Define
g = 9.8 m/s², acceleration due to gravity, positive downward.
Assume that wind resistance may be neglected.
Frog A:
u = 0.551 m/s, launch velocity, upward.
When the frog lands back on the pad, its vertical position is zero, and its vertical velocity will be 0.551 m/s downward.
If the time of flight is t, then
(0.551 m/s)*(t s) - 0.5*(9.8 m/s²)*(t s)² = 0
0.551t - 4.9t² = 0
t = 0, or t = 0.1124 s
t = 0 corresponds to launch, and t = 0.1124 s corresponds to landing.
Frog B:
Launch velocity is 1.75 m/s
When t = 0.1124 s, the position of the frog is
s = (1.75 m/s)(0.1124 s) - 0.5*(9.8 m/s²)*(0.1124 s)²
= 0.135 m
The velocity of frog B is
v = (1.75 m/s) - (9.8 m/s²)*(0.1124 s)
= 0.6485 m/s
Answer:
When frog A lands on the ground,
Frog B is 0.135 m above ground and its velocity is 0.649 m/s upward.
<h3><u>Answer;</u></h3>
= 3.73 seconds
<h3>
<u>Explanation;</u></h3>
Using the equation;
v = u + at
where v is the final velocity, u is the initial velocity and a is the acceleration due to gravity (-g) and t is the time taken.
Therefore;
v = u - gt
Thus;
v = 24.5 m/s, u = 12.1 m/s, a = -g = 9.8 m/s²
-24.5 m/s = 12.1 - 9.8 × t
36.6 = 9.8 t
t = 36.6/9.8
= 3.73 seconds
Answer:
Tension on each chain: 30.28N
Explanation:
We start by drawing a free-body diagram of all the forces acting on the flower pot. These are:
1) the weight of the flower pot which is the product of the pot's mass time the acceleration of gravity "g" (pictured in green and represented by in the diagram), and
2) the two tensions in the chains, which since they are symmetric, are of the same size, pictured in red color and represented by the letter "T" in the diagram.
We also decompose the tensions T in two components each (a vertical component V and a horizontal component h) which are pictured in blue.
All vertical forces must cancel out since the object is in static equilibrium, so:
The horizontal components h are equal, and since they act in opposite directions, they indeed cancel out.
Now to calculate the tension, we use the fact that we know the angle between T and the component V. They in fact are related via the cosine of the angle because they are the hypotenuse and the adjacent side in a right angle triangle:
The Iroquois depended on the natural resources around them to meet all of their basic needs. Because they lived in the Eastern Woodlands of North America, their food, clothing, and shelter, as well as the materials for making their tools and weapons.Hunting tools were a bow and arrow, tomahawks and clubs. They ate corn, moose, beans and deer and many other foods. Beaver skin was used for capes.
They used reeds to make baskets and they used animal hides as clothing.
The Iroquois made weapons out of bones, wood, stones, and feathers. The Iroquois used wood, stone and feathers to make arrows.Hunting tools were a bow and arrow, tomahawks and clubs.
In order to find the our own velocity with respect to land,we need to apply the theory of relative velocity.
Now consider the velocity of the ship traveling towards the north with respect to land as A.Consider our own velocity headed northwards as B.
The relative velocity is the velocity that the body A would appear to an observer on the body B and vice versa.
In this case the relative velocity would be arrived by summing up our velocity with the velocity of the ship as the object (I) is travelling in the ship.
Relative velocity = Velocity of Body A+ Velocity of Body B.
Velocity of the ship traveling towards the north with respect to land(A)= 13.0m/s. (Given)
Our own velocity headed northwards(B)= 2.8 m/s.
Relative velocity = Velocity of Body A+ Velocity of Body B.
Relative velocity= 13.0 + 2.8 = 15.8m/s.
Thus our own velocity with respect to the land is 15.8 m/s.