An incident ray that passes through the vertex of a convex lens:

Gnom [1K]

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.

3 0

2 years ago

A 3 kg model airplane is traveling at a speed of 33 m/s. The operator then increases the speed up to 45 m/s in 2 seconds. How mu

nlexa [21]

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

8 0

2 years ago

At an outdoor market, a bunch of bananas attached to the bottom of a vertical spring of force constant 16.0 N/m is set into osci

aksik [14]

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

$\omega =\sqrt{\frac{K}{m}}$

$m = \frac{K}{\omega ^{2}}$

$m = \frac{16}{1.88 ^{2}}$

m = 4.53 kg

Thus, the mass of banana bunch is 4.53 kg.

8 0

2 years ago

"a grindstone of radius 4.0 m is initially spinning with an angular speed of 8.0 rad/s. the angular speed is then increased to 1

kotegsom [21]

The average angular speed of the grindstone is 10 rad/s

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<h3>Further explanation</h3>

Let's recall Angular Speed formula as follows:

$\boxed{ \omega = \omega_o + \alpha t }$

$\boxed{ \theta = \omega_o t + \frac{1}{2} \alpha t^2 }$

$\boxed{ \omega^2 = \omega_o^2 + 2 \alpha \theta }$

$\boxed{ \theta = \frac{( \omega + \omega_o )}{2} t }$

where :

ω = final angular speed ( rad/s )

ω₀ = initial angular speed ( rad/s )

α = angular acceleration ( rad/s² )

t = elapsed time ( s )

θ = angular displacement ( rad )

$\texttt{ }$

Given:

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

Asked:

average angular speed = ?

Solution:

Firstly , we will calculate angular displacement as follows:

$\theta = \frac{( \omega + \omega_o )}{2} t$

$\theta = \frac{ ( 12 + 8.0 ) }{2} \times 4.0$

$\theta = 10 \times 4.0$

$\boxed {\theta = 40 \texttt{ rad}}$

$\texttt{ }$

Next , we could calculate the average angular speed as follows:

$\texttt{average angular speed} = \theta \div t$

$\texttt{average angular speed} = 40 \div 4.0$

$\boxed{\texttt{average angular speed} = 10 \texttt{ rad/s}}$

$\texttt{ }$

<h3>Learn more</h3>

Impacts of Gravity : brainly.com/question/5330244
Effect of Earth’s Gravity on Objects : brainly.com/question/8844454
The Acceleration Due To Gravity : brainly.com/question/4189441
Moment of Inertia : brainly.com/question/13796477
The Ratio of the Moments of Inertia : brainly.com/question/2176655

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<h3>Answer details</h3>

Grade: High School

Subject: Physics

Chapter: Rotational Dynamics

4 0

3 years ago

Read 2 more answers

Use technology and the given confidence level and sample data to find the confidence interval for the population mean muμ. Assum

Ronch [10]

Answer:

a. μ $_{95%} =$ 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 $_{95%} = x_$ ±(t*s)/sqrt(n)

where:

μ $_{95%} =$ = 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 $_{95%} =$ 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.

4 0

2 years ago