The acceleration of the box as it slides down the ramp is 4.9 m/s².
<h3>
What is the acceleration of the box down the incline?</h3>
The acceleration of the box down the incline is determined by applying Newton's second law of motion as shown below;
F - Ff = ma
where;
- F is the parallel force on the box
- Ff is the frictional force on box = 0
- m is the mass of the box
- a is the acceleration of the box
F - 0 = ma
F = ma
mg sinθ = ma
g sinθ = a
where;
- g is acceleration due to gravity
- θ is the angle of inclination of the incline
a = 9.8 x sin(30)
a = 4.9 m/s²
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Answer: B. 23 inches to the left
Explanation: A vector quantity involves both magnitude and direction.
consider the forces on mass m₁ on the incline plane :
parallel to incline , force equation is given as
T - m₁ g Sin30 = m₁ a
T = m₁ g Sin30 + m₁ a eq-1
consider the force on mass m₂ on the incline plane :
m₂ g - T = m₂ a
T = m₂ g - m₂ a eq-2
Using eq-1 and eq-2
m₂ g - m₂ a = m₁ g Sin30 + m₁ a
inserting the values
(2.3 x 9.8) - 2.3 a = (3.7 x 9.8) Sin30 + 3.7 a
a = 0.74 m/s²
Answer:
Original speed of the mess kit = 4.43 m/s at 50.67° north of east.
Explanation:
Let north represent positive y axis and east represent positive x axis.
Here momentum is conserved.
Let the initial velocity be v.
Initial momentum = 4.4 x v = 4.4v
Velocity of 2.2 kg moving at 2.9 m/s, due north = 2.9 j m/s
Velocity of 2.2 kg moving at 6.8 m/s, 35° north of east = 6.9 ( cos 35i + sin35 j ) = 5.62 i + 3.96 j m/s
Final momentum = 2.2 x 2.9 j + 2.2 x (5.62 i + 3.96 j) = 12.364 i + 15.092 j kgm/s
We have
Initial momentum = Final momentum
4.4v = 12.364 i + 15.092 j
v =2.81 i + 3.43 j
Magnitude

Direction

50.67° north of east.
Original speed of the mess kit = 4.43 m/s at 50.67° north of east.
Answer:
You should use this for work related questions :/