ANSWER:
A & C
Explanation:
Range depends upon two factors, horizontal velocity and air time of the projectile. Since velocity was constant throughout the investigation, the only factor that had an impact upon the range is air time. Increasing the height means that the projectile has a greater vertical distance to travel.
The doppler radar is used in technology in two ways;
<span>·
</span>Continuous Doppler radar – it has the capability
of receiving signals in means to provide output in velocity from the target
<span>·
</span>It may be use as radar gun in which police use
to detect speeding.
The material which is believed to explain the larger size of the outer planet in relation to the inner planet is gas while the inner is made of rock
Research on the solar system reveal to us that it is temperature of the the solar system that perfectly explain the materials both the outer and inner planets are made up of. The outer part is highly gaseous while the inner part is made of rocks
<h3>What is outer planet?</h3>
It refers to planet whose orbit lies outside the asteroid belt. Examples include Jupiter, Saturn, Uranus, or Neptune.
<h3>What is inner planet?</h3>
This simply means a planet whose orbit lies within the asteroid belt. They are planets such as Mercury, Venus, Earth, or Mars.
Learn more about planets:
brainly.com/question/11023671
Let l = Q/L = linear charge density. The semi-circle has a length L which is half the circumference of the circle. So w can relate the radius of the circle to L by
<span>C = 2L = 2*pi*R ---> R = L/pi </span>
<span>Now define the center of the semi-circle as the origin of coordinates and define a as the angle between R and the x-axis. </span>
<span>we can define a small charge dq as </span>
<span>dq = l*ds = l*R*da </span>
<span>So the electric field can be written as: </span>
<span>dE =kdq*(cos(a)/R^2 I_hat + sin(a)/R^2 j_hat) </span>
<span>dE = k*I*R*da*(cos(a)/R^2 I_hat + sin(a)/R^2 j_hat) </span>
<span>E = k*I*(sin(a)/R I_hat - cos(a)/R^2 j_hat) </span>
<span>E = pi*k*Q/L(sin(a)/L I_hat - cos(a)/L j_hat)</span>
Explanation:
For a charge concentrated nearly at a point, the electric field is directly proportional to the amount of charge; it is inversely proportional to the square of the distance radially away from the centre of the source charge and depends also upon the nature of the medium.
