Answer:
No..
Explanation:
As the bird releases the drop there is no internal force which will drive it into circular path but it will fall on tangent of the arc at the point of release because it has a tangential velocity same as bird. Path will be parabola in vertical plane.
As the person is on circular arc constantly moving it will never meet that drop.
Take east to be the positive direction. Then the resultant force from adding <em>F</em>₁ and <em>F</em>₂ is
<em>F</em>₁ + <em>F</em>₂ = (-45 N) + 63 N = 18 N
which is positive, so it's directed east.
To this we add a third force <em>F</em>₃ such that the resultant is 12 N pointing west, making it negative, so that
18 N + <em>F</em>₃ = -12 N
<em>F</em>₃ = -30 N
So <em>F</em>₃ has a magnitude of 30 N and points west.
Answer:
The smallest part of a millimeter that can be read with a digital caliper with a four digit display is 0.02mm. Thus, it has to be converted to centimetre. So, divide by 10, we then have 0.02/10= *0.002cm* not mm.
The main (and only) purpose of the turbine in the turbo jet engine is to drive the air compressor. The turbojet engine works by compressing the air using an inlet and a compressor, then mixing the fuel with the compressed air, then passing the mixture to the combustor, then passing the high pressure air through a turbine and a nozzle.
Answer:
Tp/Te = 2
Therefore, the orbital period of the planet is twice that of the earth's orbital period.
Explanation:
The orbital period of a planet around a star can be expressed mathematically as;
T = 2π√(r^3)/(Gm)
Where;
r = radius of orbit
G = gravitational constant
m = mass of the star
Given;
Let R represent radius of earth orbit and r the radius of planet orbit,
Let M represent the mass of sun and m the mass of the star.
r = 4R
m = 16M
For earth;
Te = 2π√(R^3)/(GM)
For planet;
Tp = 2π√(r^3)/(Gm)
Substituting the given values;
Tp = 2π√((4R)^3)/(16GM) = 2π√(64R^3)/(16GM)
Tp = 2π√(4R^3)/(GM)
Tp = 2 × 2π√(R^3)/(GM)
So,
Tp/Te = (2 × 2π√(R^3)/(GM))/( 2π√(R^3)/(GM))
Tp/Te = 2
Therefore, the orbital period of the planet is twice that of the earth's orbital period.
