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DIA [1.3K]
2 years ago
12

The brightness of a light bulb is most closely related to

Physics
1 answer:
vlada-n [284]2 years ago
4 0

Answer:

a ) Its resistance

Explanation:

It depends on current and resistance

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20 points please answer
Anna [14]
Another way to test your question is to build your own miniature buildings. Depending on how in-depth you go, building could get a little pricey, but if you keep it basic there shouldn't be a problem. Decide on a certain number of foundations to test [maybe 3 or so] and try simulating an earthquake. 
<span>Hope this helps! </span>
6 0
3 years ago
9.5 g bullet has a speed of 1.5 km/s what is the kinetic energy of the bullet
Amiraneli [1.4K]

The answer is  0.000824653J

You need to use the formula Mass * Velocity^2 over 2


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3 years ago
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Distance from any point on a wave to an identical point on the next wave?
lesantik [10]
It's called the "Wavelength". It corresponds to <span>the distance from any point on a wave to an identical point on the next wave and could also be from crest to crest or trough to trough.

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6 0
3 years ago
A 3.91 kg cart is moving at 5.7 m/s when it collides with a 4 kg cart which was at rest. They collide and stick together.
Nesterboy [21]

Answer:

<em>The velocity after the collision is 2.82 m/s</em>

Explanation:

<u>Law Of Conservation Of Linear Momentum </u>

It states the total momentum of a system of bodies is conserved unless an external force is applied to it. The formula for the momentum of a body with mass m and speed v is  

P=mv.  

If we have a system of two bodies, then the total momentum is the sum of the individual momentums:

P=m_1v_1+m_2v_2

If a collision occurs and the velocities change to v', the final momentum is:

P'=m_1v'_1+m_2v'_2

Since the total momentum is conserved, then:

P = P'

Or, equivalently:

m_1v_1+m_2v_2=m_1v'_1+m_2v'_2

If both masses stick together after the collision at a common speed v', then:

m_1v_1+m_2v_2=(m_1+m_2)v'

The common velocity after this situation is:

\displaystyle v'=\frac{m_1v_1+m_2v_2}{m_1+m_2}

There is an m1=3.91 kg car moving at v1=5.7 m/s that collides with an m2=4 kg cart that was at rest v2=0.

After the collision, both cars stick together. Let's compute the common speed after that:

\displaystyle v'=\frac{3.91*5.7+4*0}{3.91+4}

\displaystyle v'=\frac{22.287}{7.91}

\boxed{v' = 2.82\ m/s}

The velocity after the collision is 2.82 m/s

6 0
3 years ago
Which of the following statements is true?
Lilit [14]

A. All natural radiation is at a level low enough to be safe

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4 years ago
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