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2 years ago

15

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.

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A 1500-kg car locks its brakes and skids to a stop on a slippery horizontal road, leaving skid marks that are 15 m long. How muc

Harman [31]

Answer:

$E=88200\ J$

Explanation:

Given:

mass of car, $m=1500\ kg$

distance of skidding after the application of brakes, $d=15\ m$

coefficient of kinetic friction, $\mu_k=0.4$

<u>So, the energy dissipated during the skidding of car:</u>

<em>Frictional force:</em>

$f=\mu_k.N$

where N = normal reaction by ground on the car

$f=0.4\ties 1500\times 9.8$

$f=5880\ N$

<em>Now from the work-energy equivalence:</em>

$E=f.d$

$E=5880\times 15$

$E=88200\ J$ is the dissipated energy.

3 0

2 years ago

Calculate the acceleration of the body of mass 3kg on which a force of 42N has been applied for 5s

____ [38]

F=ma

42/3 = a
a = 14m/s^2

5 0

2 years ago

A player kicks a football into the air. It slows to a stop at its highest point in the air before falling to the ground. Which s

vivado [14]

Answer:

The ball slows down in the air due to an unbalanced force

Explanation:

When player kicks the ball, there are mainly two foces acting on this object: the force made by the player and the opposite force of gravity (which acts with a direction always to the centre of the Earth)

The force applied by the player will be decreasing, while the force of gravity is always constant, this will make that both forces will unbalance, making the football´s speed slow down

8 0

2 years ago

Read 2 more answers

A mass of 0.4 kg hangs motionless from a vertical spring whose length is 0.76 m and whose unstretched length is 0.41 m. Next the

Blizzard [7]

<h3><u>Answer;</u></h3>

= 1.256 m

<h3><u>Explanation;</u></h3>

We can start by finding the spring constant

F = k*y

Therefore; k = F/y = m*g/y

= 0.40kg*9.8m/s^2/(0.76 - 0.41)

= 11.2 N/m

Energy is conserved

Let A be the maximum displacement

Therefore; 1/2*k*A^2 = 1/2*k*(1.20 - 0.41)^2 + 1/2*m*v^2

Thus; A = sqrt((1.20 - 0.55)^2 + m/k*v^2)

= sqrt((1.20 -0.55)^2 + 0.40/9.8*1.6^2)

= 0.846 m

Thus; the length will be 0.41 + 0.846 = 1.256 m

6 0

2 years ago

Consider a spring that does not obey Hooke’s law very faithfully. One end of the spring is fixed. To keep the spring stretched o

IRINA_888 [86]

Answer:

a) $W=-0.0103125\ J$

b) $W=0.0059375\ J$

c) Compressing is easier

Explanation:

Given:

Expression of force:

$F=kx-bx^2+cx^3$

where:

$k=100\ N.m^{-1}$

$b=700\ N.m^{-2}$

$c=12000\ N.m^{-3}$

$x$ when the spring is stretched

$x$ when the spring is compressed

hence,

$F=100x-700x^2+12000x^3$

a)

From the work energy equivalence the work done is equal to the spring potential energy:

here the spring is stretched so, $x=-0.05\ m$

Now,

The spring constant at this instant:

$j=\frac{F}{x}$

$j=\frac{100\times (-0.05)-700\times (-0.05)^2+12000\times (-0.05)^3}{-0.05}$

$j=-8.25\ N.m^{-1}$

Now work done:

$W=\frac{1}{2} j.x^2$

$W=0.5\times -8.25\times (-0.05)^2$

$W=-0.0103125\ J$

b)

When compressing the spring by 0.05 m

we have, $x=0.05\ m$

<u>The spring constant at this instant:</u>

$j=\frac{F}{x}$

$j=\frac{100\times (0.05)-700\times (0.05)^2+12000\times (0.05)^3}{0.05}$

$j=4.75\ N.m^{-1}$

Now work done:

$W=\frac{1}{2} j.x^2$

$W=0.5\times 4.75\times (0.05)^2$

$W=0.0059375\ J$

c)

Since the work done in case of stretching the spring is greater in magnitude than the work done in compressing the spring through the same deflection. So, the compression of the spring is easier than its stretching.

8 0

2 years ago