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bazaltina [42]
3 years ago
13

A 5.00 Ω, a 10.0 Ω, and a 15.0 Ω resistor are connected in

Physics
1 answer:
Bad White [126]3 years ago
7 0

Answer: Option C) 3A

Explanation:

Given that,

Voltage = 90.0 Volts

Current I = ?

R1 = 5.0Ω

R2 = 10.0Ω

R3 = 15.0Ω

Since the resistors are connected in series, the total resistance (Rtotal) of the circuit is the sum of each resistance.

i.e Rtotal = R1 + R2 + R3

Rtotal = 5.0Ω + 10.0Ω + 15.0Ω = 30.0Ω

Then, apply the formula for ohms law

Voltage = Current x resistance

V = I x Rtotal

90.0V = I x 30.0Ω

I = 90.0V / 30.0Ω

I = 3 Amps (Amps is the unit of current)

Thus, the current in the circuit is 3 Amps

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First, let us derive our working equation. We all know that pressure is the force exerted on an area of space. In equation, that would be: P = F/A. From Newton's Law of Second Motion, force is equal to the product of mass and gravity: F = mg. So, we can substitute F to the first equation so that it becomes, P = mg/A. Now, pressure can also be determined as the force exerted by a fluid on an area. This fluid can be measure in terms of volume. Relating volume and mass, we use the parameter of density: ρ = m/V. Simplifying further in terms of height, Volume is the product of the cross-sectional area and the height. So, V = A*h. The working equation will then be derived to be:

P = ρgh

This type of pressure is called the hydrostatic pressure, the pressure exerted by the fluid over a known height. Next, we find the literature data of the density of seawater. From studies, seawater has a density ranging from 1,020 to 1,030 kg/m³. Let's just use 1,020 kg/m³. Substituting the values and making sure that the units are consistent:

P = (1,020 kg/m³)(9.81 m/s²)(11 km)*(1,000 m/1km)
P = 110,068,200 Pa or 110.07 MPa
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If a certain mass of mercury has a volume of 0.002 m^3 at a temperature of 20°c, what will be the volume at 50°c
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Answer:

Density of 127 I = \rm 1.79\times 10^{14}\ g/cm^3.

Also, \rm Density\ of\ ^{127}I=3.63\times 10^{13}\times Density\ of\ the\ solid\ iodine.

Explanation:

Given, the radius of a nucleus is given as

\rm r=kA^{1/3}.

where,

  • \rm k = 1.3\times 10^{-13} cm.
  • A is the mass number of the nucleus.

The density of the nucleus is defined as the mass of the nucleus M per unit volume V.

\rm \rho = \dfrac{M}{V}=\dfrac{M}{\dfrac 43 \pi r^3}=\dfrac{M}{\dfrac 43 \pi (kA^{1/3})^3}=\dfrac{M}{\dfrac 43 \pi k^3A}.

For the nucleus 127 I,

Mass, M = \rm 2.1\times 10^{-22}\ g.

Mass number, A = 127.

Therefore, the density of the 127 I nucleus is given by

\rm \rho = \dfrac{2.1\times 10^{-22}\ g}{\dfrac 43 \times \pi \times (1.3\times 10^{-13})^3\times 127}=1.79\times 10^{14}\ g/cm^3.

On comparing with the density of the solid iodine,

\rm \dfrac{Density\ of\ ^{127}I}{Density\ of\ the\ solid\ iodine}=\dfrac{1.79\times 10^{14}\ g/cm^3}{4.93\ g/cm^3}=3.63\times 10^{13}.\\\\\Rightarrow Density\ of\ ^{127}I=3.63\times 10^{13}\times Density\ of\ the\ solid\ iodine.

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3 years ago
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