Hello!
The answer is:
Luke did a work of 308N.m or 308 Joules.
<h2>
Why?</h2>
When a force is applied on a object making it to move covering a distance we call it "work". The movement caused by the force, will follow the same direction that the force.
We can calculate the work done by using the following equation:
![Work=Force*distance*Cos(\alpha)](https://tex.z-dn.net/?f=Work%3DForce%2Adistance%2ACos%28%5Calpha%29)
Where,
Work is the transferred energy.
Force, is the force applied to the object.
Distance, is the distance covered due to the applied force.
α, is the angle at the work is done.
We are given the following information to calculate the work done:
![Force=22N\\Distance=14m](https://tex.z-dn.net/?f=Force%3D22N%5C%5CDistance%3D14m)
So, substituting the given information into the equation to calculate work, we have:
![Work=Force*distance\\Work=22N*14m=308N.m](https://tex.z-dn.net/?f=Work%3DForce%2Adistance%5C%5CWork%3D22N%2A14m%3D308N.m)
Hence, we have that Luke did a work of 308N.m or 308 Jouls.
Have a nice day!
Answer:
a₁₅₇ = 1,088
Step-by-step explanation:
aₙ = a₁ + d(n-1); where 'd' if common difference
a₁₅₇ = -4 + 7(156)
= -4 + 1092
= 1,088
Convert percentages to units.
100% - 20% = 80% = 0.8
100% - 25% = 75% = 0.75
Then multiply those together:
0.8*0.75=0.6=60%
100% - 60% = 40%. That's the overall price percentage reduction.
P.S. You can convert percentages to units by diving % by 100 for example 100% divided by 100 is equal to 1.
And vice versa aka multiply units by 100 and you get %, for example 0.8 times 100 is 80%
d
Step-by-step explanation:
Answer:
![A = \frac{h\sqrt{t^{2} - h^{2}}}{2}](https://tex.z-dn.net/?f=A%20%3D%20%5Cfrac%7Bh%5Csqrt%7Bt%5E%7B2%7D%20-%20h%5E%7B2%7D%7D%7D%7B2%7D)
Step-by-step explanation:
Right triangle:
height h, width w, hypotenuse h
Area is: ![A = \frac{h*w}{2}](https://tex.z-dn.net/?f=A%20%3D%20%5Cfrac%7Bh%2Aw%7D%7B2%7D)
Pytagoras theorem: ![h^{2} + w^{2} = t^{2}](https://tex.z-dn.net/?f=h%5E%7B2%7D%20%2B%20w%5E%7B2%7D%20%3D%20t%5E%7B2%7D)
In this question:
To find the area, in terms of h and t, we have to write w as a function of h and t. So
![h^{2} + w^{2} = t^{2}](https://tex.z-dn.net/?f=h%5E%7B2%7D%20%2B%20w%5E%7B2%7D%20%3D%20t%5E%7B2%7D)
![w^{2} = t^{2} - h^{2}](https://tex.z-dn.net/?f=w%5E%7B2%7D%20%3D%20t%5E%7B2%7D%20-%20h%5E%7B2%7D)
![w = \sqrt{t^{2} - h^{2}}](https://tex.z-dn.net/?f=w%20%3D%20%5Csqrt%7Bt%5E%7B2%7D%20-%20h%5E%7B2%7D%7D)
So the area is:
![A = \frac{h*w}{2}](https://tex.z-dn.net/?f=A%20%3D%20%5Cfrac%7Bh%2Aw%7D%7B2%7D)
![A = \frac{h\sqrt{t^{2} - h^{2}}}{2}](https://tex.z-dn.net/?f=A%20%3D%20%5Cfrac%7Bh%5Csqrt%7Bt%5E%7B2%7D%20-%20h%5E%7B2%7D%7D%7D%7B2%7D)