The expression for the block's centripetal acceleration is derived as ω²r or v²/r.
<h3>
What is centripetal acceleration?</h3>
The centripetal acceleration of an object is the inward or radial acceleration of an object moving in a circular path.
The expression for the block's centripetal acceleration is derived as follows;
ω = dθ/dt
where;
- ω is the angular speed
- θ is the angular displacement
- t is the time of motion
ac = ω²r
where;
- r is the radius of the circular path
Also, ω = v/r
ac = (v/r)²r
ac = v²/r
Thus, the expression for the block's centripetal acceleration is derived as ω²r or v²/r.
Learn more about centripetal acceleration here: brainly.com/question/79801
Diagram A will result in the movement of the block to the left as a result of the forces.
<h3>What is Force?</h3>
This is referred to an influence which is capable of changing the motion of an object.
Diagram A has equal upward and downward force and left side which is 60N is higher than the right side which has 20N. The block will therefore move to the left.
Read more about Force here brainly.com/question/25239010
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Give u = start velocity
v = end velocity
v = u + at
50 = 400 + a*30
30a = -350
a = -116.67 m/
**Why the accecleration is negative number**
Because displacement, velocity, and acceleration are VECTOR QUANTITIES.
Vector Quantity must have direction.
Answer:
1058.78 ft/sec
Explanation:
Horizontal Component of Velocity; This is the velocity of a body that act on the horizontal axis. I.e Velocity along x-axis
The horizontal velocity of a body can be calculated as shown below.\
Vh = Vcos∅.......................... Equation 1
Where Vh = horizontal component of the velocity, V = The velocity acting between the horizontal and the vertical axis, ∅ = Angle the velocity make with the horizontal.
Given: V = 1178 ft/sec, ∅ = 26°
Substitute into equation 1
Vh = 1178cos26
Vh = 1178(0.8988)
Vh = 1058.78 ft/sec
Hence the horizontal component of the velocity = 1058.78 ft/sec
Answer:
108.7 V
Explanation:
Two forces are acting on the particle:
- The external force, whose work is 
- The force of the electric field, whose work is equal to the change in electric potential energy of the charge: 
where
q is the charge
is the potential difference
The variation of kinetic energy of the charge is equal to the sum of the work done by the two forces:

and since the charge starts from rest,
, so the formula becomes

In this problem, we have
is the work done by the external force
is the charge
is the final kinetic energy
Solving the formula for
, we find
