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Mkey [24]
3 years ago
5

Final velocity for 2.6 seconds

Physics
1 answer:
enyata [817]3 years ago
3 0

That's going to depend on the initial velocity, and how it changes
or doesn't change during the 2.6 seconds.  I've read the question,
and Im pretty sure you didn't tell us any of that.

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According to Wien's Law, how many times hotter is an object whose blackbody emission spectrum peaks in the blue, at a wave lengt
scZoUnD [109]

Answer:1.55 times

Explanation:

Given

First wavelength(\lambda _1)=450 nm

Second wavelength(\lambda _2)=700 nm

According wien's diplacement law

\lambda T=constant

where \lambda =wavelength

T=Temperature

Let T_1 and T_2 be the temperatures corresponding to \lambda _1 & \lambda _2 respectively.

\lambda _1\times T_1=\lambda _2\times T_2

\frac{T_1}{T_2}=\frac{\lambda _2}{\lambda _1}

\frac{T_1}{T_2}=\frac{700}{450}=1.55

Thus object with \lambda 450 nm is 1.55 times hotter than object with wavelength \lambda =700 nm

8 0
3 years ago
A 41.0 g marble moving at 2.30 m/s strikes a 25.0 g marble at rest. What is the speed of each marble immediately after the colli
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Answer:

speed of each marble after collision will be 1.728 m/sec

Explanation:

We have given mass of the marble m_1=41gram=0.041kg

Velocity of marble v_1=2.30m/sec

Its collides with other marble of mass 25 gram

So mass of other marble m_2=25gram=0.025kg

Second marble is at so v_2=0m/sec

We have to find the velocity of second marble

From momentum conservation we know that

m_1v_1+m_2v_2=(m_!+m_2)v, here v is common velocity of both marble after collision

So 0.041\times 2.30+0.025\times 0=(0.041+0.025)v

v = 1.428 m /sec

So speed of each marble after collision will be 1.728 m/sec

6 0
3 years ago
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Start with what the paragraph is about and put it basically in your own words
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Andreas93 [3]

Answer:

The combining of light nuclei is called nuclear fusion.

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