Answer:
a.After second Mr Comer's speed
b.Distance travelled by Mr.Comer in seconds
Explanation:
a. Lets recall our first equation of motion
Now we know that ,
and
Plugging the values we have.
Then Mr.Comer's speed after sec
b.
Lets find the distance and recall our third equation of motion.
So distance covered.
Dividing both sides with 2a we have.
Plugging the values.
So Mr.Comer will travel a distance of .
Answer:
The correct option is (b).
Explanation:
We need to find the work done to increase the speed of a 1 kg toy car by 5 m/s.
We know that, the work done is equal to the kinetic energy of an object i.e.
So, 12.5 J of work is done to increase the speed of a 1.0 kg toy car by 5.0 m/s.
Answer:
A. Fnx = 5.71*10⁻⁵ N , Fny= -3.67*10⁻⁵ N
B. Fn= 6.78 *10⁻⁵ N
C. α= 32.4° counterclockwise with the positive x+ axis
Explanation:
Because the particle q₃ is close to two other electrically charged particles, it will experience two electrical forces and the solution of the problem is of a vector nature.
Equivalences
1nC= 10⁻⁹C
1cm = 10⁻²m
Known data
k= 9*10⁹N*m²/C²
q₁= -2.65 nC =-2.65*10⁻⁹C
q₂= +2.00 nC = 2*10⁻⁹C
q₃= +5.00 nC= =+5*10⁻⁹C
* 10⁻²m = 4.9678* 10⁻²m
(d₁₃)² = 24.68*10⁻⁴m²
d₂₃ = 3.2 cm = 3.2*10⁻²m
Graphic attached
The directions of the individual forces exerted by q₁ and q₂ on q₃ are shown in the attached figure.
The force (F₂₃) of q₂ on q₃ is repulsive because the charges have equal signs and the forces.
The force (F₁₃) of q₁ on q₃ is attractive because the charges have opposite signs.
Magnitudes of F₁₃ and F₂₃
F₁₃ = (k*q₁*q₃)/(d₁₃)²=( 9*10⁹*2.65*10⁻⁹*5*10⁻⁹) /(24.68*10⁻⁴)
F₁₃ = 4.8 *10⁻⁵ N
F₂₃ = (k*q₂*q₃)/(d₂₃)² = ( 9*10⁹*2*10⁻⁹*5*10⁻⁹) /((3.2)²*10⁻⁴)
F₂₃ = 8.8 *10⁻⁵ N
x-y components of F₁₃ and F₂₃
F₁₃x= -4.8 *10⁻⁵ *cos β= - 4.8 *10⁻⁵(3.2/ (4.9678)= - 3.09*10⁻⁵ N
F₁₃y= -4.8 *10⁻⁵ *sin β= - 4.8 *10⁻⁵(3.8/(4.9678) = - 3.67*10⁻⁵ N
F₂₃x = F₂₃ = +8.8 *10⁻⁵ N
F₂₃y = 0
x and y components of the total force exerted on q₃ by q₁ and q₂ (Fn)
Fnx= F₁₃x+F₂₃x = - 3.09*10⁻⁵ N+8.8 *10⁻⁵ N= 5.71*10⁻⁵ N
Fny= F₁₃y+F₂₃y = - 3.67*10⁻⁵ N+0= - 3.67*10⁻⁵ N
Fn magnitude
*10⁻⁵ N
Fn= 6.78 *10⁻⁵ N
Fn direction (α)
α= -32.4°
α= 32.4° counterclockwise with the positive x+ axis
