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Allisa [31]
4 years ago
12

An electric buzzer is activated, then sealed inside a glass chamber. When all of the air is pumped out of the chamber, how is th

e sound from the buzzer affected?
Physics
1 answer:
Aleksandr-060686 [28]4 years ago
4 0
The sound is increased because sound waves are in fact mech. waves which means the  that they can't travel through empty space and thus need a medium to travel through
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which criteria are responsible for deciding whether a heterogeneous mixture is a colloid or a suspension
ad-work [718]
I think you forgot to give the options along with the question. I am answering the question based on my knowledge and research. The criteria responsible for deciding whether a heterogeneous mixture is a colloid or a suspension is whether the <span>particles remain suspended for an extended period of time. I hope it helps you.</span>
6 0
4 years ago
A cat with a mass of 5.00 kg pushes on a 25.0 kg desk with a force of 50.0N to jump off. What is the force on the desk?
olya-2409 [2.1K]

Answer:

First of all the formula is F= uR,( force= static friction× reaction)

mass= 5+25=30

F= 50

R= mg(30×10)=300

u= ?

F=UR

u= F/R

u= 50/300=0.17N

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3 years ago
Define oxidation number
Delicious77 [7]

<span>a number assigned to an element in a chemical combo that represents the number of electrons lost or gained by atom of the element in the compound.</span>

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3 years ago
Which most likely indicates a chemical change has occurred?
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green liquid becoming a red liquid

4 0
3 years ago
Read 2 more answers
Unpolarized light with intensity I0I0I_0 is incident on an ideal polarizing filter. The emerging light strikes a second ideal po
zvonat [6]

Answer:

0.293I_0

Explanation:

When the unpolarized light passes through the first polarizer, only the component of the light parallel to the axis of the polarizer passes through.

Therefore, after the first polarizer, the intensity of light passing through it is halved, so the intensity after the first polarizer is:

I_1=\frac{I_0}{2}

Then, the light passes through the second polarizer. In this case, the intensity of the light passing through the 2nd polarizer is given by Malus' law:

I_2=I_1 cos^2 \theta

where

\theta is the angle between the axes of the two polarizer

Here we have

\theta=40^{\circ}

So the intensity after the 2nd polarizer is

I_2=I_1 (cos 40^{\circ})^2=0.587I_1

And substituting the expression for I1, we find:

I_2=0.587 (\frac{I_0}{2})=0.293I_0

5 0
3 years ago
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