There is a pattern between star color and true brightness for star in the main sequence

The graph below shows the position of an ant as it crawls over a flat picnic blanket. The total time for the ant to go from the

Mice21 [21]

The average speed of the ant is 0.276 cm/s and the average velocity is 0.136 cm/s.

The correct answer is option D.

In the given graph, we can deduce the following;

the total time of the motion, = 1 mins + 45 s = 60 s + 45 s = 105 s

The average speed of the ant is calculated as;

$average \ speed = \frac{total \ distance }{total \ time }$

The total distance from the graph is calculated as follows;

first horizontal distance from 2 cm to 8 cm = 8 - 2 = 6 cm

first upward distance from 3 cm to 5 cm = 5 - 3 = 2 cm

second horizontal distance from 8 cm to 6 cm = 8 - 6 = 2 cm

second upward distance from 5 cm to 12 cm = 12 - 5 = 7 cm

third horizontal distance from 6 cm to 13 cm = 13 - 6 = 7 cm

fourth downward distance from 12 cm to 9 cm = 3 cm

final horizontal distance from 13 cm to 15 cm = 2cm

The total distance = (6 + 2 + 2 + 7 + 7 + 3 + 2) cm = 29 cm

$average \ speed = \frac{total \ distance }{total \ time } = \frac{29 \ cm}{105 \ s} = 0.276 \ cm/s$

The average velocity is calculated as the change in displacement per change in time.

The displacement is the shortest distance between the start and end positions.

This shortest distance is the straight line connecting the start and end position. Call this line P

From the end position at x = 15 cm, draw a vertical line from y = 9 cm, to y = 3 cm. The displacement = 9 cm - 3 cm = 6 cm

Also, draw a horizontal line from start at x = 2 cm to x = 15 cm. The displacement = 15 cm - 2 cm = 13 cm

Notice, you have a right triangle, now calculate the length of line P.

↓end

↓

↓ 6cm

↓

start -------------13 cm------------

Use Pythagoras theorem to solve for P.

$P^2 = 6^2 + 13^2\\\\P^2 = 36 + 169\\\\P^2 = 205\\\\P= \sqrt{205} \\\\P = 14.318 \ cm$

The average velocity of the ant is calculated as;

$average \ velocity= \frac{\Delta displacemnt }{total\ time }= \frac{14.318 \ cm}{105 \ s} = 0.136 \ cm/s \\\\$

Thus, the average speed of the ant is 0.276 cm/s and the average velocity is 0.136 cm/s.

Learn more here: brainly.com/question/589950

5 0

3 years ago

A shallow groove on the surface of the cortex is called a ________.

Juli2301 [7.4K]

Answer:

sulcus

Explanation:

A sulcus is an indentation or depression in the brain that causes it to look like it ridges or folds

Cerebral sulci and fissures are grooves between the adjacent gyri on the surface of the cerebral hemispheres.

Sulci can be basically can be divided into three basic function

limiting sulcus: This happens to develop between areas differing in structure and function, for example central sulcus

axial sulcus: This develops along the axis of a rapidly growing/developing area (e.g. calcarine sulcus)

operculated sulcus: a sulcus may be between two structurally-different areas and a third sulcus may lie in its wall and does not appear on the surface (e.g. lunate sulcus)

3 0

4 years ago

If the work function of a material is such that red light of wavelength 700 nm just barely initiates the photoelectric effect, w

marishachu [46]

Answer:

2.13 x 10^-19 J or 0.53 eV

Explanation:

cut off wavelength, λo = 700 nm = 700 x 10^-9 m

λ = 400 nm = 400 x 10^-9 m

Use the energy equation

$E = \frac{h c}{\lambda _{o}}+K$

Where, K be the work function

$\frac{h c}{\lambda} = \frac{h c}{\lambda _{o}}+K$

$K =hc\left ( \frac{1}{\lambda } -\frac{1}{\lambda _{0}}\right )$

$K =6.63\times 10^{-34}\times 3\times 10^{8}\left ( \frac{1}{4\times 10^{-7} } -\frac{1}{7\times 10^{-7}}\right )$

K = 2.13 x 10^-19 J

K = 0.53 eV

3 0

3 years ago

A horizontal 745 N merry-go-round of radius

Arturiano [62]

Answer:

The kinetic energy of the merry-goround after 3.62 s is 544J

Explanation:

Given :

Weight w = 745 N

Radius r = 1.45 m

Force = 56.3 N

To Find:

The kinetic energy of the merry-go round after 3.62 = ?

Solution:

Step 1: Finding the Mass of merry-go-round

$m = \frac{ weight}{g}$

$m = \frac{745}{9.81 }$

m = 76.02 kg

Step 2: Finding the Moment of Inertia of solid cylinder

Moment of Inertia of solid cylinder I = $0.5 \times m \times r^2$

Substituting the values

Moment of Inertia of solid cylinder I

=> $0.5 \times 76.02 \times (1.45)^2$

=> $0.5 \times 76.02\times 2.1025$

=> $79.91 kg.m^2$

Step 3: Finding the Torque applied T

Torque applied T = $F \times r$

Substituting the values

T = $56.3 \times 1.45$

T = 81.635 N.m

Step 4: Finding the Angular acceleration

Angular acceleration , $\alpha = \frac{Torque}{Inertia}$

Substituting the values,

$\alpha = \frac{81.635}{79.91}$

$\alpha = 1.021 rad/s^2$

Step 4: Finding the Final angular velocity

Final angular velocity , $\omega = \alpha \times t$

Substituting the values,

$\omega = 1.021 \times 3.62$

$\omega = 3.69 rad/s$

Now KE (100% rotational) after 3.62s is:

KE = $0.5 \times I \times \omega^2$

KE = $0.5 \times 79.91 \times 3.69^2$

KE = 544J

6 0

4 years ago

Why are aftershocks dangerous to buildings after a large earthquake

Katen [24]

because the building are weaker and they are most likley to fall

6 0

3 years ago