The answer is A because it was 2 minutes and she weighs 100 lbs.... so 2(100)=200.... 200 lbs of work
Answer:
The acceleration of a point on the wheel is 11.43 m/s² acting radially inward.
Explanation:
The centripetal acceleration acts on a body when it is performing a circular motion.
Here, a point on the bicycle is performing circular motion as the rotation of the wheel produces a circular motion.
The centripetal acceleration of a point moving with a velocity 
and at a distance of 
from the axis of rotation is given as:
Here, 
∴ 
Therefore, the acceleration of a point on the wheel is 11.43 m/s² acting radially inward.
Answer:
The acceleration is 2 m/s2.
Explanation:
We calculate the acceleration (a), with the data of mass (m) and force (F), through the formula:
F = m x a ---> a= F/m
a = 40 N/20 kg <em> 1N= 1 kg x m/s2</em>
a= 40 kgx m/s2/ 20 kg
<em>a= 2 m/s2</em>
Answer:
w = √[g /L (½ r²/L2 + 2/3 ) ]
When the mass of the cylinder changes if its external dimensions do not change the angular velocity DOES NOT CHANGE
Explanation:
We can simulate this system as a physical pendulum, which is a pendulum with a distributed mass, in this case the angular velocity is
w² = mg d / I
In this case, the distance d to the pivot point of half the length (L) of the cylinder, which we consider long and narrow
d = L / 2
The moment of inertia of a cylinder with respect to an axis at the end we can use the parallel axes theorem, it is approximately equal to that of a long bar plus the moment of inertia of the center of mass of the cylinder, this is tabulated
I = ¼ m r2 + ⅓ m L2
I = m (¼ r2 + ⅓ L2)
now let's use the concept of density to calculate the mass of the system
ρ = m / V
m = ρ V
the volume of a cylinder is
V = π r² L
m = ρ π r² L
let's substitute
w² = m g (L / 2) / m (¼ r² + ⅓ L²)
w² = g L / (½ r² + 2/3 L²)
L >> r
w = √[g /L (½ r²/L2 + 2/3 ) ]
When the mass of the cylinder changes if its external dimensions do not change the angular velocity DOES NOT CHANGE
Answer:
Explanation:
Given that,
Emf, V = 22 mV
Number of turns in the coil us 519
Rate of change of current is 10 A/s.
We need to find the magnetic flux through each turn of the coil at an instant when the current is 3.70 A.
Let's find the inductance first. So,
We have,
, 
is magnetic flux
So, the magnetic flux is 
.
