Answer:

Explanation:
As we know that 1 lb fat can be used to expend 4.1 k Calorie of energy
so here we know that we have to use 7.56 lb of body fat
so here the total energy that is to be utilized is given as

so we will have

now we know that runner expends the energy at rate of 1850 kJ/h
so we have

so it is

now we know that time to burn this energy is given as



Velocity is a vector quantity i.e. it has both magnitude and direction
Speed is a scalar quantity i.e. it has only magnitude
Answer:
75 rad/s
Explanation:
The angular acceleration is the time rate of change of angular velocity. It is given by the formula:
α(t) = d/dt[ω(t)]
Hence: ω(t) = ∫a(t) dt
Also, angular velocity is the time rate of change of displacement. It is given by:
ω(t) = d/dt[θ(t)]
θ(t) = ∫w(t) dt
θ(t) = ∫∫α(t) dtdt
Given that: α (t) = (6.0 rad/s4)t² = 6t² rad/s⁴. Hence:
θ(t) = ∫∫α(t) dtdt
θ(t) = ∫∫6t² dtdt =∫[∫6t² dt]dt
θ(t) = ∫[2t³]dt = t⁴/2 rad
θ(t) = t⁴/2 rad
At θ(t) = 10 rev = (10 * 2π) rad = 20π rad, we can find t:
20π = t⁴/2
40π = t⁴
t = ⁴√40π
t = 3.348 s
ω(t) = ∫α(t) dt = ∫6t² dt = 2t³
ω(t) = 2t³
ω(3.348) = 2(3.348)³ = 75 rad/s
The average speed of the whole travel is equal to <u>400 mph</u>.
Why?
From the statement, we know that whole travel is divided into three parts. For the first part (traveling from New York to Chicago), we have that it was 3.25 hours and the covered distance was half of the total distance (1400mi). For the second part, we have that it was 1 hour (layover time), and the covered no distance. For the third part (traveling from Chicago to Los Angeles), we have that it was 2.75 hours, and it took the other half of the total distance (1400mi).
We can calculate the average speed of the whol travel using the following formula:

Now, substituting and calculating, we have:


Hence, we have the average speed of the whole travel is equal to 400 mph.
Have a nice day!
Answer:
23. 4375 m
Explanation:
There are two parts of the rocket's motion
1 ) accelerating (assume it goes upto h1 height )
using motion equations upwards

Lets find the velocity after 2.5 seconds (V1)
V = U +at
V1 = 0 +5*2.5 = 12.5 m/s
2) motion under gravity (assume it goes upto h2 height )
now there no acceleration from the rocket. it is now subjected to the gravity
using motion equations upwards (assuming g= 10m/s² downwards)
V²= U² +2as
0 = 12.5²+2*(-10)*h2
h2 = 7.8125 m
maximum height = h1 + h2
= 15.625 + 7.8125
= 23. 4375 m