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2 years ago

6

Please answer this question I have uploaded the picture of question please answer

1 answer:

8090 [49]2 years ago

5 0

Answer:

20 mangintiude beacuse

Explanation:

mt everst is in mounatin region we eat rice we eat pizza burger sandwich go to thamel for a good reason doing lamo lamo hw so be obtidnet you 4 kaccha fail boy

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What are the three psychological processes of memory?

MariettaO [177]

1. Encoding Information
2. Storing Information
3. Retrieval Information

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3 years ago

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At the _______________ point, the particles in an object have not kinetic energy

Brums [2.3K]

The freezing point ..... :)

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2 years ago

A freshly prepared sample of radioactive isotope has an activity of 10 mCi. After 4 hours, its activity is 8 mCi. Find: (a) the

Maurinko [17]

Answer:

(a). The decay constant is $1.55\times10^{-5}\ s^{-1}$

The half life is 11.3 hr.

(b). The value of N₀ is $2.38\times10^{11}\ nuclei$

(c). The sample's activity is 1.87 mCi.

Explanation:

Given that,

Activity $R_{0}=10\ mCi$

Time $t_{1}=4\ hours$

Activity R= 8 mCi

(a). We need to calculate the decay constant

Using formula of activity

$R=R_{0}e^{-\lambda t}$

$\lambda=\dfrac{1}{t}ln(\dfrac{R_{0}}{R})$

Put the value into the formula

$\lambda=\dfrac{1}{4\times3600}ln(\dfrac{10}{8})$

$\lambda=0.0000154\ s^{-1}$

$\lambda=1.55\times10^{-5}\ s^{-1}$

We need to calculate the half life

Using formula of half life

$T_{\dfrac{1}{2}}=\dfrac{ln(2)}{\lambda}$

Put the value into the formula

$T_{\dfrac{1}{2}}=\dfrac{ln(2)}{1.55\times10^{-5}}$

$T_{\dfrac{1}{2}}=44.719\times10^{3}\ s$

$T_{\dfrac{1}{2}}=11.3\ hr$

(b). We need to calculate the value of N₀

Using formula of $N_{0}$

$N_{0}=\dfrac{3.70\times10^{6}}{\lambda}$

Put the value into the formula

$N_{0}=\dfrac{3.70\times10^{6}}{1.55\times10^{-5}}$

$N_{0}=2.38\times10^{11}\ nuclei$

(c). We need to calculate the sample's activity

Using formula of activity

$R=R_{0}e^{-\lambda\times t}$

Put the value intyo the formula

$R=10e^{-(1.55\times10^{-5}\times30\times3600)}$

$R=1.87\ mCi$

Hence, (a). The decay constant is $1.55\times10^{-5}\ s^{-1}$

The half life is 11.3 hr.

(b). The value of N₀ is $2.38\times10^{11}\ nuclei$

(c). The sample's activity is 1.87 mCi.

4 0

3 years ago

Two point charges are arranged in a line with point P, which is on the right. The

MAVERICK [17]

Answer:

chicken

Explanation:

you like chicken and want to eat it

6 0

2 years ago

A small object carrying a charge of -4.00 nC is acted upon by a downward force of 24.0 nN when placed at a certain point in an e

PIT_PIT [208]

Answer:

$E = -6 \ N/C$

Generally given that the electric field is negative it mean that its direction is opposite to that of the force

Explanation:

From the question we are told that

The charge on the small object is $Q = -4.00 \ nC = -4.00 *10^{-9} \ C$

The force is $F = 24 \ nN = 24 *10^{-9} \ N$

Generally the magnitude of the electric field is mathematically represented as

$E = \frac{F}{Q}$

=> $E = \frac{ 24 *10^{-9}} {-4 *10^{-9 }}$

=> $E = -6 \ N/C$

Generally given that the electric field is negative it mean that its direction is opposite to that of the force

6 0

2 years ago