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:
The definitive and unifying theme or idea of a story or article. It encompasses all the aspects necessary to create a coherent main idea. The central idea is typically expressed as a universal truth or theme that is built and supported by the setting and characters in a story.
Explanation:
This is the correct answer to your question.
Hope this helps!!!
Kyle.
Answer:
F₂ = -7.3 N
Explanation:
Given that,
The mass of an object, m₁ = 3.7 kg
First force, F₁ = 11 N
The net acceleration of the object is 1 m/s².
We know that,
F₁+F₂ = ma
11+F₂ = (3.7)(1)
F₂ = 3.7-11
F₂ = -7.3 N
so, the other force is 7.3 N and it is acting in west direction.
Answer:
D
Explanation:
We know the formula for Work to be:
W = f * d
Where W is work done
f is force
d is the distance
A)
Work = 50
Distance = 50
So, Force is:
Force = 50/50 = 1
B)
Work = 400
Distance = 80
Force = 400/80 = 5
C)
Work = 365
Distance = 73
Force = 365/73 = 5
D)
Work = 144
Distance = 16
Force = 144/16 = 9
Hence, D is the situation in which the force applied is the greatest.
Complete Question
A gas gun uses high pressure gas tp accelerate projectile through the gun barrel.
If the acceleration of the projective is : a = c/s m/s2
Where c is a constant that depends on the initial gas pressure behind the projectile. The initial position of the projectile is s= 1.5m and the projectile is initially at rest. The projectile accelerates until it reaches the end of the barrel at s=3m. What is the value of the constant c such that the projectile leaves the barrel with velocity of 200m/s?
Answer:
The value of the constant is 
Explanation:
From the question we are told that
The acceleration is 
The initial position of the projectile is s= 1.5m
The final position of the projectile is 
The velocity is 
Generally 
and acceleration is 
so
=> 
integrating both sides
Now for the limit
a = 200 m/s
b = 0 m/s
c = s= 3 m
d =
= 1.5 m
So we have
![[\frac{v^2}{2} ] \left | 200} \atop {0}} \right. = c [ln s]\left | 3} \atop {1.5}} \right.](https://tex.z-dn.net/?f=%5B%5Cfrac%7Bv%5E2%7D%7B2%7D%20%5D%20%5Cleft%20%7C%20200%7D%20%5Catop%20%7B0%7D%7D%20%5Cright.%20%20%3D%20c%20%5Bln%20s%5D%5Cleft%20%7C%203%7D%20%5Catop%20%7B1.5%7D%7D%20%5Cright.)
![\frac{200^2}{2} = c ln[\frac{3}{1.5} ]](https://tex.z-dn.net/?f=%5Cfrac%7B200%5E2%7D%7B2%7D%20%20%3D%20%20c%20ln%5B%5Cfrac%7B3%7D%7B1.5%7D%20%5D)
=> 
