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Alecsey [184]
3 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]3 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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A sound wave has a frequency of 250 Hz and a wavelength of 2.5 m. What is the speed of the wave?
Digiron [165]

ANSWER

625 m/s

EXPLANATION

Given:

• The frequency of the sound wave, f = 250 Hz

,

• The wavelength, λ = 2.5 m

Find:

• The speed of the wave, v

The speed of a wave of wavelength λ and frequency f is given by,

v=f\cdot\lambda

Substitute the known values and solve,

v=250Hz\cdot2.5m=625m/s

Hence, the speed of the wave is 625 m/s.

3 0
1 year ago
A video game includes an asteroid that is programmed to move in a straight line across a 17-inch monitor according to the equati
Ainat [17]

Answer:

The asteroid's acceleration at this point is 2.71\ m/s^2

Explanation:

The equation that governs the trajectory of asteroid is given by :

x=6.5t-2.3t^3

The velocity of asteroid is given by :

v=\dfrac{dx}{dt}\\\\v=\dfrac{d(6.5t-2.3t^3)}{dt}\\\\v=6.5-6.9t^2

At some point during the trip across the screen, the asteroid is at rest. It means, v = 0

So,

6.5-6.9t^2=0\\\\t=0.971\ s                      

Acceleration,

a=\dfrac{dv}{dt}\\\\a=\dfrac{d(6.5-6.9t^2)}{dt}\\\\a=-13.8t                        

Put t = 0.971 s

a=-13.8\times 0.197\\\\a=-2.71\ m/s^2

So, the asteroid's acceleration at this point is 2.71\ m/s^2 and it is decelerating.

6 0
3 years ago
A man is standing on the edge of a 20.0 m high cliff. He throws a rock horizontally with an initial velocity of 10.0 m/s. How lo
Arlecino [84]

Answer:

t = 4.08 s

R = 40.8 m

Explanation:

The question is asking us to solve for the time of flight and the range of the rock.

Let's start by finding the total time it takes for the rock to land on the ground. We can use this constant acceleration kinematic equation to solve for the displacement in the y-direction:

  • Δx = v_0 t + 1/2at²

We have these known variables:

  • (v_0)_y = 0 m/s
  • a_y = -9.8 m/s²
  • Δx_y = -20 m

And we are trying to solve for t (time). Therefore, we can plug these values into the equation and solve for t.

  • -20 = 0t + 1/2(-9.8)t²
  • -20 = 1/2(-9.8)t²
  • -20 = -4.9t²
  • t = 4.08 sec

The time it takes for the rock to reach the ground is 4.08 seconds.

Now we can use this time in order to solve for the displacement in the x-direction. We will be using the same equation, but this time it will be in terms of the x-direction.

List out known variables:

  • v_0 = 10 m/s
  • t = 4.08 s
  • a_x = 0 m/s

We are trying to solve for:

  • Δx_x = ?

By using the same equation, we can plug these known values into it and solve for Δx.

  • Δx = 10 * 4.08 + 1/2(0)(4.08)²
  • Δx = 10 * 4.08
  • Δx = 40.8 m

The rock lands 40.8 m from the base of the cliff.

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