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drek231 [11]
3 years ago
6

HELP ME PLEASE!

Physics
1 answer:
3241004551 [841]3 years ago
5 0

they have zero momentum before pushing off, is at least true

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organ

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A long, straight metal rod has a radius of 5.75 cm and a charge per unit length of 33.3 nC/m. Find the electric field at the fol
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Answer:

Explanation:

From the question;

We will make assumptions of certain values since they are not given but the process to achieve the end result will be the same thing.

We are to calculate the following task, i.e. to determine the electric field at the distances:

a)  at 4.75 cm

b)  at 20.5 cm

c) at 125.0 cm

Given that:

the charge (q) = 33.3 nC/m

= 33.3 × 10⁻⁹ c/m

radius of rod = 5.75 cm

a) from the given information, we will realize that the distance lies inside the rod. Provided that there is no charge distribution inside the rod.

Then, the electric field will be zero.

b) The electric field formula E = \dfrac{kq }{d}

E = \dfrac{9 \times 10^9 \times (33.3 \times 10^{-9}) }{0.205}

E = 1461.95 N/C

c) The electric field E is calculated as:

E = \dfrac{9 \times 10^9 \times (33.3 \times 10^{-9}) }{1.25}

E = 239.76 N/C

7 0
3 years ago
Strontium 3890Sr has a half-life of 28.5 yr. It is chemically similar to calcium, enters the body through the food chain, and co
patriot [66]

Answer:

Thus the time taken is calculated as 387.69 years

Solution:

As per the question:

Half life of ^{3890}Sr\, t_{\frac{1}{2}} = 28.5 yrs

Now,

To calculate the time, t in which the 99.99% of the release in the reactor:

By using the formula:

\frac{N}{N_{o}} = (\frac{1}{2})^{\frac{t}{t_{\frac{1}{2}}}}

where

N = No. of nuclei left after time t

N_{o} = No. of nuclei initially started with

\frac{N}{N_{o}} = 1\times 10^{- 4}

(Since, 100% - 99.99% = 0.01%)

Thus

1\times 10^{- 4} = (\frac{1}{2})^{\frac{t}{28.5}}}

Taking log on both the sides:

- 4 = \frac{t}{28.5}log\frac{1}{2}

t = \frac{-4\times 28.5}{log\frac{1}{2}}

t = 387.69 yrs

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3 years ago
Prove that s=ut+½at²​
katrin2010 [14]

Explanation:

Distance travelled = Area under the line

= ut + ½ (v-u)t

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Therefore,

Distance travelled (s) = ut + ½ (v-u)t = ut + ½ (at)t = ut + ½ at²

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