Answer:
t = 23.255 s, x = 2298.98 m, v_y = - 227.90 m / s
Explanation:
After reading your extensive writing, we are going to solve the approach.
The initial speed of the plane is 250 miles / h and it is at an altitude of 2650 m; In general, planes fly horizontally for launch, therefore this is the initial horizontal speed.
As there is a mixture of units in different systems we are going to reduce everything to the SI system.
v₀ₓ = 250 miles h (1609.34 m / 1 mile) (1 h / 3600 s) = 111.76 m / s
y₀ = 2650 m
Let's set a reference system with the x-axis parallel to the ground, the y-axis is vertical. As time is a scalar it is the same for vertical and horizontal movement
Y axis
y = y₀ + v₀ t - ½ g t²
the initial vertical velocity when the cargo is dropped is zero and when it reaches the floor the height is zero
0 = y₀ + 0 - ½ g t²
t = 
t = √(2 2650/ 9.8)
t = 23.255 s
Therefore, for the cargo to reach the desired point, it must be launched from a distance of
x = v₀ₓ t
x = 111.76 23.255
x = 2298.98 m
at the point and arrival the speed is
vₓ = v₀ₓ = 111.76
vertical speed is
v_y = v_{oy} - gt
v_y = 0 - gt
v_y = - 9.8 23.25 555
v_y = - 227.90 m / s
the negative sign indicates that the speed is down
in the attachment we have a diagram of the movement
Answer:
E = 7.99 *10^{-13} J
Explanation:
the given reaction is
we know that energy is given as
where
m_1 H^1 is mass of proton = 1.672622 *10^{-27}
m_1 H^2 is mass of deuterium = 3.344494 *10^{-27}
m_2 H^3 is mass of He = 5.008234 *10^{-27}
E = [1.672622 *10^{-27} + 3.344494 *10^{-27} - 5.008234 *10^{-27} ] *(3*10^8)^2
E = 7.99 *10^{-13} J
Answer:
go to the link quizzlet it will give you tha answer
Explanation:
Answer:
Explanation:
Given that
T₁ = 290 K
P₁ = 100 KPa
Power P =5.5 KW
mass flow rate
Lets take the exit temperature = T₂
We know that
If we assume that process inside the compressor is adiabatic then we can say that
That is why the exit pressure will be 4091 KPa.
Answer:
discrete lines are observed by the spectroscope, the emission of the lamp is of the ATOMIC source
Explanation:
Bulbs can emit light in several ways:
* When the emission is carried out by the heating of its filament, the bulb is called incandescent, in general its spectrum is similar to that of a black body, this is a continuous spectrum with a maximum dependent on the fourth power of the temperature of the filament.
* The emission can be by atomic transitions, in this case there is a discrete spectrum formed by the spectral lines of the material that forms the gas of the lamp, in general for the yellow emission the most used materials are mercury and sodium or a mixture of they.
Consequently, as discrete lines are observed by the spectroscope, the emission of the lamp is of the ATOMIC type
