Explanation: <u>Ohm's</u> <u>Law</u> is defined as the relationship between voltage and current: in a circuit, the voltage accross a conductor is directly proportional to the current.

The constant of proportionality is Resistance, whose unit is ohm (Ω).

so, Ohm's Law is

V = R.I

It is asked for the current, so, rearraging:

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

When added two more batteries, the voltage of the circuit triples:

$V_{final}=3V_{initial}$

Since they are directly proportional, if voltage triples, so does current:

$I_{final}=3I_{initial}$

$I_{final}=3(3.0)$

$I_{final}=$ 9.0

The new current, when driven by 3 batteries is 9.0 A.

3 0

3 years ago

What is the definition of the product of force and distance applied to an force

amid [387]

Answer:

that is the definition of work

Explanation:

3 0

3 years ago

The position vector of a particle of mass 1.70 kg as a function of time is given by r with arrow = (6.00 î + 5.70 t ĵ), where r

ollegr [7]

Answer:

The angular momentum of the particle is 58.14 kg m²/s along positive z- axis and is independent of time .

Explanation:

Given that,

Mass = 1.70 kg

Position vector $r=(6.00\hat{i}+5.70 t \hat{j})$

We need to calculate the angular velocity

The velocity is the rate of change of the position of the particle.

$v = \dfrac{dr}{dt}$

$v=\dfrac{d}{dt}(6.00\hat{i}+5.70 t \hat{j})$

$v=5.70\hat{j}$

We need to calculate the angular momentum of the particle

Using formula of angular momentum

$L=r\cdot p$

Where, p = mv

Put the value of p into the formula

$L=m(r\times v)$

Substitute the value into the formula

$L=1.70(6.00\hat{i}+5.70 t \hat{j}\times5.70\hat{j})$

$L=1.70\times34.2$

$L=58.14\ kgm^2/s$

Hence, The angular momentum of the particle is 58.14 kg m²/s along positive z- axis and is independent of time .

7 0

3 years ago

The battery for a certain cell phone is rated at 3.70 V. According to the manufacturer it can produce math]3.15 \times 10^{4} J[

AysviL [449]

Answer:

The average current that this cell phone draws when turned on is 0.451 A.

Explanation:

Given;

voltage of the phone, V = 3.7 V

electrical energy of the phone battery, E = 3.15 x 10⁴ J

duration of battery energy, t = 5.25 h

The power the cell phone draws when turned on, is the rate of energy consumption, and this is calculated as follows;

$P = \frac{E}{t}$

where;

P is power in watts

E is energy in Joules

t is time in seconds

$P = \frac{3.15*10^4}{5.25*3600s} = 1.667 \ W$

The average current that this cell phone draws when turned on:

P = IV

$I = \frac{P}{V} =\frac{1.667}{3.7} = 0.451 \ A$

Therefore, the average current that this cell phone draws when turned on is 0.451 A.

5 0

3 years ago