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liraira [26]
2 years ago
13

Hey guys I really really need help with this question for ASAP! Explain what chart junk is and how it differs from the kind of i

tems you should include in your graphs. Provide four examples?
the second question is Explain what chart junk is and how it differs from the kind of items you should include in your graphs. Provide four examples.
Physics
2 answers:
Step2247 [10]2 years ago
4 0

Answer:

Chartjunk refers to all visual elements in charts and graphs that are not necessary to comprehend the information represented on the graph, or that distract the viewer from this information.

Explanation:

Ahat [919]2 years ago
3 0
I found a definition: Chartjunk<span> refers to all visual elements in </span>charts<span> and graphs that are not necessary to comprehend the information represented on the graph, or that distract the viewer from this information.</span>
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Pendulum clocks are made to run at the correct rate by adjusting the pendulum’s length. Suppose you move from one city to anothe
levacccp [35]

Answer:

Obviously Lengthen...   T = 2\pi \sqrt{L/g}   or   g = 4\pi ^{2} L/g

Explanation:

As we can observe from the equation, time period of a simple pendulum depends upon the length directly. When the gravitational acceleration increases the time period of the pendulum decreases and vice versa. So, by increasing the length, the time period can be adjusted...

4 0
2 years ago
Four different observers are standing in a straight line on a street and hear a siren from a police car. Each person recorded th
Amanda [17]

Answer:

Wycleff is on block 1, Lilly is on block 4, Emilia is on block 12, and Quincy is on block 17.

Explanation:

Wycleff was at block 1 and heard a low pitch sound the whole time, so the police car must have been moving away from him.

Lilly observed was in block 4 change in pitch first.  So the car must have passed her first.

Emilia was at block 12 observed a Doppler effect after Lilly.  So the car passed her after passing Lilly

Quincy was at block 17 so she heard a high pitch sound the whole time, so the police car must have been moving toward him.

7 0
2 years ago
A tube with a cap on one end, but open at the other end, produces a standing wave whose fundamental frequency is 130.8 Hz. The s
Angelina_Jolie [31]

Answer:

A. 261.6 hz.

B. 0.656 m.

Explanation:

A.

When yhe tube is open at one end and closed at the other,

F1 = V/4*L

Where,

F1 = fundamental frequency

V = velocity

L = length of the tube

When the tube is open at both ends,

F'1 = V/2*L

Where

F'1 = the new fundamental frequency

Therefore,

V/2*L x V/4*L

F'1 = 2 * F1

= 2 * 130.8

= 261.6 hz.

B.

F1 = V/4*L

Or

F'1 = V/2*L

Given:

V = 343 m/s

F1 = 130.8

L = 343/(4 * 130.8)

= 0.656 m.

8 0
2 years ago
4. A 505-turn circular-loop coil with a diameter of 15.5 cm is initially aligned so that
Basile [38]

The strength of the magnetic field is 4.8\cdot 10^{-5} T

Explanation:

According to Faraday's Law, the magnitude of the induced emf in the coil is equal to the rate of changeof the flux linkage through the coil:

\epsilon = \frac{N\Delta \Phi}{\Delta t} (1)

where

N = 505 is the number of turns in the coil

\Delta \Phi is the change in magnetic flux through the coil

\Delta t = 2.77 ms = 2.77\cdot 10^{-3} s is the time interval

\epsilon = 0.166 V

The coil is rotated from a position perpendicular to the Earth's magnetic field to a position parallel to it, so the final flux is zero, and the magnitude of the flux change is simply equal to the initial flux:

\Delta \Phi = B A cos \theta

where

B is the strength of the magnetic field

A is the area of the coil

\theta=0^{\circ} is the angle between the normal to the coil and the field

The area of the coil can be written as

A=\pi r^2

where

r=\frac{15.5 cm}{2}=7.75 cm = 7.75\cdot 10^{-2} m is its radius

Substituting everything into eq.(1) and solving for B, we find:

\epsilon= \frac{NB\pi r^2 cos \theta}{\Delta t}\\B=\frac{\epsilon \Delta t}{\pi r^2 cos \theta}=\frac{(0.166)(2.77\cdot 10^{-3})}{(505)\pi (7.75\cdot 10^{-2})^2(cos 0^{\circ})}=4.8\cdot 10^{-5} T

Learn more about magnetic fields:

brainly.com/question/3874443

brainly.com/question/4240735

#LearnwithBrainly

8 0
2 years ago
Read 2 more answers
6. Two blocks are released from rest at the same height. Block A slides down a steeper ramp than Block B. Both ramps are frictio
Stolb23 [73]

Answer:

a. the work done by the gravitational force on Block A is <u>less than</u> the work done by the gravitational force on Block B.

b. the speed of Block A is <u>equal to</u> the speed of Block B.

c. the momentum of Block A is <u>less than</u> the momentum of Block B.

Explanation:

a. The  work done by the gravitational force is equal to:

w = m*g*h

where m is mass, g is the standard gravitational acceleration and h is height. Given that both blocks are released from rest at the same height, then, the bigger the mass, the bigger the work done.

b. With ramps frictionless, the final speed of the blocs is:

v = √(2*g*h)

which is independent of the mass of the blocks.

c. The momentum is calculated as follows:

momentum = m*v

Given that both bocks has the same speed, then, the bigger the mass, the bigger the momentum.

8 0
2 years ago
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