Answer:
Explanation:
The magnitude of the electrical force between the two point charges is
where
k is the Coulomb's constant
is the magnitude of each charge
r = 3.00 m is the separation between the two charges
Substituting the numbers into the formula, we find
imagine a sphere of radius 7x10^12m centered on the star; the surface area of this sphere is 4 pi r^2
thus, the total output of the star is distributed evenly over this sphere, so we have
15.4W/m^2 = L/(4 pi r^2) where L is the total output (luminosity)
L = 15.4 W/m^2 x 4 pi (7x10^12m)^2 = 9.48 x10^27W
<h2>
Answer: 22.05 N</h2>
Explanation:
Let's begin with the cup's free body diagram <u>(figure attached)</u>, there we are able to see the forces acting on this object:
On the X axis:
(1)
Where:
is the force of the wind
is the opposing force (that is why it has the <u>negative sign</u>)
So:
(2)
In addition we know by<u> Newton's 2nd law</u>:
(3)
Where:
is the mass of the cup
is its acceleration
Substituting (2) in (3):
(4)
Finding :
(5)
On the other hand we have the forces acting on the Y axis:
(6)
Where:
is the normal force
is the weight of the cup, being the acceleration due gravity
This means:
(7)
Substituting (5) in (7):
(8)
Finally:
This is the weight of the cup
Answer:
a. Effort = 960 Newton
b. Mechanical advantage (M.A) = 0.625
c. Velocity ratio (V.R) = 1.67
Explanation:
<u>Given the following data;</u>
- Load = 600 N
- Length of crowbar = 200 cm
- Length of load arm = 0.75 m
<u>Conversion:</u>
100 cm = 1 m
X cm = 0.75 m
Cross-multiplying, we have;
X = 0.75 * 100 = 75 cm
First of all, we would find the effort arm;
Effort arm = length of crow bar - length of load arm
Effort arm = 200 - 75
Effort arm = 125 cm
Next, we would determine the mechanical advantage (M.A) of the crow bar;
Substituting the values into the formula, we have;
<em>M.A = 0.625</em>
To find the effort of the crow bar;
Making "effort" the subject of formula, we have;
<em>Effort = 960 Newton</em>
Lastly, we would determine the velocity ratio (V.R);
<em>V.R = 1.67</em>
If you're talking about capacitor connected in series than the charge on each of the capacitor would be same since charge induced on the left plate of capacitor I.e +Q would then induce charge on the right plate of the capacitor as -Q (equal in magnitude but opposite charge) then again the other capacitor next to it would generate +Q charge and so on.