Gravity affects weight of an object
Its weight reduces as it moves away from the center as gravity is strongest near the core and reduces as you move away
Hope this helps C:
Answer:
835.29 Hz
Explanation:
When moving towards the source of sound, frequency will be given by
f*=f(vd+v)/v
Where f is the freqiency of the source, vd is the driving speed, v is the speed of sound in air, f* is the inkown frequency when moving forward.
Substituting 800 Hz for f, 340 m/s for v and 15 m/s for vd then
f*=800(15+340)/340=835.29411764704 Hz
Rounded off, the frequency is approximately 835.29 Hz
Using formula:
I=(1/2)*M*(R^2+r^2)
<span>I=0.5*0.715kg*[(12.7cm)^2+(10.7cm)^2] </span>
<span>I=98.6 kg*cm^2</span>
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.
