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.
Answer:
Explanation:
mass of charged particle m = 6 x 10⁻³ kg .
speed of particle v = 4 x 10³ m /s
speed of particle perpendicular to magnetic field = v sin37
= 4 x 10³ sin37
= 2.41 x 10³ m / s
Force on charged particle
= B q v , B is magnetic field , q is charge on particle and v is velocity perpendicular to B
Force = ma
= 6 x 10⁻³ x 8
= 48 x 10⁻³
Force = Bqv
48 x 10⁻³ = 5 x 10⁻³ q x 2.41 x 10³
q = 48 x 10⁻³ / (5 x 2.41)
= 3.98 x 10⁻³C.
Answer:
The geosphere consists of the solid Earth and the atmosphere consists of the gaseous components in the air. Thus, the answer is C.
Explanation:
when observer and source moves relative to each other then the frequency received by the observer is different from the real frequency
This apparent change in frequency due to relative motion is known as Doppler's effect.
Here we know that
here we know that
= real frequency
v = speed of sound
= speed of observer
= speed of source
so this is known as Doppler's Effect
Answer: the waves probably hits an open cavity
Explanation:
Sound waves transfers energy in form of vibration through water,the speed of sound is greatest in solids followed by liquids and gases,for we to observe such a reduction from 4000m/s to 1500m/s that means it sound had been transmitted through air like open cavity seen in intrusive rock beneath the earth.
