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2 years ago

14

The newest generation of smart phone uses three different batteries connected in a series simultaneously to extend battery life

and generate the energy necessary to power the device. If each battery is rated at 2.4 V , what is the overall voltage provided to the phone?

1 answer:

azamat2 years ago

8 0

Answer:

7.2 V

Explanation:

The three batteries are connected in series to the terminals of the phone: it means that they are connected along the same branch, so the current flowing through them is the same.

This also means that the potential difference across the phone will be equal to the sum of the voltages provided by each battery.

Here, the voltage provided by each battery is

V = 2.4 V

So, the overall voltage will be

V = 2.4 V + 2.4 V + 2.4 V = 7.2 V

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When a disaster is declared, the Federal government, led by the Federal Emergency Management Agency (FEMA), responds at the request of, and in support of, States, Tribes, Territories, and Insular Areas and local jurisdictions impacted by a disaster.

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2 years ago

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3 years ago

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Which has the most momentum?

boyakko [2]

Answer:

Both objects have the same magnitude of momentum.

Explanation:

If an object of mass $m$ is moving at a velocity of $v$ , the momentum of that object would be $m\, v$ .

The $100\; {\rm g}$ ( $0.1\; {\rm kg}$ ) object is moving at a speed of $1\; {\rm m\cdot s^{-1}}$ . The magnitude of the momentum of this object would be $0.1\; {\rm kg} \times 1\; {\rm m\cdot s^{-1}} = 0.1\; {\rm kg \cdot m \cdot s^{-1}}$ .

Similarly, the momentum of the $1\; {\rm g}$ ( $10^{-3}\; {\rm kg}$ ) object moving at a speed of $100\; {\rm m\cdot s^{-1}}$ would be $10^{-3}\; {\rm kg} \times 100\; {\rm m\cdot s^{-1}} = 0.1\; {\rm kg \cdot m \cdot s^{-1}}$ .

Hence, the magnitude of momentum is the same for the two objects.

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1 year ago

slader Question: A Model Rocket Is Launched Straight Upward With An Initial Speed Of 50m/s. Iit Accelerates With A Constant Upwa

xenn [34]

Answer:

Maximum height reached by the rocket is

$y_{max} = 308 m$

total time of the motion of rocket is given as

$T = 16.44 s$

Explanation:

Initial speed of the rocket is given as

$v_i = 50 m/s$

acceleration of the rocket is given as

$a = 2 m/s^2$

engine stops at height h = 150 m

so the final speed of the rocket at this height is given as

$v_f^2 - v_i^2 = 2 a d$

$v_f^2 - 50^2 = 2(2)(150)$

$v_f = 55.68 m/s$

so maximum height reached by the rocket is given as the height where its final speed becomes zero

so we will have

$v_f^2 - v_i^2 = 2 a d$

$0 - 55.68^2 = 2(-9.81)(y - 150)$

$y_{max} = 308 m$

Now the total time of the motion of rocket is given as

1) time to reach the height of 150 m

$v_f - v_i = at$

$55.68 - 50 = 2 t$

$t_1 = 2.84 s$

2) time to reach ground from this height

$\Delta y = v_y t + \frac{1}{2}gt^2$

$-150 = 55.68 t - \frac{1}{2}(9.81) t^2$

$t_2 = 13.6 s$

so total time of the motion of rocket is given as

$T = 13.6 + 2.84 = 16.44 s$

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3 years ago