Answer:
At a deceleration of 60g, or 60 times the acceleration due to gravity a person will travel a distance of 0.38 m before coing to a complete stop
Explanation:
The maximum acceleration of the airbag = 60 g, and the duration of the acceleration = 36 ms or 36/1000 s or 0.036 s
To find out how far (in meters) does a person travel in coming to a complete stop in 36 ms at a constant acceleration of 60g
we write out the equation of motion thus.
S = ut + 0.5at²
wgere
S = distance to come to complete stop
u = final velocoty = 0 m/s
a = acceleration = 60g = 60 × 9.81
t = time = 36 ms
as can be seen, the above equation calls up the given variable as a function of the required variable thus
S = 0×0.036 + 0.5×60×9.81×0.036² = 0.38 m
At 60g, a person will travel a distance of 0.38 m before coing to a complete stop
Answer:
E_particle = 1,129 10⁻²⁰ J / particle
T= 817.5 K
Explanation:
Energy is a scalar quantity so it is additive, let's look for the total energy of each gas
Gas a
E_a = 2 5000 = 10000 J
Gas b
E_b = 3 8000 = 24000 J
When the total system energy is mixed it is
E_total = E_a + E_b
E_total = 10000 + 24000 = 34000
The total mass is
M = m_a + m_b
M = 2 +3 = 5
The average energy among the entire mass is
E_averge = E_total / M
E_averago = 34000/5
E_average = 6800 J
One mole of matter has Avogadro's number of atoms 6,022 10²³ particles
Therefore, each particle has an energy of
E_particle = E_averag / 6.022 10²³ = 6800 /6.022 10²³
E_particle = 1,129 10⁻²⁰ J / particle
For find the temperature let's use equation
E = kT
T = E / k
T = 1,129 10⁻²⁰ / 1,381 10⁻²³
T = 8.175 102 K
T= 817.5 K
Answer:
The energy of the capacitors connected in parallel is 0.27 J
Given:
C = 
C' = 
Potential difference, V = 300 V
Solution:
Now, we know that the equivalent capacitance of the two parallel connected capacitors is given by:
The energy of the capacitor, E is given by;
Answer: 0.2 hours
Explanation: In order to solve this question we have to considerer that a recargeable battery can supply 1800 mA in one hour then we have to determine how long could this battery drive current through a long, thin wire of resistance 34 Ω .
Besides, this battery has a voltage of 12 V
so by using the Ohm law we also know that V=R*I,
Fron this we can obtain:
I= V/R= 12 V/ 34 Ω=0.35 A= 350 mA
then considering that this battery can supply 1800 mA in one hour we have this battery can supply 350 mA in x time in the form:
1hour------- 1800 mA
x hour--------350 mA
time= 350/1800= 0.2 hour
Answer:
the answer for the question is the last option
