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Answer:50
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Explanation:
Let 
be the airspeed.
Let 
be the cross wind speed.
We know that,ground speed is the vector sum of airspeed and cross wind speed and airspeed is perpendicular to cross wind speed.
If 
and 
are two perpendicular vectors,the resultant vector has the magnitude 
Given,
So,the ground speed is 
Answer:
112 m/s², 79.1°
Explanation:
In the x direction, given:
x₀ = 0 m
x = 19,500 cos 32.0° m
v₀ = 1810 cos 20.0° m/s
t = 9.20 s
Find: a
x = x₀ + v₀ t + ½ at²
19,500 cos 32.0° = 0 + (1810 cos 20.0°) (9.20) + ½ a (9.20)²
a = 21.01 m/s²
In the y direction, given:
y₀ = 0 m
y = 19,500 sin 32.0° m
v₀ = 1810 sin 20.0° m/s
t = 9.20 s
Find: a
y = y₀ + v₀ t + ½ at²
19,500 sin 32.0° = 0 + (1810 sin 20.0°) (9.20) + ½ a (9.20)²
a = 109.6 m/s²
The magnitude of the acceleration is:
a² = ax² + ay²
a² = (21.01)² + (109.6)²
a = 112 m/s²
And the direction is:
θ = atan(ay / ax)
θ = atan(109.6 / 21.01)
θ = 79.1°
Answer:
When the ejected air is moving in the downward direction then the thrust force acts in the upward direction, due to reversal thrust, the jets can take off vertically without needing a runway this way.
Explanation:
Newton’s third law motion states that for every action there will be an equal and opposite reaction.
Thrust reversal is also known as reverse thrust. It acts opposite to the motion of the aircraft by providing the deceleration.
Commercial aircraft moves the ejected air in the forward direction means that the thrust will acts opposite to the motion of the aircraft that is backward direction due to thrust reversal. This thrust force might be used to decelerate the craft.
Uses of thrust reversal in practice:
When the ejected air is moving forward direction then the thrust force moving backward direction due to reversal thrust the speed of the craft slows down.
When the ejected air is moving in the downward direction then the thrust force acts in the upward direction, due to reversal thrust, the jets can take off vertically without needing a runway this way.
We know, Mechanical Energy = K.E. + P.E.
As ball is at ground, P.E. would be zero. But as it is in motion, it must have some K.E. and that is:
K.E. = 1/2 mv²
K.E. = 1/2 * 1 * 2²
K.E. = 4/2
K.E. = 2 J
In short, Your Answer would be Option B
Hope this helps!
