Answer:
The stars are moving away from us.
Explanation:
The observed wavelengths of hydrogen transition for stars A and B (660.0 nm and 666 nm respectively) are greater than that observed in the laboratory (656.2 nm). The observed long wavelengths for the stars means that the light from the stars is red-shifted.
According to the Doppler effect, red-shifted light means that the source is moving a way from the observer; therefore, we arrive at the conclusion that the stars A and B are moving away from us.
Answer:
x2 = 0.99
Explanation:
from superheated water table
at pressure p1 = 0.6MPa and temperature 200 degree celcius
h1 = 2850.6 kJ/kg
From energy equation we have following relation
![2850.6 + [\frac{50^2}{2} * \frac{1 kJ/kg}{1000 m^2/S^2}] = h2 +[ \frac{600^2}{2} * \frac{1 kJ/kg}{1000 m^2/S^2}]](https://tex.z-dn.net/?f=2850.6%20%2B%20%5B%5Cfrac%7B50%5E2%7D%7B2%7D%20%2A%20%5Cfrac%7B1%20kJ%2Fkg%7D%7B1000%20m%5E2%2FS%5E2%7D%5D%20%3D%20h2%20%2B%5B%20%5Cfrac%7B600%5E2%7D%7B2%7D%20%2A%20%5Cfrac%7B1%20kJ%2Fkg%7D%7B1000%20m%5E2%2FS%5E2%7D%5D)
h2 = 2671.85 kJ/kg
from superheated water table
at pressure p2 = 0.15MPa
specific enthalpy of fluid hf = 467.13 kJ/kg
enthalpy change hfg = 2226.0 kJ/kg
specific enthalpy of the saturated gas hg = 2693.1 kJ/kg
as it can be seen from above value hf>h2>hg, so phase 2 is two phase region. so we have
quality of steam x2
h2 = hf + x2(hfg)
2671.85 = 467.13 +x2*2226.0
x2 = 0.99
Time before projectile hits wall
= 88.2 m / 29.4 m/s = 3 seconds
Vertical velocity of projectile after three seconds
= 3*9.8 = 29.4 m/s
Horizontal velocity of projectile after three seconds, assuming no air resistance
= 29.4 m/s (given)
Conclusion:
velocity of projectile when it hits the wall
= < 29.4, -29.4> m/s
= sqrt(29.4^2+29.4^2) m/s east-bound at 45 degrees below horizontal
= 41.58 m/s east-bound at 45 degrees below horizontal.
Arginine is a basic aminoacid, because it has two amino groups and one acid
group.
At a low pH, every ionizable group is protoned. At a little higher pH, the
acid group looses its proton. A little higher pH, one amino group looses its
proton. At a very high pH, all ionizable groups are not protoned.
Pkas
<span>
<span><span>
<span>
pka1 = 1.82
</span>
<span>
pka2 = 8.99
</span>
<span>
pka3 = 12.48
</span>
</span>
</span></span>
So 9.20 is higher tan the second pKa and lower than the third pka. This
means the acid has already lost its proton, and one of the aminos too, but the
second amino hasn’t. When an acid is not protoned, it has a negative charge.
When an amino is not protoned, it’s neutral. When an amino is protoned, it has
a positive charge. So this amnino acid has one positive charge (one of the aminos) and one negative
charge (the acid), what makes it neutral.
Answer:
a) W = - 318.26 J, b) W = 0
, c) W = 318.275 J
, d) W = 318.275 J
, e) W = 0
Explanation:
The work is defined by
W = F .ds = F ds cos θ
Bold indicate vectors
We create a reference system where the x-axis is parallel to the ramp and the axis and perpendicular, in the attached we see a scheme of the forces
Let's use trigonometry to break down weight
sin θ = Wₓ / W
Wₓ = W sin 60
cos θ = Wy / W
Wy = W cos 60
X axis
How the body is going at constant speed
fr - Wₓ = 0
fr = mg sin 60
fr = 15 9.8 sin 60
fr = 127.31 N
Y Axis
N - Wy = 0
N = mg cos 60
N = 15 9.8 cos 60
N = 73.5 N
Let's calculate the different jobs
a) The work of the force of gravity is
W = mg L cos θ
Where the angles are between the weight and the displacement is
θ = 60 + 90 = 150
W = 15 9.8 2.50 cos 150
W = - 318.26 J
b) The work of the normal force
From Newton's equations
N = Wy = W cos 60
N = mg cos 60
W = N L cos 90
W = 0
c) The work of the friction force
W = fr L cos 0
W = 127.31 2.50
W = 318.275 J
d) as the body is going at constant speed the force of the tape is equal to the force of friction
W = F L cos 0
W = 127.31 2.50
W = 318.275 J
e) the net force
F ’= fr - Wx = 0
W = F ’L cos 0
W = 0
