<span>A transverse wave is a moving wave that consists of oscillations occurring perpendicular (or right angled) to the direction of energy transfer. If a transverse wave is moving in the positive x-direction, its oscillations are in up and down directions that lie in the y–z plane. Light is an example of a transverse wave.</span>
Answer:
Force = 18N
Explanation:
Force = [ mass ( final velocity - initial velocity ) ] / time taken
using the formula, here mass is 3 kg, final velocity = 45 m/s , initial velocity = 45 m/s , time taken = 2 seconds
Force = [ 3 ( 45 - 33 ) ] / 2
Force = 18N
Answer:
Explanation:
Answer:
4.53 kg
Explanation:
spring constant, K = 16 N/m
Amplitude, A = 20 cm = 0.2 m
Maximum speed, v = 37.6 cm/s = 0.376 m/s
The formula for maximum speed is given by
v = ωA
where, ω is the angular frequency
0.376 = ω x 0.2
ω = 1.88 rad/s



m = 4.53 kg
Thus, the mass of banana bunch is 4.53 kg.
The average angular speed of the grindstone is 10 rad/s

<h3>Further explanation</h3>
<em>Let's recall </em><em>Angular Speed</em><em> formula as follows:</em>



<em>where :</em>
<em>ω = final angular speed ( rad/s )</em>
<em>ω₀ = initial angular speed ( rad/s )</em>
<em>α = angular acceleration ( rad/s² )</em>
<em>t = elapsed time ( s )</em>
<em>θ = angular displacement ( rad )</em>

<u>Given:</u>
radius of the grindstone = R = 4.0 m
initial angular speed = ω₀ = 8.0 rad/s
final angular speed = ω = 12 rad/s
elapsed time = t = 4.0 seconds
<u>Asked:</u>
average angular speed = ?
<u>Solution:</u>
<em>Firstly , we will calculate </em><em>angular displacement </em><em>as follows:</em>





<em>Next , we could calculate the </em><em>average angular speed</em><em> as follows:</em>




<h3>Learn more</h3>

<h3>Answer details</h3>
Grade: High School
Subject: Physics
Chapter: Rotational Dynamics
Answer:
a. μ
3 ± 1.8 = [1.2,4.8]
b. The correct answer is option D. No, because the sample size is large enough.
Explanation:
a. The population mean can be determined using a confidence interval which is made up of a point estimate from a given sample and the calculation error margin. Thus:
μ
±(t*s)/sqrt(n)
where:
μ
= is the 95% confidence interval estimate
x_ = mean of the sample = 3
s = standard deviation of the sample = 5.8
n = size of the sample = 41
t = the t statistic for 95% confidence and 40 (n-1) degrees of freedom = 2.021
substituting all the variable, we have:
μ
3 ± (2.021*5.8)/sqrt(41) = 3 ± 1.8 = [1.2,4.8]
b. The correct answer is option D. No, because the sample size is large enough.
Using the the Central Limit Theorem which states that regardless of the distribution shape of the underlying population, a sampling distribution of size which is ≥ 30 is normally distributed.