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3 years ago

How much current is in a circuit that includes a 9.0 volt battery and a bulb woth a resistance of 4.0 ohms

Physics

2 answers:

Elza [17]3 years ago

8 0

Answer:

Current in the circuit, I = 2.25 A

Explanation:

Given that,

Voltage of the battery, V = 9 V

Resistance of the bulb, R = 4 ohms

We need to find the current in the circuit. It can be calculated using Ohm's law. Ohm's law gives the relationship between the current, resistance and the voltage as :

$V=IR$

$I=\dfrac{V}{R}$

$I=\dfrac{9\ V}{4\ \Omega}$

I = 2.25 A

So, the current flowing in the circuit is 2.25 amperes. Hence, this is the required solution.

Vitek1552 [10]3 years ago

7 0

According to Ohm's law for a portion of the circuit we have:

U=RI=>I=U/R=9/4=2.25 A

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Calculate the molarity of a 10. 0% cacl₂ solution. The density of the solution is 1. 0835 g/cm³.

brilliants [131]

The molarity of 10% CaCl2 is 0.9%

concentration of the given salt CaCl₂ = 10%

Density of a solution = 1.0835 g/cm³

Volume = m / d

= 100 / 1.0835

= 92.29 litres

Density = mass / volume

1.0835 × 92.29 = mass

mass = 99.99 gram

Thus the molarity can be calculated by = moles of solute / volume of solution multiplied by 100

= 0.9008/ 92.29 X 100 %

= 0.009 X 100 %

= 0.9 %

The molarity of 10% CaCl2 is 0.9%

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At the equator, the radius of the Earth is approximately 6370 km. A plane flies at a very low altitude at a constant speed of v

Anna007 [38]

To solve this problem we will apply the concepts related to the kinematic equations of linear motion. For this purpose we will define the speed as the distance traveled in a given period of time. Here the distance is equivalent to the orbit traveled around the earth, that is, a circle. Approaching the height of the aircraft with the radius of the earth, we will have the following data,

$R= 6370*10^3 m$

$v = 219m/s$

$a = 17m/s^2$

The circumference of the earth would be

$\phi = 2\pi R$

Velocity is defined as,

$v = \frac{x}{t}$

$t = \frac{x}{v}$

Here $x = \phi$ , then

$t = \frac{\phi}{v} = \frac{2\pi (6370*10^3)}{219}$

$t = 1.82*10^5s$

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3 years ago

At what speed, as a fraction of c , is a particle's total energy twice its rest energy

WINSTONCH [101]

The rest energy of a particle is
$E_0=m_0 c^2$
where $m_0$ is the rest mass of the particle and c is the speed of light.

The total energy of a relativistic particle is
$E=mc^2 = \frac{m_0 c^2}{ \sqrt{1- \frac{v^2}{c^2} } }$
where v is the speed of the particle.

We want the total energy of the particle to be twice its rest energy, so that
$E=2E_0$
which means:
$\frac{m_0c^2}{ \sqrt{1- \frac{v^2}{c^2} } }=2m_0 c^2$
$\frac{1}{ \sqrt{1- \frac{v^2}{c^2} } }=2$
From which we find the ratio between the speed of the particle v and the speed of light c:
$\frac{v}{c}= \sqrt{1- (\frac{1}{2})^2 } =0.87$
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3 years ago

The minimum frequency of light needed to eject electrons from a metal is called the threshold frequency, ???? 0 ν0 . Find the mi

Allisa [31]

Answer:

<h2>E = 2.8028*10⁻¹⁹ Joules</h2>

Explanation:

The minimum energy needed to eject electrons from a metal with a threshold frequency fo is expressed as E = hfo

h = planck's constant

fo = threshold frequency

Given the threshold frequency fo = 4.23×10¹⁴ s⁻¹

h = 6.626× 10⁻³⁴ m² kg / s

Substituting this value into the formula to get the energy E

E = 4.23×10¹⁴ * 6.626 × 10⁻³⁴

E = 28.028*10¹⁴⁻³⁴

E = 28.028*10⁻²⁰

E = 2.8028*10⁻¹⁹ Joules

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