Answer:
She can swing 1.0 m high.
Explanation:
Hi there!
The mechanical energy of Jane (ME) can be calculated by adding her gravitational potential (PE) plus her kinetic energy (KE).
The kinetic energy is calculated as follows:
KE = 1/2 · m · v²
And the potential energy:
PE = m · g · h
Where:
m = mass of Jane.
v = velocity.
g = acceleration due to gravity (9.8 m/s²).
h = height.
Then:
ME = KE + PE
Initially, Jane is running on the surface on which we assume that the gravitational potential energy of Jane is zero (the height is zero). Then:
ME = KE + PE (PE = 0)
ME = KE
ME = 1/2 · m · (4.5 m/s)²
ME = m · 10.125 m²/s²
When Jane reaches the maximum height, its velocity is zero (all the kinetic energy was converted into potential energy). Then, the mechanical energy will be:
ME = KE + PE (KE = 0)
ME = PE
ME = m · 9.8 m/s² · h
Then, equallizing both expressions of ME and solving for h:
m · 10.125 m²/s² = m · 9.8 m/s² · h
10.125 m²/s² / 9.8 m/s² = h
h = 1.0 m
She can swing 1.0 m high (if we neglect dissipative forces such as air resistance).
Because field lines don’t all go in to the pole. The bottom half go outwards not inwards.
<h3><u>Answer;</u></h3>
Radius = 0.0818 m
Angular velocity = 2.775 × 10^7 rad/sec
<h3><u>Explanation;</u></h3>
The mass of proton m=1.6748 × 10^-27 kg;
Charge of electron e= 1.602 × 10^-19 C;
kinetic energy E= 2.7 MeV
= 2.7 × 10^6 × 1.602 × 10^-19 J;
= 4.32 × 10^-13 Joules
But; K.E =0.5m*v^2,
Hence v=√(2K.E/m)
Velocity = 2.27 × 10^7 m/s
Angular velocity, ω = v/r
Therefore; V = ωr
Hence; V = √(2K.E/m) = ωr
r= √(2E/m)/w = √E*√(2*m)/(eB)
= √E * √(2×1.6748×10^-27)/(1.602×10^-19 ×2.9)
but E = 4.32 × 10^-13 Joules
r = 0.0818 m
Angular speed
Angular velocity, ω = v/r , where r is the radius and v is the velocity
Therefore;
Angular velocity = 2.27 × 10^7 / 0.0818 m
= 2.775 × 10^7 rad /sec
