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2 years ago

An object has a mass of 785 g and a volume of 15 cm³. What is its density? (Give your answer in g/cm³ to 2 decimal places).

Physics

1 answer:

Anvisha [2.4K]2 years ago

4 0

Answer:

denisity = 52.33 g/c $m^{3}$

Explanation:

Density:

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

We have that m = 785 and that v = 15 c $m^{3}$ .

$d = \frac{785}{15}$

d = 52.33 $m^{3}$

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Two 5.0 mm × 5.0 mm electrodes are held 0.10 mm apart and are attached to 7.5 V battery. Without disconnecting the battery, a 0.

Musya8 [376]

Answer:

A) V = 7.5 V

B) E = 75,000 V/m

C) Q = 16.6 pC

D) V = 7.5 V

E) E = 24,000 V/m

F) Q = 52 pC

Explanation:

Given:

- The Area of plate A = ( 5 x 5 ) mm^2

- The distance between plates d = 0.10 mm

- The thickness of Mylar added t = 0.10 mm

- Voltage supplied by battery V = 7.5 V

Solution:

A) What is the capacitor's potential difference before the Mylar is inserted?

- The potential difference across the two plates is equal to the voltage provided by the battery V = 7.5 V which remains constant throughout.

B) What is the capacitor's electric field before the Mylar is inserted?

- The Electric Field E between the capacitor plates is given by:

E = V / k*d

k = 1 (air) E = 7.5 / 0.10*10^-3

E = 75,000 V/m

C) What is the capacitor's charge Q before the Mylar is inserted?

C = k*A*ε / d

k = 1 (air) C = ( 0.005^2 * 8.85*10^-12 ) / 0.0001

C = 2.213 pF

Q = C*V

Q = 7.5*(2.213)

Q = 16.6 pC

D) What is the capacitor's potential difference after the Mylar is inserted?

- The potential difference across the two plates is equal to the voltage provided by the battery V = 7.5 V which remains constant throughout.

E) What is the capacitor's electric field after the Mylar is inserted?

- The Electric Field E between the capacitor plates is given by:

E = V / k*d

k = 3.13 E = 7.5 / (3.13)0.10*10^-3

E = 24,000 V/m

F) What is the capacitor's charge after the Mylar is inserted?

C = k*A*ε / d

k = 3.13 C = 3.13*( 0.005^2 * 8.85*10^-12 ) / 0.0001

C = 6.927 pF

Q = C*V

Q = 7.5*(6.927)

Q = 52 pC

6 0

2 years ago

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

Explanation:

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

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2 years ago

Serena is a research student who has conducted an experiment on the discoloration of marble. Read about Serena’s experiment. The

sergij07 [2.7K]

The two flaws in her experiment’s design are

- She introduced at least one confounding variable. - She tried to test multiple hypotheses at a time

In the above mentioned experiment she had to have four samples to prove four hypotheses, each one separately and not to mix two hypotheses in an alone sample, that what it brings as consequence is the confusion.

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2 years ago

A scientist places a strip of hot metal on top of a block of another metal with a lower melting point. The second metal quickly

gayaneshka [121]

Answer:

Explanation:

Conduction because the heat energy is transferring directly from the separate metals convection involves "currents" like in a liquid or a gas. When you turn on the stove with a pot of water on the burner the water at the bottom of the pot gets heated first almost immediately this water rises and cooler water falls. Basically the water is "stirring" or shifting, exchanging the heat energy this is why water heats so evenly.

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2 years ago

A valuable statuette from a Greek shipwreck lies at the bottom of the Mediterranean Sea. The statuette has a mass of 10,566 g an

leonid [27]

Answer:

A) W = 103.55 N

B) mass of displaced water = 4186 g

C) W_displaced water = 41.06 N

D) Buoyant force = 41.06 N.

E) ZERO

F) 62.54 N

Explanation:

We are given;

mass of statuette;m = 10,566 g = 10.566 kg

volume = 4,064 cm³

Density of seawater;ρ = 1.03 g/mL = 1.03 g/cm³

A) The dry weight of the statuette can be calculated as;

W = mg

So;

W = 10.556 × 9.81

W = 103.55 N

B) Mass of displaced water is calculated from;

Density = mass/volume

So, mass = Density × Volume

m = 1.03 × 4,064 = 4186 g

C) Weight of displaced water is given by;

W_displaced water = (m_displaced water) × g

W_displaced water = 4.186 kg × 9.81 m/s^2 = 41.06 N

D) The buoyant force is the same as the weight of the displaced water.

Thus, Buoyant force = 41.06 N.

E) The apparent weight of the statuette is calculated from;

Apparent weight = Dry weight - Weight of displaced water

Apparent weight = 103.6 N - 41.06 N = 62.54 N. It is sitting on the bottom of the sea, so the sea floor is providing an opposite force that is equal but opposite the weight so that the net force on the statuette is zero. Since It has zero acceleration, in any direction, hence the net force on it is zero.

F. From E above, The Force required to lift the statuette = 62.54 N

4 0

2 years ago