True, physical excercise helps
Answer:
h = 2.087 m
Explanation:
Given
m₁ = 3 kg
v₁ = 20 m/s
m₂ = 2 kg
v₂ = - 14 m/s
In a completely inelastic collision the colliding objects stick together after the collision and move together as a single object.
In the given problem, lets assume that the balls of putty are initially moving along the y axis, upward direction being the positive y direction. And the collision occurs at the origin of the coordinate system.
We can apply the equation
vs = (m₁*v₁ + m₂*v₂) / (m₁ + m₂)
⇒ vs = (3 kg*20 m/s + 2 kg*(- 14 m/s)) / (3 kg + 2 kg)
⇒ vs = 6.4 m/s (↑)
To calculate the maximum height h attained by the combined system of two balls of putty after the the collision, we use the expression for linear motion under gravity:
vf² = vi² - 2*g*h
where
vf = 0 m/s
g = 9.81 m/s²
vi = vs = 6.4 m/s
finally we get h:
h = vi² / (2*g)
⇒ h = (6.4 m/s)² / (2*9.81 m/s²) = 2.087 m
Answer:
100 foot
Explanation:
Using the formula for calculating range to calculate the speed first as shown;
<h3>Range R = U²sin 2theta/g</h3>
U is the speed
theta is the observe angle
g is the acceleration due to gravity.
200 = U²sin 2(45)/9.8
Usin90 = 200 * 9.8
U² = 1960
U = √1960
U = 44.27 m/s
Get the required altitude
Altitude H = u²/2g
H = 44.27²/2(9.8)
H = 1,959.8329/19.6
<em>H = 99.99 feet</em>
<em>Hence the altitude of the rocket to the nearest foot is 100 foot</em>
Answer:
Final velocity will be 314.6 m/sec
Distance traveled = 1314.24 m
Explanation:
We have given initial velocity u = 233 m/sec
Acceleration 
Time t = 4.8 sec
From first equation of motion 
, here v is final velocity, u is initial velocity and t is time
So 
Now we have to find distance traveled
From second equation of motion
So distance traveled in given time will be 1314.24 m
Answer:
I have examined the different parts of blub and I assume the answer is (D)T and U
Hope this helps you
