Answer:
Explanation:
We shall first calculate the velocity at height h = 575 m .
acceleration a = 2.2 m /s²
v² = u² + 2 a s
u is initial velocity , v is final velocity , s is height achieved
v² = 0 + 2 x 2.2 x 575
v = 50.3 m /s
After 575 m , rocket moves under free fall so g will act on it downwards
If it travels further by height H
from the relation
v² = u² - 2 g H
v = 0 , u = 50.3 m /s
H = ?
0 = 50.3² - 2 x 9.8 H
H = 129.08 m
Total height attained by rocket
= 575 + 129.08
= 704.08 m .
<span>electromagnetic.........</span>
4. The Coyote has an initial position vector of 
.
4a. The Coyote has an initial velocity vector of 
. His position at time 
is given by the vector
where 
is the Coyote's acceleration vector at time . He experiences acceleration only in the downward direction because of gravity, and in particular 
where 
. Splitting up the position vector into components, we have 
with
The Coyote hits the ground when 
:
4b. Here we evaluate 
at the time found in (4a).
5. The shell has initial position vector 
, and we're told that after some time the bullet (now separated from the shell) has a position of 
.
5a. The vertical component of the shell's position vector is
We find the shell hits the ground at
5b. The horizontal component of the bullet's position vector is
where 
is the muzzle velocity of the bullet. It traveled 3500 m in the time it took the shell to fall to the ground, so we can solve for :
Answer:
The correct answer is B-25 V
Explanation:
We apply Ohm's Law, according to which:
V = i x R
V = 5A x 5Ω
V= 25 V
Being V the potential difference whose unit is the VOLT, i the current intensity (Ampere) and R the electrical resistance (ohm)
