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vladimir1956 [14]
3 years ago
11

A push on a 1-kg brick accelerates it. Neglecting friction, equally accelerating a 10-kg brick requires 10 times as much force.

Using your knowledge of Newton’s Second Law, explain why this is correct.
Physics
1 answer:
sergij07 [2.7K]3 years ago
6 0
It is CVM this is simple
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The half life of radium 222 is 38 seconds if you had a 12 gram sample how much would be left after 72 seconds
poizon [28]
<span>3.2 grams The first thing to do is calculate how many half lives have expired. So take the time of 72 seconds and divide by the length of a half life which is 38 seconds. So 72 / 38 = 1.894736842 So we're over 1 half life, but not quite 2 half lives. So you'll have something less than 12/2 = 6 grams, but more than 12/4 = 3 grams. The exact answer is done by dividing 12 by 2 raised to the power of 1.8947. So let's calculate 2^1.8947 power = 12 g / (e ^ ln(2)*1.8947) = 12 g / (e ^ 0.693147181 * 1.8947) = 12 g / (e ^ 1.313305964) = 12 g / 3.718446464 = 3.227154167 g So rounded to 2 significant figures gives 3.2 grams.</span>
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3 years ago
What effect does a catalyst have on a system in equilibrium?
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ANSWER:

What effect does a catalyst have on a system in equilibrium?

The system is unaffected.

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8 0
3 years ago
A car initially traveling at 27.2 m/s undergoes a constant negative acceleration of magnitude 1.90 m/s2 after its brakes are app
zubka84 [21]

Answer:

Therefore, the revolutions that each tire makes is:

\Delta \theta=22\: rev

Explanation:

We can use the following equation:

\omega_{f}^{2}=\omega_{i}^{2}-2\alpha \Delta \theta (1)

The angular acceleration is:

a_{tan}=\alpha R

\alpha=\frac{1.9}{0.325}

\alpha=5.85\: rad/s^{2}

and the initial angular velocity is:

\omega_{i}=\frac{v}{R}

\omega_{i}=\frac{27.2}{0.325}

\omega_{i}=83.69\: rad/s

Now, using equation (1) we can find the revolutions of the tire.

0=83.69^{2}-2*25.85 \Delta \theta

\Delta \theta=135.47\: rad

Therefore, the revolutions that each tire makes is:

\Delta \theta=22\: rev

I hope it helps you!

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3 years ago
Which of the following statements about the conservation of momentum is not correct? Momentum is conserved for a system of objec
Olin [163]

Answer:

wrong statement :  Momentum is not conserved for a system of objects in a head-on collision.

Explanation:

In a head on collision of two objects , two equal and opposite forces are created at the point of collision . These two forces create two impulses in opposite direction which results in equal and opposite changes in momentum in each of them . Hence net change in momentum is zero. In this way momentum is conserved in head on collision of two objects.

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3 years ago
The theoretical line perpindicular to the surface where a light ray hits a mirror is called the
valkas [14]
That's called the "normal" to the surface at that point.
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3 years ago
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