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).
The work done is by the centripetal force on mass m during an angular displacement of 2π revolutions mv²2π /r J
Centripetal force - a force acts on an moving object in circular path.
the centripetal force is given by
F= mv²/r (equation1)
Work done is given by
W = Fd (equation 2)
d = 2π
work is done by the centripetal force on mass m during an angular displacement of 2π revolutions is given by:
to calculate work done using equation 1 in 2 we get
W = mv² d/r
W = mv² × 2π /r J
The work done is by the centripetal force on mass m during an angular displacement of 2π revolutions mv²2π /r J
To know more about centripetal force :
brainly.com/question/13031430
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Answer:
80 Ω.
Explanation:
In this circuit the resistances are in series.The equivalent resistance of a series circuit is equal to the sum of the resistances. Req= 60 + 20 = 80 Ω.
Answer:
a) the maximum transverse speed of a point on the string at an antinode is 5.9899 m/s
b) the maximum transverse speed of a point on the string at x = 0.075 m is 4.2338 m/s
Explanation:
Given the data in the question;
as the equation of standing wave on a string is fixed at both ends
y = 2AsinKx cosωt
but k = 2π/λ and ω = 2πf
λ = 4 × 0.150 = 0.6 m
and f = v/λ = 260 / 0.6 = 433.33 Hz
ω = 2πf = 2π × 433.33 = 2722.69
given that A = 2.20 mm = 2.2×10⁻³
so 
= A × ω
= 2.2×10⁻³ × 2722.69 m/s
= 5.9899 m/s
therefore, the maximum transverse speed of a point on the string at an antinode is 5.9899 m/s
b)
A' = 2AsinKx
= 2.20sin( 2π/0.6 ( 0.075) rad )
= 2.20 sin( 0.7853 rad ) mm
= 2.20 × 0.706825 mm
A' = 1.555 mm = 1.555×10⁻³
so
= A' × ω
= 1.555×10⁻³ × 2722.69
= 4.2338 m/s
Therefore, the maximum transverse speed of a point on the string at x = 0.075 m is 4.2338 m/s
Answer:
oh that looks hard but sorry im not good at math ether so sorry i couldnt help
Explanation:
