Answer:
1) Q ’= 8 Q
, 2) q ’= 16 q
, 3) r ’= ¾ r
Explanation:
For this exercise we will use Coulomb's law
F = k q Q / r²
It asks us to calculate the change of any of the parameters so that the force is always F
Original values
q, Q, r
Scenario 1
q ’= 2q
r ’= 4r
F = k q ’Q’ / r’²
we substitute
F = k 2q Q ’/ (4r)²
F = k 2q Q '/ 16r²
we substitute the value of F
k q Q / r² = k q Q '/ 8r²
Q ’= 8 Q
Scenario 2
Q ’= Q
r ’= 4r
we substitute
F = k q ’Q / 16r²
k q Q / r² = k q’ Q / 16 r²
q ’= 16 q
Scenario 3
q ’= 3/2 q
Q ’= ⅜ Q
we substitute
k q Q r² = k (3/2 q) (⅜ Q) / r’²
r’² = 9/16 r²
r ’= ¾ r
Mechanical energy is the energy that is possessed by an object due to its motion or due to its position. It can either be kinetics or potential. In this problem you know it starting position so you can calculate it's potential energy (PE):
<span>PE=mass∗gravity∗height=0.3kg∗9.8m/s2∗1.8m=?
</span>The answer will typically be given in joules:
1J=kg∗m2s2 Could be wrong... But I believe it is 5.3...? as a final product.
Answer:
The y-axis should be labelled as W in Newtons (kg·m/s²)
Explanation:
The given data is presented here as follows;
Mass (kg) 
Newtons (kg·m/s²)
3.2 31.381
4.6 45.1111
6.1 59.821
7.4 72.569
9 89.241
10.4 101.989
10.9 106.892
From the table, it can be seen that there is a nearly linear relationship between the amount of Newtons and the mass, as the slope of the data has a relatively constant slope
Therefore, the data can be said to be a function of Weight in Newtons to the mass in kilograms such that the weight depends on the mass as follows;
W(m) in Newtons = Mass, m in kg × g
Where;
g is the constant of proportionality
Therefore, the y-axis component which is the dependent variable is the function, W(m) = Weight of the body while the x-axis component which is the independent variable is the mass. m
The graph of the data is created with Microsoft Excel give the slope which is the constant of proportionality, g = 9.8379, which is the acceleration due to gravity g ≈ 9.8 m/s²
We therefore label the y-axis as W in Newtons (kg·m/s²)
Answer:
The minimum possible coefficient of static friction between the tires and the ground is 0.64.
Explanation:
if the μ is the coefficient of static friction and R is radius of the curve and v is the speed of the car then, one thing we know is that along the curve, the frictional force, f will be equal to the centripedal force, Fc and this relation is :
Fc = f
m×(v^2)/(R) = μ×m×g
(v^2)/(R) = g×μ
μ = (v^2)/(R×g)
= ((25)^2)/((100)×(9.8))
= 0.64
Therefore, the minimum possible coefficient of static friction between the tires and the ground is 0.64.
Making a wire thicker has the same effect as making a road wider. It makes it easier for the electron traffic to flow. The resistance decreases, and the current (traffic) increases.
