Here in crash test the two forces are acting on the dummy in two different directions
As we know that force is a vector quantity so we need to use vector addition laws in order to find the resultant force on it.
So here two forces are given in perpendicular direction with each other so as per vector addition law we need to use Pythagoras theorem to find the resultant of two vectors
so we can say
here given that
now we will plug in all data in the above equation
so it will have net force 4501.9 N which will be reported by sensor
Answer:
energy that is stagnant and cannot be changed
Answer:
221.17 kJ
Explanation: Note the heat of vaporization is in kJ/mol,then to determine the number of moles of water: divide the mass by 18. Then multiply the number of moles by the molar heat of vaporization of water.
N = 97.6 ÷ 18
Q=molar heat *moles
Q = (40.79) * (97.6 ÷ 18)
This is approximately 221.17 kJ
Answer:
t₁ = 3 s
Explanation:
In this exercise, the vertical displacement equation is not given
y = 240 t + 16 t²
Where y is the displacement, 240 is the initial velocity and 16 is half the value of the acceleration
Let's replace
864 = 240 t + 16 t²
Let's solve the second degree equation
16 t² + 240 t - 864 = 0
Let's divide by 16
t² + 15 t - 54 = 0
The solution of this equation is
t = [-15 ± √(15 2 - 4 1 (-54)) ] / 2 1
t = [-15 ±√(225 +216)] / 2
t = [-15 + - 21] / 2
We have two solutions.
t₁ = [-15 +21] / 2
t₁ = 3 s
t₂ = -18 s
Since time cannot have negative values, the correct t₁ = 3s
<h3>Answer;</h3>
<u>It would make the lens stronger. </u>
<h3>Explanation;</h3>
- The focal length is the distance between the optical center or the center of the lens to the focal point of a convex or concave lens.
- The power of the convex lens is lens ability to undertake refraction or bend light. It is given as the reciprocal of focal length.
- Power of the lens = 1/ f; therefore the smaller the focal length the higher the power and the larger the focal length the lower the power.
- Thus; decreasing the focal length of a convex lens makes the lens stronger.
