<h2>
The gravitational acceleration on planet X is 32.7 m/s²</h2>
Explanation:
The acceleration due to gravity is given by
![g=\frac{GM}{R^2}](https://tex.z-dn.net/?f=g%3D%5Cfrac%7BGM%7D%7BR%5E2%7D)
where G is gravitational constant, M is mass and R is radius.
For earth we have
g = 9.81 m/s²
That is
![\frac{GM_e}{R_e^2}=9.81m/s^2](https://tex.z-dn.net/?f=%5Cfrac%7BGM_e%7D%7BR_e%5E2%7D%3D9.81m%2Fs%5E2)
For planet X we have
![R=3R_e\\\\M=30M_e](https://tex.z-dn.net/?f=R%3D3R_e%5C%5C%5C%5CM%3D30M_e)
Substituting
![g=\frac{GM}{R^2}=\frac{G\times 30M_e}{(3R_e)^2}=\frac{30}{9}\times 9.81\\\\g=32.7m/s^2](https://tex.z-dn.net/?f=g%3D%5Cfrac%7BGM%7D%7BR%5E2%7D%3D%5Cfrac%7BG%5Ctimes%2030M_e%7D%7B%283R_e%29%5E2%7D%3D%5Cfrac%7B30%7D%7B9%7D%5Ctimes%209.81%5C%5C%5C%5Cg%3D32.7m%2Fs%5E2)
The gravitational acceleration on planet X is 32.7 m/s²
The rotational effect of a force is called torque.it is the cause of rotation or angular deceleration
<span>τ=rXF
where
</span>τ = F r sin @
hope it helps
Answer:
accleraion normal ski 0.0735 m/s², new ski 0.219 m/s²
Explanation:
This exercise should work with the one-dimensional kinetic equations, specifically with the equation
x = v₀ t + ½ a t²
When the skier is at the exit they are at rest, so their initial speed is zero
(v₀ = 0)
x = ½ a t²
Let's calculate the acceleration for normal skiing
a₁ = 2 x / t²
a₁ = 2 175/69²
a₁ = 0.0735 m/s²
a₁ = 7.35 10⁻² m/s²
Let's calculate the acceleration with the new plastic ski
a₂ = 2 x / t₂²
a₂ = 2 175/40²
a₂ = 0.219 m/s²
a₂= 2.19 10⁻¹ m/s²
Answer:
a.3.20m
b.0.45cm
Explanation:
a. Equation for minima is defined as: ![sin \theta=\frac{m\lambda}{\alpha}](https://tex.z-dn.net/?f=sin%20%5Ctheta%3D%5Cfrac%7Bm%5Clambda%7D%7B%5Calpha%7D)
Given
,
and
:
#Substitute our variable values in the minima equation to obtain
:
![\theta=sin^-^1 (\frac{3\times 6.33\times 10^-^7}{0.00015})\\\\\theta=0.01266rad](https://tex.z-dn.net/?f=%5Ctheta%3Dsin%5E-%5E1%20%28%5Cfrac%7B3%5Ctimes%206.33%5Ctimes%2010%5E-%5E7%7D%7B0.00015%7D%29%5C%5C%5C%5C%5Ctheta%3D0.01266rad)
#draw a triangle to find the relationship between
and
.
#where ![y=4.05cm](https://tex.z-dn.net/?f=y%3D4.05cm)
![L=y/tan(\theta)=3.20](https://tex.z-dn.net/?f=L%3Dy%2Ftan%28%5Ctheta%29%3D3.20)
Hence the screen is 3.20m from the split.
b. To find the closest minima for green(the fourth min will give you the smallest distance)
#Like with a above, the minima equation will be defined as:
, where
given that it's the minima with the smallest distance.
![sin \theta=\frac{4\lambda}{\alpha}\\\theta=sin^-^1 (\frac{4\times 6.33\times 10^-^7}{0.00015})\\\\\theta=0.01688rad](https://tex.z-dn.net/?f=sin%20%5Ctheta%3D%5Cfrac%7B4%5Clambda%7D%7B%5Calpha%7D%5C%5C%5Ctheta%3Dsin%5E-%5E1%20%28%5Cfrac%7B4%5Ctimes%206.33%5Ctimes%2010%5E-%5E7%7D%7B0.00015%7D%29%5C%5C%5C%5C%5Ctheta%3D0.01688rad)
#we then use
to calculate
=4.5cm
Then from the equation subtract
from
:
![4.50cm-4.05cm=0.45cm](https://tex.z-dn.net/?f=4.50cm-4.05cm%3D0.45cm)
Hence, the distance
is 0.45cm