Answer:
The velocity of the other fragment immediately following the explosion is v .
Explanation:
Given :
Mass of original shell , m .
Velocity of shell , + v .
Now , the particle explodes into two half parts , i.e .
Since , no eternal force is applied in the particle .
Therefore , its momentum will be conserved .
So , Final momentum = Initial momentum
The velocity of the other fragment immediately following the explosion is v .
We have that the instantaneous velocity of the shuttlecock when it hits the ground is
From the question we are told
Assuming the acceleration is still -9.81 m/s2, what is the instantaneous velocity of the
shuttlecock when it hits the ground? Show your work below.
Generally the equation for acceleration is mathematically given as
Where
acceleration is still -9.81 m/s2,
Hence,
Therefore
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Answer:
The displacement of the volleyball is 2.62 m
Explanation:
Given;
initial velocity of the volleyball, u = 7.5 m/s
final velocity of the volleyball, v = 2.2 m/s
displacement of the volleyball, d = ?
Apply the following kinematic equation;
v² = u² - 2gd
2gd = u² - v²
Therefore, the displacement of the volleyball is 2.62 m
Answer:
Acceleration of the crate is 0.362 m/s^2.
Explanation:
Given:
Mass of the box, m = 40 kg
Applied force, F = 15 N
Angle at which the force is applied, = 15°
We have to find the magnitude of the acceleration.
Let the acceleration be "a".
FBD is attached with where we can see the horizontal and vertical component of force.
⇒ and ⇒
⇒ ⇒
⇒ Applying concept of forces.
⇒
⇒
⇒ <em> ...Newtons second law Fnet = ma</em>
⇒
⇒ Plugging the values.
⇒ <em>...f is the friction which is zero here.</em>
⇒
⇒
Magnitude of the acceleration of the crate is 0.362 m/s^2.
