Answer:
See Explanation
Explanation:
According to Newton's second law of motion, the acceleration of a body is proportional to the net external force that acts on the body.
A body accelerated when it is acted upon by an unbalanced net external force.
When the external forces acting on a body are balanced, the effect of each force is cancelled by the other hence the body is not accelerated according to Newton's second law.
Answer:
a) v = 0.9167 m / s, b) A = 0.350 m, c) v = 0.9167 m / s, d) A = 0.250 m
Explanation:
a) to find the velocity of the wave let us use the relation
v = λ f
the wavelength is the length that is needed for a complete wave, in this case x = 5.50 m corresponds to a wavelength
λ = x
λ = x
the period is the time for the wave to repeat itself, in this case t = 3.00 s corresponds to half a period
T / 2 = t
T = 2t
period and frequency are related
f = 1 / T
f = 1 / 2t
we substitute
v = x / 2t
v = 5.50 / 2 3
v = 0.9167 m / s
b) the amplitude is the distance from a maximum to zero
2A = y
A = y / 2
A = 0.700 / 2
A = 0.350 m
c) The horizontal speed of the traveling wave (waves) is independent of the vertical oscillation of the particles, therefore the speed is the same
v = 0.9167 m / s
d) the amplitude is
A = 0.500 / 2
A = 0.250 m
Answer:
24m/s²
Explanation:
Given
Distance S = 3m
Time of fall = 0.5sec
Required
Acceleration due to gravity
Using the equation of motion
S = ut+1/2gt²
Substitute the given values
3 = 0+1/2g(0.5)²
3 = 1/2(0.25)g
3 = 0.125g
g = 3/0.125
g = 24
Hence the value for the acceleration of gravity on this new planet is 24m/s²
The question just basically explained what happens
To solve the problem it is necessary to use Newton's second law and statistical equilibrium equations.
According to Newton's second law we have to

where,
m= mass
g = gravitational acceleration
For the balance to break, there must be a mass M located at the right end.
We will define the mass m as the mass of the body, located in an equidistant center of the corners equal to 4m.
In this way, applying the static equilibrium equations, we have to sum up torques at point B,

Regarding the forces we have,

Re-arrange to find M,



Therefore the maximum additional mass you could place on the right hand end of the plank and have the plank still be at rest is 16.67Kg