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3 years ago

6

Use the drop-down menu to complete the statement. Based on the field lines, the electric charges indicated by the question marks

are .

2 answers:

velikii [3]3 years ago

8 0

Answer:

the same

Explanation: got it right on edge

zheka24 [161]3 years ago

5 0

Answer: The same

Explanation: I just did it on Edg

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What is the name of the specification that indicates the frequencies that are dedicated for data transmission over cable lines?

mihalych1998 [28]

The answer to this question is the term
DOCSIS. A DOCSIS or Data Over Cable Service Interface Specification is a telecommunication standard or interface where in an internet was being provided by the use of cables. The advantage of using DOCSIS is that the speed of the internet is faster using this kind of interface.

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3 years ago

Having a good range of motion is super important because that is what allows you to ________________.

kotykmax [81]

A is the correct answer I think

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2 years ago

The fulcrum is between the effort and the load

Inessa05 [86]

Class 1 lever

Explanation:

In a class 1 lever, the fulcrum is placed between the effort and the load. This lever systems is the most common.

The effort is the force input and the load is the force output
The fulcrum is a hinge between the load and effort.
Movement of the effort and load are in opposite directions.
There are other classes of lever like the class 2 and 3.
They all have different load, fulcrum and effort configurations

learn more:

Load related problems brainly.com/question/9202964

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3 years ago

Pllssss answer thiss ...<br> its about the reading of VOLTMETER AND AMMETER

Afina-wow [57]

Answer:

The answer is C 1.8V and 0.38A

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2 years ago

At the intersection of Texas Avenue and University Drive,

Zielflug [23.3K]

Answer:

The initial speed of the truck is 21.93 m/s, and the initial speed of the car is 19.524 m/s

Explanation:

We can use conservation of momentum to find the initial velocities.

Taking the unit vector $\hat{i}$ pointing north and $\hat{j}$ pointing east, the final velocity will be

$\vec{V}_f = 16.0 \frac{m}{s} \ ( \ cos(24.0 \°) \ , \ sin (24.0 \°) \ )$

$\vec{V}_f = ( \ 14.617 \frac{m}{s} \ , \ 6.508 \frac{m}{s} \ )$

The final linear momentum will be:

$\vec{P}_f = (m_{car}+ m_{truck}) * V_f$

$\vec{P}_f = (950 \ kg \ + 1900 \ kg \ ) * ( \ 14.617 \frac{m}{s} \ , \ 6.508 \frac{m}{s} \ )$

$\vec{P}_f = (2.850 \ kg \ ) * ( \ 14.617 \frac{m}{s} \ , \ 6.508 \frac{m}{s} \ )$

$\vec{P}_f = ( \ 41,658.45 \frac{ kg \ m}{s} \ , \ 18,547.8 \frac{kg \ m}{s} \ )$

As there are not external forces, the total linear momentum must be constant.

So:

$\vec{P}_0= \vec{P}_f$

As initially the car is travelling east, and the truck is travelling north, the initial linear momentum must be

$\vec{P}_0= ( m_{truck} * v_{truck}, m_{car}* v_{car} )$

so:

$\vec{P}_0= \vec{P}_f$

$( m_{truck} * v_{truck}, m_{car}* v_{car} ) = ( \ 41,658.45 \frac{ kg \ m}{s} \ , \ 18,547.8 \frac{kg \ m}{s} \ )$

so

$\left \{ {{m_{truck} \ v_{truck} = 41,658.45 \frac{ kg \ m}{s} } \atop {m_{car} \ v_{car}=18,547.8 \frac{kg \ m}{s} }} \right.$

So, for the truck

$m_{truck} \ v_{truck} = 41,658.45 \frac{ kg \ m}{s}$

$1900 \ kg \ v_{truck} = 41,658.45 \frac{ kg \ m}{s}$

$v_{truck} = \frac{41,658.45 \frac{ kg \ m}{s}}{1900 \ kg}$

$v_{truck} = \frac{41,658.45 \frac{ kg \ m}{s}}{1900 \ kg}$

$v_{truck} = 21.93 \frac{m}{s}$

And, for the car

$950 \ kg \ v_{car}=18,547.8 \frac{kg \ m}{s}$

$v_{car}=\frac{18,547.8 \frac{kg \ m}{s}}{950 \ kg}$

$v_{car}=19.524 \frac{m}{s}$

5 0

2 years ago