Answer:
Kf > Ka = Kb > Kc > Kd > Ke
Explanation:
We can apply
E₀ = E₁
where
E₀: Mechanical energy at the beginning of the motion (top of the incline)
E₁: Mechanical energy at the end (bottom of the incline)
then
K₀ + U₀ = K₁ + U₁
If v₀ = 0 ⇒ K₀
and h₁ = 0 ⇒ U₁ = 0
we get
U₀ = K₁
U₀ = m*g*h₀ = K₁
we apply the same equation in each case
a) U₀ = K₁ = m*g*h₀ = 70 Kg*9.81 m/s²*8m = 5493.60 J
b) U₀ = K₁ = m*g*h₀ = 70 Kg*9.81 m/s²*8m = 5493.60 J
c) U₀ = K₁ = m*g*h₀ = 35 Kg*9.81 m/s²*4m = 1373.40 J
d) U₀ = K₁ = m*g*h₀ = 7 Kg*9.81 m/s²*16m = 1098.72 J
e) U₀ = K₁ = m*g*h₀ = 7 Kg*9.81 m/s²*4m = 274.68 J
f) U₀ = K₁ = m*g*h₀ = 105 Kg*9.81 m/s²*6m = 6180.30 J
finally, we can say that
Kf > Ka = Kb > Kc > Kd > Ke
Answer:
121Sb=57.2%
123Sb=42.8%
Explanation:
We are given that
Atomic mass of 121Sb=120.904 amu
Atomic mass of 123Sb=122.904 amu
Average atomic mass of antimony=121.760 amu
We have to find the percent of each of the isotope.
Let x be the percent of 121Sb and 1-x be the percent of 123Sb.
Using formula of average atomic weight
Average atomic weight=atomic weight of 121Sb
percentage abundance of isotope 121Sb+atomic weight of 123Sb
percentage abundance of isotope 123Sb
Substitute the values






Percentage of 121Sb=
57.2%
Abundance of isotope 123Sb=1-0.572=0.428
Percentage of isotope 123Sb=
42.8%
Answer:
t = 0.85[s]
Explanation:
To solve this problem we must make a complete description of this. By doing an internet search, we find the description of this problem as if of the question.
<u>Description</u>
<u />
"An alligator swims to the left with a constant velocity of 5 m s when the alligator season a bird straight ahead the alligator speeds up with a constant acceleration of 3 m/s^2 leftward until it reaches a final velocity of 35 Ms left work how many seconds does it take the alligator to speed up from 5 m/s to 35 m/s".
To solve this problem we must identify the initial data:
v0 = initial velocity = 5 [m/s]
a = acceleration = 3 [m/s^2]
vf = final velocity = 35[m/s]
t = time = ?
Using the following kinematic equation, we can find the time that is required.
![v_{f}=v_{0}+a*t\\35=5+35*t\\t=\frac{35-5}{35} \\t=0.85[s]](https://tex.z-dn.net/?f=v_%7Bf%7D%3Dv_%7B0%7D%2Ba%2At%5C%5C35%3D5%2B35%2At%5C%5Ct%3D%5Cfrac%7B35-5%7D%7B35%7D%20%5C%5Ct%3D0.85%5Bs%5D)
The force exerted by student A with his scooter is 306 N and that of student B is 204 N.
<h3>
Force applied by each student</h3>
The force exerted by each student is calculated from Newton's second law of motion.
F = ma
where;
- m is mass
- a is acceleration
F(A) = 127.5 x 2.4
F(A) = 306 N
F(B) = 120 x 1.7
F(B) = 204 N
Thus, the force exerted by student A with his scooter is 306 N and that of student B is 204 N.
Learn more about force here: brainly.com/question/12970081
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Answer:
35 kg
Explanation: Newton's Second Law of Motion reinterpreted as
or 