Answer:
Magnitude of net force will be 432.758 N
Explanation:
We have given x component of acceleration 
And vertical component of acceleration 
Mass of the ball m = 0.40 kg
So net acceleration 
Now according to second law of motion
Force = mass × acceleration
So F = 0.40×1081.896 = 432.758 N
The answer for the following problem is explained below.
Therefore the volume charge density of a substance (ρ) is 0.04 ×
C.
Explanation:
Given:
radius (r) =2.1 cm = 2.1 ×
m
height (h) =8.8 cm = 8.8 ×
m
total charge (q) =6.1×
C
To solve:
volume charge density (ρ)
We know;
<u> ρ =q ÷ v</u>
volume of cylinder = π ×r × r × h
volume of cylinder =3.14 × 2.1 × 2.1 ×
× 8.8 ×
volume of cylinder (v) = 122.23 ×
<u> ρ =q ÷ v</u>
ρ = 6.1×
÷ 122.23 ×
<u>ρ = 0.04 × </u>
<u> C</u>
Therefore the volume charge density of a substance (ρ) is 0.04 ×
C.
Hello!
Answer: 7918 J
Explanation:
We are assuming that the floor (field) is completely horizontal since there's no information about that in the statement.
We are going to use the following formula:

Where:


º

Then, by substituting we have:

Answer:
A. The bomb will take <em>17.5 seconds </em>to hit the ground
B. The bomb will land <em>12040 meters </em>on the ground ahead from where they released it
Explanation:
Maverick and Goose are flying at an initial height of
, and their speed is v=688 m/s
When they release the bomb, it will initially have the same height and speed as the plane. Then it will describe a free fall horizontal movement
The equation for the height y with respect to ground in a horizontal movement (no friction) is
[1]
With g equal to the acceleration of gravity of our planet and t the time measured with respect to the moment the bomb was released
The height will be zero when the bomb lands on ground, so if we set y=0 we can find the flight time
The range (horizontal displacement) of the bomb x is
[2]
Since the bomb won't have any friction, its horizontal component of the speed won't change. We need to find t from the equation [1] and replace it in equation [2]:
Setting y=0 and isolating t we get

Since we have 


Replacing in [2]


A. The bomb will take 17.5 seconds to hit the ground
B. The bomb will land 12040 meters on the ground ahead from where they released it