Answer:
Explanation:
For this problem we use the translational equilibrium condition. Our reference frame for block 1 is one axis parallel to the plane and the other perpendicular to the plane.
X axis
-Aₓ - f_e +T = 0 (1)
Y axis
N₁ - W_y = 0 ( 2)
let's use trigonometry for the weight components
sin θ = Wₓ / W
cos θ = W_y / W
Wₓ = W sin θ
W_y = W cos θ
We write the diagram for the second body.
Note that in the block the positive direction rd upwards, therefore for block 2 the positive direction must be downwards
W₂ -T = 0 (3)
we add the equations is 1 and 3
- W₁ sin θ - μ N₁ + W₂ = 0
from equation 2
N₁ = W₁ cos θ
we substitute
-W₁ sin θ - μ (W₁ cos θ) + W₂ = 0
W₂ = m₁ g (without ea - very expensive)
This is the smallest value that supports the equilibrium system
The formula should be (2πr)/T. r=radius T=time. if time is in seconds the formula would look like this
Answer:
The largest to smallest change in momentum with respect to magnitude of change in momentum is as follows;
1st- Collision (1) & Collision (2 or b in question)
2nd- Collision (4)
3rd- Collision (3)
Explanation:
This is because momentum is mass times into velocity. i.e.
P=m.v (kg.m/s - S.I unit)
(where p is momentum, m is mass of object and v is velocity or speed object)
If mass remains constant(real life scenario) then change in momentum is directly related to change in speed. i.e
Δp=m⋅(Δv)=m⋅(vf−vi) where vf is final velocity and vi is initial velocity.
By using above formula ;
we can calculate change in momentum for different collisions with respect to cart B.
m= mass of cart B
Collision (1) Δp=m⋅(Δv)=m⋅(vf−vi)=m.(0-1.0)=-m kg.m/s (where "-" indicates deceleration or stopping of object.)
Collision (2 or b in question ) Δp=m⋅(Δv)=m⋅(vf−vi)=m.(0-1.0)=-m kg.m/s
Collision (3) Δp=m⋅(Δv)=m⋅(vf−vi)=m.(0-0)=0 (which indicates that object remains stationary before and after collision and momentum for cart B is 0)
Collision (4) Δp=m⋅(Δv)=m⋅(vf−vi)=m.(0.67-0)=0.67m kg.m/s
Therefore collisions (1) and (2 or b) are ranked 1st, collision (4) ranked 2nd and collision (3) ranked 3rd.
Answer:
55mg/kg * 0.02kg which gives 1.1 mg pesticides
Explanation:
(Options are missing)
If the average mass or weight of a roach is 0.02kg
And 55mg is needed per 1 kg roaches
The amount or size of pesticide required for 0.02kg roach is 55mg/kg * 0.02kg
Size = 55mg/kg * 0.02 kg
Size = 1.1mgkg/kg
Size = 1.1 mg
So, the size of pesticide required is 1.1mg
resultant force = thrust – weight
acceleration = resultant force (newtons, N) divided by mass (kilograms, kg).
Acceleration = resultant force divided by mass
53N/0.56
=94.64 approximately 95
= 95m/s^2
This means that, every second, the speed of the rocket increases by 95m/s2
the S.I unit of Acceleration is meter per second square.
