Answer:
270 m
Explanation:
Given:
v₀ = 63 m/s
a = 2.8 m/s²
t = 4.0 s
Find: Δx
Δx = v₀ t + ½ at²
Δx = (63 m/s) (4.0 s) + ½ (2.8 m/s²) (4.0 s)²
Δx = 274.4 m
Rounded to two significant figures, the displacement is 270 meters.
Answer:
The tension in the string at that point is 90.75 N
Explanation:
Given;
mass of the object, m = 3 kg
length of string, r = 0.25 m
the angular velocity, ω = 11 rad/s
The tension on string can be equated to the centrifugal force on the object;
T = mω²r
Where;
T is the tension in the string
m is mass of the object
ω is the angular velocity
r is the radius of the circular path
T = 3 x (11)² x 0.25
T = 90.75 N
Therefore, the tension in the string at that point is 90.75 N
Hi there!
We can use the following kinematic equation:
The initial velocity is 0 m/s, so:
vf = final velocity (? m/s)
a = acceleration due to gravity (g)
d = vertical height (m)
Plug in the givens and solve:
<h2>Answer:</h2>
Acceleration= -1.11 m/sec²
<h2>Explanation:</h2>
Date Given to us is
Mass = 150 kg
Time = 1.5 minutes = 90 seconds
Distance = 2500 meters
Initial Velocity = 120 m/s
Final Velocity = 20 m/s
Acceleration = ?
<u>Solution:</u>
By using First Equation of motion
Vf = Vi + at
Putting the values
20 = 120 + a (90)
Subtracting 120 on both sides
20-120 = 120 + a(90) - 120
-100 = 90 a
Dividing both sides by 90
-100/90 = 90 a / 90
-1.11 = a
So the acceleration is -1.11 m/s²