Sound level at distance of 15 m is given as 20 dB
so intensity at this distance is given as
now if we move closer to some some distance the sound level is now 50 dB
now the intensity is given as
now we know that
so now the distance from friend must be 47 cm
The spring is initially stretched, and the mass released from rest (v=0). The next time the speed becomes zero again is when the spring is fully compressed, and the mass is on the opposite side of the spring with respect to its equilibrium position, after a time t=0.100 s. This corresponds to half oscillation of the system. Therefore, the period of a full oscillation of the system is
Which means that the frequency is
and the angular frequency is
In a spring-mass system, the maximum velocity of the object is given by
where A is the amplitude of the oscillation. In our problem, the amplitude of the motion corresponds to the initial displacement of the object (A=0.500 m), therefore the maximum velocity is
My very eager mother just served us nine pizzas
Answer:
The minimum total speed is 11.2km/s
Explanation:
We are been asked to find the escape velocity.
Escape velocity is defined as the minimum initial velocity that will take a body(projectile)away above the surface of a planet(earth) when it's projected vertically upwards.
The formula to calculate the escape velocity is Ve = √2gR
For the earth g = 9.8m/s2 , R = 6.4*10^6
Substituting into the equation Ve = √2*9.8*6.4*10^6 = 11.2*10^3m/s
=11.2km/s