Answer:
Thus, the initial velocity of the bullet is 507.5 m/s.
Explanation:
mass of bullet, m = 0.0085 kg
mass of block, M = 0.99 kg
Height raised, h = 0.95 m
Let the initial velocity of bullet is u and the final velocity of block and bullet system is v.
Use conservation of energy
Potential energy of bullet bock system = kinetic energy of bullet block system
Now use conservation of linear momentum
mu = (M+m) v
0.0085 x u = (0.99 + 0.0085) x 4.32
0.0085 u = 4.314
u = 507.5 m/s
Thus, the initial velocity of the bullet is 507.5 m/s.
Answer:
m = 0.164 kg
Explanation:
T (period)
k (force/spring constant)
m (mass)
T = 2*Pi*sqrt(m/k)
T/(2*Pi) = sqrt(m)/sqrt(k)
(T/(2*Pi))*sqrt(k) = sqrt(m)
m = ((T/(2*Pi))*sqrt(k))^2
m = 4.5*((1.2/(2*Pi)))^2
m = 0.1641403175
Answer:
41.6666667 m / s
Explanation:
Esa es la respuesta. Esto es en metros.
The correct answer for the given question above would be option B. Air flowing from the equator to the poles rises and falls in looping patterns and these patterns are called convection cells. Convection cells<span> is an important process in the formation of landforms and the movement of winds. Hope this answers your question.</span>
use the formula: v^2=(3kT)/m
Where:
<em>v is the velocity of a molecule</em>
<em>k is the Boltzmann constant (1.38064852e-23 J/K)</em>
<em>T is the temperature of the molecule in the air</em>
<em>m is the mass of the molecule</em>
For an H2 molecule at 20.0°C (293 K):
v^2 = 3 × 1.38e-23 J/K × 293 K / (2.00 u × 1.66e-27 kg/u)
v^2 = 3.65e+6 m^2/s^2
v = 1.91e+3 m/s
For an O2 molecule at same temp.:
v^2 = 3 × 1.38e-23 J/K × 293 K / (32.00 u × 1.66e-27 kg/u)
v^2 = 2.28e+5 m^2/s^2
v = 478 m/s
Therefore, the ratio of H2:O2 velocities is:
1.91e+3 / 478 = 4.00
