Vector quantity is defined by direction as well as magnitude both
so now lets discuss all option
1) distance = it is total path length between two points so there is no direction needed in it so it is scalar
2) Speed = it is ratio of total distance covered and total time, so it also do not require any direction.
<em>3)velocity = it is ratio of displacement and time, and displacement is always given with direction so velocity is a vector quantity</em>
4) time = time is the measurement of the interval of two events and we do not require any direction in it so its a scalar quantity.
Answer:
In this case, the horizontal velocity of the rocket starts from the acceleration, so if its velocity drops to zero,
Explanation:
When a satellite is in orbit the most important force is the docking of gravity with the Earth
F = m a
where the acceleration is centripetal and F is the force of universal attraction
centripetal acceleration is
a = v² / r
F = m v² / r
In this case, the horizontal velocity of the rocket starts from the acceleration, so if its velocity drops to zero, the force also drops to serious and the satellite steels to Earth.
The speed of the satellite is provides the speed, by local for smaller speeds in satellite, it descends in its orbits and when the speed is amate you have the energy to stop an orb to go to a higher orbit.
<span> Weight = mass x acceleration
Earths acceleration is 9.8 m/s*2
1 kg = 2.2 lbs, so 2.0 lbs x 1 kg/2.2 lbs = 0.91 kg
The bag would have a weight of 9.8 x 0.91 = 8.9 N
1. 8.9 x 1/6 = 1.5 N
2. 8.9 x 2.64 = 23.5 N
The mass of the bag at all three locations is 0.91 kg. Mass does not change, the different locations only change its weight. </span>
Answer:
The heavier piece acquired 2800 J kinetic energy
Explanation:
From the principle of conservation of linear momentum:
0 = M₁v₁ - M₂v₂
M₁v₁ = M₂v₂
let the second piece be the heavier mass, then
M₁v₁ = (2M₁)v₂
v₁ = 2v₂ and v₂ = ¹/₂ v₁
From the principle of conservation of kinetic energy:
¹/₂ K.E₁ + ¹/₂ K.E₂ = 8400 J
¹/₂ M₁(v₁)² + ¹/₂ (2M₁)(¹/₂v₁)² = 8400
¹/₂ M₁(v₁)² + ¹/₄M₁(v₁)² = 8400
K.E₁ + ¹/₂K.E₁ = 8400
Now, we determine K.E₁ and note that K.E₂ = ¹/₂K.E₁
1.5 K.E₁ = 8400
K.E₁ = 8400/1.5
K.E₁ = 5600 J
K.E₂ = ¹/₂K.E₁ = 0.5*5600 J = 2800 J
Therefore, the heavier piece acquired 2800 J kinetic energy
