Answer:
earth
sorry I need points but I don’t know the answer again I’m really sorry
Answer:
The moment arm is 0.6 m
Explanation:
Given that,
First force 
Second force 
Distance r = 0.2 m
We need to calculate the moment arm
Using formula of torque

So, Here,

We know that,
The torque is the product of the force and distance.
Put the value of torque in the equation


Where,
=First force
=First force
=Second force
= distance
Put the value into the formula


Hence, The moment arm is 0.6 m
Answer: 0.258 N
Explanation:
As the density of the object is much less than the density of water, it’s clear that the buoyant force, is greater than the weight of the object, which means that in normal conditions, it would float in water.
So, in order to get the ball submerged in water, we need to add a downward force, that add to the weight, in order to compensate the buoyant force, as follows:
F = Fb – Fg
Fb= δH20* 4/3*π*(d/2)³ * g
Fg = δb* 4/3*π*(d/2)³ *g
F= (δH20- δb) * 4/3*π*(d/2)³*g
Replacing by the values of the densities, and the ball diameter, we finally get:
F= 0.258 N
Answer:
a) > x<-c(1,2,3,4,5)
> y<-c(1.9,3.5,3.7,5.1,6)
> linearmodel<-lm(y~x)
And the output is given by:
> linearmodel
Call:
lm(formula = y ~ x)
Coefficients:
(Intercept) x
1.10 0.98
b) 
And if we compare this with the general model 
We see that the slope is m= 0.98 and the intercept b = 1.10
Explanation:
Part a
For this case we have the following data:
x: 1,2,3,4,5
y: 1.9,3.5,3.7,5.1, 6
For this case we can use the following R code:
> x<-c(1,2,3,4,5)
> y<-c(1.9,3.5,3.7,5.1,6)
> linearmodel<-lm(y~x)
And the output is given by:
> linearmodel
Call:
lm(formula = y ~ x)
Coefficients:
(Intercept) x
1.10 0.98
Part b
For this case we have the following trend equation given:

And if we compare this with the general model 
We see that the slope is m= 0.98 and the intercept b = 1.10