B) atmosphere pressure, i believe
X=1/2 at^2
3.1=1/2 a *0.64
a=9.68
v=at
v=0.8*9.6875=7.75
Answers:
a) -171.402 m/s
b) 17.49 s
c) 1700.99 m
Explanation:
We can solve this problem with the following equations:
(1)
(2)
(3)
Where:
is the bomb's final height
is the bomb's initial height
is the bomb's initial vertical velocity, since the airplane was moving horizontally
is the time
is the acceleration due gravity
is the bomb's range
is the bomb's initial horizontal velocity
is the bomb's final velocity
Knowing this, let's begin with the answers:
<h3>b) Time
</h3>
With the conditions given above, equation (1) is now written as:
(4)
Isolating :
(5)
(6)
(7)
<h3>a) Final velocity
</h3>
Since , equation (3) is written as:
(8)
(9)
(10) The negative sign only indicates the direction is downwards
<h3>c) Range
</h3>
Substituting (7) in (2):
(11)
(12)
Answer:
6.1 × 10^9 Nm-1
Explanation:
The electric field is given by
E= Kq/d^2
Where;
K= Coulombs constant = 9.0 × 10^9 C
q = magnitude of charge = 1.62×10−6 C
d = distance of separation = 1.53 mm = 1.55 × 10^-3 m
E= 9.0 × 10^9 × 1.62×10−6/(1.55 × 10^-3 )^2
E= 14.58 × 10^3/2.4 × 10^-6
E= 6.1 × 10^9 Nm-1
<span>The majority of the asteroids in our solar system are found
in the space between the orbits of Mars and Jupiter. (B)</span>
