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2 years ago

Air is colorless, odorless, and tasteless. describe one way that air can be shown to exist.

Physics

1 answer:

miskamm [114]2 years ago

3 0

Light a match or make a sound. It will tell us if there is any air present in the atmosphere or not

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Why did Rita's hands get hot when she rubbed them ?

Setler79 [48]

Answer:

due to production of heat through friction

4 0

3 years ago

Read 2 more answers

31. Draw a free body diagram for a 15.5N box that is being pushed to the right with a 18. N force while experiencing 4.30 N of r

posledela

Answer:

See answers below

Explanation:

F = mg,

15.5 N = m(9.8 m/s²)

m = 1.58 kg

Fnet = Applied force - resistance,

Fnet = 18 N - 4.30 N,

Fnet = 13.70 N

Fnet = ma

13.70 N = (1.58 kg)a

a = 8.67 m/s²

For the free body diagram, draw a box with an upward arrow labeled 15.5 N, a downward label labeled 15.5 N, a right label labeled 18 N, and a left label labeled 4.30 N.

7 0

2 years ago

a 2100-kg car drives with a speed of 18 m/s onb a flat road around a curve that has a radius of curvature of 83m. The coefficien

Law Incorporation [45]

Answer:

<u>The magnitude of the friction force is 8197.60 N</u>

Explanation:

Using the definition of the centripetal force we have:

$\Sigma F=ma_{c}=m\frac{v^{2}}{R}$

Where:

m is the mass of the car
v is the speed
R is the radius of the curvature

Now, the force acting in the motion is just the friction force, so we have:

$F_{f}=m\frac{v^{2}}{R}$

$F_{f}=2100\frac{18^{2}}{83}$

$F_{f}=8197.60 \: N$

<u>Therefore the magnitude of the friction force is 8197.60 N</u>

I hope it helps you!

7 0

3 years ago

What are 5 wave interactions with matter ?

vichka [17]

An echo
Refraction
Diffraction
Transmission
reflection

7 0

2 years ago

Equations to use: v= λ ∙ f v=d/t

Margarita [4]

b. 460.8 m/s

Explanation:

The relationship between the speed of the wave along the string, the length of the string and the frequency of the note is

$f=\frac{v}{2L}$

where v is the speed of the wave, L is the length of the string and f is the frequency. Re-arranging the equation and substituting the data of the problem (L=0.90 m and f=256 Hz), we can find v:

$v=2Lf=2(0.90 m)(256 Hz)=460.8 m/s$

c. 18,000 m

Explanation:

The relationship between speed of the wave, distance travelled and time taken is

$v=\frac{d}{t}$

where

v = 6,000 m/s is the speed of the wave

d = ? is the distance travelled

t = 3 s is the time taken

Re-arranging the formula and substituting the numbers into it, we find:

$d=vt=(6,000 m/s)(3 s)=18,000 m$

3 0

2 years ago