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Nadya [2.5K]
3 years ago
9

Using a series circuit, remove one battery from the circuit by re-wiring. How did it change the brightness of the bulbs? How doe

s the voltage of the batteries affect the brightness of the bulbs?
Physics
1 answer:
morpeh [17]3 years ago
8 0
The higher the voltage in series, the brighter the bulb gets. so removing one battery from the circuit causes a reduction in the brightness of the bulb.
hope this helps
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suppose a ball had a potential energy of 5 j when you dropped it. What would be its kinetic energy just as it hit the ground. (i
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Since we are ignoring air resistance which is a non-conservative force, the potential energy will be completely converted into kinetic energy, resulting in a final kinetic energy of 5J.
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Please select the word from the list that best fits the definition Visualizes where items are located
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Answer:I’m pretty sure it’s spatial

Explanation:

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If the force that propels the cannonball forward is 500N, how much force will move the cannon backward?
Licemer1 [7]

Answer:

Well it would be equal to 500N because pushing forward the ball (or whatever maybe a body) would push the canon back an even 500N backwards...

Explanation:

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3 years ago
Light from a laser of wavelength λ1 shines normally on a pair of narrow slits separated by distance D. This results in a differe
lana66690 [7]

Answer:

Explanation:

The angular position of first intensity maxima is given by the expression

= λ / d where λ is wave length of light and d is slit separation

putting the values given in the problem. ,

.02 = λ₁ / D

Now wavelength has been increased to 2λ₁

angular separation of first maxima

=  2λ₁ /D

= 2 x .02

= .04 radians.

6 0
3 years ago
An offshore oil well is 2 kilometers off the coast. The refinery is 4 kilometers down the coast. Laying pipe in the ocean is twi
shusha [124]

Answer:

Rectangular path

Solution:

As per the question:

Length, a = 4 km

Height, h = 2 km

In order to minimize the cost let us denote the side of the square bottom be 'a'

Thus the area of the bottom of the square, A = a^{2}

Let the height of the bin be 'h'

Therefore the total area, A_{t} = 4ah

The cost is:

C = 2sh

Volume of the box, V = a^{2}h = 4^{2}\times 2 = 128            (1)

Total cost, C_{t} = 2a^{2} + 2ah            (2)

From eqn (1):

h = \frac{128}{a^{2}}

Using the above value in eqn (1):

C(a) = 2a^{2} + 2a\frac{128}{a^{2}} = 2a^{2} + \frac{256}{a}

C(a) = 2a^{2} + \frac{256}{a}

Differentiating the above eqn w.r.t 'a':

C'(a) = 4a - \frac{256}{a^{2}} = \frac{4a^{3} - 256}{a^{2}}

For the required solution equating the above eqn to zero:

\frac{4a^{3} - 256}{a^{2}} = 0

\frac{4a^{3} - 256}{a^{2}} = 0

a = 4

Also

h = \frac{128}{4^{2}} = 8

The path in order to minimize the cost must be a rectangle.

8 0
3 years ago
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