Answer:
The values is 
Explanation:
From the question we are told that
The speed of the fire engine is 
The frequency of the tone is 
The speed of sound in air is 
The beat frequency is mathematically represented as

Where
is the frequency of sound heard by the people in the fire engine and is is mathematically evaluated as
![f_a = [\frac{v_s + v }{v_s -v} ]* f](https://tex.z-dn.net/?f=f_a%20%20%3D%20%20%5B%5Cfrac%7Bv_s%20%2B%20v%20%7D%7Bv_s%20%20-v%7D%20%5D%2A%20f)
substituting values
![f_a = [\frac{340 + 5 }{340 -5} ]* 500](https://tex.z-dn.net/?f=f_a%20%20%3D%20%20%5B%5Cfrac%7B340%20%2B%205%20%7D%7B340%20%20-5%7D%20%5D%2A%20500)

Thus


Answer:
Vi = 8.28 m/s
Explanation:
This problem is related to the projectile motion.
As we know there are two components of motion associated with this, the horizontal component and vertical component.
The horizontal distance covered by the ball is
Vx*t = x
Vx*t = 5.3
Vx = 5.3/t eq. 1
Also we know that
Vx = Vicos(60)
Vx = Vi*0.5 eq. 2
equate eq. 1 and eq. 2
5.3/t = Vi*0.5
5.3/0.5 = Vi*t
Vi*t = 10.6 eq. 3
The vertical distance is
Vy = y1 + Vyi*t - 0.5gt²
also we know that
Vyi = Visin(60)
Vyi = Vi*0.866
It is given that V1 = 1.9 m and and Vy = 3 m is the vertical distance
3 = 1.9 + Vi*0.866*t - 0.5gt²
3 = 1.9 + Vi*0.866*t - 0.5(9.8)t²
3 = 1.9 + 0.866(Vi*t) - 0.5(9.8)t²
3 = 1.9 + 0.866(Vi*t) - 0.5(9.8)t²
1.1 = 0.866(Vi*t) - 4.9t²
0.866(Vi*t) = 4.9t² + 1.1
substitute Vi*t = 10.6 in above equation
0.866(10.6) = 4.9t² + 1.1
9.18 = 4.9t² + 1.1
4.9t² = 8.08
t² = 8.08/4.9
t² = 1.648
t = 1.28 sec
Finally, initial speed can be found by substituting the value of t into eq. 3
Vi*t = 10.6
Vi = 10.6/t
Vi = 10.6/1.28
Vi = 8.28 m/s
Answer:
In the table there we always eat.
It just based on my own...
Answer:
Explanation:
Height covered = 12m
time to fall by 12 m
s = 1/2 gt²
12 = 1/2 g t²
t = 1.565 s
Horizontal distance of throw
= 8.5 x 1.565
= 13.3 m
This distance is to be covered by dog during the time ball falls ie 1.565 s
Speed of dog required = 13.3 / 1.565
= 8.5 m /s
b ) dog will catch the ball at a distance of 13.3 m .
True if you have proper stance and use your body the right way then the ball will be below your waist to allow for more control.