Answer:
60 m/s
Explanation:
From the law of conservation of energy,
The kinetic energy of the plane = Energy of store in the spring when the plane lands.
1/2mv² = 1/2ke²
making v the subject of the equation.
v = √(ke²/m).................... Equation 1
Where v = the plane landing speed, k = spring constant, e = extension. m = mass of the plane.
Given: m = 15000 kg, k = 60000 N/m, e = 30 m.
Substitute into equation 1
v = √(60000×30²/15000)
v = √(4×900)
v = √(3600)
v = 60 m/s.
Hence the plane's landing speed = 60 m/s
Answer:
True
Explanation:
Velocity is a vector quantity, which means that it carries both magnitude and direction. Hence when direction of a particle changes, although magnitude (speed) may remain same, it's velocity changes due to direction change. For ex. A particle is m... A particle is moving along x axis with speed 1m/s, it's velocity will be represented as 1i (i represents unit vector along x)
But if it now starts moving along y axis, it's velocity is 1j (j represents unit vector along y axis). Hence velocity changes with direction.
brainllest pls .
Answer:
Cat and a dog
Explanation:
The cat might bark instead of meow because it has certain traits from the dog. The dog however may still bark because it is a dog!
Answer:
A
Explanation:
if the man doubles his force to 40 and the box was yet to move that means acceleration also doubled so your answer would be A
Answer: the minimum spacing that must be there between two objects on the earth's surface if they are to be resolved as distinct objects by this telescope 6.45 cm
Explanation:
Given that;
diameter of the mirror d = 1.7 m
height h = 180 km = 180 × 10³ m
wavelength λ = 500 nm = 5 × 10⁻⁹ m
Now Angular separation from the peak of the central maximum is expressed as;
sin∅= 1.22 λ / d
sin∅ = (1.22 × 5 × 10⁻⁹) / 1.7
sin∅ = 3.588 × 10⁻⁷
we know that;
sin∅ = object separation / distance from telescope
object separation =
sin∅ × distance from telescope
object separation = 3.588 × 10⁻⁷ × 180 × 10³
object separation =6.45 × 10⁻² m
then we convert to centimeter
object separation = 6.45 cm
Therefore the minimum spacing that must be there between two objects on the earth's surface if they are to be resolved as distinct objects by this telescope 6.45 cm
