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swat32
2 years ago
9

You need to determine the density of an unknown liquid and decide to perform an experiment. You notice that a wooden block float

s in this liquid with 4.6 cm of the side of the block submerged. Then you place the same block in water and notice that it also floats but with 5.8 cm submerged. What is the density of the unknown liquid.
Physics
1 answer:
Allushta [10]2 years ago
5 0

Answer:

pu = 1260.9kg/m^3

the density of the unknown liquid is 1260.9kg/m^3

Explanation:

The density of a liquid is inversely proportional to the volume (height) of object submerged in it.

High density liquid possess higher buoyant force preventing objects from submerging.

p ∝ 1/V ∝ 1/h

since V = Ah

pu/pw = hw/hu

pu = pwhw/hu

Where;

p = density

h = height submerged

pu and pw is the density of unknown liquid and water respectively

hu and hw is the height of object submerged in unknown liquid and water respectively

pw = 1000kg/m^3

hu = 4.6cm = 0.046m

hw = 5.8cm = 0.058m

Substituting the given values;

pu = 1000×0.058/0.046

pu = 1260.9kg/m^3

the density of the unknown liquid is 1260.9kg/m^3

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gaseous h2 and br2 are added to an evacuated 1.15L container kept at 298K. The intial partial pressurre of H2(g) is 0.782 atm an
Nastasia [14]

The partial pressures of HBr when the system reaches equilibrium is 2.4 X 10⁻¹¹ atm

<u>Explanation:</u>

H₂ + Br₂ ⇒ 2HBr

PH₂ = 0.782atm

PBr₂ = 0.493atm

Kp = (PHBr)²/ (PH₂) (PBr₂) = 1.4 X 10⁻²¹

At equilibrium:

Let 2x = pressure of HBr

PH₂ = 0.782 -x

PBr₂ = 0.493 - x

Kp = (2x)^2 / (0.782-x)(0.493-x)

Now, because Kp is very small, x will be very small compared to 0.782 and 0.493.

Then,

Kp = 1.4X10⁻²¹ = (4x²) / (0.782)(0.493)

x = 1.2X10⁻¹¹

PHBr = 2x = 2.4 X 10⁻¹¹ atm

Therefore, the partial pressures of HBr when the system reaches equilibrium is 2.4 X 10⁻¹¹ atm

3 0
2 years ago
A person is working on a steel structure while standing on the ground. An accident occurred where 5 A pass through the structure
matrenka [14]

Answer:

35mA

Explanation:

Hello!

To solve this problem we must use the following steps

1. Find the electrical resistance of the metal rod using the following equation

R=\alpha  \frac{l}{a}

WHERE

α=

metal rod resistivity=2x10^-4 Ωm

l=leght=2m

A=  Cross-sectional area

A=\frac{\pi }{4} d^2=\frac{\pi }{4} (0.06)^2=0.00283

solving

R=(2x10^-4)\frac{2}{0.00283} =0.14

2. Now we model the system as a circuit with parallel resistors, where we will call 1 the metal rod and 2 the man(see attached image)

3.we know that the sum of the currents in 1 and 2 must be equal to 5A, by the law of conservation of energy

I1+I2=5

4.as the voltage on both nodes is the same we can use ohm's law in resitance 1 and 2 (V=IR)

V1=V2

(0.14I1)=2000(i2)

solving for i1

I1=14285.7i2

5.Now we use the equation found in step 3

14285.7i2+i2=5

i2=\frac{5}{14285.7+1} =3.5x10^-4A=35mA

6 0
2 years ago
Select the correct answer from each drop-down menu.
bagirrra123 [75]

Answer:

(1) An object that’s negatively charged has more electrons than protons.

(2) An object that’s positively charged has fewer electrons than protons.

(3) An object that’s not charged has the same number of electrons than protons.

Explanation :

Objects have three subatomic particles that are Electrons, protons, and neutrons.

Protons and neutrons are found in the nucleus and electrons rotate or move outside the nucleus. Naturally, protons are positively charged, neutrons have no charge, and electrons are negatively charged.

Therefore, an object that is negatively charged has more electrons than protons.  An object that is not charged has the same number of electrons than protons. An object that is positively charged has fewer electrons than protons.

8 0
2 years ago
A 4.50-kg wheel that is 34.5 cm in diameter rotates through an angle of 13.8 rad as it slows down uniformly from 22.0 rad/s to 1
Mila [183]

Answer:

-10.9 rad/s²

Explanation:

ω² = ω₀² + 2α(θ - θ₀)

Given:

ω = 13.5 rad/s

ω₀ = 22.0 rad/s

θ - θ₀ = 13.8 rad

(13.5)² = (22.0)² + 2α (13.8)

α = -10.9 rad/s²

6 0
2 years ago
Read 2 more answers
A small object carrying a charge of -3.00 nc is acted upon by a downward force of 30.0 nn when placed at a certain point in an e
arlik [135]
The working equation for this one is:

E = F/Q, where E is the strength of the electric field, F is the electric force and Q is the charge. Substituting the corresponding values, the strength of the electric field is equal to

E = -30 nN/-3 nC
E = 10 nN/nC


3 0
2 years ago
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