Answer:
1.11 m/s
Explanation:
The motion of the boat is an example of accelerated motion, since the velocity is not constant. However, we don't need to find the acceleration, because we are only interested in the average velocity of the boat, which is given by:

where d is the total distance covered and t the time taken. In this problem, the boat covered a distance of d = 20 m and it takes t = 18 s, therefore the average velocity is

Are you answering a question or asking? You have already seemed to get the answer, A is the correct answer :I
<span>Now that you know the time to reach its maximum height, you have enough information to find out the initial velocity of the second arrow. Here's what you know about it: its final velocity is 0 m/s (at the maximum height), its time to reach that is 2.8 seconds, but wait! it was fired 1.05 seconds later, so take off 1.05 seconds so that its time is 1.75 seconds, and of course gravity is still the same at -9.8 m/s^2. Plug those numbers into the kinematic equation (Vf=Vi+a*t, remember?) for 0=Vi+-9.8*1.75 and solve for Vi to get.......
17.15 m/s</span>
Answer:
the boat would be deeped by 3200 m
Explanation:
Given that
The boat arrives back after 4 seconds
And, the speed of the sound in water is 1,600 m/s
We need to find out how much deep is the water
So,
As we know that
Distance = ( speed × time) ÷ 2
Here we divided by 2 because the boat arrives back
= (1600 × 4) ÷ 2
= 3200 m
Therefore the boat would be deeped by 3200 m
Answer:
y = 77.74 10⁻⁵ m
Explanation:
For this exercise we can use Newton's second law
F = m a
a = F / m
a = 4.9 10⁻¹⁶ / 9.1 10⁻³¹
a = 0.538 10¹⁵ m / s
This is the vertical acceleration of the electron.
Now let's use kinematics to find the time it takes to move the
x= 29 mm = 29 10⁻³ m
On the x axis
v = x / t
t = x / v
t = 29 10⁻³ / 1.7 10⁷
t = 17 10⁻¹⁰ s
Now we can look for vertical distance at this time.
y =
t + ½ a t²
y = 0 + ½ 0.538 10¹⁵ (17 10⁻¹⁰)²
y = 77.74 10⁻⁵ m