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Alecsey [184]
4 years ago
6

A battery does work on electric charges to bring them to a position of higher electric potential energy so that they can flow th

rough a circuit to a lower potential energy. the potential difference between the terminals of a battery, when no current flows to an external circuit, is referred to as the terminal voltage. the internal resistance of a battery decreases with decreasing temperature. a battery is a device that produces electricity by transforming chemical energy into electrical energy.\
Physics
1 answer:
Ipatiy [6.2K]4 years ago
7 0
(Missing question is: which of the following statements are true?)

a) <span>A battery does work on electric charges to bring them to a position of higher electric potential energy so that they can flow through a circuit to a lower potential energy --> TRUE
That's true: the battery does the work to move the charges to one end of its terminal thus creating a potential difference between the two terminals. Then, when it is connected to the circuit, charges start to flow to the terminal at lower potential (through the circuit)

b) </span><span>the potential difference between the terminals of a battery, when no current flows to an external circuit, is referred to as the terminal voltage --> FALSE
This is false: when no current flows, it is called e.m.f. (electromotive force)

c) </span><span>the internal resistance of a battery decreases with decreasing temperature 
--> TRUE
In fact, the dependence of a resistance with the temperature is:
</span>R=R_0[1+\alpha (T-T_0)]
with \alpha being generally positive, therefore the value of the resistance is proportional to T, and when T decreases, R decreases as well.

<span>d) a battery is a device that produces electricity by transforming chemical energy into electrical energy --> TRUE
</span><span>That's true: a battery uses chemical reactions to create a potential difference between the two terminals that can be exploited to make charges flowing through a circuit.


</span>
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It is recommended that drinking water contain 1.6 ppm fluoride (F−) for prevention of tooth decay. Consider a reservoir with a d
icang [17]

Answer:

total amount of fluoride 3.6 *10^6 g

Explanation:

we know that

volume of reservoir = \pi r^2 h

                                  =\pi* [\frac{4.70*10^2}{2}]^2 *14.2

                                  = 2.25*10^6 m^3

denso=ity of water = 1000 kg/m3

mass of water = 1000*2.25*10^6

                         = 2.25*10^9 kg = 2.25*10^12 g

fluoride 1.6 ppm

total amount of fluoride = \frac{1.6}{10^6} * 2.25*10^{12} g

                                         = 3.6 *10^6 g

7 0
3 years ago
An earthquake emits both S-waves and P-waves which travel at different speeds through the Earth. A P-wave travels at 9000 m/s an
Sergio [31]

Answer:

The distance between earthquake center and the measuring station is 1350 kilometers.

Explanation:

Let the earthquake center be at a distance of 'S' meters from the recording station.

Now from the basic relation of distance, speed and time we know that

Distance=Speed\times Time

For a Primary wave (P wave) let us assume that it appraoches the measuring station after t_{1} minutes

Thus making use of the above relation we have

Distance=V_{p}\times Time\\\\\therefore D=9000\times t_{1}.......(i)

Now since it is given that the secondary wave (S wave) reaches the measuring spot after 2 minutes or 120 seconds thus the time taken by secondary waves to reach recorder equals t_{1}+120 making use of the same relation we get

Distance=V_{s}\times Time\\\\\therefore D=5000\times (t_{1}+120).......(ii)

Solving equation 'i' and 'ii' we get

D=5000\times (\frac{D}{9000}+120)\\\\\therefore \frac{D}{5000}=\frac{D}{9000}+120\\\\\frac{D}{5000}-\frac{D}{9000}=120\\\\\therefore D=\frac{120}{\frac{1}{5000}-\frac{1}{9000}}=1350000meters=1350kilomerers

8 0
3 years ago
Plz help idk and it's due tomorrow
____ [38]
1) G. The continents are now farther apart. 2) A. P waves. They usually travel 60% faster than S waves. 3) H. A hot spot caused the Hawaiian Islands to form.
7 0
4 years ago
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While at the county fair, you decide to ride the Ferris wheel.Having eaten too many candy apples and elephant ears, you find the
Gwar [14]

Explanation:

ride. You estimate the radius of thebig wheel to be 15 {\rm m}, and you use your watch tofind that each loop around takes 25 {\rm s}.a.What is your speed

3 0
3 years ago
At t=0 a grinding wheel has an angular velocity of 25.0 rad/s. It has a constant angular acceleration of 26.0 rad/s2 until a cir
Agata [3.3K]

Answer:

a) The total angle of the grinding wheel is 569.88 radians, b) The grinding wheel stop at t = 12.354 seconds, c) The deceleration experimented by the grinding wheel was 8.780 radians per square second.

