W = Fd = 4(2100) = 8400 J
So the answer is A) 8400 J
I was just rewriting my notes on the work lesson I did in class today, so I saw this question at the perfect time!! :)
Hope it helps!! :)
Answer:
Newton's first law: An object at rest remains at rest, or if in motion, remains in motion at a constant velocity unless acted on by a net external force. ... An object sliding across a table or floor slows down due to the net force of friction acting on the object.
Explanation:
please give me a heart
Answer:
Explanation:
Given:
- mass of vehicle,
- radius of curvature,
- coefficient of friction,
<u>During the turn to prevent the skidding of the vehicle its centripetal force must be equal to the opposite balancing frictional force:</u>
where:
coefficient of friction
normal reaction force due to weight of the car
velocity of the car
is the maximum velocity at which the vehicle can turn without skidding.
False... I hope that helps ;)
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.
