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2 years ago

If you comb your hair and the comb becomes negatively charged, your hair becomes?

2 answers:

7 0

The negative charges transfer from your hair to the comb, this leads the comb to be negatively charged and your hair to become positively charged.

slamgirl [31]2 years ago

5 0

A positive charge cause dosent your hair static up when that happens or no?

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Difference between rest and motion

creativ13 [48]

Rest - it is the state in which body doesn’t move from it’s place

motion - it is the state in which body moves from it’s place

3 0

2 years ago

An astronaut is standing on the surface of a planetary satellite that has a radius of 1.74 × 10^6 m and a mass of 7.35 × 10^22 k

ExtremeBDS [4]

Answer:

2.87 km/s

Explanation:

radius of planet, R = 1.74 x 10^6 m

Mass of planet, M = 7.35 x 10^22 kg

height, h = 2.55 x 10^6 m

G = 6.67 x 106-11 Nm^2/kg^2

Use teh formula for acceleration due to gravity

$g=\frac{GM}{R^{2}}$

$g=\frac{6.67\times 10^{-11}\times 7.35\times 10^{22}}{1.74^{2}\times 10^{12}}$

g = 1.62 m/s^2

initial velocity, u = ?, h = 2.55 x 10^6 m , final velocity, v = 0

Use third equation of motion

$v^{2}=u^{2}-2gh$

0 = v² - 2 x 1.62 x 2.55 x 10^6

v² = 8262000

v = 2874.37 m/s

v = 2.87 km/s

Thus, the initial speed should be 2.87 km/s.

6 0

2 years ago

A violin string vibrates at 260 Hz when unfingered. At what frequency will it vibrate if it is fingered one fourth of the way do

Advocard [28]

Answer:

346.66 Hz

Explanation:

$l_1$ = Length of string which is unfingered = l

$l_2$ = Length of string which is vibrate when fingered = $l-\dfrac{1}{4}l=\dfrac{3}{4}l$

$f_1$ = Unfingered frequency = 260 Hz

$f_2$ = Fingered frequency

Frequency is inversely proportional to length

$f=\dfrac{1}{l}$

So,

$\dfrac{f_1}{f_2}=\dfrac{l_2}{l_1}\\\Rightarrow \dfrac{f_1}{f_2}=\dfrac{\dfrac{3}{4}l}{l}\\\Rightarrow \dfrac{f_1}{f_2}=\dfrac{3}{4}\\\Rightarrow f_2=\dfrac{4}{3}f_1\\\Rightarrow f_2=\dfrac{4}{3}260\\\Rightarrow f_2=346.66\ Hz$

The frequency of the fingered string is 346.66 Hz

3 0

2 years ago

What is the magnitude of the magnetic field at a point midway between them if the top one carries a current of 19.5 A and the bo

Phantasy [73]

Answer:

The magnetic field will be $\large{\dfrac{1.4 \times 10^{-4}}{d}} T$ , '2d' being the distance the wires.

Explanation:

From Biot-Savart's law, the magnetic field ( $\large{\overrightarrow{B}}$ ) at a distance ' $r$ ' due to a current carrying conductor carrying current ' $I$ ' is given by

$\large{\overrightarrow{B} = \dfrac{\mu_{0}I}{4 \pi}} \int \dfrac{\overrightarrow{dl} \times \hat{r}}{r^{2}}}$

where ' $\overrightarrow{dl}$ ' is an elemental length along the direction of the current flow through the conductor.

Using this law, the magnetic field due to straight current carrying conductor having current ' $I$ ', at a distance ' $d$ ' is given by

$\large{\overrightarrow{B}} = \dfrac{\mu_{0}I}{2 \pi d}$

According to the figure if ' $I_{t}$ ' be the current carried by the top wire, ' $I_{b}$ ' be the current carried by the bottom wire and ' $2d$ ' be the distance between them, then the direction of the magnetic field at 'P', which is midway between them, will be perpendicular towards the plane of the screen, shown by the $\bigotimes$ symbol and that due to the bottom wire at 'P' will be perpendicular away from the plane of the screen, shown by $\bigodot$ symbol.

Given $\large{I_{t} = 19.5 A}$ and $\large{I_{B} = 12.5 A}$

Therefore, the magnetic field ( $\large{B_{t}}$ ) at 'P' due to the top wire

$B_{t} = \dfrac{\mu_{0}I_{t}}{2 \pi d}$

and the magnetic field ( $\large{B_{b}}$ ) at 'P' due to the bottom wire

$B_{b} = \dfrac{\mu_{0}I_{b}}{2 \pi d}$

Therefore taking the value of $\mu_{0} = 4\pi \times 10^{-7}$ the net magnetic field ( $\large{B_{M}}$ ) at the midway between the wires will be

$\large{B_{M} = \dfrac{4 \pi \times 10^{-7}}{2 \pi d} (I_{t} - I_{b}) = \dfrac{2 \times 10^{-7}}{d} = \dfrac{41.4 \times 10 ^{-4}}{d}} T$

5 0

2 years ago

Sally and Sam are in a spaceship that comes to within 14,000 km of the asteroid Ceres. Determine the force Sally experiences, in

tatiyna

Sam and Sally are traveling aboard a spacecraft that approaches the asteroid Ceres within 14,000 kilometers. Sally will experience 1.989 × 10⁻¹¹ N of force.

<h3>What is the gravitational force?</h3>

Newton's law of gravity states that each particle having mass in the universe attracts each other particle with a force known as the gravitational force.

The gravitational force is proportional to the product of the masses of the two bodies and inversely proportional to the square of their distance.

Given data

Mass of asteroid ,m₁ = 8.7 1020 kg

Mass of sally,m₂ = 67 kg

Gravitational constant,G = 6.6 × 10⁻¹¹ kg⁻² m²

Distance of seperation,R = 14,000 km

$\rm F = G\frac{m_1m_2}{R^ 2} \\\\ F = 6.6 \times 10^{-11 }\times \frac{8.71020 \times 67 }{(14,000)^2} \\\\F = 1.989 \times 10^{-11 } \ N$

Hence, the force Sally experiences will be 1.989 × 10⁻¹¹ N.

To learn more about the gravitational force, refer to the link;

brainly.com/question/24783651

#SPJ1

4 0

1 year ago