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Marysya12 [62]
3 years ago
15

When connected to a battery, a lightbulb glows brightly. If the battery is reversed and reconnected to the bulb, the bulb will g

low (a) brighter (b) dimmer (c) with the same brightness (d) not at all
Physics
1 answer:
attashe74 [19]3 years ago
8 0

Answer: c) with the same brightness

Explanation: The load in this case the bulb, is not polarized ( it has no positive and negative points) thus any connection relative to the battery (source) will have no effect on it brightness.

Also, brightness is a function of current and in this case the voltage ( from battery) and resistance of load (bulb) is constant, and according to ohms law (V=IR) if the current is constant at the first connection, it will be the same at the reversed connection.

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<h2>Answer:</h2>

<em>Hello, </em>

<h3><u>QUESTION)</u></h3>

<em>✔ We have: KE = PE (potential energy) </em>

<em>PE = m x g x h </em>

The potential energy that the pebble of mass 1 has is called PE1 and the potential energy that the pebble of mass 2 has is called PE2  

PE1 = PE2 ⇔ PE1/PE2 = 1

\frac{m_1\times g\times h}{m_2\times g\times 4h} = 1 \\ \\  \frac{m_1}{m_2\times 4} = 1 \\ \\  \frac{m_1}{m_2} = 4

The mass m1 is therefore 4 times greater than that of the stone of mass m2.

 

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2 years ago
The circumference of a sphere was measured to be
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To solve this problem we will apply the concepts related to the calculation of the surface, volume and error through the differentiation of the formulas given for the calculation of these values in a circle. Our values given at the beginning are

\phi = 76cm

Error (dr) = 0.5cm

The radius then would be

\phi = 2\pi r \\76cm = 2\pi r\\r = \frac{38}{\pi} cm

And

\frac{d\phi}{dr} = 2\pi \\d\phi = 2\pi dr \\0.5 = 2\pi dr

PART A ) For the Surface Area we have that,

A = 4\pi r^2 \\A = 4\pi (\frac{38}{\pi})^2\\A = \frac{5776}{\pi}

Deriving we have that the change in the Area is equivalent to the maximum error, therefore

\frac{dA}{dr} = 4\pi (2r) \\dA = 4r (2\pi dr)

Maximum error:

dA = 4(\frac{38}{\pi})(0.5)

dA = \frac{76}{\pi}cm^2

The relative error is that between the value of the Area and the maximum error, therefore:

\frac{dA}{A} = \frac{\frac{76}{\pi}}{\frac{5776}{\pi}}

\frac{dA}{A} = 0.01315 = 1.31\%

PART B) For the volume we repeat the same process but now with the formula for the calculation of the volume in a sphere, so

V = \frac{4}{3} \pi r^3

V = \frac{4}{3} \pi (\frac{38}{\pi})^3

V = \frac{219488}{3\pi^2}

Therefore the Maximum Error would be,

\frac{dV}{dr} = \frac{4}{3} 3\pi r^2

dV = 2r^2 (2\pi dr)

dV = 4r^2 (\pi dr)

Replacing the value for the radius

dV = 4(\frac{38}{\pi})^2(0.5)

dV = \frac{2888}{\pi^2} cm^3

And the relative Error

\frac{dV}{V} = \frac{ \frac{2888}{\pi^2}}{ \frac{219488}{3\pi^2} }

\frac{dV}{V} = 0.03947

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h = height of pellet = 100 m

Vf = final velocity of pellet = 0 m/s (since, pellet will momentarily stop at highest point)

Vi = Initial Velocity of Pellet = ?

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Vi = √(1960 m²/s²)

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2(1.625 m/s²)h' = (44.27 m/s)² - (0 m/s)²

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