The answer for APEX Learning is Ocean
Answer:
Explanation:
λ = wave length = 632 x 10⁻⁹
slit width a = 2 x 10⁻³ m
angular separation of central maxima
= 2 x λ /a
= 2 x 632 x 10⁻⁹ / 2 x 10⁻³
= 632 x 10⁻⁶ rad
width in m of light spot.
= 632 x 10⁻⁶ x 376000 km
= 237.632 km
1) The total mechanical energy of the rock is:
where U is the gravitational potential energy and K the kinetic energy.
Initially, the kinetic energy is zero (because the rock starts from rest, so its speed is zero), and the total mechanical energy of the rock is just gravitational potential energy. This is equal to
where
is the mass,
is the gravitational acceleration and
is the height.
Putting the numbers in, we find the potential energy
2) Just before hitting the ground, the potential energy U is zero (because now h=0), and all the potential energy of the rock converted into kinetic energy, which is equal to:
where v is the speed of the rock just before hitting the ground. Since the mechanical energy of the rock must be conserved, then the kinetic energy K before hitting the ground must be equal to the initial potential energy U of the rock:
3) For the work-energy theorem, the work W done by the gravitational force on the rock is equal to the variation of kinetic energy of the rock, which is:
Answer:
<h3>The answer is 1600 kgm/s</h3>
Explanation:
The momentum of an object can be found by using the formula
<h3>momentum = mass × velocity</h3>
From the question
mass = 200 kg
velocity / speed = 8m/s
We have
momentum = 200 × 8
We have the final answer as
<h3>1600 kgm/s</h3>
Hope this helps you
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.