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.
Using the equation v(average)=x traveled/time
v = 100/2.5
You get 40 kilometers per hour
Hope this helped!
Answer: 0.42watts
Explanation:
Energy of the battery = 500 J
Time = 1200 seconds
Power of the instrument = ?
Recall that power is the rate of energy expended in doing work. Thus, power is energy expended divided by time taken.
i.e Power = (energy/time)
Power = 500J/1200 seconds
Power = 0.416 watts (Round to the nearest hundredth which is 0.42 watts)
Thus, the power of the instrument is 0.42watts