Answer:
v = 719.2 m / s and a = 83.33 m / s²
Explanation:
This is a rocket propulsion system where the system is made up of the rocket plus the ejected mass, where the final velocity is
v - v₀ = 
ln (M₀ / M)
where v₀ is the initial velocity, v_{e} the velocity of the gases with respect to the rocket and M₀ and M the initial and final masses of the rocket
In this case, if fuel burns at 75 kg / s, we can calculate the fuel burned for the 10 s
m_fuel = 75 10
m_fuel = 750 kg
As the rocket initially had a mass of 3000 kg including 1000 kg of fuel, there are still 250 kg, so the mass of the rocket minus the fuel burned is
M = 3000 -750 = 2250 kg
let's calculate
v - 0 = 2500 ln (3000/2250)
v = 719.2 m / s
To calculate the acceleration, let's use the concept of the rocket thrust, which is the force of the gases on it. In the case of the rocket, it is
Push = v_{e} dM / dt
let's calculate
Push = 2500 75
Push = 187500 N
If we use Newton's second law
F = m a
a = F / m
let's calculate
a = 187500/2250
a = 83.33 m / s²
<span>When reading a buret, the initial reading should be taken from the top of the glassware and the final volume should still taken at the top. If the buret is completely, the initial volume for most buret would be zero. though, there are some where their initial starts at 50 decreasing to zero.</span>
Answer:
v = -v₀ / 2
Explanation:
For this exercise let's use kinematics relations.
Let's use the initial conditions to find the acceleration of the electron
v² = v₀² - 2a y
when the initial velocity is vo it reaches just the negative plate so v = 0
a = v₀² / 2y
now they tell us that the initial velocity is half
v’² = v₀’² - 2 a y’
v₀ ’= v₀ / 2
at the point where turn v = 0
0 = v₀² /4 - 2 a y '
v₀² /4 = 2 (v₀² / 2y) y’
y = 4 y'
y ’= y / 4
We can see that when the velocity is half, advance only ¼ of the distance between the plates, now let's calculate the velocity if it leaves this position with zero velocity.
v² = v₀² -2a y’
v² = 0 - 2 (v₀² / 2y) y / 4
v² = -v₀² / 4
v = -v₀ / 2
We can see that as the system has no friction, the arrival speed is the same as the exit speed, but with the opposite direction.
Explanation:
Given that,
Mass if the rock, m = 1 kg
It is suspended from the tip of a horizontal meter stick at the 0-cm mark so that the meter stick barely balances like a seesaw when its fulcrum is at the 12.5-cm mark.
We need to find the mass of the meter stick. The force acting by the stone is
F = 1 × 9.8 = 9.8 N
Let W be the weight of the meter stick. If the net torque is zero on the stick then the stick does not move and it remains in equilibrium condition. So, taking torque about the pivot.
W = 3.266 N
The mass of the meters stick is :
So, the mass of the meter stick is 0.333 kg.
