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jeyben [28]
3 years ago
10

An ideal battery would produce an extraordinarily large current if "shorted" by connecting the positive and negative terminals w

ith a short wire of very low resistance. Real batteries do not. The current of a real battery is limited by the fact that the battery itself has resistance.
What is the resistance of a 9.0 V battery that produces a 17 A current when shorted by a wire of negligible resistance?

R=____
Physics
1 answer:
liubo4ka [24]3 years ago
6 0

Answer:

Resistance, R=0.529\ \Omega

Explanation:

Given that,

Voltage of the battery, V = 9 volts

Current produced in the circuit, I = 17 A

We need to find the resistance when shorted by a wire of negligible resistance. It is a case of Ohm's law. The voltage is given by :

V=IR

R=\dfrac{V}{I}

R=\dfrac{9\ V}{17\ A}

R=0.529\ \Omega

So, the resistance in the circuit is 0.529 ohms. Hence, this is the required solution.

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5 0
3 years ago
Calculate the amount of energy produced in joules by 100- watt light bulb lit for 2.5 hours.
9966 [12]
100/2.5 is 40.
40 is the energy that is being produced 
4 0
3 years ago
A block of mass 4 kilograms is initially moving at 5m/s on a horizontal surface. There is friction between the block and the sur
Dmitriy789 [7]

Answer:

d = 2.54 [m]

Explanation:

Through the theorem of work and energy conservation, we can find the work that is done. Considering that the energy in the initial state is only kinetic energy, while the energy in the final state is also kinetic, however, this is zero since the body stops.

E_{k1}+W=E_{k2}\\

where:

W = work [J]

Ek1 = kinetic energy at initial state [J]

Ek2 = kinetic energy at the final state = 0.

We must remember that kinetic energy can be calculated by means of the following expression.

\frac{1}{2} *m*v^{2}-W=0\\W= \frac{1}{2} *4*(5)^{2}\\W= 50 [J]

We know that work is defined as the product of force by distance.

W=F*d

where:

F = force [N]

d = distance [m]

But the friction force is equal to the product of the normal force (body weight) by the coefficient of friction.

f=m*g*0.5\\f = 4*9.81*0.5\\f = 19.62 [N]

Now solving the equation for the work.

d=W/F\\d = 50/19.62\\d = 2.54[m]

4 0
3 years ago
The Hi line of the Balmer series is emitted in the transition from n = 3 to n = 2. Compute the wavelength of this line for l H a
irina1246 [14]

Answer:

\lambda=550\ nm

Explanation:

The Hi line of the Balmer series is emitted in the transition from n = 3 to n = 2 i.e. n_i=3 and n_f=2

The wavelength of Hi line of the Balmer series is given by :

\dfrac{1}{\lambda}=R(\dfrac{1}{n_f}-\dfrac{1}{n_i})

\dfrac{1}{\lambda}=1.09\times 10^7\times (\dfrac{1}{2}-\dfrac{1}{3})

\lambda=5.50\times 10^{-7}\ m

\lambda=550\ nm

So, the wavelength for this line is 550 nm. Hence, this is the required solution.

6 0
3 years ago
135g of an unknown substance gains 9133 J of heat as it is heated from 25⁰C to 100⁰C. Using the chart below, determine the ident
telo118 [61]

Answer:

The unknown substance is Aluminum.

Explanation:

We'll begin by calculating the change in the temperature of substance. This can be obtained as follow:

Initial temperature (T₁) = 25 ⁰C

Final temperature (T₂) = 100 ⁰C

Change in temperature (ΔT) =?

ΔT = T₂ – T₁

ΔT = 100 – 25

ΔT = 75 ⁰C

Finally, we shall determine the specific heat capacity of the substance. This can be obtained as follow:

Change in temperature (ΔT) = 75 ⁰C

Mass of the substance (M) = 135 g

Heat (Q) gained = 9133 J

Specific heat capacity (C) of substance =?

Q = MCΔT

9133 = 135 × C × 75

9133 = 10125 × C

Divide both side by 10125

C = 9133 / 10125

C = 0.902 J/gºC

Thus, the specific heat capacity of substance is 0.902 J/gºC

Comparing the specific heat capacity (i.e 0.902 J/gºC) of substance to those given in the table above, we can see clearly that the unknown substance is aluminum.

7 0
3 years ago
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