Newton's 2nd law of motion:
Force = (mass) x (acceleration)
= (1,127 kg) x (6 m/s² forward)
= (1,127 x 6) newtons forward
= 6,762 newtons forward
______________________________
Momentum = (mass) x (speed)
= (69 kg) x (6 m/s)
= 414 kg-m/s
<span>Inertia keeps us orbiting because any object with mass has the tendency to resist changes to their direction and speed of movement. Combine that with the interaction of the gravitational attraction of the sun, and that is what keeps Earth in orbit. The sun’s gravitational force is one that is proportional to Earth’s mass, and it acts in a way that is almost exactly perpendicular to Earth’s motion. This keeps Earth from spinning into the sun or far away from it.</span>
Answer:
Explanation:
As the path is straight, so the speed is equivalent to velocity. Now. assuming that the acceleration and deceleration of the train are constant. So, change of velocity with respect to time for acceleration as well as deceleration is constant. Hence, the slope of the speed-time graph is constant for the time of acceleration as well as deceleration. The speed for the time from 
to 
is constant, so slope for this interval of time is zero. The speed-time graph is shown in the figure.
The total distance covered by the train during the entire journey is the area of the speed-time graph.
Area
As velocity is in 
and time is in 
so the unit of area is 
Hence, the total distance is 
.
An object that has kinetic energy must be <em>moving</em>.
The formula for an object's kinetic energy is
KE = (1/2) · (the object's mass) · <u><em>(the object's speed)²</em></u>
As you can see from the formula, if the object has no speed, then its kinetic energy is zero. That's why kinetic energy is usually called the "energy of motion", and if an object HAS kinetic energy, then that tells you right away that it must be moving.
Answer:
D) 15s
Explanation:
let Te be the period of the block-spring system on earth and Tm be the period of the same system on the moon.let g1 be the gravitational acceleration on earth and g2 be the gravitational acceleration on the moon.
the period of a pendulum is given by:
T = 2π√(L/g)
so on earth:
Te = 2π√(L/g1)
= 6s
on the moon;
Tm = 2π√(L/g2)
since g2 = 1/6 g1 then:
Tm = 2π√(L/(1/6×g1))
= √(6)×2π√(L/(g1))
and 2π√(L/(g1)) = Te = 6s
Tm = (√(6))×6 = 14.7s ≈ 15s
Therefore, the period of the block-spring system on the moon is 15s.
