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.
In 1912, Bohr<span> was working for the Nobel laureate J.J. Thompson in England when he was introduced to Ernest Rutherford, whose </span>discovery<span> of the nucleus and development of an atomic model had earned him a Nobel Prize in chemistry in 1908. Under Rutherford's tutelage, </span>Bohr<span> began studying the properties of atoms.
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I believe the answer is B)
Hope this helps*
Answer:
44.3 m/s
Explanation:
Given that a ball is thrown horizontally from the top of a building 100m high. The ball strikes the ground at a point 120 m horizontally away from and below the point of release.
What is the magnitude of its velocity just before it strikes the ground ?
The parameters given are:
Height H = 100m
Since the ball is thrown from a top of a building, initial velocity U = 0
Let g = 9.8m/s^2
Using third equation of motion
V^2 = U^2 + 2gH
Substitute all the parameters into the formula
V^2 = 2 × 9.8 × 100
V^2 = 200 × 9.8
V^2 = 1960
V = 44.27 m/s
Therefore, the magnitude of its velocity just before it strikes the ground is 44.3 m/s approximately
Answer:
the object is stopped at B.
