Answer:
a) a = 3.72 m / s², b) a = -18.75 m / s²
Explanation:
a) Let's use kinematics to find the acceleration before the collision
v = v₀ + at
as part of rest the v₀ = 0
a = v / t
Let's reduce the magnitudes to the SI system
v = 115 km / h (1000 m / 1km) (1h / 3600s)
v = 31.94 m / s
v₂ = 60 km / h = 16.66 m / s
l
et's calculate
a = 31.94 / 8.58
a = 3.72 m / s²
b) For the operational average during the collision let's use the relationship between momentum and momentum
I = Δp
F Δt = m v_f - m v₀
F =
F = m [16.66 - 31.94] / 0.815
F = m (-18.75)
Having the force let's use Newton's second law
F = m a
-18.75 m = m a
a = -18.75 m / s²
Answer:
Explanation:
fundamental frequency, f = 250 Hz
Let T be the tension in the string and length of the string is l ans m be the mass of the string initially.
the formula for the frequency is given by
.... (1)
Now the length is doubled ans the tension is four times but the mass remains same.
let the frequency is f'
.... (2)
Divide equation (2) by equation (1)
f' = √2 x f
f' = 1.414 x 250
f' = 353.5 Hz
Answer:
<h3>Newton's 2nd law states acceleration is proportional to the net force acting on an object. The net force is the vector sum of all the forces applied to the object. ... In this case the acceleration (slowing down) of the puck is proportional to the amount of friction.</h3>
Explanation:
<h3>mark as brainliast</h3>
Explanation:
The two postulates of special theory of relativity
Postulate 1: The law of physics are invariant under any of inertial frame of reference.
Postulate 2: The velocity of light is remains same in each ans every frame of reference and independent of relativity.
They are differ from classical mechanics that in classical mechanics there is no change in mass and length in relative velocity but in relativistic mechanics it changes.
These two postulates implements in phenomenon like time dilation , length contraction etc.
Thanks