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kenny6666 [7]
3 years ago
12

The electrical charge of an atom as a whole is?

Physics
2 answers:
gayaneshka [121]3 years ago
8 0
<span>The electrical charge of an atom as a whole is?

Answer:</span><span>C. Neutral</span>

Natali5045456 [20]3 years ago
5 0
Im pretty sure its <span>C. Neutral.</span>
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A hydrometer is made of a tube of diameter 2.3cm.The mass of the tube and it's content is 80g. If it floats in a liquid density
iris [78.8K]

Answer:

The depth to which the hydrometer sinks is approximately 24.07 cm

Explanation:

The given parameters are;

The diameter of the hydrometer tube, d = 2.3 cm

The mass of the content of the tube, m = 80 g

The density of the liquid in which the tube floats, ρ = 800 kg/m³

By Archimedes' principle, the up thrust (buoyancy) force acting on the hydrometer = The weight of the displaced liquid

When the hydrometer floats, the up-thrust is equal to the weight of the hydrometer which by Archimedes' principle, is equal to the weight of the volume of the liquid displaced by the hydrometer

Therefore;

The weight of the liquid displaced = The weight of the hydrometer, W = m·g

Where;

g = The acceleration due to gravity ≈ 9.81 m/s²

∴ W = 80 g × g

The volume of the liquid that has a mass of 80 g (0.08 kg), V = m/ρ

V = 0.08 kg/(800 kg/m³) = 0.0001 m³ = 0.0001 m³ × 1 × 10⁶ cm³/m³ = 100 cm³

The volume of the liquid displaced = 100 cm³ = The volume of the hydrometer submerged, V_h

V_h = A × h

Where;

A = The cross-sectional area of the tube = π·d²/4

h = The depth to which the hydrometer sinks

h = V_h/A

∴ h = 100 cm³/( π × 2.3²/4 cm²) ≈ 24.07 cm

The depth to which the tube sinks, h ≈ 24.07 cm.

3 0
2 years ago
A computer is connected across a 110 V power supply. The computer dissipates 123 W of power in the form of electromagnetic radia
nexus9112 [7]

Answer:

81.3ohms

Explanation:

Resistance is known to provide opposition to the flow of electric current in an electric circuit.

Power dissipated by the computer is expressed as;

Power = current (I) × Voltage(V)

P = IV... (1)

Note that from ohms law, V = IR

I = V/R ... (2)

Substituting equation 2 into 1, we will have;

P = (V/R)×V

P = V²/R.. (3)

Given source voltage = 100V, Power dissipated = 123W

To get resistance R of the computer, we will substitute the given value into equation 3 to have

123 = 100²/R

R = 100²/123

R = 10,000/123

R = 81.3ohms

The resistance of the computer is 81.3ohms

3 0
3 years ago
Read 2 more answers
Given three different locations on Earth's surface, where will the weight of a person be greatest?
Feliz [49]

Answer:

Explanation:

In order to answer this question, we simply have to refer to the laws of the equations of gravitational mechanics.

The equation given by Newton tells us that  

F = \frac{Gm_{1} m_{2}  }{r^{2} }

In the case where we compare a specific place where the Force of Gravity is greater or lesser, we focus on the term assigned to the Planet's Radius.

In the case of G, m_{1} ,m_{2}, we understand that they are constant.

We can easily notice that the more the Radius (Height seen from a viewer on the ground), the lower the force will be.

In other words, the smaller the radius in which the measurement is made with respect to the center of the earth, the greater the gravitational force.

In that order of ideas the smallest radio has South Pole, which is about 6356 km from the center of the Earth on the Equator line

4 0
2 years ago
Which of the following is a correct equation for total energy?
Liula [17]

Answer:

kinetic energy + potential energy

6 0
3 years ago
Read 2 more answers
A current of 5 A is flowing in a 20 mH inductor. The energy stored in the magnetic field of this inductor is:_______
Kipish [7]

Answer:

C. 0.25J

Explanation:

Energy stored in the magnetic field of the inductor is expressed as E = 1/2LI² where;

L is the inductance

I is the current flowing in the inductor

Given parameters

L = 20mH = 20×10^-3H

I = 5A

Required

Energy stored in the magnetic field.

E = 1/2 × 20×10^-3 × 5²

E = 1/2 × 20×10^-3 × 25

E = 10×10^-3 × 25

E = 0.01 × 25

E = 0.25Joules.

Hence the energy stored in the magnetic field of this inductor is 0.25Joules

7 0
3 years ago
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