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>.
Answer:
Speed has no effect on voltage.
Explanation:
The voltage of the battery in your car is always 12 to 13.5 volts. It makes no difference whether the car is home in the garage or zooming down the interstate.
a. As sun peeks, it creates a straight line- sun to tree to shadow, angle=180°
b. During morning, the sun-tree-shadow changes from 180° to 90°, during noon, angle =90°.During sunset the angle changes from 90 to 0°.
c. Acute angles created in afternoon.
d. A right angle would be created at noon.
e. Obtuse angle would be created in morning.
f. Yes, a straight angle would be created at sunrise.
g. When sun is at mid-morning, the angle=45°+90°= 135°
Answer:
the maximum height reached by the object from where it was launched is 0.4591 m
Explanation:
initial speed of the object, u= 3 m/s
The velocity at the maximum height will always be 0.
Therefore, final velocity, v= 0 m/s
Using the Newton's equation of motion,
v^2 - u^2 = 2*g*h(max)
0 - u^2= 2*g*h(max)
h(max) = -u^2 /2* g
where g is the gravitational acceleration.
g= - 9.8 m/s^2
substituting the values in equation,
h(max)= - (3*3) / 2*(-9.8)
h(max) = 0.4591 m
the maximum height reached by the object from where it was launched is 0.4591 m
learn more about maximum height reached here:
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