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REY [17]
2 years ago
8

Two 1.50-V batteries-with their positive terminals in the same direction-are inserted in series into a flashlight. One battery h

as an internal resistance of 0.255ω , and the other has an internal resistance of 0.15ω . When the switch is closed, the bulb carries a current of 600mA .(a) What is the bulb's resistance?
Physics
1 answer:
DIA [1.3K]2 years ago
8 0

Based on the voltage of the batteries and the internal resistance, the bulb's resistance is 4.595 Ω

<h3>What is the resistance?</h3>

First, find the total emf of the circuit:

= 2 x 1.5V

= 3.0V

The internal resistance is:

= 0.255 + 0.15

= 0.405 Ω

The resistance in the bulb is:

= 3.0 / 0.600 - 0.405

= 4.595 Ω

Find out more on resistance at brainly.com/question/24974191

#SPJ4

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Answer: Base units

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Number of waves that pass a given point in one second
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A 25 kg block is held against a compressed spring and then the spring is allowed to decompress giving the block a velocity. The
Alex787 [66]

Answer:

h=18.05 cm

Explanation:

Given that

m= 25 kg

K= 1300 N/m

x=26.4 cm

θ= 19.5 ∘

When the block just leave the spring then the speed of block = v m/s

From energy conservation

\dfrac{1}{2}Kx^2=\dfrac{1}{2}mv^2

Kx^2=mv^2

v=\sqrt{\dfrac{kx^2}{m}}

By putting the values

v=\sqrt{\dfrac{kx^2}{m}}

v=\sqrt{\dfrac{1300\times 0.264^2}{25}}

v=1.9 m/s

When block reach at the maximum height(h) position then the final speed of the block will be zero.

We know that

V_f^2=V_i^2-2gh

By putting the values

0^2=1.9^2-2\times 10\times h

h=0.1805 m

h=18.05 cm

4 0
3 years ago
A 2498 kg car is moving at 17.1 m/s slams on its brakes and slows to 2.6 m/s. What is the magnitude (absolute value) of the impu
bija089 [108]

Answer:

<em>J=36221 Kg.m/s</em>

Explanation:

<u>Impulse-Momentum Theorem</u>

These two magnitudes are related in the following way. Suppose an object is moving at a certain speed v_1 and changes it to v_2. The impulse is numerically equivalent to the change of linear momentum. Let's recall the momentum is given by

p=mv

The initial and final momentums are, respectively

p_1=mv_1,\ p_2=mv_2

The change of momentum is

\Delta p=p_2-p_1=m(v_2-v_1)

It is numerically equal to the Impulse J

J=\Delta p

J=m(v_2-v_1)

We are given

m=2498\ kg,\ v_1=17.1\ m/s,\ v_2=2.6\ m/s

The impulse the car experiences during that time is

J=2498(2.6-17.1)=2498(-14.5)

J=-36221 Kg.m/s

The magnitude of J is

J=36221 Kg.m/s

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Andreas93 [3]

Answer:

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