Answer:
Yes the frequency of the angular simple harmonic motion (SHM) of the balance wheel increases three times if the dimensions of the balance wheel reduced to one-third of original dimensions.
Explanation:
Considering the complete question attached in figure below.
Time period for balance wheel is:


m = mass of balance wheel
R = radius of balance wheel.
Angular frequency is related to Time period as:

As dimensions of new balance wheel are one-third of their original values


a = ( V2 - V1)/( t2 - t1)
3.2 = ( 23.5m/s - 15.2m/s)/(t - 0)
3.2m/s = 8.3/t
t(3.2) = 8.3
t = 8.3/3.2
t = 2.59 seconds
Answer:
It says about the motion and the graph of the object is stationary, basically travelling at the same speed at any time of the graph. It will never change.
Explanation:
To draw a diagram:
1. Draw an object and represent the speed as stationary and constant at any time.
Mechanical advantage is defined as the ratio of output load to the input load. The mechanical advantage of the machine will be 0.1.
<h3>What is
mechanical advantage?</h3>
Mechanical advantage is a measure of the ratio of output force to input force in a system,
It is used to obtain the efficiency of forces in levers and pulleys. It is an effective way of amplifying the force in simple machines like levers.
The theoretical mechanical advantage is defined as the ratio of the force responsible for the useful work in the system to the applied force.
Given
applied force = 250 N
Output force = 25
Mechanical advantage = work output / work input



Hence the mechanical advantage of the machine will be 0.1
To learn more about the mechanical advantage refer to the link;
brainly.com/question/7638820
Answer:
Explanation:
Given that,
Mass m = 6.64×10^-27kg
Charge q = 3.2×10^-19C
Potential difference V =2.45×10^6V
Magnetic field B =1.6T
The force in a magnetic field is given as Force = q•(V×B)
Since V and B are perpendicular i.e 90°
Force =q•V•BSin90
F=q•V•B
So we need to find the velocity
Then, K•E is equal to work done by charge I.e K•E=U
K•E =½mV²
K•E =½ ×6.64×10^-27 V²
K•E = 3.32×10^-27 V²
U = q•V
U = 3.2×10^-19 × 2.45×10^6
U =7.84×10^-13
Then, K•E = U
3.32×10^-27V² = 7.84×10^-13
V² = 7.84×10^-13 / 3.32×10^-27
V² = 2.36×10^14
V=√2.36×10^14
V = 1.537×10^7 m/s
So, applying this to force in magnetic field
F=q•V•B
F= 3.2×10^-19 × 1.537×10^7 ×1.6
F = 7.87×10^-12 N