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3 years ago

The potential difference across the terminals of a battery is 8. 4v when there is a current of 1. 50 a in the ba

Physics

1 answer:

Vikentia [17]3 years ago

7 0

Answer:

The potential difference across the terminals of a battery is 8. 4v when there is a current of 1. 50 a in the ba

Explanation:

The potential difference across the terminals of a battery is `8.4 V` when there is a current of `1.50 A` in the battery from the negative to the positive terminal. When the current is `3.50 Air`, the reverse direction, the potential difference becomes `9.4 V`. <br>(a) What is the internal resistance of the battery? <br>(b) What is the emf of the battery?

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Two bullets have masses of 3.0 g and 6.0 g respectively. Both are fired with a speed of 40.0 m/s.(a) Which bullet has more kinet

Stolb23 [73]

Answer:

(a) The bullet with a mass of 6.0 g

(b) $\frac{K_2}{K_1}=2$

Explanation:

(a) Kinetic energy is defined as:

$K=\frac{mv^2}{2}$

So, we have:

$K_1=\frac{3*10^{-3}kg(40\frac{m}{s})^2}{2}\\K_1=2.4J$

$K_2=\frac{6*10^{-3}kg(40\frac{m}{s})^2}{2}\\K_2=4.8J$

The bullet with a mass of 6.0 g have more kinetic energy

(b) We have $m_2=2m_1(1)$ and $v_1=v_2(2)$ . So:

$K_1=\frac{m_1v_1^2}{2}\\K_2=\frac{m_2v_2^2}{2}\\\frac{K_2}{K_1}=\frac{\frac{m_2v_2^2}{2}}{\frac{m_1v_1^2}{2}}(3)$

Replacing (1) and (2) in (3):

$\frac{K_2}{K_1}=\frac{\frac{2m_1v_1^2}{2}}{\frac{m_1v_1^2}{2}}\\\frac{K_2}{K_1}=2$

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3 years ago

100 kw of power is delivered to the other side of a city by a pair of power lines with the voltage difference of 13014.1 v.

FinnZ [79.3K]

A) The power delivered to the lines is
$P_{in}= 100 kW=1 \cdot 10^5 W$
And the voltage at which the lines work is
$V=13014.1 V$
Since the power delivered is the product between the voltage and the current:
$P=VI$
We can find the current flowing in the lines:
$I= \frac{P}{V}= \frac{1 \cdot 10^5 W}{13014.1 V}=7.68 A$

b) The voltage change along each line can be found by using Ohm's law:
$\Delta V = IR = (7.68 A)(10 \Omega)=76.8 V$

c) The power wasted as heat along each line is given by:
$P_d = I^2 R = (7.68 A)^2 (10 \Omega) = 590 W$
And since we have 2 lines, the total power wasted as heat in both lines is
$P_d = 2 \cdot 590 W=1180 W$

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3 years ago

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3 years ago

When you stretch a spring 33 cm past its natural length, it exerts a force of 6 N. What is the spring constant of this spring?

PilotLPTM [1.2K]

The spring formula:

   Force = - (spring constant) · (distance stretched or squashed)

   6 N = - (spring constant) · (33 cm)

   = - (0.33 m) · (spring constant)

Spring constant = - (6 N) / (0.33 m)

=    - 18.18 N/meter .

The negative signs means that the spring always wants to go in the
opposite direction from where you forced it, and return to its relaxed
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Harrizon [31]

Answer:

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Explanation:

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