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2 years ago

5

A plastic rod is rubbed against a wool shirt, thereby acquiring a charge of −4.9 µc. how many electrons are transferred from the

wool shirt to the plastic rod? the elemental charge is 1.6 × 10−19 c

1 answer:

creativ13 [48]2 years ago

5 0

To find the number of electrons transferred, we should divide the total charge acquired by the rod $Q=-4.9 \mu C=-4.9 \cdot 10^{-6}C$ by the charge of a single electron ( $e=-1.6 \cdot 10^{-19}C$ ), and we find:
$N= \frac{Q}{e}= \frac{-4.9 \cdot 10^{-6}C}{-1.6 \cdot 10^{-19}C} =3.1 \cdot 10^{13}$

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Carbon is allowed to diffuse through a steel plate 15 mm thick. The concentrations of carbon at the two faces are 0.65 and 0.30

beks73 [17]

Answer:

T=575.16K

Explanation:

To solve the problem we proceed to use the 1 law of diffusion of flow,

Here,

$J=-D\frac{\Delta C}{\Delta x}$

$\Delta C$ is the rate in concentration

$\Delta x$ is the rate in thickness

D is the diffusion coefficient, where,

$D= D_0 exp(\frac{Q_d}{RT})$

Replacing D in the first law,

$J=-(D_0 exp(\frac{-Q_D}{RT}))\frac{\Delta }{\Delta x}$

clearing T,

$T=\frac{Q_d}{R*ln(\frac{J*\Delta x}{D_0*\Delta C})}$

Replacing our values

$T=-\frac{80000}{8.31*ln(\frac{(6.2*10^{-7})(-15*10^{-3})}{(1.43*10^{-9})(0.65-0.30)})}$

$T=-\frac{80000}{-138.09}$

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4 0

2 years ago

A 10 cm wide box is held between two springs in a 1 m gap on a frictionless surface. The left spring has a natural length of 80

Y_Kistochka [10]

Answer:

The distance is $x =0.291 \ m$

Explanation:

From the question we are told that

The width of the box is $b = 10 \ cm = \frac{10}{100} = 0.10 \ m$

The gap is of length $l = 1\ m$

The natural length of the first spring is $y = 80 \ cm = \frac{80 }{100} = 0.8 \ m$

The spring constant of the first spring is $k_1 = 200 N/m$

The second spring has a natural length of $z = 90 \ cm = \frac{90}{100} = 0.9 \ m$

The spring constant of the first spring is $k_2 = 350 \ N/m$

Let the distance from the center of the box to the left edge be x

So at equilibrium

The force applied by the first spring is

$F_1 = k_1 * (0.8 -x)$

and

The force applied by the second spring is

$F_2 = k_2 * [ 0.9 - (0.9 -x)]$

Now at equilibrium

$F_1 = F_2$

So

$k_1 * (0.8 -x) = k_2 * [ 0.9 - (0.9 -x)]$

substituting values

$200 * (0.8 -x) = 350 * [ 0.9 - (0.9 -x)]$

=> $160 -200x) = 350x$

=> $160 =550x$

=> $x =0.291 \ m$

6 0

2 years ago

A twig from a tree drops from a 200m high cliff on to a beach below.during its fall 40% twig's energy is converted into thermal

kolezko [41]

<span>Calculating for potential energy P.E=mgh
m=mass
g=acceleration due to gravity
h=height
</span>Let the twig's mass be M<span>
</span><span>
therefore at the start, the twig has 200*9.8*M potential energy.

</span>During its fall 40% twig's energy is converted into thermal energy and the remaining <span>60% of this will turn into kinetic energy during the fall, giving us 1176M
Now the equation for kinetic energy, K.E is 0.5*m*v^2
m is mass
v is velocity

Therefore 1176M=0.5*M*v^2 (mass will divide out)
velocity is </span><span>48.49 m/s</span>

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2 years ago