Answer:
(a) W= 44N
(b)W= 31.65 N
Explanation:
Data
T=44 N : Maximum force that the rope can withstand without breaking
Newton's second law:
∑F = m*a Formula (1)
∑F : algebraic sum of the forces in Newton (N)
m : mass in kilograms (kg)
a : acceleration in meters over second square (m/s²)
(a) We apply the formula (1) at constant speed , then, a=0
W: heaviest fish that can be pulled up vertically
∑F = 0
T-W =0
W = T
W= 44N
(b) We apply the formula (1) , a= 1.26 m/s²
W: heaviest fish that can be pulled up vertically
W= m*g
m= W/g
g= 9.8 m/s² : acceleration due to gravity
∑F = 0
T-W = m*a
T= W+(W/g)*a
44=W*(1+1/9.8)* (1.26 )
44= W* 1.39
W= 44/1.39
W= 31.65 N
Answer:
The energy of motion increases
Explanation:
The reason the energy of motion increases is because as the speed increases there is more motion and more energy
Answer:
the frequency is the fundamental and distance is L = ¼ λ
Explanation:
This problem is a phenomenon of resonance between the frequency of the tuning fork and the tube with one end open and the other end closed, in this case at the closed end you have a node and the open end a belly, so the wavelength is the basis is
λ = 4 L
In this case L = 19.4 cm = 0.194 m
let's use the relationship between wave speed and wavelength frequency and
v = λ f
where the frequency is f = 440 Hz
v = 4 L f
let's calculate
v = 4 0.194 440
v = 341.44 m / s
so the frequency is the fundamental and distance is
L = ¼ λ
The work done on the mass is approximately 1J
<h3>How to calculate work done on mass</h3>
From the question, we are to determine the work done on the mass
The work done can be calculated from the formula for Potential energy
Work done = P.E = mgh
Where m is the mass
g is the acceleration due to gravity (g = 10 m/s²)
h is the height
From the question,
m = 2.0 kg
h = 0.050 m
Putting the values into the question, we get
Work done = 2.0 × 10 × 0.050
Work done = 1 J
Hence, the work done on the mass is approximately 1J
Learn more on how to calculate work done here: brainly.com/question/14460830
Answer:
y = [tg α] *x - [1/2 g/ (v0 ²cos²(α))]* x² , which describes a parabolic path
Explanation:
since
x = (v0 cos(α))t
where x represents horizontal distance covered by the projectile
and
y = (v0 sin(α))t −(1/2) gt^2
where y represents vertical distance, we can replace the parameter t in order to have a function of coordinates, then
x = (v0 cos(α))t → x /(v0 cos(α))= t
replacing t in the equation of y
y = (v0 sin(α))t −(1/2) gt^2 = (v0 sin(α))x/(v0 cos(α)) −(1/2) g[x/(v0 cos(α))]² =
[sin(α) /cos(α)]* x - 1/2 g x²/(v0 cos(α))² = [tg α] *x - [1/2 g/ (v0 ²cos²(α))]* x²
therefore
y = [tg α] *x - [1/2 g/ (v0 ²cos²(α))]* x²
since tg α= constant=C1 , [-1/2 g/ (v0 ²cos²(α))]= constant=C2 , then
y = C1*x + C2* x²
which describes a parabolic path