Answer:
t = 1.4[s]
Explanation:
To solve this problem we must use the principle of conservation of linear momentum, which tells us that momentum is conserved before and after applying a force to a body. We must remember that the impulse can be calculated by means of the following equation.
where:
P = impulse or lineal momentum [kg*m/s]
m = mass = 50 [kg]
v = velocity [m/s]
F = force = 200[N]
t = time = [s]
Now we must be clear that the final linear momentum must be equal to the original linear momentum plus the applied momentum. In this way we can deduce the following equation.
where:
m₁ = mass of the object = 50 [kg]
v₁ = velocity of the object before the impulse = 18.2 [m/s]
v₂ = velocity of the object after the impulse = 12.6 [m/s]
![(50*18.2)-200*t=50*12.6\\910-200*t=630\\200*t=910-630\\200*t=280\\t=1.4[s]](https://tex.z-dn.net/?f=%2850%2A18.2%29-200%2At%3D50%2A12.6%5C%5C910-200%2At%3D630%5C%5C200%2At%3D910-630%5C%5C200%2At%3D280%5C%5Ct%3D1.4%5Bs%5D)
Answer:
a. get warmer.
Explanation:
When the water vaporous reach the upper layer of the atmosphere they get a cooler air to which they loose their temperature and condense to form clouds as a the temperature of the air increases.
It may be noted that the water looses its high amount of latent heat of vaporization to condense into water this significantly increases the temperature of the air in contact.
Answer:
17. NADH has a molar extinction coefficient of 6200 M2 cm at 340 nm. Calculate the molar concentration of NADH required to obtain an absorbance of 0.1 at 340 nm in a 1-cm path length cuvette. 18. A sample with a path length of 1 cm absorbs 99.0% of the incident light at a wavelength of 274 nm, measured with respect to an appropriate solvent blank. Tyrosine is known to be the only chromophore present in the sample that has significant absorption at 274 nm. Calculate the molar concentration of tyrosine in the sample.
Explanation:
Answer:
The maximum altitude reached with respect to the ground = 0.5 + 0.510 = 1.01 km
Explanation:
Using the equations of motion,
When the rocket is fired from the ground,
u = initial velocity = 0 m/s (since it was initially at rest)
a = 10 m/s²
The engine cuts off at y = 0.5 km = 500 m
The velocity at that point = v
v² = u² + 2ay
v² = 0² + 2(10)(500) = 10000
v = 100 m/s
The velocity at this point is the initial velocity for the next phase of the motion
u = 100 m/s
v = final velocity = 0 m/s (at maximum height, velocity = 0)
y = vertical distance travelled after the engine shuts off beyond 0.5 km = ?
g = acceleration due to gravity = - 9.8 m/s²
v² = u² + 2gy
0 = 100² + 2(-9.8)(y)
- 19.6 y = - 10000
y = 510.2 m = 0.510 km
So, the maximum altitude reached with respect to the ground = 0.5 + 0.510 = 1.01 km
Hope this helps!!!
