Answer:
Yes
Explanation:
In the International System of Units (SI), energy is measured in joules. One joule is equal to the work done by a one- newton force acting over a one- metre distance
Answer:
12.32 m/s
Explanation:
Using the formula of maximum height of a projectile,
H = u²sin²Ф/2g................... Equation 1
Where H = maximum height, u = initial velocity, Ф = angle of projection, g = acceleration due to gravity
make u the subject of the equation
u = √(2Hg/sin²Ф)............ Equation 2
Given: H = 2.3 m, Ф = 33°, g = 9.8 m/s²
Substitute into equation 2
u = √[(2×2.3×9.8)/sin²33°]
u =√ [45.08/(0.545)²]
u = 45.08/0.297
u = √(151.785)
u = 12.32 m/s
Answer:
V=I×R
<em>4</em><em>.</em><em>5</em><em> </em><em>=</em><em> </em><em>I×</em><em>9</em>
<em> </em><em> </em><em> </em><em>I</em><em>.</em><em> </em><em>=</em><em> </em><em>4</em><em>.</em><em>5</em><em>/</em><em>9</em>
<em> </em><em> </em><em> </em><em>I</em><em>. </em><em>=</em><em> </em><em>0</em><em>.</em><em>5</em><em> </em><em>A</em>
<em>curre</em><em>nt</em><em> </em><em>is</em><em> </em><em>0</em><em>.</em><em>5</em><em> </em><em>A</em>
Answer:652.05 J
Explanation:
Given
Weight of lifter ![W=345 N](https://tex.z-dn.net/?f=W%3D345%20N)
vertical distance move ![h=1.89 m](https://tex.z-dn.net/?f=h%3D1.89%20m)
Work done in lifting the weight is equal change in Potential Energy of weight
Change in Potential Energy ![=m g h](https://tex.z-dn.net/?f=%3Dm%20g%20h)
![\Delta PE=345 \times 1.89=652.05 J](https://tex.z-dn.net/?f=%5CDelta%20PE%3D345%20%5Ctimes%201.89%3D652.05%20J)
therefore work done is equal to
To solve this problem we must rely on the equations of the simple harmonic movement that define the period as a function of length and gravity as
![T = 2\pi \sqrt{\frac{l}{g}}](https://tex.z-dn.net/?f=T%20%3D%202%5Cpi%20%5Csqrt%7B%5Cfrac%7Bl%7D%7Bg%7D%7D)
Where
l = Length
g = Gravity
Re-arrange to find L,
![L = g (\frac{T}{2\pi})^2](https://tex.z-dn.net/?f=L%20%3D%20g%20%28%5Cfrac%7BT%7D%7B2%5Cpi%7D%29%5E2)
Our values are given as
![g = 9.81m/s](https://tex.z-dn.net/?f=g%20%3D%209.81m%2Fs)
![T = 10.1s](https://tex.z-dn.net/?f=T%20%3D%2010.1s)
Replacing,
![L = g (\frac{T}{2\pi})^2](https://tex.z-dn.net/?f=L%20%3D%20g%20%28%5Cfrac%7BT%7D%7B2%5Cpi%7D%29%5E2)
![L = (9.81) (\frac{10.1}{2\pi})^2](https://tex.z-dn.net/?f=L%20%3D%20%289.81%29%20%28%5Cfrac%7B10.1%7D%7B2%5Cpi%7D%29%5E2)
![L = 25.348m](https://tex.z-dn.net/?f=L%20%3D%2025.348m)
Therefore the height would be 25.348m