Answer:
Explanation:
Consider the initial position of the frog (20 m above ground) as the reference position. All measurements are positive measured upward.
Therefore,
u = 10 m/s, initial upward velocity.
H = - 20 m, position of the ground.
g = 9.8 m/s², acceleration due to gravity.
Part (a)
When the frog reaches a maximum height of h from the reference position, its velocity is zero. Therefore
u² - 2gh = 0
h = u²/(2g) = 10²/(2*9.8) = 5.102 m
At maximum height, the frog will be 20 + 5.102 = 25.102 m above ground.
Answer: 25.1 m above ground
Part (b)
Let v = the velocity when the frog hits the ground. Then
v² = u² - 2gH
v² = 10² - 2*9.8*(-20) = 492
v = 22.18 m/s
Answer: The frog hits the ground with a velocity of 22.2 m/s
Answer:
x = 400 [m]
Explanation:
To solve this problem we must use the following kinematics equations, first, we find the final speed, and then we proceed to find the distance traveled.
where:
Vf = final velocity [m/s]
Vi = initial velocity = 15 [m/s]
a = acceleration = 5 [m/s^2]
t = time = 10 [s]
Note: the positive sign in the Equation indicates that the car is accelerating, i.e. its speed is increasing.
<u>Now replacing</u>
Vf = 15 + (5*10)
Vf = 65 [m/s]
Now using the second equation:
where:
x = distance traveled [m]
x = (65^2 - 15^2)/ (2*5)
x = 400 [m]
If everyone died then the too people were not on the plane they were near by when the plane crashed and were not onbard.
<u>Answer</u>
5) b-c
6) a-b and
e-f
7) f-g
9) a-b = 0 m/s
c-d = 0.6667 m/s
e-f = 0 m/s
f-g = -3 m/s
10) b-c ⇒ The cart is acceleration.
e-f ⇒ The cart is moving backwards with a constant velocity.
<u>Explanation</u>
Answer
5) b-c
In the section b-c the cart is accelerating because the slope of the graph is changing. The gradient that represent velocity is increasing.
6) a-b and e-f
At this sections the distance is not changing at all. This can only mean that the cart is not moving. It is at rest.
7) f-g
At this section the slope is negative meaning the cart is moving back to where it came from.
9) a-b = 0 m/s
At a-b the cart is not moving. So the velocity is zero.
<u> c-d = 0.66667 m/s</u>
Velocity = distance / time
=(50-40)/(40-25)
= 10/15
= 0.6667 m/s
<u> e-f = 0 m/s</u>
At e-f the cart is not moving. So the velocity is zero.
<u> f-g = -3 m/s</u>
Velocity = distance / time
= (60-30)/(65-75)
= 30/-10
= - 3 m/s
10) b-c ⇒ The cart is acceleration.
e-f ⇒ The cart is moving backwards with a constant velocity.
You would be in an editing mode when you can see a blinking insertion point on a field. In addition, in any word processing software, it would basically signify that you can already type on the document. Another name for the insertion point would be the I-beam as displayed on the screen.
