The position-time graphs show the relationship between the position of an object (shown on the y-axis) and the time (shown on the x-axis) to show velocity.
<h3>What is velocity?</h3>
Velocity is a vector quantity that tells the distance an object has traveled over a period of time.
Displacement is a vector quality showing total length of an area traveled by a particular object.
Imagine a time-position graph where the velocity of an object is constant. What will be observed on the graph concerning the slope of the line segment as well as the velocity of the object?
The slope of the line is equal to zero and the object will be stationary.
The position-time graphs show the relationship between the position of an object (shown on the y-axis) and the time (shown on the x-axis) to show velocity.
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Answer:
2.2 x 10-19
Explanation:
Kinetic Energy = 1/2 m v ^2
Answer:
4.42 x 10⁷ W/m²
Explanation:
A = energy absorbed = 500 J
η = efficiency = 0.90
E = Total energy
Total energy is given as
E = A/η
E = 500/0.90
E = 555.55 J
t = time = 4.00 s
Power of the beam is given as
P = E /t
P = 555.55/4.00
P = 138.88 Watt
d = diameter of the circular spot = 2.00 mm = 2 x 10⁻³ m
Area of the circular spot is given as
A = (0.25) πd²
A = (0.25) (3.14) (2 x 10⁻³)²
A = 3.14 x 10⁻⁶ m²
Intensity of the beam is given as
I = P /A
I = 138.88 / (3.14 x 10⁻⁶)
I = 4.42 x 10⁷ W/m²
Answer:
E = 31.329 N/C.
Explanation:
The differential electric field 
at the center of curvature of the arc is
<em>(we have a cosine because vertical components cancel, leaving only horizontal cosine components of E. )</em>
where 
is the radius of curvature.
Now
,
where 
is the charge per unit length, and it has the value
Thus, the electric field at the center of the curvature of the arc is:
Now, we find 
and 
. To do this we ask ourselves what fraction is the arc length 3.0 of the circumference of the circle:
and this is
radians.
Therefore,
evaluating the integral, and putting in the numerical values we get:
Answer:
The required angle is (90-25)° = 65°
Explanation:
The given motion is an example of projectile motion.
Let 'v' be the initial velocity and '∅' be the angle of projection.
Let 't' be the time taken for complete motion.
Let 'g' be the acceleration due to gravity
Taking components of velocity in horizontal(x) and vertical(y) direction.
= v cos(∅)
= v sin(∅)
We know that for a projectile motion,
t =
Since there is no force acting on the golf ball in horizonal direction.
Total distance(d) covered in horizontal direction is -
d = ×t = vcos(∅)× = 
.
If the golf ball has to travel the same distance 'd' for same initital velocity v = 23m/s , then the above equation should have 2 solutions of initial angle 'α' and 'β' such that -
α +β = 90° as-
d = 
= ![\frac{v^{2}sin(2[90-β]) }{g}](https://tex.z-dn.net/?f=%5Cfrac%7Bv%5E%7B2%7Dsin%282%5B90-%CE%B2%5D%29%20%7D%7Bg%7D)
=
= 
.
∴ For the initial angles 'α' or 'β' , total horizontal distance 'd' travelled remains the same.
∴ If α = 25° , then
β = 90-25 = 65°
∴ The required angle is 65°.
