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nevsk [136]
3 years ago
6

If a ball is rolling at a velocity of 1.5 m/sec and has a momentum of 10.0 kg·m/sec, what is the mass of the 

Physics
1 answer:
AlladinOne [14]3 years ago
4 0
Momentum = (mass) x (speed)

10 kg-m/sec = (mass) x (1.5 m/s)

Divide each side by  1.5 m/s :

Mass =  10 kg-m/sec / 1.5 m/s  =  <em>(6 and 2/3) kg</em>
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Answer:

Aerobic exercise includes?

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(B) anaerobic exercise

Explanation:

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3 years ago
Can someone please tell me if this sentence makes sense?
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41. Electric and magnetic forces can both make certain objects move. For example, a positively charged particle will repel anoth
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Answer:

Place the north pole of a magnet next to the north pole of another magnet.

Explanation:

Looking at the comments, we can see that the options are:

Place the south pole of a magnet next to the north pole of another magnet.

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First, we know that a positively charged particle will repel another positively charged particle.

The same thing happens for magnetic forces (usually we define a magnetic flow from the south pole to the north pole, so we can define the south pole as the "positive" and the north pole as the "negative", but this is only notation and do not really matter), a south pole of a magnet will repel another south pole of a magnet (and the same happens for the north poles)

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8 0
3 years ago
Your spaceship lands on an unknown planet. To determine the characteristics of this planet, you drop a wrench from 4.50 m above
Virty [35]

8.98×10^6\:\text{m}

Explanation:

First we need to find the acceleration due to gravity on the planet. The wrench took 0.809 s to fall from a height of 4.50 m so we can use the equation

y = -\frac{1}{2}gt^2

Solving for g, we get

g = -\dfrac{2y}{t^2} = -\dfrac{2(-4.50\:\text{m})}{(0.809\:\text{s})} = 13.8\:\text{m/s}^2

Recall that the acceleration due to gravity on a planet's surface can be written as

g = G\dfrac{M_p}{R_p^2}

We can express the mass of the planet M_p in terms of its density \rho as follows:

M_p = \rho \left(\dfrac{4\pi}{3}R_p^3\right) = \dfrac{4\pi}{3}\rho R_p^3

The expression for g then becomes

g = \dfrac{G}{R_p^2} \left(\dfrac{4\pi}{3}\rho R_p^3\right) = \dfrac{4\pi G}{3}\rho R_p

Solving for R_p, we get

R_p = \dfrac{3g}{4\pi G\rho}

\:\:\:\:\:\:\:= \left[\dfrac{3(13.8\:\text{m/s}^2)}{4\pi (6.674×10^{-11}\:\text{Nm}^2\text{/kg}^2)(5500\:\text{kg/m}^3)}\right]

\:\:\:\:\:\:\:= 8.98×10^6\:\text{m}

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vredina [299]
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