The team monitoring a space probe exploring the outer solar system fi nds that radio transmissions from the probe take 2.53 hours
1 answer:
You might be interested in
Answer:
The period of motion of new mass T = 0.637 sec
Explanation:
Given data
Mass of object (m) = 9 gm = 0.009 kg
Δx = 3.5 cm = 0.035 m
We know that spring force is given by
F = m g
F = 0.009 × 9.81 = 0.08829 N
Spring constant
k = 2.522
New mass = 26 gm = 0.026 kg
Now the period of motion is given by
T = 0.637 sec
This is the period of motion of new mass.
Answer:
A
B
C
D
Explanation:
Considering the first question
From the question we are told that
The spring constant is
The potential energy is
Generally the potential energy stored in spring is mathematically represented as
=>
=>
=>
Considering the second question
From the question we are told that
The mass of the dart is m = 0.050 kg
Generally from the law of energy conservation
=>
=>
Considering the third question
The height at which the dart was fired horizontally is
Generally from the law of energy conservation
Here KE is kinetic energy of the dart which is mathematical represented as
=>
=>
=>
Considering the fourth question
Generally the total time of flight of the dart is mathematically represented as
=>
=>
Generally the horizontal distance from the equilibrium position to the ground is mathematically represented as
=>
=>
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