Answer:
The new speed of the aircraft with respect to the ground is 1414.3 mph.
Explanation:
Given that,
Angle = 37°
Velocity of jet airliner = 693 mph
Velocity of wind = 798 mph
We know that,
The new velocity of the aircraft with respect to the ground
We need to calculate the new speed of the aircraft with respect to the ground
Using formula for velocity
Put the value into the formula
Hence, The new speed of the aircraft with respect to the ground is 1414.3 mph.
Answer:
Explanation:
Given data
Terminal velocity for spread eagle position vt=130 km/h
Terminal velocity for nosedive position vt=326 km/h
The terminal speed of the diver is given by
Therefore the area is given by
Since everything else is constant in the two dives except for the terminal velocity, the ratio between the area in the slow position to the area in the fast position is
Explanation:
From Newton's second law:
F = ma
Given that m = 4 kg and a = 8 m/s²:
F = (4 kg) (8 m/s²)
F = 32 N
If m is reduced to 1 kg and F stays at 32 N:
32 N = (1 kg) a
a = 32 m/s²
So the acceleration increases by a factor of 4.
No, speed is not a vector. it is scalar.
because it doesn't need direction but need magnitude.
