Answer:
A. plot an H-R diagram for the stars in the cluster.
Explanation:
A star cluster can be defined as a constellation of stars, due to gravitational force, which has the same origin.
The astronomy student would have to plot an H-R diagram for the stars in the cluster and determine the age of the cluster by observing the turn-off point. The turn-off is majorly as a result of gradual depletion of the source of energy of the star. Thus, it projects off the constellation.
115 divided by 2 equals 57.5
Answer:
V₀y = 0 m/s
t = 2.47 s
V₀ₓ = 61.86 m/s
Vₓ = 61.86 m/s
Explanation:
Since, the ball is hit horizontally, there is no vertical component of velocity at initial point. So, the initial vertical velocity (V₀y) will beL
<u>V₀y = 0 m/s</u>
For the initial vertical velocity of golf ball we consider the vertical motion and apply 2nd equation of motion:
Y = V₀y*t + (0.5)gt²
where,
Y = Height = 30 m
g = 9.8 m/s²
t = time to hit the ground = ?
Therefore,
30 m = (0 m/s)(t) + (0.5)(9.8 m/s²)t²
t² = 30 m/4.9 m/s²
t = √6.122 s²
<u>t = 2.47 s</u>
For initial vertical velocity we analyze the horizontal motion of the ball. We neglect the frictional effects in horizontal motion thus the speed remains uniform. Hence,
V₀ₓ = Xt
where,
V₀ₓ = Initial vertical Velocity = ?
X = Horizontal Distance = 25 m
Therefore,
V₀ₓ = (25 m)(2.47 s)
<u>V₀ₓ = 61.86 m/s</u>
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Due, to uniform motion in horizontal direction:
Final Vertical Velocity = Vₓ = V₀ₓ
Vₓ = 61.86 m/s
Answer:
a) v = 88.54 m/s
b) vf = 26.4 m/s
Explanation:
Given that;
m = 1400.0 kg
a)
by using the energy conservation
loss in potential energy is equal to gain in kinetic energy
mg × ( 3200-2800) = 1/2 ×m×v²
so
1400 × 9.8 × 400 = 0.5 × 1400 × v²
5488000 = 700v²
v² = 5488000 / 700
v² = 7840
v = √7840
v = 88.54 m/s
b)
Work done by all forces is equal to change in KE
W_gravity + W_non - conservative = 1/2×m×(vf² - vi²)
we substitute
1400 × 9.8 × ( 3200-2800) - (5 × 10⁶) = 1/2 × 1400 × (vf² -0 )
488000 = 700 vf²
vf² = 488000 / 700
vf² = 697.1428
vf = √697.1428
vf = 26.4 m/s
Given that Oxygen has an oxidation state of 2 -, you can combine Mn 3+ with O 2- ions to form Mn2O3, and you can combine Mn 2+ with O 2- to form Mn2O2 which is MnO.
The other compounds imply oxidation states of N, Br and Cl that does not exist.
Therefore, the answer is the option D. MnO