Hi there!
We can use the following kinematic equation:
vf = final velocity (? m/s)
vi = intial velocity (0 m/s)
a = acceleration (5 m/s²)
d = displacement (8 m)
Plug in the givens and solve.
Explanation:
Kepler’s third law states that for all objects orbiting a given body, the cube of the semimajor axis (A) is proportional to the square of the orbital period (P).
For each of our planets orbiting the Sun, the relationship between the orbital period and semimajor axis can be represented by the equation as:
k is constant of proportionality
It is required to solve the above equation for k
Answer:
(A) The speed just as it left the ground is 30.25 m/s
(B) The maximum height of the rock is 46.69 m
Explanation:
Given;
weight of rock, w = mg = 20 N
speed of the rock at 14.8 m, u = 25 m/s
(a) Apply work energy theorem to find its speed just as it left the ground
work = Δ kinetic energy
F x d = ¹/₂mv² - ¹/₂mu²
mg x d = ¹/₂m(v² - u²)
g x d = ¹/₂(v² - u²)
gd = ¹/₂(v² - u²)
2gd = v² - u²
v² = 2gd + u²
v² = 2(9.8)(14.8) + (25)²
v² = 915.05
v = √915.05
v = 30.25 m/s
B) Use the work-energy theorem to find its maximum height
the initial velocity of the rock = 30.25 m/s
at maximum height, the final velocity = 0
- mg x H = ¹/₂mv² - ¹/₂mu²
- mg x H = ¹/₂m(0) - ¹/₂mu²
- mg x H = - ¹/₂mu²
2g x H = u²
H = u² / 2g
H = (30.25)² / 2(9.8)
H = 46.69 m
Answer:
-30°C
Explanation:
F-32/180 =C-0/100
or, -22-32/180=C/100
or, -54/180*100=C
or, -0.3*100=C
therefore, C= -30
-22°F = -30°C
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C₂H₃O₂⁻ is an anion.
<u>Explanation:</u>
NaC₂H₃O₂(s) → Na⁺(aq) + C₂H₃O₂⁻(aq)
NaC₂H₃O₂ when dissociated, yields Na⁺ and C₂H₃O₂⁻.
Anion is a negatively charged ion.
In this case, C₂H₃O₂⁻ is an anion.
