Dressage. It’s an event in horseback riding.
Answer:
Variation
Explanation:
This explains variation because theres many different species in the question