Answer:
3m/s
Explanation:
Time=5s
Distance =15m
Speed=distance/time
Putting the values
Speed=15m/5s
Speed=3m/s is the answer
Hope it will help you. :)
Answer:
<em>The balloon is 66.62 m high</em>
Explanation:
<u>Combined Motion
</u>
The problem has a combination of constant-speed motion and vertical launch. The hot-air balloon is rising at a constant speed of 14 m/s. When the camera is dropped, it initially has the same speed as the balloon (vo=14 m/s). The camera has an upward movement for some time until it runs out of speed. Then, it falls to the ground. The height of an object that was launched from an initial height yo and speed vo is

The values are


We must find the values of t such that the height of the camera is 0 (when it hits the ground)


Multiplying by 2

Clearing the coefficient of 

Plugging in the given values, we reach to a second-degree equation

The equation has two roots, but we only keep the positive root

Once we know the time of flight of the camera, we use it to know the height of the balloon. The balloon has a constant speed vr and it already was 15 m high, thus the new height is



It started that the present is the key to the past. The process that we see in operation today are the same ones that have operated in the geologic past.
Answer:
A) Concentration of A left at equilibrium of we started the reaction with [A] = 2.00 M and [B] = 2.00 M is 0.55 M.
B) Final concentration of D at equilibrium if the initial concentrations are [A] = 1.00 M and [B] = 2.00 M is 0.90 M.
[D] = 0.90 M
Explanation:
With the first assumption that the volume of reacting mixture doesn't change throughout the reaction.
This allows us to use concentration in mol/L interchangeably with number of moles in stoichiometric calculations.
- The first attached image contains the correct question.
- The solution to part A is presented in the second attached image.
- The solution to part B is presented in the third attached image.