Answer:
The drop time ball 1 is less than the drop time of ball 2. A further explanation is provided below.
Explanation:
The net force acting on the ball will be:
⇒ 
Here,
F = Force
m = mass
g = acceleration
Now,
According to the Newton's 2nd law of motion, we get
⇒ 
To find the value of "a", we have to substitute "
" in the above equation,
⇒ 
⇒ 
We can see that, the acceleration is greater for the greater mass of less for the lesser mass. Thus the above is the appropriate solution.
Answer:
to make calculation more easy to get
Explanation:
if you are using chart or calculate Thermodynamic problems you will not never solve this problem with out using data table for thermodynamic
<span>The offspring will have the exact same genetic makeup as the parent. This is because there is no other parent involved other than the one parent.</span>
Answer:
0.143 m
Explanation:
The relationship between force applied on a string and stretching of the spring is given by Hooke's law:
where
F is the force exerted on the spring
k is the spring constant of the spring
x is the stretching of the spring from its equilibrium position
In this problem, we have:
F = 20 N is the force applied on the spring
k = 140 N/m is the spring constant
Solving for x, we find how far the spring will stretch:
Answer:
xf = 5.68 × 10³ m
yf = 8.57 × 10³ m
Explanation:
given data
vi = 290 m/s
θ = 57.0°
t = 36.0 s
solution
firsa we get here origin (0,0) to where the shell is launched
xi = 0 yi = 0
xf = ? yf = ?
vxi = vicosθ vyi = visinθ
ax = 0 ay = −9.8 m/s
now we solve x motion: that is
xf = xi + vxi × t + 0.5 × ax × t² ............1
simplfy it we get
xf = 0 + vicosθ × t + 0
put here value and we get
xf = 0 + (290 m/s) cos(57) (36.0 s)
xf = 5.68 × 10³ m
and
now we solve for y motion: that is
yf = yi + vyi × t + 0.5 × ay × t
² ............2
put here value and we get
yf = 0 + (290 m/s) × sin(57) × (36.0 s) + 0.5 × (−9.8 m/s2) × (36.0 s) ²
yf = 8.57 × 10³ m
