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
Force (N) = mass (kg) x velocity (m/s) / time (s)
Answer:
(a) 19.62 N
(b) Box moves down the slope
(c) 24.43 N
Explanation:
(a)
2 Kg box causes tension
hence
where m is mass and g is gravitational force
T'=4*9.81 sin 35= 22.5071 N
Since T' is greater than
, then the box moves down the slope
(c)
Acceleration a=

When moving, the box will exert force T"=
T"= 4*9.81 sin 35 +(4*0.48)= 24.43 N
Answer:
a) Velocity = 4.2m/s
b) Acceleration = 2.94m/s^2
c) Force exerted on the floor= 1401.4×10^3N
Explanation:
a) Velocity,V=sqrt(2×9.8×0.900)
V= 4.2m/s
b) Vf2= V^2+2ay2
a= 4.2^2 - 0/2×3
a= 17.64/6= 2.94m/s^2
c) Newton's 2nd law indicates:
Fnet= F - mg=ma
F= m(g+a)
F=110(9.8+2.94)
F=110×12.94
F= 1401.4N
Answer: One canister contains 1.03 Kg of fuel,
Explanation:
The density is defined as the relation between the mass and the volume 
First of all you need to have the same units for volume and density, so:
For the volume,

For the density,

From the density equation we have
, so:
