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Arte-miy333 [17]

2 years ago

11

Suppose you performed the experiment in atmosphere of Argon at 25 deg. C, (viscosity of argon is 2.26X10^-5 N.s/m^2 at that temp

erature), and measured terminal speed of 2.64X10^-5 m/s and weight-neutralizing voltage of 35.0 V. How many electrons have been attached to, or detached from the initially neutral plastic sphere

1 answer:

yuradex [85]2 years ago

7 0

Answer:

2*10^9electrons

Explanation:

Remember that the net force will be zero at terminal voltege so

Mg = 6πrng

At 35v

We have

qvr = 6πrng

q= 6 x 3.142* nx 2.6*10^-5/35

q,= 3.2x 10^ - 10C

So using n= q/e

= 3.2x 10^ - 10C/1.6*10-19

= 2*10^9electrons

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If a nucleus decays by gamma decay to a daughter nucleus, which of the following statements about this decay are correct? (There

GarryVolchara [31]

Answer: Option (b) is the correct answer.

Explanation:

A gamma particle is basically a photon of electromagnetic radiation with a short wavelength.

Symbol of a gamma particle is $^{0}_{0}\gamma$ . Hence, charge on a gamma particle is also 0.

For example, $^{234}_{91}Pa \rightarrow ^{234}_{91}Pa + ^{0}_{0}\gamma + Energy$

So, when a nucleus decays by gamma decay to a daughter nucleus then there will occur no change in the number of protons and neutrons of the parent atom but there will be loss of energy as a nuclear reaction has occurred.

Thus, we can conclude that the statement daughter nucleus has the same number of nucleons as the original nucleus., is correct about if a nucleus decays by gamma decay to a daughter nucleus.

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3 years ago

The horizontal force exerted between the tires of a 500kg car and the ground is 980N. if the car starts from rest, how far will

Maksim231197 [3]

Answer:

Explanation:

d = ½at²

d = ½(F/m)t²

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6 0

1 year ago

A uniform beam with mass M and length L is attached to the wall by a hinge, and supported by a cable. A mass of value 3M is susp

Jobisdone [24]

Answer:

The tension is $T= \frac{11}{2\sqrt{3} } Mg$

The horizontal force provided by hinge $Fx= \frac{11}{4\sqrt{3} } Mg$

Explanation:

From the question we are told that

The mass of the beam is $m_b =M$

The length of the beam is $l = L$

The hanging mass is $m_h = 3M$

The length of the hannging mass is $l_h = \frac{3}{4} l$

The angle the cable makes with the wall is $\theta = 60^o$

The free body diagram of this setup is shown on the first uploaded image

The force $F_x \ \ and \ \ F_y$ are the forces experienced by the beam due to the hinges

Looking at the diagram we ca see that the moment of the force about the fixed end of the beam along both the x-axis and the y- axis is zero

So

$\sum F =0$

Now about the x-axis the moment is

$F_x -T cos \theta = 0$

=> $F_x = Tcos \theta$

Substituting values

$F_x =T cos (60)$

$F_x= \frac{T}{2} ---(1)$

Now about the y-axis the moment is

$F_y + Tsin \theta = M *g + 3M *g ----(2)$

Now the torque on the system is zero because their is no rotation

So the torque above point 0 is

$M* g * \frac{L}{2} + 3M * g \frac{3L}{2} - T sin(60) * L = 0$

$\frac{Mg}{2} + \frac{9 Mg}{4} - T * \frac{\sqrt{3} }{2} = 0$

$\frac{2Mg + 9Mg}{4} = T * \frac{\sqrt{3} }{2}$

$T = \frac{11Mg}{4} * \frac{2}{\sqrt{3} }$

$T= \frac{11}{2\sqrt{3} } Mg$

The horizontal force provided by the hinge is

$F_x= \frac{T}{2} ---(1)$

Now substituting for T

$F_{x} = \frac{11}{2\sqrt{3} } * \frac{1}{2}$

$Fx= \frac{11}{4\sqrt{3} } Mg$

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liq [111]

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Valentin [98]

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