After looking at the transverse waves in the diagram you listed above, the one diagram that does represent the direction of particle X at the instant show in diagram number 3. The direction of the wave motion is up. The correct answer choice will be 3.
Answer:
The value of the spring constant of this spring is 1000 N/m
Explanation:
Given;
equilibrium length of the spring, L = 10.0 cm
new length of the spring, L₀ = 14 cm
applied force on the spring, F = 40 N
extension of the spring due to applied force, e = L₀ - L = 14 cm - 10 cm = 4 cm
From Hook's law
Force applied to a spring is directly proportional to the extension produced, provided the elastic limit is not exceeded.
F ∝ e
F = ke
where;
k is the spring constant
k = F / e
k = 40 / 0.04
k = 1000 N/m
Therefore, the value of the spring constant of this spring is 1000 N/m
We have that for the Question "the acceleration of the object at time t = 0.7 s is most nearly equal to which of the following?"
- it can be said that the acceleration of the object at time t = 0.7 s is most nearly equal to the slope of the line connecting the origin and the point where the graph where the graph crosses the 0.7s grid line
From the question we are told
the acceleration of the object at time t = 0.7 s is most nearly equal to which of the following?
Generally the equation for the Force is mathematically given as
F=\frac{F}{dx}
Therefore
F=-kdx
k=600Nm^{-1}
now
K.E=0.5x ds^2
K.E=600*(-0.1^2)
K.E=3J
Therefore
the acceleration of the object at time t = 0.7 s is most nearly equal to the slope of the line connecting the origin and the point where the graph where the graph crosses the 0.7s grid line
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Answer: 13.2 seconds.
Explanation: using equation of motion; S= ut +1/2at² where u = initial velocity=0
S= distance travelled
a = acceleration due gravity
t= time.
1 foot = 0.305m so,
S= 2860 feet =872.3m
S= ut+1/2 at²
872.3 = 0×t + 1/2×10 × t²
872.3 =0 + 5t²
T²= 872.3/5
T²= 174.46
Take the square root of T we then have;
t = 13.2 seconds to one decimal place.
Answer:
Because the wavelengths of macroscopic objects are too short for them to be detectable.
Explanation:
Wavelength of an object is given by de Broglie wavelength as:
Where, 'h' is Planck's constant, 'm' is mass of object and 'v' is its velocity.
So, for macroscopic objects, the mass is very large compared to microscopic objects. As we can observe from the above formula, there is an inverse relationship between the mass and wavelength of the object.
So, for vary larger masses, the wavelength would be too short and one will find it undetectable. Therefore, we don't observe wave properties in macroscopic objects.
