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3 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]3 years ago

5 0

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

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Please help,,, question on image

MA_775_DIABLO [31]

Answer:

first you have to

Explanation:

3 0

3 years ago

A physics student swings a tennis ball connected to a rope in a vertical circle with a constant speed of 6.29 m/s. The ball has

Alex777 [14]

Answer:

r = 0.5 m

Explanation:

First we find the angular speed of the ball by using its period:

ω = θ/t

For the time period:

ω = angular speed = ?

θ = angular displacement = 2π rad

t = time period = 0.5 s

Therefore,

ω = 2π rad/0.5 s

ω = 12.56 rad/s

Now, for the radius:

v = rω

r = v/ω

where,

v = linear speed = 6.29 m/s

r = radius = ?

r = (6.29 m/s)/(12.56 rad/s)

8 0

3 years ago

Based on Archimedes' principle, the greatest buoyant force an object can experience in water is determined by which quantity?

kicyunya [14]

I think it the objects mass

6 0

3 years ago

Una partícula se mueve en el plano XY efectúa un desplazamiento mientras actúa sobre ella una fuerza constante. X= (4i + 3j) m,

dsp73

Answer:

a) La magnitud del desplazamiento es de 5 m

La magnitud de la fuerza es 20 N

b) El trabajo realizado por la fuerza es de 100 J

c) El ángulo entre la fuerza y el plano es 0 °

Explanation:

a) La magnitud del desplazamiento se encuentra por la relación;

$\left | X \right | = \sqrt{X_{x}^{2}+X_{y}^{2}}$

Lo que da;

$\left | X \right | = \sqrt{4^{2}+3^{2}} = 5 \ m$

De manera similar, la magnitud de la fuerza, F, se encuentra como sigue;

$\left | F \right | = \sqrt{F_{x}^{2}+F_{y}^{2}}$

Lo que da;

$\left | F \right | = \sqrt{16^{2}+12^{2}} = 20 \ N$

b) El trabajo, W, realizado por la fuerza = Fuerza, F × Distancia, X

∴ Ancho = 20 N × 5 m = 100 N · m = 100 J

c) La dirección de la fuerza viene dada por la siguiente fórmula;

$tan^{-1} \left (\dfrac{F_y}{F_x} \right ) = tan^{-1} \left (\dfrac{12}{16} \right ) = 38.9^{\circ}$

La dirección del plano viene dada por la siguiente fórmula;

$tan^{-1} \left (\dfrac{X_y}{X_x} \right ) = tan^{-1} \left (\dfrac{3}{4} \right ) = 38.9^{\circ}$

Por tanto, el ángulo entre la fuerza y el plano = 0 °

La fuerza actúa a lo largo del plano.

6 0

3 years ago

A charged particle moving along the x-axis enters a uniform magnetic field pointing along the z-axis. Because of an electric fie

Furkat [3]

Answer:

Explanation:

Let the charge particle have charge equal to +q .

force due to electric field will be along the field that is along y - axis . To balance it force by magnetic force must be along - y axis. ( negative of y axis )

force due to magnetic field = q ( v x B ) , v is velocity and B is magnetic field.

F = q ( v i x B k ) , ( velocity is along x direction and magnetic field is along z axis. )

= (Bqv) - j

= - Bqv j

The force will be along - ve y - direction .

If we take charge as negative or - q

force due to electric field will be along - y axis .

magnetic force = F = -q ( v i x B k )

= + Bqv j

magnetic force will be along + y axis

So it is difficult to find out the nature of charge on the particle from this experiment.

5 0

3 years ago