B. only its velocity should change
Answer:
D. Dylan is incorrect because a 90-degree launch angle results in the largest vertical range
Explanation:
Projectile is the motion of an object thrown into space. When an object is thrown into space, the only force which acts on it is the acceleration due to gravity.
An object thrown into space would reach maximum height (vertical range) if it is launched at an angle of 90 degrees. For maximum horizontal range, the object needs to be launched at an angle of 45 degrees.
Therefore Dylan is incorrect because a 90-degree launch angle results in the largest vertical range
Answer:
35 mph
Explanation:
The key of this problem lies in understanding the way that projectile motion works as we are told to neglect the height of the javelin thrower and wind resistance.
When the javelin is thown, its velocity will have two components: a x component and a y component. The only acceleration that will interact with the javelin after it was thown will be the gravety, which has a -y direction. This means that the x component of the velocity will remain constant, and only the y component will be affected, and can be described with the constant acceleration motion properties.
When an object that moves in constant acceleration motion, the time neccesary for it to desaccelerate from a velocity v to 0, will be the same to accelerate the object from 0 to v. And the distance that the object will travel in both desaceleration and acceleration will be exactly the same.
So, when the javelin its thrown, it willgo up until its velocity in the y component reaches 0. Then it will go down, and it will reach reach the ground in the same amount of time it took to go up and, therefore, with the same velocity.
Force on the particle is defined as the application of the force field of one particle on another particle. The magnitude and direction of the electrical force will be 4.05×10⁴N towards the north.
<h3>What is electrical force?</h3>
Force on the particle is defined as the application of the force field of one particle on another particle. It is a type of virtual force.
The given data in the problem is
q₁ is the negative charge = 6 µC=6×10⁻⁶ C
q₂ is the positive charge = 3 µC=3×10⁻⁶ C
r is the distance between the charges=0.002 m
is the electric force =?
The value of electric force will be;
![\rm F_E= \frac{Kq_1q_2}{r^2} \\\\ F_E= \frac{9\times 10^9\times 6\times 10^{-6}\times3\times10^{-6}}{(0.002)^2}\\\\ \rm F_E=4.05\times10^4\;N](https://tex.z-dn.net/?f=%5Crm%20F_E%3D%20%5Cfrac%7BKq_1q_2%7D%7Br%5E2%7D%20%5C%5C%5C%5C%20F_E%3D%20%5Cfrac%7B9%5Ctimes%2010%5E9%5Ctimes%206%5Ctimes%2010%5E%7B-6%7D%5Ctimes3%5Ctimes10%5E%7B-6%7D%7D%7B%280.002%29%5E2%7D%5C%5C%5C%5C%20%5Crm%20F_E%3D4.05%5Ctimes10%5E4%5C%3BN)
Hence the magnitude and direction of the electrical force will be 4.05×10⁴N towards the north.
To learn more about the electrical force refer to the link;
brainly.com/question/1076352