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3 years ago

A 0.25 kg ball is rolled across the floor with a velocity of 3.5 m/s. What is the ball's momentum?

Physics

1 answer:

solniwko [45]3 years ago

7 0

<h2>Answer:0.875 $Kgms^{-1}$ </h2>

Explanation:

Momentum is defined as the product of mass of an object and the velocity of that object.

Let $m$ be the mass of the object.

Let $v$ be the velocity of the object.

Let $p$ be the momentum of the object.

Given that $m=0.25Kg$

$v=3.5ms^{-1}$

$p=mv$

$p=0.25\times 3.5=0.875Kgms^{-1}$

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sergij07 [2.7K]

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$\boxed{\mathfrak{Power(P)=\frac{Voltage(V)^2}{Resistance(R)} }}$

V = 12 V
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$\boxed{\mathfrak{Power(P)=\frac{Voltage(V)^2}{Resistance(R)} }}$

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<em>these lights operate at the usual 240 volts direct from the main electricity supply. Therefore,</em>

V = 240 V

$\implies \mathsf{100=\frac{240^2}{R} }$

<em>R and 100 can interchange places</em>

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The brightness of the bulb in series is <em><u>less than</u></em> when they're placed individually.

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Their resistances are nothing but mere numbers and the sum of two numbers(positive of course) is greater than the numbers.

So, the effective resistance of some bulbs in series <u>is more</u> than the individual resistance.

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<em>Brightness, i. e., Power</em>

$\boxed{\mathfrak{Power \propto \frac{1}{Resistance} }}$

If resistance increases, Power decreases.

Here, the effective resistance was for sure larger, therefore resistance was increasing, hence power decreased taking brightness along with it.

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