The forward force you exert on the fish and your backward action will allow you to reach the shore.
<h3>
Newton's third law of motion</h3>
Newton's third law of motion states that for every action, there is an equal and opposite reaction.
Fa = -Fb
Let's assume the fish is held in the hook, this will give you the opportunity to throw the fish forward while still holding it.
When the the fish is thrown forward, you will move backwards with an equal force based on Newton's third law. Your backward momentum towards the shore will help to maintain equal linear momentum between you and the fish.
Thus, this forward force of the fish and your backward action will allow you to reach the shore.
Learn more Newton's third law of motion here: brainly.com/question/25998091
Centripetal force is equal to (mv^2)/r
The way I use to answer these question is to set every variable to 1
m=1
v=1
r=1
so centripetal force =1
then change the variable we're looking at
and since we're find when it's half we could either change it to 1/2 or 2, but 2 is easier to use
m=1
v=2
r=1
((1)×(2)^2)/1=4
So the velocity in the 1st part is half the velocity in the 2nd part and the centripetal force is 4× less
The answer is the centripetal force is 1/4 as big the second time around
Answer:
Take-off velocity = v = 81.39[m/s]
Explanation:
We can calculate the takeoff speed easily, using the following kinematic equation.
where:
a = acceleration = 4[m/s^2]
x = distance = 750[m]
vi = initial velocity = 25 [m/s]
vf = final velocity
![v_{f}=\sqrt{(25)^{2}+(2*4*750) } \\v_{f}=81.39[m/s]](https://tex.z-dn.net/?f=v_%7Bf%7D%3D%5Csqrt%7B%2825%29%5E%7B2%7D%2B%282%2A4%2A750%29%20%7D%20%5C%5Cv_%7Bf%7D%3D81.39%5Bm%2Fs%5D)
Answer:
v' = 1.5 m/s
Explanation:
given,
mass of the bullet, m = 10 g
initial speed of the bullet, v = 300 m/s
final speed of the bullet after collision, v' = 300/2 = 150 m/s
Mass of the block, M = 1 Kg
initial speed of the block, u = 0 m/s
velocity of the block after collision, u' = ?
using conservation of momentum
m v + Mu = m v' + M u'
0.01 x 300 + 0 = 0.01 x 150 + 1 x v'
v' = 0.01 x 150
v' = 1.5 m/s
Speed of the block after collision is equal to v' = 1.5 m/s
