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 .
I think it is B the rise of the land
Draw a velocity-time diagram as shown below.
Because a velocity of 26.82 m/s is attained in 4.00 s from rest, the average acceleration is
a = 26.82/4 = 6.705 m/s²
The time required to reach maximum velocity of 82.1 m/s is
t₁ = (82.1 m/s)/(6.705 m/s²) = 12.2446 s
The distance traveled during the acceleration phase is
s₁ = (1/2)at₁²
= (1/2)*(6.705 m/s²)*(12.2446 s)²
= 502.64 m
Answer:
The time required to reach maximum speed is 12.245 s
The distance traveled during the acceleration phase is 502.6 m
Answer:
Option (1), option (4) and option (5)
Explanation:
The main observations of Ernest Rutherford's experiment are given below:
1. most of the positively charged particles pass straight, it means there is an empty space in the atom.
2. Very few positively charged particles retraces their path.
So,
The positively charged particles were deflected because like charges repel, that means they are deflected by protons.
Almost all the positively charge concentrate in a very small part which is called nucleus.
Answer:
Michael's final velocity is 19.62 m/s.
Explanation:
We can find the final velocity of Michael by using the following kinematic equation:
(1)
Where:
: is the final velocity =?
: is the initial velocity = 1.62 m/s
a: is the acceleration = 1.2 m/s²
t: is the time = 15 s
By entering the above values into equation (1) we have:


Therefore, Michael's final velocity is 19.62 m/s.
I hope it helps you!