Answer:
2.43J
Explanation:
Given parameters:
Mass of the arrow = 0.155kg
Velocity = 31.4m /s
Unknown:
Kinetic energy when it leaves the bow = ?
Solution:
The kinetic energy of a body is the energy in motion of the body;
it can be derived using the expression below:
K.E = 
m v²
m is the mass
v is the velocity
Solve for K.E;
K.E = x 0.155 x 31.4 = 2.43J
To solve this you must set up what is called a proportion. A proportion is a way of comparing two comparing values where one of the four values is missing. In your problem the missing value is the height of the smallest tree in the model.
To set up a proportion, you need all of your values. The easiest way to do this is to list them:
Highest tree in real life: 40ft
Highest tree in model: 10ft
Smallest tree in real life: 4ft
Smallest tree in model: x
So know you can set your proportion like this:
40/4 = 4/x
(When setting up a proportion, you always want to have the values belong to each other. For example don't put the height of the small tree in the model underneath the value of the highest tree in real life.)
So know to find what the x values equals, we need to cross multiply. And then all that's left after that is to solve for x.
40 times x = 4 times 4
40x = 16
x = 2.5
The smallest tree in the model should equal 2.5 feet.
Hope this helps! :)
Answer:
<u>The transformation of energy in a torch light is as follows:</u>
1) When the torch is turned ON, the chemical energy in the batteries is converted into electrical energy.
2) The electrical energy is converted into heat and light energy. (We feel the torch to be hot after some time and we can see the light energy)
Hope this helped!
<h2>~AnonymousHelper1807</h2>
Answer: a switch can do A, B and E
Explanation:
V = t^2 - 9t + 18
position, s
s = t^3 /3 - 4.5t^2 +18t + C
t = 0, s = 1 => 1=C => s = t^3/3 -4.5t^2 + 18t + 1
Average velocity: distance / time
distance: t = 8 => s = 8^3 / 3 - 4.5 (8)^2 + 18(8) + 1 = 27.67 m
Average velocity = 27.67 / 8 = 3.46 m/s
t = 5 s
v = t^2 - 9t + 18 = 5^2 - 9(5) + 18 = -2 m/s
speed = |-2| m/s = 2 m/s
Moving right
V > 0 => t^2 - 9t + 18 > 0
(t - 6)(t - 3) > 0
=> t > 6 and t > 3 => t > 6 s => Interval (6,8)
=> t < 6 and t <3 => t <3 s => interval (0,3)
Going faster and slowing dowm
acceleration, a = v' = 2t - 9
a > 0 => 2t - 9 > 0 => 2t > 9 => t > 4.5 s
Then, going faster in the interval (4.5 , 8) and slowing down in (0, 4.5)