Explanation:

Since the grinding wheel accelerates and decelerates at constant rate, motion can be represented by the following kinematic equations:

\theta = \theta_{o} + \omega_{o}\cdot t + \frac{1}{2}\cdot \alpha \cdot t^{2}

\omega = \omega_{o} + \alpha \cdot t

\omega^{2} = \omega_{o}^{2} + 2 \cdot \alpha \cdot (\theta-\theta_{o})

Where:

\theta_{o}, \theta - Initial and final angular position, measured in radians.

\omega_{o}, \omega - Initial and final angular speed, measured in radians per second.

\alpha - Angular acceleration, measured in radians per square second.

t - Time, measured in seconds.

Likewise, the grinding wheel experiments two different regimes:

1) The grinding wheel accelerates during 2.40 seconds.

2) The grinding wheel decelerates until rest is reached.

a) The change in angular position during the Acceleration Stage can be obtained of the following expression:

\theta - \theta_{o} = \omega_{o}\cdot t + \frac{1}{2}\cdot \alpha \cdot t^{2}

If \omega_{o} = 25\,\frac{rad}{s}, t = 2.40\,s and \alpha = 26\,\frac{rad}{s^{2}}, then:

\theta-\theta_{o} = \left(25\,\frac{rad}{s} \right)\cdot (2.40\,s) + \frac{1}{2}\cdot \left(26\,\frac{rad}{s^{2}} \right)\cdot (2.40\,s)^{2}

\theta-\theta_{o} = 134.88\,rad

The final angular angular speed can be found by the equation:

\omega = \omega_{o} + \alpha \cdot t

If  \omega_{o} = 25\,\frac{rad}{s}, t = 2.40\,s and \alpha = 26\,\frac{rad}{s^{2}}, then:

\omega = 25\,\frac{rad}{s} + \left(26\,\frac{rad}{s^{2}} \right)\cdot (2.40\,s)

\omega = 87.4\,\frac{rad}{s}

The total angle that grinding wheel did from t = 0 s and the time it stopped is:

\Delta \theta = 134.88\,rad + 435\,rad

\Delta \theta = 569.88\,rad

The total angle of the grinding wheel is 569.88 radians.

b) Before finding the instant when the grinding wheel stops, it is needed to find the value of angular deceleration, which can be determined from the following kinematic expression:

\omega^{2} = \omega_{o}^{2} + 2 \cdot \alpha \cdot (\theta-\theta_{o})

The angular acceleration is now cleared:

\alpha = \frac{\omega^{2}-\omega_{o}^{2}}{2\cdot (\theta-\theta_{o})}

Given that \omega_{o} = 87.4\,\frac{rad}{s}, \omega = 0\,\frac{rad}{s} and \theta-\theta_{o} = 435\,rad, the angular deceleration is:

\alpha = \frac{ \left(0\,\frac{rad}{s}\right)^{2}-\left(87.4\,\frac{rad}{s} \right)^{2}}{2\cdot \left(435\,rad\right)}

\alpha = -8.780\,\frac{rad}{s^{2}}

Now, the time interval of the Deceleration Phase is obtained from this formula:

\omega = \omega_{o} + \alpha \cdot t

t = \frac{\omega - \omega_{o}}{\alpha}

If \omega_{o} = 87.4\,\frac{rad}{s}, \omega = 0\,\frac{rad}{s}  and \alpha = -8.780\,\frac{rad}{s^{2}}, the time interval is:

t = \frac{0\,\frac{rad}{s} - 87.4\,\frac{rad}{s} }{-8.780\,\frac{rad}{s^{2}} }

t = 9.954\,s

The total time needed for the grinding wheel before stopping is:

t_{T} = 2.40\,s + 9.954\,s

t_{T} = 12.354\,s

The grinding wheel stop at t = 12.354 seconds.

c) The deceleration experimented by the grinding wheel was 8.780 radians per square second.

4 0
4 years ago
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