Answer:
F = 800 [N]
Explanation:
To be able to calculate this problem we must use the principle of momentum before and after the impact of the hammer.
We must summarize that after the impact the hammer does not move, therefore its speed is zero. In this way, we can propose the following equation.
ΣPbefore = ΣPafter
where:
m₁ = mass of the hammer = 0.15 [m/s]
v₁ = velocity of the hammer = 8 [m/s]
F = force [N] (units of Newtons)
t = time = 0.0015 [s]
v₂ = velocity of the hammer after the impact = 0
![(0.15*8)-(F*0.0015) = (0.15*0)\\F*0.0015 = 0.15*8\\F = 1.2/(0.0015)\\F = 800 [N]](https://tex.z-dn.net/?f=%280.15%2A8%29-%28F%2A0.0015%29%20%3D%20%280.15%2A0%29%5C%5CF%2A0.0015%20%3D%200.15%2A8%5C%5CF%20%3D%201.2%2F%280.0015%29%5C%5CF%20%3D%20800%20%5BN%5D)
Note: The force is taken as negative since it is exerted by the nail on the hammer and this force is directed in the opposite direction to the movement of the hammer.
Lungs is what helps u breath
<u>Answer :</u>
(a) d = 0.25 m
(b) d = 0.5 m
<u>Explanation :</u>
It is given that,
Frequency of sound waves, f = 686 Hz
Speed of sound wave at 
is, v = 343 m/s
(1) Perfectly destructive interference occurs when the path difference is half integral multiple of wavelength i.e.
........(1)
Velocity of sound wave is given by :
Hence, when the speakers are in phase the smallest distance between the speakers for which the interference of the sound waves is perfectly destructive is 0.25 m.
(2) For constructive interference, the path difference is integral multiple of wavelengths i.e.
( n = integers )
Let n = 1
So, 
Hence, the smallest distance between the speakers for which the interference of the sound waves is maximum constructive is 0.5 m.
Answer
given,
v = 128 ft/s
angle made with horizontal = 30°
now,
horizontal component of velocity
vx = v cos θ = 128 x cos 30° = 110.85 ft/s
vertical component of velocity
vy = v sin θ = 128 x sin 30° = 64 m/s
time taken to strike the ground
using equation of motion
v = u + at
0 =-64 -32 x t
t = 2 s
total time of flight is equal to
T = 2 t = 2 x 2 = 4 s
b) maximum height
using equation of motion
v² = u² + 2 a h
0 = 64² - 2 x 32 x h
64 h = 64²
h = 64 ft
c) range
R = v_x × time of flight
R = 110.85 × 4
R = 443.4 ft
