Answer:
Work done, W = 6 J
Explanation:
It is given that,
Force of gravity acting on the book, weight of the book is 15 N
We need to find the work done in lifting the book straight up for a distance of 0.4 meters.
The weight of the book is acting in downward direction and the book is lifted straight up, it means angle between them is 180 degrees. Work done is given by :
So, the magnitude of work done in lifting the book is 6 joules.
D. Velocity because it describes a speed and direction
The applied force is different for the two cases
The case A with a greater force involves the greatest momentum change
The case A involves the greatest force.
<h3>What is collision?</h3>
- This is the head-on impact between two object moving in opposite or same direction.
The initial momentum of the two ball is the same.
P = mv
where;
- m is the mass of each
- v is the initial velocity of each ball
Since the force applied by the arm is different, the final velocity of the balls before stopping will be different.
Thus, the final momentum of each ball will be different
The impulse experienced by each ball is different since impulse is the change in momentum of the balls.
J = ΔP
The force applied by the rigid arm is greater than the force applied by the relaxed arm because the force applied by the rigid arm will cause the ball to be brought to rest faster.
Thus, we can conclude the following;
- The applied force is different for the two cases
- The case A with a greater force involves the greatest momentum change
- The case A involves the greatest force.
Learn more about impulse here: brainly.com/question/25700778
Answer:
The answer is below
Explanation:
Well, a hurricane is a storm with violent winds (more less with a force of 12 in the scale that measures winds) and it happens especially in the Caribbean sea. While a typhoon is also a strong storm but it happens especially in the Indian region and the western pacific oceans.
In conclusion, their difference is the area in which they happen.
Answer:
x = 0.775m
Explanation:
Conceptual analysis
In the attached figure we see the locations of the charges. We place the charge q₃ at a distance x from the origin. The forces F₂₃ and F₁₃ are attractive forces because the charges have an opposite sign, and these forces must be equal so that the net force on the charge q₃ is zero.
We apply Coulomb's law to calculate the electrical forces on q₃:
(Electric force of q₂ over q₃)
(Electric force of q₁ over q₃)
Known data
q₁ = 15 μC = 15*10⁻⁶ C
q₂ = 6 μC = 6*10⁻⁶ C
Problem development
F₂₃ = F₁₃
(We cancel k and q₃)
q₂(2-x)² = q₁x²
6×10⁻⁶(2-x)² = 15×10⁻⁶(x)² (We cancel 10⁻⁶)
6(2-x)² = 15(x)²
6(4-4x+x²) = 15x²
24 - 24x + 6x² = 15x²
9x² + 24x - 24 = 0
The solution of the quadratic equation is:
x₁ = 0.775m
x₂ = -3.44m
x₁ meets the conditions for the forces to cancel in q₃
x₂ does not meet the conditions because the forces would remain in the same direction and would not cancel
The negative charge q₃ must be placed on x = 0.775 so that the net force is equal to zero.
