Answer:
19.8 m/s
Explanation:
Given:
Maximum vertical displacement of the object (H) = 20 m
Acceleration due to gravity (g) = 9.8 m/s²
At maximum height, the velocity of the object is 0 m/s for a moment. So, final velocity (v) at the maximum height is 0 m/s.
Now, let the initial velocity or velocity at launch be 'u' m/s.
Now, using the following equation of motion for vertical motion:

Rewriting in terms of 'u', we get:

Plug in the given values and solve for 'u'. This gives,

Therefore, the vertical velocity at the launch is 19.8 m/s.
<span>Most objects tend to contain the same numbers of positive and negative charge because this is the most stable situation. In fact, if an object has an excess of positive charge, it tends to attract an equal number of negative charges to balance this effect and restore neutrality: the attracted negative charges combine with the excess of positive charges, leaving the object electrically neutral.</span>
Answer:
Explanation:
Given that,
Height of the bridge is 20m
Initial before he throws the rock
The height is hi = 20 m
Then, final height hitting the water
hf = 0 m
Initial speed the rock is throw
Vi = 15m/s
The final speed at which the rock hits the water
Vf = 24.8 m/s
Using conservation of energy given by the question hint
Ki + Ui = Kf + Uf
Where
Ki is initial kinetic energy
Ui is initial potential energy
Kf is final kinetic energy
Uf is final potential energy
Then,
Ki + Ui = Kf + Uf
Where
Ei = Ki + Ui
Where Ei is initial energy
Ei = ½mVi² + m•g•hi
Ei = ½m × 15² + m × 9.8 × 20
Ei = 112.5m + 196m
Ei = 308.5m J
Now,
Ef = Kf + Uf
Ef = ½mVf² + m•g•hf
Ef = ½m × 24.8² + m × 9.8 × 0
Ef = 307.52m + 0
Ef = 307.52m J
Since Ef ≈ Ei, then the rock thrown from the tip of a bridge is independent of the direction of throw
Radio waves are what cell phone company rely on
Answer:
The distance away the center of the earthquake is 1083.24 km.
Explanation:
Given that,
Speed of transverse wave = 9.1\ km/s
Speed of longitudinal wave = 5.7 km/s
Time = 71 sec
We need to calculate the distance of transverse wave
Using formula of distance

....(I)
The distance of longitudinal wave
....(II)
From the first equation

Put the value of t in equation (II)




Hence, The distance away the center of the earthquake is 1083.24 km.