Answer:
1.68m/s^2
Explanation:
using V^2=U^2+2×a×s formula.
If,
(Final Velocity)V=45
(Initial Velocity)U=0 because it starts from rest
(Distance)S=600m
(acceleration)a= ?
now,
using formula,
45^2=0^2+2×a×600
2025= 1200a
a=2025÷1200
a = 1.68m/s^2(The unit of acceleration is m/s^2)
Therefore the acceleration is 1.68m/s^2
Hope it works !!!
Let's be clear: The plane's "395 km/hr" is speed relative to the
air, and the wind's "55 km/hr" is speed relative to the ground.
Before the wind hits, the plane moves east at 395 km/hr relative
to both the air AND the ground.
After the wind hits, the plane still maintains the same air-speed.
That is, its velocity relative to the air is still 395 km/hr east.
But the wind vector is added to the air-speed vector, and the
plane's velocity <span>relative to the ground drops to 340 km/hr east</span>.
When the contracting gas and dust from nebula become so dense and hot that nuclear fusion starts
<span>When an observer is moving towards a stationary object making a sound,
he runs the risk of startling the stationary object.
On the other hand, if the stationary object is the one making the sound,
then the pitch of the sound perceived by the observer is higher than the
sound that the object is actually making.</span>
The answer is solar flare ; an eruption of high - energy radiation from the sun.
