Answer:
a) 24.918 m
b) 4.5078 seconds
c) [0.18, 0.39]
Explanation:
t = Time taken
u = Initial velocity
v = Final velocity
s = Displacement
a = Acceleration due to gravity = 9.81 m/s²
At the beginning of the last 1 second the velocity will be 15.095 m/s
Taking that as the final velocity of the path from the maximum height we find time
So, the ball traveled for 1+1.539 = 2.2539 seconds from the maximum height to the ground.
The maximum height the ball traveled is 24.918 m.
The time the ball will take to go up is 2.2539 seconds
So, total time the ball will be in flight is 2.2539+2.2539 = 4.5078 seconds
Solving the above equation we get
t = 0.18 seconds
Solving the equation we get
t = 0.39 seconds
So, the ball will traveling in the [0.18, 0.39] seconds of the flight
Answer:
DNA identical to the DNA of the parent
Explanation:
Answer:
The position of the image = 7.2 cm.
Explanation:
Given:
A Converging lens:
Object distance = 12 cm = -12 cm (sign convention)
Focal length = 18 cm
We have to find the position of the image.
Let the position of the image be .
Sign convention:
The focal length of converging (convex) lens is always positive,while object distance is negative.
Using lens formula:
⇒
⇒
⇒
⇒
⇒
⇒
⇒
⇒
⇒ cm
So the position of the image = 7.2 cm.
Use the formula n = c / v, where c is the speed of light and v is the speed of light in the environment, that is, use the speed in brackets and the correct answer will be the one that will be the smallest result
At t =0, the velocity of A is greater than the velocity of B.
We are told in the question that the spacecrafts fly parallel to each other and that for the both spacecrafts, the velocities are described as follows;
A: vA (t) = ť^2 – 5t + 20
B: vB (t) = t^2+ 3t + 10
Given that t = 0 in both cases;
vA (0) = 0^2 – 5(0) + 20
vA = 20 m/s
For vB
vB (0) = 0^2+ 3(0) + 10
vB = 10 m/s
We can see that at t =0, the velocity of A is greater than the velocity of B.
Learn more: brainly.com/question/24857760
Read each question carefully. Show all your work for each part of the question. The parts within the question may not have equal weight. Spacecrafts A and B are flying parallel to each other through space and are next to each other at time t= 0. For the interval 0 <t< 6 s, spacecraft A's velocity v A and spacecraft B's velocity vB as functions of t are given by the equations va (t) = ť^2 – 5t + 20 and VB (t) = t^2+ 3t + 10, respectively, where both velocities are in units of meters per second. At t = 6 s, the spacecrafts both turn off their engines and travel at a constant speed. (a) At t = 0, is the speed of spacecraft A greater than, less than, or equal to the speed of spacecraft B?