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Nastasia [14]
3 years ago
9

Question in pic.. thanks!

Physics
1 answer:
11Alexandr11 [23.1K]3 years ago
3 0

That's false. Displacement would be (r2 - r1) .

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You walk from your bedroom, 25m to the mailbox, and then walk 25m back to your bedroom. What is your total distance? What is you
Yuki888 [10]

Answer:

Distance is 50m

Displacement is 0m

Explanation:

Distance is based on the amount of length you covered, regardless of where you end.

Displacement only considered where you started and where you ended, which is at the same spot in this case. Therefore, no displacement.

7 0
3 years ago
A tuba may be treated like a tube closed at one end. If a tuba has a fundamental frequency of 88.4 Hz, determine the first three
yulyashka [42]

Answer with Explanation:

We are given that

Fundamental frequency,f=88.4 Hz

Speed of sound,v=343 m/s

We have to find the first three overtones.

Tube is closed

The overtone of closed pipe  is equal to odd number of fundamental frequency.

Therefore, the overtone of tuba

f'=nf

Where n=3,5,7,..

Substitute n=3

f'=3\times 88.4=265.2Hz

For second overtone

f'=5\times 88.4=442Hz

For third overtone

f'=7\times 88.4=618.8Hz

4 0
3 years ago
Identify Label these cell parts
Kruka [31]

Answer:

I don't know this answer because I am not American

8 0
3 years ago
Read 2 more answers
You are trying to determine the specific gravity of a solid object that floats in water. If m is the mass of your object, mS is
Alisiya [41]

Answer:

Specific Gravity = m/[m(s)-m(os)]

Explanation:

Specific gravity, also called relative density, is the ratio of the density of a substance to the density of a reference substance. By this definition we need to find out the ratio of density of the object of mass m to the density of the surrounding liquid.

m = mass of the object

<u>Weight in air</u>

W (air) = mg, where g is the gravitational acceleration

<u>Weight with submerged with only one mass</u>

m(s)g + Fb = mg + m(b)g, <em>consider this to be equation 1</em>

where Fb is the buoyancy force

Weight with submerged with both masses

m(os)g + Fb’ = mg + m(b)g, <em>consider this to be equation 2</em>

<u>equation 1 – equation 2 would give us</u>

m(s)g – m(os)g = Fb’ – Fb

where Fb = D x V x g, where D is the density of the liquid the object is submerged in, g is the force of gravity and V is the submerged volume of the object

m(s)g – m(os)g = D(l) x V x g

m(s) – m(os) = D(l) x V

we know that Mass = Density x V, which in our case would be, D(b) x V, which also means

V = Mass/D(b), where D(b) is the density of the mass

<u>Substituting V into the above equation we get</u>

m(s) – m(os) = [D(l) x m)/ D(b)]

Rearranging to get the ratio of density of object to the density of liquid

D(b)/D(l) = m/[m(s)-m(os)], where D(b)/D(l) denotes the specific gravity

8 0
3 years ago
Frank has a sample of steel that has a mass of 80 grams if the density is 8g/cm3 what is the volume
nikklg [1K]
Density is mass divided by volume. rho=m/v. So, v=m/rho. In frank's case this is 80/8 = 10 cm^3.
7 0
3 years ago
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