A run though an open field during a thunderstorm is the answer
Answer:
45.89m/s²
Explanation:
Given
Distance S = 305m
Time t = 3.64s
To get the acceleration during this run, we will apply the equation of motion:
S = ut+1/2at²
Substitute the given parameters into the formula and calculate the value of a
305 = 0+1/2 a(3.64)²
304 = 1/2(13.2496)a
304 = 6.6248a
a = 304/6.6248
a = 45.89m/s²
Hence the average acceleration during this run is 45.89m/s²
B) 7.87 m/s
The gravitational pull is the rate of change of velocity which is the acceleration. Formula for acceleration is;

Given:
• Initial velocity = 0m/s; I dropped the ball, and didn't throw it, so it was at rest firstly
• Time taken = 2.40s
• Acceleration = 3.28m/s^2
We're require to find the final velocity, at which the ball hit the ground with. Ignoring air resistance, keep in mind that the velocity of an object increases as it comes closer to the ground.


Answer:
- 0.6
Explanation:
Given that angle between normal y axis is 62° so angle between normal
and x axis will be 90- 62 = 28 °. Since incident ray is along x axis , 28 ° will be the angle between incident ray and normal ie it will be angle of incidence
Angle of incidence = 28 °
angle of reflection = 28°
Angle between incident ray and reflected ray = 28 + 28 = 56 °
Angle between x axis and reflected ray = 56 °
x component of reflected ray
= - cos 56 ( it will be towards - ve x axis. )
- 0.6
<em>The answer is </em>Ninth <em>and </em>Tenth <em>grade so the answer would be</em> B
<em>I hope this helps you </em>