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timama [110]
2 years ago
10

What keeps an inflated balloon from falling down if you rub it against your hair and place it against a wall? 1. rubbing distort

s the atoms inside the ballon and polarizes it. 2. rubbing leaves a balloon electrically charged; the charged balloon polarizes the wall. 3. when you rub the balloon against your hair, it will remove some m?
Physics
1 answer:
Keith_Richards [23]2 years ago
3 0
2 because when you are doing this it causes friction Which then cause the balloon to stick
~dany-ley
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A rigid, uniform bar with mass mmm and length bbb rotates about the axis passing through the midpoint of the bar perpendicular t
Pie

Answer:

I = \frac{mvb}{6}

Explanation:

we know angular velocity in terms of moment of inertia and angular speed

       L = Iω ....                        (1)

moment of inertia of rod rotating about its center of length b

 

      I = \frac{ mb^2}{12}  ........               .(2)  

using         v = ωr  

where w is angular velocity

and r is radius of  rod which is equal to b

        so we get  2v =  ωb  

                            ω  = 2v/b  .................            (3)    

here velocity is two time because two opposite ends  are moving opposite with a velocity v so net velocity will be 2v

put second and third equation in ist equation

                 L   =   \frac{mb^2}{12}×\frac{2v}{b}

              so final answer will be      L  =   \frac{mvb}{6}

7 0
2 years ago
A source charge generates an electric field of 4286 N/C at a distance of 2.5 m. What is the magnitude of the source charge?
mrs_skeptik [129]
The Answer is 3.0uc. I took the quiz.
8 0
2 years ago
Read 2 more answers
A long thin uniform rod of length 1.50 m is to be suspended from a frictionless pivot located at some point along the rod so tha
Dvinal [7]

Answer:

0.087 m

Explanation:

Length of the rod, L = 1.5 m

Let the mass of the rod is m and d is the distance between the pivot point and the centre of mass.

time period, T = 3  s

the formula for the time period of the pendulum is given by

T = 2\pi \sqrt{\frac{I}{mgd}}    .... (1)

where, I is the moment of inertia of the rod about the pivot point and g is the acceleration due to gravity.

Moment of inertia of the rod about the centre of mass, Ic = mL²/12

By using the parallel axis theorem, the moment of inertia of the rod about the pivot is

I = Ic + md²

I = \frac{mL^{2}}{12}+ md^{2}

Substituting the values in equation (1)

3 = 2 \pi \sqrt{\frac{\frac{mL^{2}}{12}+ md^{2}}{mgd}}

9=4\pi^{2}\times \left ( \frac{\frac{L^{2}}{12}+d^{2}}{gd} \right )

12d² -26.84 d + 2.25 =  0

d=\frac{26.84\pm \sqrt{26.84^{2}-4\times 12\times 2.25}}{24}

d=\frac{26.84\pm 24.75}{24}

d = 2.15 m , 0.087 m

d cannot be more than L/2, so the value of d is 0.087 m.

Thus, the distance between the pivot and the centre of mass of the rod is 0.087 m.

3 0
1 year ago
PLEASE HELP!!!!!! WILL GIVE BRAINLIEST TO WHOEVER ANSWERS WITH THE RIGHT ANSWER !!!!!!!! 
Solnce55 [7]

It would be B and D your welcome


7 0
2 years ago
Read 2 more answers
A 10N force pulls to the right and friction opposes 2N. If the object is 20kg,find the acceleraton.
zmey [24]

Force = mass * acceleration

10 N - 2 N = 20 kg * acceleration

8 N = 20 kg * acceleration

8 / 20 = acceleration

2/5 m/s^2 = acceleration

8 0
1 year ago
Read 2 more answers
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