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3 years ago

5

The electric potential 1.34 m from a charge is 580 V. What is the value of the charge? Include the sign, + or -. (The answer is

___ *10^-8 c. Just fill in the number, not the power.)

1 answer:

blagie [28]3 years ago

5 0

Answer: 8.6

Explanation:

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A 7.8-kg solid sphere, made of metal whose density is 2500 kg/m3, is suspended by a cord. When the sphere is immersed in water (

Klio2033 [76]

Answer:

Density of the liquid = 1470.43 kg/m³

Explanation:

Given:

Mass of solid sphere(m) = 6.1 kg

Density of the metal = 2600 kg/m³

Thus volume of the liquid :

Volume of the sphere = 6.1 kg/2600 kg/m³ = 0.002346 m³

The volume of water displaced is equal to the volume of sphere (Archimedes' principle)

Volume displaced = 0.002346 m³

Buoyant force =

Where

is the density of the fluid

g is the acceleration due to gravity

V is the volume displaced

The free body diagram of the sphere is shown in image.

According to image:

Acceleration due to gravity = 9.81 ms⁻²

Tension force = 26 N

Applying in the equation to find the density of the liquid as:

Thus, the density of the liquid = 1470.43 kg/m³

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2 years ago

A throttle position sensor waveform is going to be observed. At what setting should the volts per division be set to see the ent

Mars2501 [29]

Answer:

1 V / div

Explanation:

Solution:

- The vertical scale has eight divisions.

- If each division is set to equal 1 volt, the display will show 0 to 8 volts.

- This is okay in a 0 to 5 volt variable sensor such as a throttle position (TP) sensor.

- The volts per division (V/div) should be set so that the entire anticipated waveform can be viewed.

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3 years ago

The two types of decomposers

ycow [4]

Bacteria,Worms, Slugs, Snails, and Fungi are all types of decomposers

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3 years ago

The electric field at the distance of 3.5 meters from an infinite wall of charges is 125 N/C. What is the magnitude of the elect

RideAnS [48]

Explanation:

It is given that,

Distance, r = 3.5 m

Electric field due to an infinite wall of charges, E = 125 N/C

We need to find the electric field 1.5 meters from the wall, r' = 1.5 m. Let it is equal to E'. For an infinite wall of charge the electric field is given by :

$E=\dfrac{\lambda}{2\pi \epsilon_o r}$

It is clear that the electric field is inversely proportional to the distance. So,

$\dfrac{E}{E'}=\dfrac{r'}{r}$

$E'=\dfrac{Er}{r'}$

$E'=\dfrac{125\times 3.5}{1.5}$

E' = 291.67 N/C

So, the magnitude of the electric field 1.5 meters from the wall is 291.67 N/C. Hence, this is the required solution.

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4 years ago

Which of the following is not a current use of radioisotopes?

andrey2020 [161]

Answer;

light fixtures

Explanation;

Radioactive isotopes have a variety of applications. They are useful because either we can detect their radioactivity or we can use the energy they release.

These uses includes;

Radioactive Dating; Radioactive isotopes are useful for establishing the ages of various objects.
Medical Applications; Radioactive isotopes have numerous medical applications; sterilization of equipment, diagnosing and treating illness and diseases.
Detection of leaks in building structures
Irradiation of Food; the radiation emitted by some radioactive substances can be used to kill microorganisms on a variety of foodstuffs, extending the shelf life of these products. etc.

5 0

3 years ago

Read 2 more answers