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2 years ago

Like magnetism, static electricity can attract and repel. True False

Physics

2 answers:

earnstyle [38]2 years ago

5 0

The answer to the question is True

sweet-ann [11.9K]2 years ago

4 0

True because my hair demonstrates it xD

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Calculate the amount of heat transferred when 710 grams of water warms from an initial temperature of 4.0 ºC to a final temperat

tigry1 [53]

Answer:

Q = 62383.44 Joules

Explanation:

Given that,

Mass of water, m = 710 gm

Initial temperature of water, $T_i=4^{\circ} C$

Final temperature of water, $T_f=25^{\circ} C$

The specific heat capacity of liquid water is, $c=4.184\ J/g\ ^oC$

Heat transferred is given by :

$Q=mc(T_f-T_i)$

$Q=710\times 4.184\times (25-4)$

Q = 62383.44 Joules

So, the amount of heat transferred is 62383.44 Joules. Hence, this is the required solution.

3 0

3 years ago

Do all objects have inertia <br><br> a) true<br> b) false

Natasha2012 [34]

Answer:

The answer is A. Which is true

4 0

2 years ago

The Lamborghini Huracan has an initial acceleration of 0.75g. Its mass, with a driver, is 1510 kg.

Cerrena [4.2K]

Answer:

11109.825 N

Explanation:

Given Data:

total mass =m=1510 kg

initial acceleration (a) =0.75g ( g=9.81 m/s² )

F=ma

= (1510)*( 0.75*9.81)

= 11109.825 N

4 0

3 years ago

Over the past 150 years, what has happened to the amount of forest cover in Minnesota?

Airida [17]

I think D. None of the above

I think that it

4 0

3 years ago

Charge q is accelerated starting from rest up to speed v through the potential difference V. What speed will charge q have after

adoni [48]

Answer: v = $1.19 * 10^{6} m/s$

Explanation: q = magnitude of electronic charge = $1.609 * 10^{-19} c$

mass of an electronic charge = $9.10 * 10^{-31} kg$

V= potential difference = 4V

v = velocity of electron

by using the work- energy theorem which states that the kinetic energy of the the electron must equal the work done use in accelerating the electron.

kinetic energy = $\frac{mv^{2} }{2}$ , potential energy = qV

hence, $\frac{mv^{2} }{2} = qV$

$\frac{9.10 *10^{-31} * v^{2} }{2} = 1.609 * 10^{-16} * 4\\\\\\\\9.10*10^{-31} * v^{2} = 2 * 1.609 *10^{-16} * 4\\\\\\9.10 *10^{-31} * v^{2} = 1.287 *10^{-15} \\\\v^{2} = \frac{1.287 *10^{-15} }{9.10 *10^{31} } \\\\v^{2} = 1.414*10^{15} \\\\v = \sqrt{1.414*10^{15} } \\\\v = 1.19 * 10^{6} m/s$

7 0

2 years ago