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3 years ago

Can charging by friction occur only in solids? explain using an example.

1 answer:

8 0

The answer to the question can charging by friction occur only in solids is no. not at all. the static charge generated by following fluids can be hazardous. just as the static discharge when you touch a doorknob produces a spark, lightening is the spark from the static charge built up in the atmosphere.

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As part of your daily workout, you lie on your back and push with your feet against a platform attached to two stiff springs arr

zubka84 [21]

Answer:

$894.12\ \text{N}$

$284.013\ \text{J}$

Explanation:

U = Energy = 76 J

x = Displacement of spring = 0.17 m

k = Spring constant

Energy is given by

$U=\dfrac{1}{2}kx^2\\\Rightarrow k=\dfrac{2U}{x^2}\\\Rightarrow k=\dfrac{2\times 76}{0.17^2}\\\Rightarrow k=5259.51\ \text{Nm}$

Force is given by

$F=kx\\\Rightarrow F=5259.51\times 0.17\\\Rightarrow F=894.12\ \text{N}$

The magnitude of force must you apply to hold the platform is $894.12\ \text{N}$ .

Now $x=0.17+0.2=0.37\ \text{m}$

$U=\dfrac{1}{2}kx^2\\\Rightarrow U=\dfrac{1}{2}\times 5259.51\times 0.37^2\\\Rightarrow U=360.013\ \text{J}$

Additional energy

$360.013-76=284.013\ \text{J}$

8 0

2 years ago

The accepted standard measurement for mass is the A) liter. B) pound. C) Newton. D) kilogram.

andreev551 [17]

D) Kilograms is the standard unit of mass

6 0

3 years ago

An electric ceiling fan is rotating about a fixed axis with an initial angular velocity magnitude of 0.300 rev/s . The magnitude

Salsk061 [2.6K]

1) 1.2 m/s

First of all, we need to find the angular velocity of the blade at time t = 0.200 s. This is given by

$\omega_f = \omega_i + \alpha t$

where

$\omega_i = 0.300 rev/s$ is the initial angular velocity

$\alpha = 0.895 rev/s^2$ is the angular acceleration

Substituting t = 0.200 s, we find

$\omega_f = 0.300 + (0.895)(0.200)=0.479 rev/s$

Let's now convert it into rad/s:

$\omega_f = 2\pi \cdot 0.479 rev/s=3.01 rad/s$

The distance of a point on the tip of the blade is equal to the radius of the blade, so half the diameter:

$r=\frac{0.800}{2}=0.400 m$

And so now we can find the tangential speed at t = 0.200 s:

$v=\omega_f r =(3.01)(0.400)=1.2 m/s$

2) $2.25 m/s^2$

The tangential acceleration of a point rotating at a distance r from the centre of the circle is

$a_t = \alpha r$

where $\alpha$ is the angular acceleration.

First of all, we need to convert the angular acceleration into rad/s^2:

$\alpha = 0.895 rev/s^ \cdot 2 \pi =5.62 rad/s^2$

A point on the tip of the blade has a distance of

r = 0.400 m

From the centre; so, the tangential acceleration is

$a_t = (5.62)(0.400)=2.25 m/s^2$

3) $3.6 m/s^2$

The centripetal acceleration is given by

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

where

v is the tangential speed

r is the distance from the centre of the circle

We already calculate the tangential speed at point a):

v = 1.2 m/s

while the distance of a point at the end of the blade from the centre is

r = 0.400 m

Therefore, the centripetal acceleration is

$a=\frac{1.2^2}{0.400}=3.6 m/s^2$

7 0

3 years ago

A helicopter flies over the arctic ice pack at a constant altitude, towing an airborne 130-kg laser sensor that measures the thi

umka21 [38]

Answer:T=1316.21 N

Explanation:

The tension has two components: Vertical and Horizontal. The

horizontal component is ma, the vertical component is mg. Using

Pythagoras theorem, we can find the tension as:

T=((ma)^2 (mg)^2)^(1/2)

T=((129*2.84)^2 (129*9.8)^2)^(1/2)

T=1316.21 N

8 0

2 years ago

A 0.14-MIN baseball is dropped from rest. It has a momentum of 0.90 kg⋅m/skg⋅m/s just before it lands on the ground.

nikitadnepr [17]

The time spent in the air by the ball at the given momentum is 6.43 s.

The given parameters;

<em>momentum of the ball, P = 0.9 kgm/s</em>
<em>weight of the ball, W = 0.14 N</em>

The impulse experienced by the ball is calculated as follows;

$Ft = \Delta P$

where;

$Ft$ is impulse

$\Delta P$ is change in momentum

The time of motion of the ball is calculated as follows;

$t = \frac{\Delta P}{F} \\\\t = \frac{0.9 - 0}{0.14} \\\\t = 6.43 \ s$

Thus, the time spent in the air by the ball at the given momentum is 6.43 s.

Learn more here:brainly.com/question/13468390

7 0

1 year ago