Answer:
A) Average speed = 18.75 m/s
B) More time is spent at 15 m/s than at 25 m/s.
Explanation:
Let the first distance be d1 and the second distance be d2.
We are given;
d1 = 10 km = 10000 m
d2 = 10 km = 10000 m
Speed; v1 = 15 m/s
Speed; v2 = 25 m/s
Now, the formula for distance is; Distance = speed x time
Thus:
d1 = v1 x t1
t1 = d1/v1 = 10000/15 = 666.67 seconds
Also,
d2 = v2 x t2
t2 = d2/v2 = 10000/25 = 400 seconds
Average speed = total distance/total time = (10000 + 10000)/(666.67 + 400) = 18.75 m/s
From earlier, since t1 = 666.67 seconds and t2 = 400 seconds, then;
More time at 15 m/s than at 25 m/s.
Answer:
8.91 J
Explanation:
mass, m = 8.20 kg
radius, r = 0.22 m
Moment of inertia of the shell, I = 2/3 mr^2
= 2/3 x 8.2 x 0.22 x 0.22 = 0.265 kgm^2
n = 6 revolutions
Angular displacement, θ = 6 x 2 x π = 37.68 rad
angular acceleration, α = 0.890 rad/s^2
initial angular velocity, ωo = 0 rad/s
Let the final angular velocity is ω.
Use third equation of motion
ω² = ωo² + 2αθ
ω² = 0 + 2 x 0.890 x 37.68
ω = 8.2 rad/s
Kinetic energy,
K = 0.5 x 0.265 x 8.2 x 8.2
K = 8.91 J
Answer: The ratio of atoms of potassium to ratio of atoms of oxygen is 4:2
Explanation:
According to the law of conservation of mass, mass can neither be created nor be destroyed, and remains conserved. The mass of products must be same as that of the reactants.
Thus the number of atoms of each element must be same on both sides of the equation so as to keep the mass same and thus balanced chemical equations are written.
K exists as atoms and oxygen exist as molecule which consists of 2 atoms. The ratio of number of atoms on both sides of the reaction are same and thus the ratio of atoms of potassium to ratio of atoms of oxygen is 4:2.
The International System Units or the SI units is scientific method of expressing the magnitudes or quantities of important natural phenomena. There are seven base units in the system, from which other units are derived. This system was formerly called the meter-kilogram-second (MKS) system.
Answer:
Explanation:
a )
Time to reach the speed of 20 m/s with an acceleration of 2 m/s² can be calculated as follows .
v = u + a t
20 = 0 + 2 t
t = 20 /2 = 10 s .
Total time = 10 s + 20 s + 5 s = 35 s .
b) Average velocity = Total distance travelled / total time
Distance travelled in first 10 s
S₁ = ut + 1/2 a t²
= 0 + .5 x 2 x 10²
= 100 m
Distance travelled in next 20 s
S₂= 20s x 20 m/s = 400 m
Distance travelled in last 5 s .
deceleration in last 5 s
v = u + at
0 = 20 m/s + a x 5
a = - 4 m/s²
v² = u² - 2 a s
0 = (20 m/s)² - 2 x 4 m/s² x s
s = 50 m
S₃ = 50 m
Total distance = S₁ + S₂ + S₃
= 100 m + 400 m + 50 m
= 550 m .
Average velocity = 550 m / 35 s
= 15.71 m /s .
