For the answer to the question above, first find out the gradient.
<span>m = rise/run </span>
<span>=(y2-y1)/(x2-x1) </span>
<span>the x's and y's are the points given: "After three hours, the velocity of the car is 53 km/h. After six hours, the velocity of the car is 62 km/h" </span>
<span>(x1,y1) = (3,53) </span>
<span>(x2,y2) = (6,62) </span>
<span>sub values back into the equation </span>
<span>m = (62-53)/(6-3) </span>
<span>m = 9/3 </span>
<span>m = 3 </span>
<span>now we use a point-slope form to find the the standard form </span>
<span>y-y1 = m(x-x1) </span>
<span>where x1 and y1 are any set of point given </span>
<span>y-53 = 3(x-3) </span>
<span>y-53 = 3x - 9 </span>
<span>y = 3x - 9 + 53 </span>
<span>y = 3x + 44 </span>
<span>y is the velocity of the car, x is the time.
</span>I hope this helps.
We first calculate the acceleration on the ball using:
2as = v² - u²; u = 0 because ball is initially at rest
a = (36)²/(2 x 0.35)
a = 1850 m/s²
F = ma
F = 0.058 x 1850
= 107.3 Newtons
Answer:
The torque on the loop is 
Nm
Explanation:
Given:
Current 
A
Magnetic field 
T
Area of loop 
Angle between magnetic field and area vector 
21°
Form the formula of torque in case of magnetic field,
г 
Where 
magnetic moment
г 
г 
г 
Nm
Therefore, the torque on the loop is Nm
Answer:
18 Ω
Explanation:
As K and F are at the same voltage, we can redraw the diagram as in figure 2
Series resistances add directly, so we get figure 3
Adding parallel resistances gets us to figure 4
Now we can move two 6Ω resistances for clarification in figure 5
As the voltage between C and J will be identically split between D and H, there will be no voltage drop across the middle 6Ω resister and no current through it, identical to an infinite resistance, so that 6Ω can be eliminated as in figure 6
Add series resistances to get to figure 7
Add parallel resistances to get to figure 8
Add series resistances to get to figure 9
<span>The electric force is given by:
F = [ k*(q1)*(q2) ] / d^2
F = Electric force
k = Coulomb's constant
q1 = Charge of one proton
q2 = Charge of second proton
d = Distance between centers of mass
Values:
F = unknown
k = 8.98E 9 N-m^2/C^2
q1 = 1.6E-19
q2 = 1.6E-19
d = 1.0E-15 m
Insert values into F = [ k*(q1)*(q2) ] / d^2
F = [ (8.98E 9 N-m^2/C^2) * (1.6E-19) * (1.6E-19) ] / (1.0E-15 m)^2
F = </span>229.888 N
answer
the electric force of repulsion between nuclear protons is 229.888 N
