Answer:
v = 4.4271 m/s
Explanation:
Given
m = 3 Kg
R = 0.2 m
∅ = 15°
h = 1.5 m
g = 9.8 m/s²
v = ?
Ignoring frictional losses, at the bottom of the plane
Total kinetic energy is = Potential Energy at the top of plane
Using Law of conservation of energy we have
U = Kt + Kr
m*g*h = 0.5*m*v² + 0.5*I*ω²
knowing that
Icylinder = 0.5*m*R²
ω = v/R
we have
m*g*h = 0.5*m*v² + 0.5*(0.5*mR²)*(v/R)² = 0.75*m*v²
⇒ v = √(g*h/0.75) = √(9.8 m/s²*1.5 m/0.75)
⇒ v = 4.4271 m/s
Answer:
The wheel spins for 11.43s. The number cannot be determined.
Explanation:
We know that initial velocity is 124.4 rpm = 13.027137522 rad/s, acceleration is -1.14 rad/s^2, and final velocity is assumed to be 0 rad/s. We are asked to find t and displacement. We use the equation
where
,
,
, and
.
Rearrange the equation to obtain
. Plug in the numbers and solve to obtain
.
The number of the wheel cannot be determined as we do not know the placement of numbers on the wheel.
Answer:
3000J
Explanation:
The gravitational energy for any object is given by E=mgh
where m is mass, g is gravitational field strength anf h is height above ground
When you reverse the direction of the current, the current loop generated by the magentic field is revered.
Hi there!
(a)
Recall that:

W = Work (J)
F = Force (N)
d = Displacement (m)
Since this is a dot product, we only use the component of force that is IN the direction of the displacement. We can use the horizontal component of the given force to solve for the work.

To the nearest multiple of ten:

(b)
The object is not being displaced vertically. Since the displacement (horizontal) is perpendicular to the force of gravity (vertical), cos(90°) = 0, and there is NO work done by gravity.
Thus:

(c)
Similarly, the normal force is perpendicular to the displacement, so:

(d)
Recall that the force of kinetic friction is given by:

Since the force of friction resists the applied force (assigned the positive direction), the work due to friction is NEGATIVE because energy is being LOST. Thus:

In multiples of ten:

(e)
Simply add up the above values of work to find the net work.

Nearest multiple of ten:

(f)
Similarly, we can use a summation of forces in the HORIZONTAL direction. (cosine of the applied force)



Nearest multiple of ten:
