Differentiate scalar & vector quantity?
1 answer:
Scalar Quantity :-
→ These are the quantities with magnitude only . These quantities doesn't have to be mentioned with direction
eg.)=> Mass , Temprature .
Vector Quantity :-
→ These quantities are described with both Magnitude and Direction . These quantities follow special type of algebra called Vector algebra .
eg.)=> Force , Displacement
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Answer:
The coefficient of friction is (F/(19.6·m)
Explanation:
The given parameters are;
The force applied to the immovable body = F
The time duration the force acts = t
The time the body spends in motion = 3·t
The acceleration due to gravity, g = 9.8 m/s²
From Newton's second law of motion, we have;
The impulse of the force = F × t = m × Δv₁
Where;
Δv₁ = v₁ - 0 = v₁
The impulse applied by the force of friction, is
× (3·t - t) =
× (2·t)
Given that the motion of the object is stopped by the frictional force, we have;
The impulse due to the frictional force = Momentum change = m × Δv₂ = × (2·t)
Where;
Δv₂ = v₂ - 0 = v₂
Given that the velocity, v₂, at the start of the deceleration = The velocity at the point the force ceased to act, v₁, we have;
m × Δv₂ = × (2·t) = m × Δv₁ = F × t
∴ × (2·t) = F × t
= F × t/(2·t) = F/2
The coefficient of dynamic friction, = Frictional force/(The weight of the body) = (F/2)/(9.8 × m)
= (F/(19.6·m)
The coefficient of friction, = (F/(19.6·m)
Answer:
A) The acceleration is zero
<em>B) The total distance is 112 m</em>
Explanation:
<u>Velocity vs Time Graph</u>
It shows the behavior of the velocity as time increases. If the velocity increases, then the acceleration is positive, if the velocity decreases, the acceleration is negative, and if the velocity is constant, then the acceleration is zero.
The graph shows a horizontal line between points A and B. It means the velocity didn't change in that interval. Thus the acceleration in that zone is zero.
A. To calculate the acceleration, we use the formula:
Let's pick the extremes of the region AB: (0,8) and (12,8). The acceleration is:
This confirms the previous conclusion.
B. The distance covered by the body can be calculated as the area behind the graph. Since the velocity behaves differently after t=12 s, we'll split the total area into a rectangle and a triangle.
Area of rectangle= base*height=12 s * 8 m/s = 96 m
Area of triangle= base*height/2 = 4 s * 8 m/s /2= 16 m
The total distance is: 96 m + 16 m = 112 m