Answer:
high tension: 4.2 × 1.5 = 6.3 cm/s
medium tension: 2.8 ×1.5 = 4.2 cm/s
low tension: 0.8 × 1.5 = 1.2 cm/s
Explanation: Given Settings:
amplitude: 0.75 cm
damping: zero
Using
Speed = frequency ×wavelength
Where
Wavelength = 0.75 × 2 = 1.5 cm
Therefore:
high tension: 4.2 × 1.5 = 6.3 cm/s
medium tension: 2.8 ×1.5 = 4.2 cm/s
low tension: 0.8 × 1.5 = 1.2 cm/s
The dimension of K is M/ T^2
according to the question T=2π square root ofm/k here 2 pi is constant so
T= root of m /k and root of k = root of m/ T now by squaring on both the sides we get the answer k= M/ T^2
complete question :
A spring is hanging down from the ceiling, and an object of mass m is attached to the free end. The object is pulled down, thereby stretching the spring, and then released. The object oscillates up and down, and the time T required for one complete up-and-down oscillation is given by the equation T=√2πm/k, where k is known as the spring constant. What must be the dimension of k for this equation to be dimensionally correct?
To learn more about dimension:
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Your answer would be
D. Gasoline-powered cars are better for past acceleration and traveling at high speeds.
I just took this quiz and can confirm this is the correct answer!
Hope this helps :)
Answer:
distance stop 1.52m,
velocity 4.0 m/s y^
Explanation:
The movement of the particle is two-dimensional since it has acceleration in the x and y axes, the way to solve it is by working each axis independently.
a) At the point where the particle begins to return its velocity must be zero (Vfx = 0)
Vfₓ = V₀ₓ + aₓ t
t = - V₀ₓ/aₓ
t = - 2.4/(-1.9)
t= 1.26 s
At this time the particle stops, let's find his position
X1 = V₀ₓ t + ½ aₓ t²
X1= 2.4 1.26 + ½ (-1.9) 1.26²
X1= 1.52 m
At this point the particle begins its return
b) The velocity has component x and y
As a section, the X axis x Vₓ = 0 m/s is stopped, but has a speed on the y axis
Vfy= Voy + ay t
Vfy= 0 + 3.2 1.26
Vfy = 4.0 m/s
the velocity is
V = (0 x^ + 4.0 y^) m/s
c) In order to make the graph we create a table of the position x and y for each time, let's start by writing the equations
X = V₀ₓ t+ ½ aₓ t²
Y = Voy t + ½ ay t²
X= 2.4 t + ½ (-1.9) t²
Y= 0 + ½ 3.2 t²
X= 2.4 t – 0.95 t²
Y= 1.6 t²
With these equations we build the table to graph, for clarity we are going to make two distance graph with time, one for the x axis and another for the y axis
Chart to graph
Time (s) x(m) y(m)
0 0 0
0.5 0.960 0.4
1 1.45 1.6
1.50 1.46 3.6
2.00 1.00 6.4