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2 years ago

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If you're ever standing on a mountaintop when a dark cloud passes overhead and your hair stands up, get off the mountain fast. H

ow would your hair have acquired the charge to make it stand up?

1 answer:

OleMash [197]2 years ago

8 0

Answer:

The hairs would have acquired charge by the passing of dry winds resulting in the loss of electron.

Explanation:

While standing on the top of a mountain if a person gets its hairs stand up after a cloud passes over, this might happen due to the static electric charges on the lower surface of the cloud are opposite in nature to that of hairs which the hairs would have acquired by the passing of dry winds which would have resulted in the loss of electron from the hair tip.

Similar case happens when we rub a dry plastic ruler or a dry plastic comb on our hairs.

You might be interested in

Angular and Linear Quantities: A child is riding a merry-go-round that has an instantaneous angular speed of 1.25 rad/s and an a

serious [3.7K]

To solve this problem we will use the kinematic equations of angular motion in relation to those of linear / tangential motion.

We will proceed to find the centripetal acceleration (From the ratio of the radius and angular velocity to the linear velocity) and the tangential acceleration to finally find the total acceleration of the body.

Our data is given as:

$\omega = 1.25 rad/s \rightarrow$ The angular speed

$\alpha = 0.745 rad/s2 \rightarrow$ The angular acceleration

$r = 4.65 m \rightarrow$ The distance

The relation between the linear velocity and angular velocity is

$v = r\omega$

Where,

r = Radius

$\omega =$ Angular velocity

At the same time we have that the centripetal acceleration is

$a_c = \frac{v^2}{r}$

$a_c = \frac{(r\omega)^2}{r}$

$a_c = \frac{r^2\omega^2}{r}$

$a_c = r \omega^2$

$a_c = (4.65 )(1.25 rad/s)^2$

$a_c = 7.265625 m/s^2$

Now the tangential acceleration is given as,

$a_t = \alpha r$

Here,

$\alpha =$ Angular acceleration

r = Radius

$\alpha = (0.745)(4.65)$

$\alpha = 3.46425 m/s^2$

Finally using the properties of the vectors, we will have that the resulting component of the acceleration would be

$|a| = \sqrt{a_c^2+a_t^2}$

$|a| = \sqrt{(7.265625)^2+(3.46425)^2}$

$|a| = 8.049 m/s^2 \approx 8.05 m/s2$

Therefore the correct answer is C.

7 0

2 years ago

A fighter jet is catapulted off an aircraft carrier from rest to 75 m/s. If the aircraft carrier deck is 100 m long, what is the

egoroff_w [7]

The acceleration of the jet is $28.1 m/s^2$

Explanation:

Since the motion of the jet is a uniformly accelerated motion, we can use the following suvat equation:

$v^2-u^2=2as$

where

v is the final velocity

u is the initial velocity

a is the acceleration

s is the displacement

For the jet in this problem, we have

u = 0

v = 75 m/s

s = 100 m

Solving for a, we find the acceleration:

$a=\frac{v^2-u^2}{2s}=\frac{75^2-0}{2(100)}=28.1 m/s^2$

Learn more about acceleration:

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#LearnwithBrainly

4 0

2 years ago

A 15-g bullet moving at 300 m/s passes through a 2.0 cm thick sheet of foam plastic and emerges with a speed of 90 m/s. Let's as

Shkiper50 [21]

Answer:

Explanation:

a) Change in momentum, Δp = mΔv = m(v - u) = (15 * 10-3) * (90 - 300) = -3.15 kg-m2

b) Acceleration of the bullet, a = (v2 - u2) / 2s = (902 - 3002) / (2 * 0.02) = -2047500 m/s2

So, the bullet is in contact with the plastic for the time, $\bigtriangleupt = \frac{(v - u)}{a} =\frac{(90 - 300)}{(-2047500)} = 1.03 \times 10^{-4} s$

c) Average force, $F_{avg} =\frac{\bigtriangleup p}{\bigtriangleup t} =\frac{(-3.15)}{(1.03 \times 10^{-4})} = 30.6 kN$

8 0

2 years ago

Consider a turntable to be a circular disk of moment of inertia It rotating at a constant angular velocity ωi around an axis thr

Rom4ik [11]

Answer:

Note: Angular momentum is always conserved in a collision.

The initial angular momentum of the system is

L = ( It ) ( ωi )

where It = moment of inertia of the rotating circular disc,

ωi = angular velocity of the rotating circular disc

The final angular momentum is

L = ( It + Ir ) ( ωf )

where ωf is the final angular velocity of the system.

Since the two angular momenta are equal, we see that

( It ) ( ωi ) = ( It + Ir ) ( ωf )

so making ωf the subject of the formula

ωf = [ ( It ) / ( It + Ir ) ] ωi

Explanation:

7 0

2 years ago

A uniform electric field with a magnitude of 125 000 N/C passes through a rectangle with sides of 2.50 m and 5.00 m. The angle b

jeyben [28]

Answer:

$6.60\cdot 10^5 Nm^2/C$

Explanation:

The electric flux through the rectangle is given by

$\Phi = E A cos \theta$

where

E is the electric field strength

A is the area of the rectange

$\theta$ is the angle between the direction of the electric field and of the vector normal to the plane of the rectangle

In this problem we have

E = 125 000 N/C

The area of the rectangle is

$A=2.50 m \cdot 5.00 m=12.5 m^2$

and the angle is

$\theta=65.0^{\circ}$

so, the electric flux is

$\Phi = (125,000 N/C)(12.5 m^2)(cos 65^{\circ})=6.60\cdot 10^5 Nm^2/C$

8 0

2 years ago