Answer:
Option D
Explanation:
<u><em>Given:</em></u>
Mass = m = 110 kg
Acceleration due to gravity = g = 9.8 m/s
<u><em>Required:</em></u>
Weight = W = ?
<u><em>Formula</em></u>
W = mg
<u><em>Solution:</em></u>
W = (110)(9.8)
W = 1078 N
The dog’s speed is
A) 0.61 m/s
Explanation:
Given formula:
ME=
mv²+mgh
To make height the subject of the formula, follow the following procedures;
Subtract
mv² from both side of equation
M.E -
mv² =
mv² -
mv²+mgh
This gives:
M.E -
mv² = mgh
Multiply both sides of the expression by 
( M.E -
mv² ) x
=
x mgh
h = ( M.E -
mv² ) x 
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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)
=> 
