For this, you need the v-squared equation, which is v(final)² = v(initial)² + 2aΔx
The averate acceleration is thus a = (v(final)² - v(initial)²) / 2Δx = (20² - 15²) / 2(50) = 175 / 100 = 1.75 m/s²
So the average acceleration is 1.75 m/s²
20 ohms in parallel with 16 ohm= 8.89
20x16/20+16. Product over sum
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:
0.976 c
Explanation:
= velocity of object 1 relative to earth = 0.80 c
= velocity of object 2 relative to object 1 = 0.80 c
= velocity of object 2 relative to earth
Velocity of object 2 relative to earth is given as
![v_{2e}= \frac{v_{1e} + v_{21}}{1 + \frac{v_{1e}v_{21}}{c^{2}}}](https://tex.z-dn.net/?f=v_%7B2e%7D%3D%20%5Cfrac%7Bv_%7B1e%7D%20%2B%20v_%7B21%7D%7D%7B1%20%2B%20%5Cfrac%7Bv_%7B1e%7Dv_%7B21%7D%7D%7Bc%5E%7B2%7D%7D%7D)
![v_{2e}= \frac{0.80 c + 0.80 c}{1 + \frac{(0.80c)(0.80c)}{c^{2}}}](https://tex.z-dn.net/?f=v_%7B2e%7D%3D%20%5Cfrac%7B0.80%20c%20%2B%200.80%20c%7D%7B1%20%2B%20%5Cfrac%7B%280.80c%29%280.80c%29%7D%7Bc%5E%7B2%7D%7D%7D)
= 0.976 c
Answer:
B, It reacts with oxygen to form a new substance
Explanation:
B