Answer: When rubbing a balloon with a wool cloth, it puts negative charges on the balloon. Negative charges attract to positive charges. If a balloon is not rubbed with the wool cloth, it has an equal amount of negative to positive charges, so it will attract to a rubbed balloon.

8 0

3 years ago

Find the volume of a rectangular prism that is 8m long, 4m wide, and 300cm high

konstantin123 [22]

L=8m
B=4m
H=300cm=3m

$\\ \bull\tt\longmapsto Volume=LBH$

$\\ \bull\tt\longmapsto Volume=8(4)(3)$

$\\ \bull\tt\longmapsto Volume=96m^3$

6 0

2 years ago

Read 2 more answers

I stretch a rubber band and "plunk" it to make it vibrate in its fundamental frequency. I then stretch it to twice its length an

Nikitich [7]

Answer:

The new frequency (F₂ ) will be related to the old frequency by a factor of one (1)

Explanation:

Fundamental frequency = wave velocity/2L

where;

L is the length of the stretched rubber

Wave velocity = $\sqrt{\frac{T}{\frac{M}{L}}}$

Frequency (F₁) = $\frac{\sqrt{\frac{T}{\frac{M}{L}}}}{2*L}$

To obtain the new frequency with respect to the old frequency, we consider the conditions stated in the question.

Given:

L₂ =2L₁ = 2L

T₂ = 2T₁ = 2T

(M/L)₂ = 0.5(M/L)₁ = 0.5(M/L)

F₂ = $\frac{\sqrt{\frac{2T}{0.5(\frac{M}{L})}}}{4*L} = \frac{\sqrt{4(\frac{T}{\frac{M}{L}}})}{4*L} = \frac{2}{2} [\frac{\sqrt{\frac{T}{\frac{M}{L}}}}{2*L}] = F_1$

Therefore, the new frequency (F₂ ) will be related to the old frequency by a factor of one (1).

7 0

3 years ago