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Vinvika [58]
3 years ago
9

A friend says that the reason one's hair stands out while touching a charged Van de Graaff generator is simply that the hair str

ands become charged and are light enough so that the repulsion between strands is visible. Do you agree or disagree
Physics
1 answer:
Llana [10]3 years ago
5 0

Answer:

Explanation:

Yes I agree with the statement .

When a person who is perfectly insulated from the earth , touches a Van de Graaff , his body acquires charge . when the hair acquires it, it stands out due to mutual repulsion . It is to be noted here that at pointed areas on a surface , there is larger accumulation of charge. Accumulation of charge is greater at hair tops .

It is also a general observation that when a bird sits on high tension wire , his feather stands out due to the same reason.

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A −3.0 nC charge is on the x-axis at x=−9 cm and a +4.0 nC charge is on the x-axis at x=16 cm. At what point or points on the y-
alexdok [17]

Answer:

y = 10.2 m

Explanation:

It is given that,

Charge, q_1=-3\ nC

It is placed at a distance of 9 cm at x axis

Charge, q_2=+4\ nC

It is placed at a distance of 16 cm at x axis

We need to find the point on the y-axis where the electric potential zero. The net potential on y-axis is equal to 0. So,

\dfrac{kq_1}{r_1}+\dfrac{kq_2}{r_2}=0

Here,

r_1=\sqrt{y^2+9^2} \\\\r_2=\sqrt{y^2+15^2}

So,

\dfrac{kq_1}{r_1}=-\dfrac{kq_2}{r_2}\\\\\dfrac{q_1}{r_1}=-\dfrac{q_2}{r_2}\\\\\dfrac{-3\ nC}{\sqrt{y^2+81} }=-\dfrac{4\ nC}{\sqrt{y^2+225} }\\\\3\times \sqrt{y^2+225}=4\times \sqrt{y^2+81}

Squaring both sides,

3\times \sqrt{y^2+225}=4\times \sqrt{y^2+81}\\\\9(y^2+225)=16\times (y^2+81)\\\\9y^2+2025=16y^2-+1296\\\\2025-1296=7y^2\\\\7y^2=729\\\\y=10.2\ m

So, at a distance of 10.2 m on the y axis the electric potential equals 0.

8 0
3 years ago
Question 5 of 10
muminat
The correct answer to this question is D
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3 years ago
What dose nuclear reactions produce?
aleksandrvk [35]

Answer:

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A 360-N child is in a swing that is attached to a pair of ropes 1.90 m long. Find the gravitational potential energy of the chil
Igoryamba

Answer:

684J

Explanation:

So basically the formula for gravitational potential energy is Mass X Gravity X height. That is G.p.e = mgh

We don't have the mass but since we have the height, we multiply directly with the height since the quantity of weight is already given.

so G.p.e = 360 X 1.9 = 684J

Note that; The answer is in joules because g.p.e is work done.

Hope that was helpful!!

7 0
3 years ago
The pendulum consists of two slender rods AB and OC which have a mass of 3 kg/m. The thin plate has a mass of 12 kg/m2 . a) Dete
jeka57 [31]

Answer:

The answer is below

Explanation:

a) The location ӯ of the center of mass G of the pendulum is given as:

y=\frac{0+(\pi*(0.3\ m) ^2*12kg/m^2*1.8\ m-\pi*(0.1\ m) ^2*12kg/m^2*1.8\ m)+0.75\ m*1.5\ m *3\ kg/m}{(\pi*(0.3\ m) ^2*12kg/m^2-\pi*(0.1\ m) ^2*12kg/m^2)+3\ kg/m^2*0.8\ m+3\ kg/m^2*1.5\ m} \\\\y=0.88\ m

b)  the mass moment of inertia about z axis passing the rotation center O is:

I_G=\frac{1}{12}*3(0.8)(0.8)^2+ 3(0.8)(0.888)^2-\frac{1}{2}*(12)(\pi)(0.1)^2(0.1)^2 -(12)(\pi)(0.1)^2(1.8-\\0.888)^2+\frac{1}{2}*(12)(\pi)(0.3)^2(0.3)^2 +(12)(\pi)(0.3)^2(1.8-0.888)^2+\frac{1}{12}*3(1.5)(1.5)^2+\\3(1.5)(0.888-0.75)^2\\\\I_G=13.4\ kgm^2

c) The mass moment of inertia about z axis passing the rotation center O is:

I_o=\frac{1}{12}*3(0.8)(0.8)^2+ \frac{1}{3}* 3(1.5)(1.5)^2+\frac{1}{2}*(12)(\pi)(0.3)^2(0.3)^2 +(12)(\pi)(0.3)^2(1.8)^2-\\\frac{1}{2}*(12)(\pi)(0.1)^2(0.1)^2 -(12)(\pi)(0.1)^2(1.8)^2\\\\I_o=13.4\ kgm^2

3 0
3 years ago
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