Answer:
Net displacement = 0
Distance traveled = 2PQ <_up and down
Explanation:
At point x = 0, the particle accelerates. Since there will be change of velocity at that point. The the force of the particle will change from negative sign to positive sign according to the given figure, we can therefore conclude that the particle will have a turning point at point x = 0.
Given that a 2.0 kg particle moving along the z-axis experiences the force shown in a given figure.
Force is the product of mass and acceleration. While acceleration is the rate of change of velocity. Both the force and acceleration are vector quantities. They have both magnitude and direction.
If the particle's velocity is 3.0 m/s at x = 0 m, that mean that the particle experience change of velocity at point x = 0. Since the the force of the particle will change from negative sign to positive sign according to the given figure, we can therefore conclude that the particle will have a turning point at point x = 0.
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Answer:
the magnitude of the velocity of the block just after impact is 2.598 m/s and the original speed of the bullect is 324.76m/s.
Explanation:
a) Kinetic energy of block = potential energy in spring
½ mv² = ½ kx²
Here m stands for combined mass (block + bullet),
which is just 1 kg. Spring constant k is unknown, but you can find it from given data:
k = 0.75 N / 0.25 cm
= 3 N/cm, or 300 N/m.
From the energy equation above, solve for v,
v = v √(k/m)
= 0.15 √(300/1)
= 2.598 m/s.
b) Momentum before impact = momentum after impact.
Since m = 1 kg,
v = 2.598 m/s,
p = 2.598 kg m/s.
This is the same momentum carried by bullet as it strikes the block. Therefore, if u is bullet speed,
u = 2.598 kg m/s / 8 × 10⁻³ kg
= 324.76 m/s.
Hence, the magnitude of the velocity of the block just after impact is 2.598 m/s and the original speed of the bullect is 324.76m/s.
Answer:
as it travel through the space it behave like a wave and has an oscillating electric field components and an oscillating magnetic field
Answer:
Explanation:
The momentum of the first piece = m v =- m x 31 i kg m/s in - x direction direction
The momentum of the second piece = -m x 31 j kg m /s in Y - direction
Total momentum = - 31 m( i + j )
To conserve momentum , the third piece must have momentum equal to this
and opposite to it
So momentum of the third piece = 3m x V = 31 m ( i +j )
V = 31/3 ( i + j ) =
Magnitude of velocity V = √2 x 31/ 3 = 14.6 m / s
Its direction will be towards the vector i + j ie 45° from x - axis in positive direction