Answer:
1.) Magnitude = 5596 N
2.) Direction = 60 degrees
Explanation: You are given that the breakdown vehicle A is exerting a force of 4000 N at angle 45 degree to the vertical and breakdown vehicle B is exerting a force of 2000 N
Let us resolve the two forces into X and Y component
Sum of the forces in the X - component will be 4000 × cos 45 = 2828.43 N
Sum of the forces in the Y - component will be 2000 + ( 4000 × sin 45 )
= 2000 + 2828.43
= 4828.43 N
The resultant force R will be
R = sqrt ( X^2 + Y^2 )
Substitutes the forces at X component and Y component into the formula
R = sqrt ( 2828.43^2 + 4828.43^2 )
R = sqrt ( 31313752.53 )
R = 5595.87 N
The direction will be
Tan Ø = Y/X
Substitute Y and X into the formula
Tan Ø = 4828.43 / 2828.43
Tan Ø = 1.707106
Ø = tan^-1( 1.707106 )
Ø = 59.64 degree
Therefore, approximately, the magnitude and direction of the resultant force on the truck are 5596 N and 60 degree respectively.
The answer for this question, If I am correct, should be answer "D".
<h3>Question:</h3>
A 2.0-cm length of wire centered on the origin carries a 20-A current directed in the positive y direction. Determine the magnetic field at the point x = 5.0m on the x-axis.
<h3>
Answer:</h3>
1.6nT [in the negative z direction]
<h2>
Explanation:</h2>
The magnetic field, B, due to a distance of finite value b, is given by;
B = (μ₀IL) / (4πb
) -----------(i)
Where;
I = current on the wire
L = length of the wire
μ₀ = magnetic constant = 4π × 10⁻⁷ H/m
From the question,
I = 20A
L = 2.0cm = 0.02m
b = 5.0m
Substitute the necessary values into equation (i)
B = (4π × 10⁻⁷ x 20 x 0.02) / (4π x 5.0 
)
B = (10⁻⁷ x 20 x 0.02) / (5.0 )
B = (10⁻⁷ x 20 x 0.02) / (5.0 
)
B = (10⁻⁷ x 20 x 0.02) / (25.0)
B = 1.6 x 10⁻⁹T
B = 1.6nT
Therefore, the magnetic field at the point x = 5.0m on the x-axis is 1.6nT.
PS: Since the current is directed in the positive y direction, from the right hand rule, the magnetic field is directed in the negative z-direction.
When the metals touch
together, half the charge of the charged metal flows to the other because the
electrons all repel each other. Therefore this also means that each metal ball
contains the same amount of electrons. Each ball has 5^10 electrons, this is
equivalent to a total charge of:
Q1 = Q2 = (1.602 * 10^-19
coulombs / electron) 5^10 electrons = Q
Q = 1.564 * 10^-12 C
Now using the Coulombs
law to find for the electric force:
F = k q1 q2 / r^2 = k (Q)^2
/ r^2
where k is a contant = 9
* 10^9 N m^2 / C^2
r = the distance of the
two metals = 0.2 m
So,
F = (9 * 10^9 N m^2 /
C^2) (1.564 * 10^-12 C)^2 / (0.2 m)^2
F = 5.51 * 10^-13 N
Since the two metals
repel therefore they are the one which exerts the force hence the magnitude
must be negative:
<span>F = - 5.51 * 10^-13 N</span>
Answer:
the vertical distance between the two object will increase uniformly when they are dropped after a fixed interval of time
Explanation:
Since airplane is moving horizontally with constant speed v
so when object is dropped from the plane then the speed of the object will be same as that of the speed of the airplane
so we can say that two object when dropped after some interval of time then they always lie in same vertical line
now we know that they both have same acceleration in vertical line so the motion of two objects relative to each other in vertical direction is always uniform motion because they have no acceleration with respect to each other
So the vertical distance between the two object will increase uniformly when they are dropped after a fixed interval of time
