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2 years ago

Help asap I will give brainleast

Physics

1 answer:

NARA [144]2 years ago

5 0

3rd option. Plant D and E did not have enough radiant energy

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For the following elementary reaction 2br• -> br2-. The rate of consumption of the reaction and the rate of formation of prod

Scorpion4ik [409]

Answer: $-\frac{1}{2}\times \frac{d[Br^.]}{dt}=+\frac{d[Br_2]}{dt}$

Explanation:

Rate of a reaction is defined as the rate of change of concentration per unit time.

Thus for reaction:

$2Br^.\rightarrow Br_2$

The rate in terms of reactants is given as negative as the concentration of reactants is decreasing with time whereas the rate in terms of products is given as positive as the concentration of products is increasing with time.

$Rate=-\frac{d[Br^.]}{2dt}$

or $Rate=+\frac{d[Br_2]}{dt}$

Thus $-\frac{d[Br^.]}{2dt}=+\frac{d[Br_2]}{dt}$

4 0

2 years ago

Read the list of phrases from the article.

ICE Princess25 [194]

Answer:

(D)

to establish an understanding of key concepts relating to population biology

Explanation:

Thats what I would go with but I didn't read the article so I don't know what context was used. Good luck! :)

8 0

2 years ago

If you get too many fouls, you may not be able to play in the rest of the game.

trapecia [35]

Answer:

In the NBA that's true

4 0

2 years ago

Read 2 more answers

When beryllium-7 ions (m = 11.65 × 10-27 kg) pass through a mass spectrometer, a uniform magnetic field of 0.205 T curves their

AysviL [449]

Answer:

ratio =0.3075 T

Explanation:

The magnetic field B creates a force on a moving charge such that

$F = qvB$

Now this causes a centripetal acceleration

$F = = mv^2/r$

$qvB = mv^2/r$ ...........(i)

$B = mv/(rq)$ ...............(ii)

If accelerating potential V is same and then kinetic energy equals the potential energy difference

$\frac{1}{2} mv^2 = Vq$

$v = \sqrt{(2Vq/m)}$ put these value in equation (ii)

$B = m\frac{\sqrt{(2Vq/m)} }{rq}$

simplifying we get

$B =m \frac{(\sqrt{ 2Vm/q})}{r}$

for same location r will be same in both case

$B_{7} = \frac{ \sqrt{(m_{7})(2V/q) }}{r}$ ..............(iii)

$B_{10} = \frac{ \sqrt{(m_{10})(2V/q) }}{r}$ ..........(iv)

dividing (iv) and (iii) equation we get

$\frac{B_{10}}{B_{7}} = \sqrt{\frac{m_{10}}{{m_7}} }$

${B_{10}} = B_{7} \sqrt{\frac{m_{10}}{{m_7}} }$

$B_{10} = 0.2574T\sqrt{\frac{ (1.663x10^-26}{(1.165x10^-26)}$

so on solving we get

=0.3075 T

5 0

2 years ago

A cyclist has a constant speed of 12 m/s. What is the magnitude of the displacement of the cyclist after 18 seconds?

Katyanochek1 [597]

Answer:

216 m

Explanation:

Assuming a straight line:

Δx = vt

Δx = (12 m/s) (18 s)

Δx = 216 m

7 0

2 years ago