Answer:
The maximum height the box will reach is 1.72 m
Explanation:
F = k·x
Where
F = Force of the spring
k = The spring constant = 300 N/m
x = Spring compression or stretch = 0.15 m
Therefore the force, F of the spring = 300 N/m×0.15 m = 45 N
Mass of box = 0.2 kg
Work, W, done by the spring = 
and the kinetic energy gained by the box is given by KE = 
Since work done by the spring = kinetic energy gained by the box we have
= therefore we have v = 
= 
= 
= 5.81 m/s
Therefore the maximum height is given by
v² = 2·g·h or h = 
= 
= 1.72 m
Incomplete question.The Complete question is here
A flat uniform circular disk (radius = 2.00 m, mass = 1.00 ✕ 102 kg) is initially stationary. The disk is free to rotate in the horizontal plane about a friction less axis perpendicular to the center of the disk. A 40.0-kg person, standing 1.25 m from the axis, begins to run on the disk in a circular path and has a tangential speed of 2.00 m/s relative to the ground.
a.) Find the resulting angular speed of the disk (in rad/s) and describe the direction of the rotation.
b.) Determine the time it takes for a spot marking the starting point to pass again beneath the runner's feet.
Answer:
(a)ω = 1 rad/s
(b)t = 2.41 s
Explanation:
(a) initial angular momentum = final angular momentum
0 = L for disk + L............... for runner
0 = Iω² - mv²r ...................they're opposite in direction
0 = (MR²/2)(ω²) - mv²r
................where is ω is angular speed which is required in part (a) of question
0 = [(1.00×10²kg)(2.00 m)² / 2](ω²) - (40.0 kg)(2.00 m/s)²(1.25 m)
0=200ω²-200
200=200ω²
ω = 1 rad/s
b.)
lets assume the "starting point" is a point marked on the disk.
The person's angular speed is
v/r = (2.00 m/s) / (1.25 m) = 1.6 rad/s
As the person and the disk are moving in opposite directions, the person will run part of a revolution and the turning disk would complete the whole revolution.
(angle) + (angle disk turns) = 2π
(1.6 rad/s)(t) + ωt = 2π
t[1.6 rad/s + 1 rad/s] = 2π
t = 2.41 s
Answer:
B = 0.546 T, F = 2.59 10⁻¹² N
Explanation:
The magnetic force is
F = q v x B
We can calculate the magnitude of the force and find the direction by the right hand rule
F = q v B sin θ
Let's use Newton's second law
F = m a
Acceleration is centripetal
a = v² / r
We substitute
q v B sin θ = m v² / r
The angle between the field and the radius of the circle is 90º so sin 90 = 1
q B = m v / r
B = m v / q r
Let's calculate ’
B = 1.67 10⁻²⁷ 2.97 10⁷ / (1.60 10⁻¹⁹ 0.568)
B = 0.546 T
The foce is
F = q v B
F = 1.60 10⁻¹⁹ 2.97 10⁷ 0.546
F = 2.59 10⁻¹² N
We can calculate the acceleration of Cole due to friction using Newton's second law of motion:
where
is the frictional force (with a negative sign, since the force acts against the direction of motion) and m=100 kg is the mass of Cole and the sled. By rearranging the equation, we find
Now we can use the following formula to calculate the distance covered by Cole and the sled before stopping:
where
is the final speed of the sled
is the initial speed
is the distance covered
By rearranging the equation, we find d:
Answer:
Im pretty sure its number 2
Explanation:
