Answer:
4 m/
Explanation:
From Equilibrium of forces, The Tension in string is cancelled by the Weight (product of mass and acceleration due to gravity) of the body acting downwards.
The Net force = Mass * Acceleration.
Since Net Force = 20 Newton, Mass = 5kg, therefore;
20 = 5kg * acceleration. Dividing the RHS and LHS of the equation by 5, we have;
Acceleration = which gives 4.
Note: RHS means Right Hand Side.
LHS means Left Hand Side.
Answer:
a = - 9.8 j ^ m/s²
Explanation:
This is a projectile launch problem, they give us the initial velocity in the two components
v₀ₓ = 17.1 m / s
= 14.7 m / s
They indicate that the only acceleration that exists is the acceleration of gravity, which acts in the direction towards the center of the Earth, in general in a coordinate system it coincides with the direction of the y axis.
a = - g j ^
a = - 9.8 j ^ m /s²
Answer:
The smallest distance the student that the student could be possibly be from the starting point is 6.5 meters.
Explanation:
For 2 quantities A and B represented as
and
The sum is represented as
For the the values given to us the sum is calculated as
Now the since the uncertainity inthe sum is
The closest possible distance at which the student can be is obtained by taking the negative sign in the uncertainity
Thus closest distance equals meters