Answer:
the answer is b
Explanation:
Second and third class levers are differentiated by <u>the location of the </u><u>load.</u>
<em>Hope</em><em> </em><em>this</em><em> </em><em>help</em><em> </em><em>you</em><em> </em><em>out </em><em>and have</em><em> </em><em>a </em><em>nice</em><em> </em><em>day </em><em>=</em><em>)</em>
Given :
A 0.50-kg mass is attached to a spring of spring constant 20 N/m along a horizontal, frictionless surface.
The object oscillates in simple harmonic motion and has a speed of 1.5 m/s at the equilibrium position.
To Find :
At what location are the kinetic energy and the potential energy the same.
Solution :
Let, at location x from the equilibrium position the kinetic energy and the potential energy the same.
So,
Hence, this is the required solution.
1. To solve this problem you can use the formula:
Force = mass × acceleration
16 = 5 × a
16/5 = a
3.2 m/sec^2 = acceleration
2. To solve this formula use the same formula that was given above:
F = ma
F= 1239 kg × 4m/sec
F = 4956 N
3. To solve this problem use the formula:
Weight = mass × gravity
W = 35 × 9.8
W = 343 N
4. To solve this problem use the formula:
Momentum = mass × velocity
M = 95 kg × 8 m/s
M = 760 kg × m/sec
hope this helps :)
Wavelength = (speed) / (frequency)
= (30 m/s) / (60/sec) =
= 0.5 meter .