B. opposite charge and smaller mass
The highest point<span> of the </span>pendulums<span> swing is when the potential energy is at its </span>highest<span> and the </span>kinetic energy<span> is at its lowest.</span>
A car of mass 1535 kg collides head-on with a parked truck of mass 2000 kg. Spring mounted bumpers ensure that the collision is essentially elastic. If the velocity of the truck is 17 km/h (in the same direction as the car's initial velocity) after the collision, what was the initial speed of the car <u>20kmh</u>
<h3>What is
collision ?</h3>
A collision in physics is any situation in which two or more bodies quickly exert forces on one another. Despite the fact that the most common usage of the word "collision" refers to situations in which two or more objects clash violently, the scientific usage of the word makes no such assumptions.
The following are a few instances of physical encounters that scientists might classify as collisions:
- Legs of an insect are said to collide with a leaf when it falls on one.
- Every contact of a cat's paws with the ground while it strides across a lawn is seen as a collision, as is every brush of its fur with a blade of grass.
To learn more about collision from the given link:
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The range of the projectile is 188 m
Explanation:
The motion of the arrow in this problem is a projectile motion, so it follows a parabolic path which consists of two independent motions:
- A uniform motion (constant velocity) along the horizontal direction
- An accelerated motion with constant acceleration (acceleration of gravity) in the vertical direction
The path of a projectile is the combination of these two motions: see figure in attachment.
In order to find the horizontal range of the projectile, we just need to calculate the horizontal distance travelled.
We have:
t = 5.0 s (time of fligth of the projectile)
and the horizontal velocity is constant, and it is given by
where
is the initial velocity
is the angle of projection
Substituting,
And therefore, the range of the projectile is:
Learn more about projectile motion:
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Answer:
Work= -7.68×10⁻¹⁴J
Explanation:
Given data
q₁=q₂=1.6×10⁻¹⁹C
r₁=2.00×10⁻¹⁰m
r₂=3.00×10⁻¹⁵m
To find
Work
Solution
The work done on the charge is equal to difference in potential energy
W=ΔU
![Work=U_{1}-U_{2}\\ Work=-kq_{1}q_{2}[\frac{1}{r_{2}}-\frac{1}{r_{1}} ]\\Work=(-9*10^{9})*(1.6*10^{-19} )^{2}[\frac{1}{3.0*10^{-15} }-\frac{1}{2*10^{-10} } ]\\ Work=-7.68*10^{-14}J](https://tex.z-dn.net/?f=Work%3DU_%7B1%7D-U_%7B2%7D%5C%5C%20Work%3D-kq_%7B1%7Dq_%7B2%7D%5B%5Cfrac%7B1%7D%7Br_%7B2%7D%7D-%5Cfrac%7B1%7D%7Br_%7B1%7D%7D%20%5D%5C%5CWork%3D%28-9%2A10%5E%7B9%7D%29%2A%281.6%2A10%5E%7B-19%7D%20%29%5E%7B2%7D%5B%5Cfrac%7B1%7D%7B3.0%2A10%5E%7B-15%7D%20%7D-%5Cfrac%7B1%7D%7B2%2A10%5E%7B-10%7D%20%7D%20%5D%5C%5C%20%20Work%3D-7.68%2A10%5E%7B-14%7DJ)
