Answer:
Vertical component of velocity is 9.29 m/s
Explanation:
Given that,
Velocity of projection of a projectile, v = 22 m/s
It is fired at an angle of 22°
The horizontal component of velocity is v cosθ
The vertical component of velocity is v sinθ
So, vertical component is given by :



Hence, the vertical component of the velocity is 9.29 m/s
Answer. Explanation: Frequency of the sound decreases and the speed of sound becomes 346m/s from near about 1500 m/s.
Answer:
Car H
Explanation:
Frictional force is a resistant force. It is given as:
F = u*m*g
Where u = coefficient of friction
m = mass
g = acceleration due to gravity
From the formula above, we see that frictional force is dependent on the mass of object and the coefficient of friction.
Since they all have the same tires, the coefficient of friction between the tire and the floor is the same for each car. Acceleration due to gravity, g, is constant.
The only factor that determines the frictional force of each car is the mass. Hence, the more the mass, the more the frictional force.
So, the most massive car will have the most frictional force and hence, will come to a stop quicker than the others. The least massive car will have the least frictional force and so, will take a longer time to stop.
a) KE=0.5*mv^2==0.5*145*25^2=45312.5 J
b) PE=mgh=145*9.8*3.5=4973.5 J
c) ME=KE+PE=m(0.5v^2+gh)=62524 J
Answer:
Wavelength of the sound wave that reaches your ear is 1.15 m
Explanation:
The speed of the wave in string is

where T= 200 N is tension in the string ,
=1.0 g/m is the linear mass density


Wavelength of the wave in the string is

The frequency is

The required wavelength pf the sound wave that reaches the ear is( take velocity of air v=344 m/s)
