The ball X has a known position at 3 different times, each 1 second apart. The positions of the ball are (1.8,2.2)m, (4.4,4.8)m,
1 answer:
Answer:
Explanation:
From the position coordinates given , it appears that the ball moves simultaneously along x and y direction.
Displacement along x direction in one second = 4.4 - 1.8 = 2.6 m
So velocity along x direction V_x =
Similarly velocity along y direction V_y(1) =
In the next phase velocity changes both in x and y direction.
velocity in x - direction V_x(2) = [tex]\frac{2}{s}[/tex
Velocity in Y- direction V_y(2) = [tex]\frac{3.1}{s}[/tex
Acceleration in x direction = change of velocity in x direction
= ( 2 - 2.6 ) = -.6 m s⁻²
Acceleration in y direction = ( 3.1 - 2.6 ) = 0.5 m s⁻²
Total acceleration =\sqrt{( -.6 )² + ( .5 )²}
= .78 ms⁻²
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Answer:
1250 N
Explanation:
length of wrench = 20 cm = 0.2 m
Write the length of wrench in vector form
r = 0.2 j metre
Direction of force=
Let the magnitude of force is F
So, write the force in vector form
F = F
torque, T = 150 Nm
Torque = T = r x F
F = 1250 N
Thus, the magnitude of force is 1250 N.
Answer:
(a) 7.315 x 10^(-14) N
(b) - 7.315 x 10^(-14) N
Explanation:
As you referred at the final remark, the electron and proton undergo a magnetic force of same magnitude but opposite direction. Using the definition of magnetic force, a cross product must be done. One technique is either calculate the magnitude of the velocity and magnetic field and multiplying by sin (90°), but it is necessary to assure both vectors are perpendicular between each other ( which is not the case) or do directly the cross product dealing with a determinant (which is the most convenient approach), thus,
(a) The electron has a velocity defined as:
In respect to the magnetic field;
The magnetic force can be written as;
Bear in mind
thus,
Note: The cross product is operated as a determinant. Likewise, the product of the unit vector k is squared and that is operated as dot product whose value is equal to one, i.e,
(b) Considering the proton charge has the same magnitude as electron does, but the sign is positive, thus
Note: The cross product is operated as a determinant. Likewise, the product of the unit vector k is squared and that is operated as dot product whose value is equal to one, i.e,
Final remarks: The cross product was performed in R3 due to the geometrical conditions of the problem.