You are moving a desk that has a mass of 36 kg; its acceleration is 0.5 m / s 2. What is the force being applied
1 answer:
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Answer:
Diagrams in pictures
Explanation:
Using energy I can get
m g h = 1/2 m v^2
So the velocity at the end of the ramp is the squareroot of two times the initial height of the box times the gravity constant.
(H= 1,3m sin25)
V=2,32m/s
V= a t
And
X= v t +1/2 a t^2
Knowing v=2,32 m/s and x= 1,3 m
I can get
a= 6,21m/s2
F= m a
I can get the force of the box when it collides with the spring
F= 12, 425 N
The force the spring makes on the box then is
F = -12,425N = -k d
Then the spring's constant is k= 51,75N/m
To make the two diagrams I need the functions of time when the box slows down
I use the same two equations
V= a t
And
X= v t + 1/2 a t^2
Being now 2,32 my initial velocity and 0 my final velocity, and my distance 0,24 m.
I get there the time t=0,0689 seconds and the acceleration a= -33,67 m/s2 (negative because it's slowing down).
Then,
V(t)= - 33,67 m/s2 t for time between 0 and 0,689 sec
X(t)= 2,32 m/s t + 1/2 33,67 m/s2 t^2.
for time between 0 and 0,689 sec
Diagrams and equations are in the pictures
Answer:
Magnitude of the magnetic field inside the solenoid near its centre is 1.293 x 10⁻³ T
Explanation:
Given;
number of turns of solenoid, N = 269 turn
length of the solenoid, L = 102 cm = 1.02 m
radius of the solenoid, r = 2.3 cm = 0.023 m
current in the solenoid, I = 3.9 A
Magnitude of the magnetic field inside the solenoid near its centre is calculated as;
Where;
μ₀ is permeability of free space = 4π x 10⁻⁷ m/A
Therefore, magnitude of the magnetic field inside the solenoid near its centre is 1.293 x 10⁻³ T