All categories

3 years ago

True or False? A mechanical wave needs a medium to transfer energy. True False

Physics

1 answer:

Vlada [557]3 years ago

6 0

Answer:

true I think

Explanation:

sorry if I'm wrong, have a good day:)

You might be interested in

A helicopter lifts 35m vertically into the air and then comes back down to the ground.

Phoenix [80]

Explanation:

Distance = Total path length = 35m + 35m = 70m

Displacement = Final - Initial position = 0m

5 0

3 years ago

I MARK BRAINLIEST, PLEASE ANSWER ASAP!!! PHYSICS 20 POINTS!!

irina [24]

The first one is C. Hope this helps!! If you need anymore help just message me and I will try to get back to you quickly and help in any way i can!

5 0

3 years ago

Read 2 more answers

The density of Mercury is 1.36 × 10 by 4 Kgm - 3 at 0 degrees. Calculate its value at 100 degrees and at 22 degrees. Take cubic

Drupady [299]

a) Density at 100 degrees: $1.34\cdot 10^4 kg/m^3$

Explanation:

The density of mercury at 0 degrees is $d=1.36\cdot 10^4 kg/m^3$

Let's take 1 kg of mercury. Its volume at 0 degrees is

$V=\frac{m}{d}=\frac{1 kg}{1.36\cdot 10^4 kg/m^3}=7.35\cdot 10^{-5} m^3$

The formula to calculate the volumetric expansion of the mercury is:

$\Delta V= \alpha V \Delta T$

where

$\alpha=180\cdot 10^{-6} K^{-1}$ is the cubic expansivity of mercury

V is the initial volume

$\Delta T$ is the increase in temperature

In this part of the problem, $\Delta T=100 C-0 C=100 C=100 K$

So, the expansion is

$\Delta V= \alpha V \Delta T=(180\cdot 10^{-6} K^{-1})(7.35\cdot 10^{-5} m^3)(100 K)=1.3\cdot 10^{-6} m^3$

So, the new density is

$d'=\frac{m}{V+\Delta V}=\frac{1 kg}{7.35\cdot 10^{-5} m^3+1.3\cdot 10^{-6} m^3}=1.34\cdot 10^4 kg/m^3$

b) Density at 22 degrees: $1.355\cdot 10^4 kg/m^3$

We can apply the same formula we used before, the only difference here is that the increase in temperature is

$\Delta T=22 C-0 C=22 C=22 K$

And the volumetric expansion is

$\Delta V= \alpha V \Delta T=(180\cdot 10^{-6} K^{-1})(7.35\cdot 10^{-5} m^3)(22 K)=2.9\cdot 10^{-7} m^3$

So, the new density is

$d'=\frac{m}{V+\Delta V}=\frac{1 kg}{7.35\cdot 10^{-5} m^3+2.9\cdot 10^{-7} m^3}=1.355\cdot 10^4 kg/m^3$

8 0

3 years ago

An image cannot be projected and forms where light rays appear to originate.

irakobra [83]

An VIRTUAL image cannot be projected, and

forms where light rays appear to originate.

3 0

3 years ago

How does an oxygen concentrator machine work?

zaharov [31]

-It takes air from it's surroundings.
-Compresses the air
-Removes Nitrogen
-Delivers an electric Interface
-Then it delivers the purified oxygen

3 0

4 years ago