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4 years ago

13

Which statements describe the factors affecting the strength of an electrical force? Check all that apply.

2 answers:

mote1985 [20]4 years ago

4 0

Answer:

The answer for me is B

Explanation:

Assuming we hv two charged papers after it has been rubbed in the hair when we bring these two charged papers together we get a stronger electrical force between them

sergeinik [125]4 years ago

4 0

Answer:

The stronger the charge on two objects, the stronger the electrical force between them.

The closer together two objects are, the stronger the electrical force between them.

Explanation:

answered it on edge

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Which is not a scale used to measure temperature?

garik1379 [7]

<span>There are three temperature scales in use today, Fahrenheit, Celsius and Kelvin. Fahrenheit temperature scale is a scale based on 32 for the freezing point of water and 212 for the boiling point of water, the interval between the two being divided into 180 parts.</span>

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4 years ago

Read 2 more answers

a shot putter accelerates a 7.3kg shot from rest to 14m/s in 1.5r seconds. what average power was developed?

Sedaia [141]

Explanation:

$power = \frac{energy \: expended}{time}$

m = 7.3kg

u = 0

v = 14m/s

t = 1.5sec

P = (0.5×7.3×14²) ÷ 1.5

P = 476.93

P = 477 watt

3 0

4 years ago

A spring with a constant of 92N/m is compressed 2.8 cm. How much potential energy is stored in the spring?

Arada [10]

Answer:

0.036J

Explanation:

Given parameters:

Spring constant , K = 92N/m

Compression = 2.8cm = 0.028m

Unknown:

Potential energy = ?

Solution:

To solve this problem;

P.E = $\frac{1}{2}$ K e²

K is the spring constant

e is the compression

so;

P.E = $\frac{1}{2}$ x 92 x 0.028² = 0.036J

6 0

3 years ago

which of the following is not used to protect us from the possibility of receiving an electric shock

Juli2301 [7.4K]

Answer:water

Explanation:

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3 years ago

Find the elongation produced in a copper wire of length 2m and radius 5mm, when suspended by a block

igomit [66]

Answer:

0.104 m

Explanation:

Stress, $\sigma=\frac {F}{A}$

Where F is force and A is area. Also, F=mg where m is mass and g is acceleration due to gravity

$Area= \pi r^{2}$

$Strain=\frac {\triangle l}{l}$ where $\triangle l$ is the elongation and l is the original length

$E=\frac {stress}{strain}=\frac {\frac {F}{A}}{\frac {\triangle l}{l}}=\frac {Fl}{A\triangle l}$

Making $\triangle l$ the subject then

$\triangle l=\frac {Fl}{AE}=\frac{mg l}{\pi r^{2} E}$

By substituting the given values and taking g as 9.81 then

$\triangle l=\frac {Fl}{AE}=\frac{500\times 9.81\times 2 m}{\pi \times 0.005^{2}\times 1.2\times 10^{9}}=0.104087333 m\approx 0.104 m$

3 0

3 years ago